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http://math.stackexchange.com/questions/268206/lower-semicontinuity-and-closed-graph	# lower semicontinuity and closed graph I would like to prove that for a LSC function, its epigraph is closed. I saw some longer proof here, but why would the following not hold ? : $f LSC := \liminf f(x_n) \ge f(x)$ when $x_n \rightarrow x$ then $x_n,a_n \rightarrow x,a \implies f(x) \le \liminf f(x_n) \le \liminf a_n = a$ so $x,a \in epi(f)$ - ## 1 Answer This is a streamlined version of the proof in wikipedia. Perhaps it's a bit too streamlined (write it up using quantifiers!). Otherwise it appears perfectly valid to me. - yeah, i spotted other weird stuff in the article in reference. not to be taken as a reference ! – nicolas Dec 31 '12 at 17:44	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 4, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9654937386512756, "perplexity_flag": "middle"}
http://psychology.wikia.com/wiki/Semantic_holism	# Semantic holism Talk0 31,725pages on this wiki Assessment \| Biopsychology \| Comparative \| Cognitive \| Developmental \| Language \| Individual differences \| Personality \| Philosophy \| Social \| Methods \| Statistics \| Clinical \| Educational \| Industrial \| Professional items \| World psychology \| Philosophy Index: Aesthetics · Epistemology · Ethics · Logic · Metaphysics · Consciousness · Philosophy of Language · Philosophy of Mind · Philosophy of Science · Social and Political philosophy · Philosophies · Philosophers · List of lists Semantic holism is a doctrine in the philosophy of language to the effect that a certain part of language, be it a term or a complete sentence, can only be understood through its relations to a (previously understood) larger segment of language. There is substantial controversy, however, as to exactly what the larger segment of language in question consists of. In recent years, the debate surrounding semantic holism, which is one among the many forms of holism that are debated and discussed in modern philosophy, has tended to centre around the view that the "whole" in question consists of an entire language. ## Background Since the use of a linguistic expression is only possible if the speaker who uses it understands its meaning, one of the central problems for analytic philosophers has always been the question of meaning. What is it? Where does it come from? How is it communicated? And, among these questions, what is the smallest unit of meaning, the smallest fragment of language with which it is possible to communicate something? At the end of the 18th and beginning of the 19th century, Gottlob Frege and his followers abandoned the view, common at the time, that a word gets its meaning in isolation, independently from all the rest of the words in a language. Frege, as an alternative, formulated his famous Context principle, according to which it is only within the context of an entire sentence that a word acquires its meaning. In the 1950's, the agreement that seemed to have been reached regarding the primacy of sentences in semantic questions began to unravel with the collapse of the movement of logical positivism and the powerful influence exercised by the philosophical investigations of the later Wittgenstein. Wittgenstein wrote in the Philosophical Investigations, in fact, that "comprehending a proposition means comprehending a language." About the same time or shortly after, W.V.O. Quine wrote that "the unit of measure of empirical meaning is all of science in its globality"; and Donald Davidson, in 1967, put it even more sharply by saying that "a sentence (and therefore a word) has meaning only in the context of a (whole) language." ## Problems with semantic holism If semantic holism is interpreted as the thesis that any linguistic expression E (a word, a phrase or sentence) of some natural language L cannot be understood in isolation and that there are inevitably many ties between the expressions of L, it follows that in order to understand E one must understand a set K of expressions to which E is related. If, in addition, no limits are placed on the size of K (as in the cases of Davidson, Quine and, perhaps, Wittgenstein), then K coincides with the "whole" of L. The many and substantial problems with this position have been described by Michael Dummett, Jerry Fodor, Ernest Lepore and others. In the first place, it is impossible to understand how a speaker of L can acquire knowledge of (learn) the meaning of E, for any expression E of the language. Given the limits of our cognitive abilities, we will never be able to master the whole of the English (or Italian or German) language, even on the assumption that languages are static and immutable entities (which is false). Therefore, if one must understand all of a natural language L in order to understand the single word or expression E, then language learning is simply impossible. Semantic holism, in this sense, also fails to explain how two speakers can mean the same thing when using the same linguistic expression, and therefore how communication is even possible between them. Given a sentence P, since Fred and Mary have each mastered different parts of the English language and P is related differently to the sentences in each part, the result is that P means one thing for Fred and something else for Mary. Moreover, if a sentence P derives its meaning from the relations which it entertains with the totality of sentences of a language, as soon as the vocabulary of an individual changes by the addition or elimination of a sentence $P'$, the totality of relations changes, and therefore also the meaning of P. As this is a very common phenomenon, the result is that P has two different meanings in two different moments during the life of the same person. Consequently, if I accept the truth of a sentence and then reject it later on, the meaning of that which I rejected and that which I accepted are completely different and therefore I cannot change my opinions with regard to the same sentences. ## Semantic holism and holism of mental content These sorts of counterintuitive consequences of semantic holism also affect another form of holism, often identified with but, in fact, distinct from semantic holism: the holism of mental content. This is the thesis that the meaning of a particular propositional attitude (thought, desire, belief) acquires its content by virtue of the role that it plays within the web which connects it to all the other propositional attitudes of an individual. Since there is a very tight relationship between the content of a mental state M and the sentence P which expresses it and makes it publicly communicable, the tendency in recent discussion is to consider the term "content" to apply indifferently to both linguistic expressions and mental states, regardless of the extremely controversial question of which category (the mental or the linguistic) has priority over the other and which, instead, possesses only a derived meaning. So, it would seem that semantic holism ties the philosopher’s hands. By making it impossible to explain language learning and to provide a unique and consistent description of the meanings of linguistic expressions, it blocks off any possibility of formulating a theory of meaning; and, by making it impossible to individuate the exact contents of any propositional attitude—given the necessity of considering a potentially infinite and continuously evolving set of mental states—it blocks off the possibility of formulating a theory of the mind. Given this situation one may rightly wonder why semantic holism has received such serious attention from the philosophical community at all. ## Semantic holism and confirmational holism The key to answering this question lies in going back to Quine and his attack on logical positivism. The logical positivists, who dominated the philosophical scene for almost the entire first half of the twentieth century, maintained that genuine knowledge consisted in all and only such knowledge as was capable of manifesting a strict relationship with empirical experience. Therefore, they believed, the only linguistic expressions (manifestations of knowledge) which possessed meaning were those which either directly referred to observable entities or which could be reduced to a vocabulary which directly referred to such entities. A sentence S contained knowledge only if it possessed a meaning, and it possessed a meaning only if it was possible to refer to a set of experiences which could, at least potentially, verify it and to another set which could potentially falsify it. Underlying all this, there is an implicit and powerful connection between epistemological and semantic questions. This connection carries over into the work of Quine in Two Dogmas of Empiricism. Quine’s holistic argument against the neo-positivists set out to demolish the assumption that every sentence of a language is bound univocally to its own set of potential verifiers and falsifiers and the result was that the epistemological value of every sentence must depend on the entire language. Since the epistemological value of every sentence, for Quine just as for the positivists, was the meaning of that sentence, then the meaning of every sentence must depend on every other. As Quine states it: All of our so-called knowledge or convictions, from questions of geography and history to the most profound laws of atomic physics or even mathematics and logic, are an edifice made by man which touches experience only at the margins. Or, to change images, science in its globality is like a force field whose limit points are experiences…a particular experience is never tied to any proposition inside the field except indirectly, for the needs of equilibrium which affect the field in its globality. For Quine then (although Fodor and Lepore have maintained the contrary), and for many of his followers, confirmation holism and semantic holism are inextricably linked. Since confirmation holism is widely accepted among philosophers, a serious question for them has been to determine whether and how the two holisms can be distinguished or how the undesirable consequences of unbuttoned holism, as Michael Dummett has called it, can be limited. ## Moderate holism Numerous philosophers of language have taken the latter avenue, abandoning the early Quinean holism in favour of what Michael Dummett has labelled semantic molecularism. These philosophers generally deny that the meaning of an expression E depends on the meanings of the words of the entire language L of which it is part and sustain, instead, that the meaning of E depends on some subset of L. These positions, notwithstanding the fact that many of their proponents continue to call themselves holists, are actually intermediate between holism and atomism. Dummett, for example, after rejecting Quinean holism (holism tout court in his sense), takes precisely this approach. But those who would opt for some version of moderate holism need to make the distinction between the parts of a language which are "constitutive" of the meaning of an expression E and those which are not without falling into the extraordinarily problematic analytic/synthetic distinction. Fodor and Lepore (1992) present several arguments to demonstrate that this is impossible. ### Arguments against molecularism According to Fodor and Lepore, there is a quantificational ambiguity in the molecularist’s typical formulation of his thesis: someone can believe P only if she believes a sufficient number of other propositions. They propose to disambiguate this assertion into a strong and a weak version: (S) $\forall p \exists q \ne p \Box (B(x,p) \rightarrow B(x,q))$ (W) $\Box (B(x,p) \rightarrow \exists q \ne p (B(x,q))$ The first statement asserts that there are other propositions, besides p, that one must believe in order to believe that p. The second says that one cannot believe that p unless there are other propositions in which one believes. If one accepts the first reading, then one must accept the existence of a set of sentences which are necessarily believed and hence fall into the analytic/synthetic distinction. The second reading is useless (too weak) to serve the molecularist’s needs since it only requires that if, say, two people believe the same proposition p, they also believe in at least one other proposition. But, in this way, each one will connect to p his own inferences and communication will remain impossible. Carlo Penco criticizes this argument by pointing out that there is an intermediate reading which Fodor and Lepore have left out of count: (I) $\Box (B(x,p) \land B(y,p) \rightarrow \exists q \ne p (B(x,q) \land B(y,q))$ This says that two people cannot believe the same proposition unless they also both believe a proposition different from p. This helps to some extent but there is still a problem in terms of identifying how the different propositions shared by the two speakers are specifically related to each other. Dummett’s proposal is based on an analogy from logic. In order to understand a logically complex sentence it is necessary to understand one which is logically less complex. In this manner, the distinction between logically less complex sentences which are constitutive of the meaning of a logical constant and logically more complex sentences which are not takes on the role of the old analytic/synthetic distinction. "The comprehension of a sentence in which the logical constant does not figure as a principal operator depends on the comprehension of the constant, but does not contribute to its constitution." For example, one can explain the use of the conditional in $(a \lor \lnot b) \rightarrow c$ by stating that the whole sentence is false if the part before the arrow is true and c is false. But in order to understand $a \lor \lnot b$ one must already know the meaning of "not" and "or." This is, in turn, explained by giving the rules of introduction for simple schemes such as $P \lor Q$ and $\lnot Q$. To comprehend a sentence is to comprehend all and only the sentences of less logical complexity than the sentence that one is trying to comprehend. However, there is still a problem with extending this approach to natural languages. If I understand the word "hot" because I have understood the phrase "this stove is hot", it seems that I am defining the term by reference to a set of stereotypical objects with the property of being hot. If I don’t know what it means for these objects to be "hot", such a set or listing of objects is not helpful. ## Holism and compositionality The relationship between compositionality and semantic holism has also been of interest to many philosophers of language. On the surface it would seem that these two ideas are in complete and irremediable contradiction. Compositionality is the principle that states that the meaning of a complex expression depends on the meaning of its parts and on its mode of composition. As stated before, holism, on the other hand, is the thesis that the meanings of expressions of a language are determined by their relations with the other expressions of the language as a whole. Peter Pagin, in an essay called Are Compositionality and Holism Compatible identifies three points of incompatibility between these two hypotheses. The first consists in the simple observation that while, for holism, the meaning of the whole would seems to precede that of its parts in terms of priority, for compositionality, the reverse is true, the meaning of the parts precedes that of the whole. The second incoherence consists in the fact that "strange" meanings must be attributed to the components of larger expressions would apparently result from any attempt to reconcile compositionality and holism. Pagin takes a specific holistic theory of meaning – inferential role semantics, the theory according to which the meaning of an expression is determined by the inferences that it involves – as his paradigm of holism. If we interpret this theory holistically, the result will be that every accepted inference which involves some expression will enter into the meaning of that expression. Suppose, for example, that Fred believes that "Brown cows are dangerous". That is, he accepts the inference from "brown cows" to "dangerous." This entails that this inference is now part of the meaning of "brown cow." According to compositionality then, "cow implies dangerous" and "brown implies dangerous" are both true because they are the constituents of the expression "brown cow." But is this really an inevitable consequence of the acceptance of the holism of inferential role semantics? To see why it's not assume the existence of a relation of inference I between two expressions x and y and that the relation applies just in case F accepts the inference from x to y. Suppose that in the extension of I, there are the following pairs of expressions ("The sky is blue and leaves are green", "the sky is blue") and ("brown cow", "dangerous"). There is also a second relation P which applies to two expressions just in case the first is part of the second. So, ("brown, "brown cow") belongs to the extension of P. Two more relations, "Left" and "Right", are required: $L(\alpha,\beta,\gamma) \leftrightarrow P(\alpha,\beta) \land I(\beta,\gamma)$ $R(\alpha,\beta,\gamma) \leftrightarrow P(\alpha,\gamma) \land I(\beta,\gamma)$ The first relation means that L applies between α,β and γ just in case α is a part of β and F accepts the inference between β and γ. The relation R applies between α, β, and γ just in case α is a part of γ and F accepts the inference from β to γ. The Global Role, G(α), of a simple expression α can then be defined as: $G(\alpha) = ([(\beta,\gamma) : L(\alpha,\beta,\gamma)], [(\beta,\gamma) : R(\alpha,\beta,\gamma)]) \;$ The global role of $\alpha$ consists in a pair of sets each one of which is composed of a pair of sets of expressions. If F accepts the inference from $\beta$ to $\gamma$ and $\alpha$ is a part of $\gamma$, then the couple $(\beta,\gamma)$ is an element of the set which is an element of the right side of the Global Role of α. This makes Global Roles for simple expressions sensitive to changes in the acceptance of inferences by F. The Global Role for complex expressions can be defined as: $G(\beta) = [G(\alpha_1)...G(\alpha_n)] \;$ The Global Role of the complex expression β is the n- tuple of the global roles of its constituent parts. The next problem is to develop a function which assigns meanings to Global Roles. This function is generally called a homomorphism and says that for every syntactic function G which assigns to simple expressions α1...αn some complex expression β, there exists a function F from meanings to meanings: $h(G(\beta)) = h([G(\alpha_1)...G(\alpha_n)]) = F(h(G(\alpha_1))...h(G(\alpha_n))). \;$ This function is one to one in that it assigns exactly one meaning to every Global Role. According to Fodor and Lepore, holistic inferential role semantics leads to the absurd conclusion that part of the meaning of “brown cow” is constituted by the inference “Brown cow implies dangerous.” This is true if the function from meanings to Global Roles is one to one. In this case, in fact, the meanings of “brown”, “cow” and “dangerous” all contain the inference “Brown cows are dangerous”!! But this only true if the relation is one to one. Since it is one to one, “brown” would not have the meaning it has unless it had the global role that it has. If we change the relation so that it is many to one (h), many global roles can share the same meaning. So suppose that the meaning of “brown “is given by M(“brown”). It does not follow from this that L(“brown”, “brown cow”, “dangerous”) is true unless all of the global roles that h assigns to M(“brown”) contain (“brown cow”, “dangerous”). And this is not necessary for holism. In fact, with this many to one relation from Global Roles meanings, it is possible to change opinions with respect to an inference consistently. Suppose that B and C initially accept all of the same inferences, speak the same language and they both accept that “brown cows imply dangerous.” Suddenly, B has changes his mind and rejects the inference. If the function from meanings to Global Role is one to one, then many of B’s Global Roles have changed and therefore their meanings. But if there is no one to one assignment, then B’s change in belief in the inference about brown cows does not necessarily imply a difference is the meaning of the terms he uses. Therefore, it is not intrinsic to holism that communication or change of opinion is impossible. ## Holism and Externalism Since the concept of semantic holism, as explained above, is often used to refer to theories of meaning in natural languages but also to theories of mental content such as the hypothesis of a language of thought, the question often arises as to how to reconcile the idea of semantic holism (in the sense of the meanings of expressions in mental languages) with the phenomenon called externalism in philosophy of mind. Externalism is the thesis that the propositional attitudes of an individual are determined, at least in part, by her relations with her environment (both social and natural). Hilary Putnam formulated the thesis of the natural externalism of mental states in his The Meaning of "Meaning". In it, he described his famous thought experiment involving Twin Earths: two individuals, Calvin and Carvin, live, respectively, on the real earth (E) of our everyday experience and on an exact copy (E') with the only difference being that on E "water" stands for the substance $H_2O$ while on E' it stands for some substance macroscopically identical to water but which is actually composed of XYZ. According to Putnam, only Calvin has genuine experiences which involve water and so only his term "water" really refers to water. Tyler Burge, in Individualism and the Mental, describes a different thought experiment which led to the notion of the social externalism of mental contents. In Burge's experiment, a person named Jeffray believes that he has arthritis in his thighs and we can correctly attribute to him the (mistaken) belief that he has arthritis in his thighs because he is ignorant of the fact that arthritis is a disease of the articulation of the joints. In another society, there is an individual named Goodfrey who also believes that he has arthritis in the thighs. But in the case of Goodfrey the belief is correct because in the counterfactual society in which he lives "arthritis" is defined as a disease which can include the thighs. The question then arises of the possibility of reconciling externalism with holism. The one seems to be saying that meanings are determined by the external relations (with society or the world), while the other suggests that meaning is determined by the relation of words (or beliefs) to all the other words (or beliefs). Frederik Stjernberg identifies at least three possible ways to reconcile them and then points out some objections. The first approach is to insist that there is no conflict because holists do not mean the phrase "determine beliefs" in the sense of individuation but rather of attribution. But the problem with this is that if one is not a "realist" about mental states, then all we are left with is the attributions themselves and, if these are holistic, then we really have a form of hidden constitutive holism rather than a genuine attributive holism. But if one is a "realist" about mental states, then why not say that we can actually individuate them and therefore that instrumentalist attributions are just a short-term strategy? Another approach is to say that externalism is valid only for certain beliefs and that holism only suggests that beliefs are determined only in part by their relations with other beliefs. In this way, it is possible to say that externalism applies only to those beliefs which are not determined by their relations with other beliefs (or for the part of a belief that is not determined by its relations with other parts of other beliefs), and holism is valid to the extent that beliefs (or parts of beliefs) are not determined externally. The problem here is that the whole scheme is based on the idea that certain relations are constitutive (i.e. necessary) for the determination of the beliefs and others are not. Thus, we have reintroduced the idea of an analytic/synthetic distinction with all of the problems that that carries with it. A third possibility is to insist that there are two distinct types of belief: those determined holistically and those determined externally. Perhaps the external beliefs are those that are determined by their relations with the external world through observation and the holistic ones are the theoretical statements. But this implies the abandonment of a central pillar of holism: the idea that there can be no one to one correspondence between behavior and beliefs. There will be cases in which the beliefs that are determined externally correspond one to one with perceptual states of the subject. One last proposal is to carefully distinguish between so-called narrow content states and broad content states. The first would be determined in a holistic manner and the second non-holistically and externalistically. But how to distinguish between the two notions of content while providing a justification of the possibility of formulating an idea of narrow content that does not depend on a prior notion of broad content? These are some of the problems and questions that have still to be resolved by those who would adopt a position of "holist externalism" or "externalist holism". ## References • Dummett, Michael. The Logical Basis of Metaphysics. Harvard University Press. Cambridge (MA). 1978. • Quine, W.V.O.. From a Logical Point of View. Harvard University Press. Cambridge (MA). 1953. • Davidson, Donald. Inquiries into Truth and Interpretation. Clarendon Press. Oxford. 1984. • Wittgenstein, Ludwig. Philosophical Investigations. Basil Blackwell. Oxford. 1967. • Fodor, J. and Lepore, E. Holism: A Shopper's Guide. Blackwell. Oxford. 1992. • Penco, Carlo. Olismo e Molecularismo in Olismo ed. Massimo Dell'Utri. Quodlibet. Macerata. 2002. • Peter Pagin. Are Compositionality and Holism Compatible? in Olismo ed. Massimo dell'Utri. Quodlibet. Macerata. 2002. • Putnam, Hilary. The Mind is not only Computation in Olismo ed. Massimo dell'Utri. Quodlibet. Macerata. 2002. • Stjernberg, Fredrik. On the Combination of Holism and Externalism in Olismo ed. Massimo dell'Utri. Quodlibet. Macerata. 2002. • Putnam, Hilary. Mind, Language and Reality. Cambridge University Press. Cambridge. 1975. • Burge, Tyler. Individualism and the Mental in Midwest Studies in Philosophy. 4. pp. 73-121. 1979. ## Read more • Assessment \| Biopsychology \| Comparative \| Cognitive \| Developmental \| Language \| Individual... Connotation and denotation • Assessment \| Biopsychology \| Comparative \| Cognitive \| Developmental \| Language \| Individual... Norm (philosophy) • Assessment \| Biopsychology \| Comparative \| Cognitive \| Developmental \| Language \| Individual... Objectivity (philosophy) # Photos Add a Photo 6,465photos on this wiki • by Dr9855 2013-05-14T02:10:22Z • by PARANOiA 12 2013-05-11T19:25:04Z Posted in more... • by Addyrocker 2013-04-04T18:59:14Z • by Psymba 2013-03-24T20:27:47Z Posted in Mike Abrams • by Omaspiter 2013-03-14T09:55:55Z • by Omaspiter 2013-03-14T09:28:22Z • by Bigkellyna 2013-03-14T04:00:48Z Posted in User talk:Bigkellyna • by Preggo 2013-02-15T05:10:37Z • by Preggo 2013-02-15T05:10:17Z • by Preggo 2013-02-15T05:09:48Z • by Preggo 2013-02-15T05:09:35Z • See all photos See all photos >	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 20, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9525458812713623, "perplexity_flag": "middle"}
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Some notation: $\mathbb Z_n$ denotes the set of integers up to n i.e. $\mathbb Z_n=\{i│i ∈ \mathbb Z,0≤i < n \}$. $\mathbb Z_n^*$ denotes invertible elements of $\mathbb Z_n$ i.e. \$\mathbb ... 0answers 49 views ### Finding the random $r$ in a Paillier encrypted message with knowledge of the private key. In the Paillier cryptosystem, suppose that I know a Ciphertext encrypted with some unknown random $r$ i.e. $$C = (g^m r^n) \bmod n^2$$ I know $g, n$, the prime factorization of $n$, i.e., $pq$. I ... 1answer 147 views ### Multiplicative inverse trouble in RSA Wikipedia entry I'm having a bit of trouble working through an example in the RSA entry on Wikipedia. At step 5, $d$ is calculated as $2753$. However, $d$, which is the multiplicative inverse of $e$, can be ... 2answers 144 views ### Diffie–Hellman Problem I've been reading the Wikipedia Article on the Diffie–Hellman Problem and I've been wondering whether I understood it correctly, because if I did it seems fairly easy to solve. 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Of course you could brute force it but I'm interested in ...	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 65, "mathjax_display_tex": 3, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9258318543434143, "perplexity_flag": "middle"}
http://mathoverflow.net/questions/55442/why-can-we-define-the-moment-map-in-this-way-i-e-why-is-this-form-exact/55451	## Why can we define the moment map in this way (i.e. why is this form exact)? ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) Given a symplectic manifold $(X, \omega)$ and a group $G$ acting on $X$ preserving the symplectic form, we define the moment map `$\mu : X \to \mathfrak{g}^$` so that $$\langle d\mu(v), \xi\rangle = \omega\big(\xi^(x), v\big)$$ where $\xi \in \mathfrak{g}$, `$\xi^$` is the vector field generated by $\xi$, and $v \in T_x(X)$. Why can we do this? Why is $\mu$ defined? I read the above definition as follows. Define the `$\mathfrak{g}^$`-valued 1-form $\theta$ via $$\theta(v) = \omega(\xi^*,v).$$ Then this form is exact, and so we define $\mu$ so that $d\mu = \theta$. Why can we do this? Having done some simple computations, the fact that $G$ preserves the symplectic form seems to be imply that exactness holds (and from the above definition it is necessary), but I don't know why this should be the case. My question is: Why does the fact that $G$ preserves the symplectic form imply that the 1-form $\theta$ defined above is exact? Also: Is the preservation of the symplectic form equivalent to the exactness of $\theta$? - ## 3 Answers Both answers are "No." There are well-known obstructions to the existence of an equivariant momentum mapping arising from the action by symplectomorphisms of a group $G$ on a symplectic manifold. They can be phrased in many ways, but if $G$ is connected and its Lie algebra is semisimple, for example, the obstructions vanish. A nice treatment can be found in the classic paper by Atiyah and Bott: "The moment map and equivariant cohomology" in Topology 1984. After-dinner update: Let me try to add some more details, since it seems I was a little too quick both reading the question and also the comments! (In my defense, I was getting really hungry and wanted to get home!) Let $(M,\omega)$ be a symplectic manifold and let $G$ be a connected Lie group acting on $M$ via symplectomorphisms. Let $\mathfrak{g}$ be the Lie algebra of $G$ and for every $X \in \mathfrak{g}$ let $\xi_X$ denote the corresponding vector field on $M$. Since $G$ acts symplectomorphically, $\xi_X$ is a symplectic vector field; that is, $\mathcal{L}_{\xi_X} \omega = 0$, which, as Ben points out, implies that $d i_{\xi_X}\omega = 0$. Let $\mathrm{Sym}(M)$ denote the Lie algebra of symplectic vector fields on $M$. We have a Lie algebra homomorphism $\mathfrak{g} \to \mathrm{Sym}(M)$. A symplectic vector field $\xi$ is said to be hamiltonian if $i_\xi\omega$ is not merely closed, but also exact. The $G$-action is hamiltonian if the $\xi_X$ are hamiltonian for all $X \in \mathfrak{g}$. Let $\mathrm{Ham}(M)$ denote the Lie algebra of hamiltonian vector fields. It is not hard to show that the Lie bracket of two symplectic vector fields is hamiltonian, so $\mathrm{Ham}(M)$ is an ideal in $\mathrm{Sym}(M)$ with abelian quotient. This gives rise to a short exact sequence of Lie algebras $$0 \longrightarrow \mathrm{Ham}(M) \longrightarrow \mathrm{Sym}(M) \longrightarrow H^1(M) \longrightarrow 0$$ where $H^1(M)$ is the first de Rham cohomology group thought of as an abelian Lie algebra. If $\xi \in \mathrm{Ham}(M)$, then $i_\xi \omega = df$ for some $f \in C^\infty(M)$. This defines a map $C^\infty(M) \to \mathrm{Ham}(M)$ which is also a Lie algebra homomorphism if we make $C^\infty(M)$ into a Lie algebra via the Poisson bracket. The kernel of this map consists of the locally constant functions, whence we have another short exact sequence of Lie algebras $$0 \longrightarrow H^0(M) \longrightarrow C^\infty(M) \longrightarrow \mathrm{Ham}(M) \longrightarrow 0$$ Putting these two sequences together we have a four-term exact sequence starting and ending at the first two de Rham cohomology groups: $$0 \longrightarrow H^0(M) \longrightarrow C^\infty(M) \longrightarrow \mathrm{Sym}(M) \longrightarrow H^1(M) \longrightarrow 0$$ Now the symplectic $G$-action defines a Lie algebra homomorphism $\mathfrak{g} \to \mathrm{Sym}(M)$ and for the existence of a momentum map, we want this to lift to a Lie algebra morphism $\mathfrak{g} \to C^\infty(M)$. There is an immediate obstruction for the map to lift at the level of vector spaces, namely the map $\mathfrak{g} \to \mathrm{Sym}(M) \to H^1(M)$ is a Lie algebra cocycle with values in the trivial module $H^1(M)$, and so defines a class in $H^1(\mathfrak{g};H^1(M))$. If this class vanishes, we do get a map $\mathfrak{g} \to C^\infty(M)$ which may only be a homomorphism modulo $H^0(M)$. In other words, it defines a Lie algebra homomorphism to a central extension of $C^\infty(M)$ defined by a 2-cocycle with values in the trivial module $H^0(M)$. The moment map will exist if the class of this cocycle in $H^2(\mathfrak{g}; H^0(M))$ vanishes. For example, if $\mathfrak{g}$ is semisimple, then both $H^1$ and $H^2$ vanish and the momentum map exists. The beautiful observation of Atiyah and Bott is that the obstruction can be reinterpreted in terms of the Cartan model for the equivariant de Rham cohomology of $M$, where it becomes simply the obstruction to extending $\omega$ to an equivariant cocycle. - 2 How can "No" answer a question that begines with "Why..."? :) – Mariano Suárez-Alvarez Feb 14 2011 at 21:02 1 Good point! The answer is "No" if you omit the "Why" :) Otherwise the question is ill-posed. – José Figueroa-O'Farrill Feb 14 2011 at 21:05 Jose, I'm really confused by your answer. At least what I learned in my symplectic geometry class is that the answer is "yes." – Ben Webster♦ Feb 14 2011 at 21:45 To be fair, my question conflates the two, so your answer isn't actually that far off. – Simon Rose Feb 14 2011 at 22:23 Well, as comments show, I wasn't thinking with the highest level of clarity myself. That's the beauty of MO, you get to iteratively converge on the right answer. – Ben Webster♦ Feb 14 2011 at 22:44 ### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you. One thing that seems to be confusing you is that there is no "the moment map." There often several different moment maps for the same action, and as Jose says, sometimes none. On the other hand, your questions seem to really be If X is a vector field, are the conditions • $X$ preserves $\omega$ and • $i_X\omega=\omega(X,-)$ is an exact 1-form equivalent?" The answer is "no", but it's "yes" if you replace "exact" by "closed". The preservation of $\omega$ by $X$ is the same as saying that the Lie derivative $\mathcal{L}_X\omega=0$. By Cartan's magic formula, $$\mathcal{L}_X\omega=d(i_X\omega)+i_X(d\omega)=d(i_X\omega)$$ since the second term is zero by the closedness of $\omega$. This distinction is important, since lots of symplectic actions don't have moment maps because of the existence of $H^1$. Think about, for example, the 2-torus $T^2$ acting on itself; any invariant volume form can also be thought of as an invariant symplectic form, bu the map $\mathfrak t^2\to H^1(T^2)$ given by $X\mapsto [i_X\omega]$ is an isomorphism. No non-trivial element of the Lie algebra acts by a Hamiltonian vector field, even though they all act by symplectic ones. - I'm a bit confused here. Isn't this just saying that $i_X \omega$ is closed? – Simon Rose Feb 14 2011 at 22:25 Yes, my bad. Just writing a little too fast. – Ben Webster♦ Feb 14 2011 at 22:35 OK, it's fixed now. As Jose said, there are obstructions to the existence for a moment map for a symplectic action, and one of them is that $i_X\omega$ must be exact, not just closed. – Ben Webster♦ Feb 14 2011 at 22:38 So I probably have to look more at José's answer then. – Simon Rose Feb 14 2011 at 22:39 ... or what you just said. – Simon Rose Feb 14 2011 at 22:39 There is yet another interpretation of the already mentioned obstructions for the existence and uniqueness of momentum maps (moment mappings, moment maps, momentum mappings....) in terms of Poisson geometry. This even generalizes then to Poisson manifolds and makes things perhaps a little bit more transparent: For the existence of a (non-equivariant) momentum map you ask Poisson vector fields to be Hamiltonian, the obstruction therefore lies in the first Poisson cohomology, which is precisely the quotient of Poisson vector field modulo Hamiltonian vector fields. In the symplectic case this quotient becomes canonically isomorphic to the first deRham cohomology as symplecticl vector fields correspond to closed one-forms while Hamiltonian ones correspond to exact one-forms via the musical isomorphism induced by $\omega$. The uniqueness (and also the equivariance) is then controlled by the zeroth Poisson cohomology which are th functions with trivial Hamiltonian vector field, also called the Poisson center. In the symplectic and connected case, these are just the constant functions, but in the general Poisson case this might be a much more interesting cohomology. The advantage of this point of view is, perhaps, the fact that now all obstructions are somehow arising from the same complex: the canonical complex associated to a Poisson structure $\pi$. As a vector space this is just the sections of the exterior algebra of the tangent bundle (i.e. the multivector fields) and the differential is the Schouten bracet with $\pi$. In the symplectic case, this boils down to the deRham complex via the musical isomorphim. You can find a detailed discussion of this in the (by the way very: nice) book of Ana Cannas da Silva and Alan Weinstein on "Geometric Models for Noncommutative Algebras". - 1 A small nitpick is that I would not say that the obstruction lies in the Poisson cohomology. This is an equivariant problem, hence the Lie algebra has to come in somewhere. The obstructions in the da Silva and Weinstein are, respectively, a Lie algebra homomorphism from $\mathfrak{g}$ to the first Poisson cohomology (which, unless I misunderstand, need not be a 1-cocycle in general) and a class in the second cohomology of $\mathfrak{g}$ with values in the zeroth Poisson cohomology. I wonder whether there is perhaps an equivariant Poisson cohomology theory where the obstruction does lie. – José Figueroa-O'Farrill Feb 16 2011 at 10:49 Sorry, I was a little bit sloppy. Concerning the equivariant Poisson cohomology, I think I have seen something like that somewhere. Don't remember, though... :( In any case, this should be rather straightforward to cook up a reasonable Cartan like model for it. – Stefan Waldmann Feb 16 2011 at 14:19 Thanks -- googling I just found this paper by Viktor Ginzburg: arxiv.org/abs/dg-ga/9611002. Having glanced at it very quickly, he seems to apply this to the case of Poisson Lie groups acting on a Poisson manifold and the resulting momentum mapping. – José Figueroa-O'Farrill Feb 16 2011 at 20:13 Oh, yes. That is the one I had in mind. Sorry for not googling myself :) – Stefan Waldmann Feb 17 2011 at 8:27	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 79, "mathjax_display_tex": 6, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9314595460891724, "perplexity_flag": "head"}
http://math.stackexchange.com/questions/142000/bijection-proof?answertab=votes	# Bijection proof I got this on an exam and struggled to complete it, could anyone offer a proof? Thanks! Let $X$ be a finite set. Let $f: X \longrightarrow X$ be a bijection. For $n \in \mathbb{Z}^+$, set $$f^n = \underbrace{f \circ f \circ \cdots \circ f}_{n \text{ times}}.$$ Prove that there exists $m \in \mathbb{Z}^+$ such that $f^m = \mathrm{Id}_X$. - ## 5 Answers There are finitely many bijections, so you must have $f_j=f_k$ for some $j\lt k$. Then consider $f_{k-j}$. - Wouldn't it be better to use superscripts, $f^j$ instead of $f_j$ for the $j$-fold iteration of a function? – Michael Hardy May 7 '12 at 18:28 @Michael, yes. When I answered the question, it had not been posted in TeX, and it had fm and fn. On that shaky basis, I thought OP would be more comfortable with $f_m$ than with $f^m$. Perhaps I should have used $f^m$ anyway. If it bothers you, feel free to edit. – Gerry Myerson May 8 '12 at 3:33 HINT: 1. For each $n$, $f^n=\underbrace{f\circ f\circ\ldots \circ f}_{n}$ is a bijection from $X$ to $X$. 2. Are there infinitely many bijections from a finite set to itself? 3. Pigeonhole principle. - Hint: If you wanted to prove that there exists $m\in\mathbb{N}$ such that $f^m (x)=x$ for just one $x\in X$, how would you do it (remember that $X$ is finite)? Next, repeat the argument with $f^m$ taking the place of $f$ (remember that $(f^m)^n=f^{mn}$). An easy induction on the number of elements in $X$. For me, this is the easiest way to see it. - A simple proof is to use Lagrange's theorem: the set of bijections $X\to X$ is a finite group and hence $f^m=\operatorname{id}$ for say $m=n!$, where $n=\|X\|$. - 1 Gerry's answer is much more elementary. – lhf May 7 '12 at 0:44 In my exam i got to the point where i figured out that there was a limited number of bijections n! but how do you skip from there to the fact that f^m=id?? – Katie May 7 '12 at 0:51 The cyclic subgroup generated by f contains the inverse of f. – Shahab May 7 '12 at 0:57 @Katie, that's what Pete Clark calls Lagrange's Little Theorem: in a finite group of order N, we have $x^N=1$ for every $x$. – lhf May 7 '12 at 1:01 @lhf: Thanks for the plug. Actually though I reserve LLT for the case of a finite commutative group, because in this case one really can carry over the argument from the standard proof of Fermat's Little Theorem. So far as I know -- and I think we've had a good discussion of this point on this site -- there is not a similarly easy combinatorial proof in the non-commutative case. (Right?) – Pete L. Clark May 7 '12 at 4:25 show 1 more comment $$x, f(x), f(f(x)), f(f(f(x), \ldots$$ This list cannot go on forever, since $X$ is finite. So it must reach one that has appeared in the list earlier. Suppose $$f(f(f(f(f(f(f(x))))))),$$ the seventh one, appeared earlier as the third one. Then $$f(f(f(x))) = f(f(f(f(f(f(f(x))))))).$$ Since $f$ is one-to-one, we can cancel: $$x = f(f(f(f(x)))).$$ But that doesn't mean $f\circ f\circ f\circ f$ is the identity. It only means that if you apply it to this one element, $x$, you get back $x$, NOT that if you apply it to every element $y$, you'll get back $y$. So suppose $x$ re-appears in four steps $$x \mapsto f(x) \mapsto f(f(x)) \mapsto f(f(f(x))) \mapsto f(f(f(f(x))))$$ and some other element, $w$, re-appears in six steps: $$f(f(f(f(f(f(w)))))) = w.$$ The smallest common multiple of $4$ and $6$ is $12$, so after $12$ steps, both $w$ and $x$ will re-appear simultaneously. There are finitely many elements of $X$. Find the smallest common multiple of the numbers of steps it takes to get them to re-appear. After that many steps, the all re-appear simultaneously. So $f$ iterated that many times is the identity function on $X$. -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 50, "mathjax_display_tex": 7, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9586036801338196, "perplexity_flag": "head"}
http://mathoverflow.net/questions/7787/intuition-behind-moduli-space-of-curves/7795	Intuition behind moduli space of curves Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) For a genus g compact smooth surface $M$, an algebraic structure is the same as a complex structure is the same as a conformal structure. So the moduli space of smooth curves should be the same as the moduli space of conformal structures on $M$. A conformal structure is an equivalence class of Riemmanian metrics that give the same angle measurements. If I embed $M$ in $\mathbb{R}^3$, I get a metric on $M$. This gives me a conformal structure, hence a point in the moduli space. The moduli space is known to have complex dimension 3g-3 (except for g=0, where the moduli space is a point, and g=1, where the moduli space is 1-dimensional). My question is: can we visualize the 6g-6 real dimensions as deformations of the embedding of $M$ in $\mathbb{R}^3$. In particular, how can we see that small deformations of a 2-sphere are conformally equivalent to the original 2-sphere. - 3 Great question! – Sam Nead Dec 4 2009 at 17:23 1 Although the space of deformations is 6g-6 for g>1, the machinery used to prove that is quite different from the machinery used to prove that all 2-spheres are conformally equivalent. In particular, the latter is one branch of the uniformization theorem and intuition about how that applies to the specific case you ask about, 2-spheres embedded in 3D, probably won't give intuition about the g>1 case without quite a bit of further work. – Dan Piponi Dec 4 2009 at 19:01 Is it even clear what subspace of the space of conformal structures can be realized by surfaces embedded in $\mathbb R^3$? Do all hyperbolic metrics admit conformal embeddings in $\mathbb R^3$? – Ryan Budney Dec 5 2009 at 3:28 1 My intuition is that if you wrinkle the surface, you can basically decouple its conformal class from the macroscopic geometry of the embedding. It might not be easy to develop this idea, and I also have a vague recollection that there has been a paper on this. – Greg Kuperberg Dec 5 2009 at 3:38 5 Answers There is one thing that you can "see" when you have a surface embedded in $R^3$. Namelly, the moduli space of conformal (complex) structures on a surface of genus $g$, $g>0$ is non-compact. If you take in $R^3$ a surface that has a very long cylinder inside of it, such that the circle generated by it on the surface is not contractible, then in the moduli space of complex curves you get a point "close to its boundary". The longer the cylinder close you get to the boundary. En example will be a long and thin torus. Considering such pictures you will be able to viualise at least points of the moduli space that are the most close to the boundary in the case of arbirtary genus. Over thing you could destinguish are complex curves that admit an anti-holomorphic involution - just take a surface in $R^3$ symmetric with respect the $x,y$ plane. As for small deformation of the sphere, unfotrunatelly I don't think it will be any easier to "see" that it is the same as on any other sphere. You will need PDE to prove it. Even to prove that it has a holomorphic sturcture you will need to know that for any metric on a 2-dimesnional surface you have local isotermal coordinates, where it looks as $f(x,y)(dx^2+dy^2)$. ADDED. By Pogorelov's theorem every sphere with positive Gauss curvature can be emdedded isometrically in R^3. So a sphere close to a unite sphere is just a sphere whose curvature is approximatively 1. Knowing this does not make it any easier for you to know that the conformal structure on this sphere is standard, according to what I understand. - You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you. I think sigfpe's comment is correct -- the formula $3g-3$ is only true for $g \geq 2$, and however you can "visualize" or understand these $3g-3$ dimensions for $g \geq 2$ probably won't work or help for $g=0,1$. I'm not sure if this will help, but one way (the only way that I know, but there are others) to "visualize" the $6g-6$ dimensions is using the Fenchel-Nielsen coordinates. Roughly it works like this: You can take your genus $g \geq 2$ surface and cut it along $3g-3$ circles to get a bunch of pairs of pants ($2g-2$ of them). Then $3g-3$ of the $6g-6$ dimensions come from the lengths of the circles; the remaining $3g-3$ dimensions come from how much we "twist" when we identify the pants along the circles. A key fact which makes this work is that, for each specification of the lengths of the boundary circles, there is a unique pair of pants (unique complex structure). If you google "Fenchel-Nielsen coordinates" I am sure that you will find lots of papers with nice pictures illustrating this, and explaining this in more detail. Anyway, you'll notice that you can't do this for $g=0,1$. A sphere can't be cut into pairs of pants, nor can a torus. - This is related to Kevin's answer but goes back to the 19th century. A Riemann surface is planar if a simple closed curve separates it. A Riemann surface of genus g becomes planar after cutting g handles along simple closed curves. Then the classical uniformaization theorem states that it is equivalent to a sphere with 2g slits parallel to the "x axis". Then the location of the slits is determine by the 2g centers (4g real coordinates) and the 2g lengths (2g real coordinates). This give 6g real coordinates, which are then reduced by the 6 real dimensions of the conformal automorphisms of the sphere. Thus Riemann surfaces of genus g have 6g-6 real parameters, plus the number of parameters of the conformal automorphisms of the generic surface of genus g, which adds 6 for g=0 and 2 for g=1. (Alan Mayer explained it this way in a colloquium at Brandeis about 1967.) As usual, induction is the easiest way to see anything, and here one degenerates a Riemann surface to one of lower genus by acquiring a node (shrinking off a loop). Then the lower genus Riemann surface has 3(g-1)-3 complex parameters. the two identified points have 2 parameters, and we add one more for the degeneration direction. (Such degeneration methods occur in Wirtinger's book on theta functions in 1895.) - (EDIT 1: Replaced hand-waving argument in third paragraph with a hopefully less incorrect version) (EDIT 2: Added final paragraph about obtaining all conformal deformations for surfaces other than sphere.) I think it is possible to see the infinitesimal rigidity of the sphere, even if it does involve a PDE as Dmitri says. I think you can also try and see if for other embedded surfaces, all infinitesimal deformations of conformal structure are accounted for by deformations of the embedding in a similar way. For the case of S2, what you want is to do is take a normal vector field V (i.e. infinitesimal change of embedding) and produce a tangent vector field X such that flowing along X gives the same infinitesimal change in conformal structure as flowing along V. This should amount to solving a linear PDE, so as Dmitri says a PDE is definitely involved, but probably not as hard as proving the existence of isothermal coordinates (which from memory is non-linear). For the standard embedding of S2 there can't be too many choices for this linear differential operator given that it has to respect the SO(3)-symmetry. I guess we're looking for a first-order equivariant linear operator from normal vector fields to tangent vector fields. If we identify normal fields with functions then two possible candidates are to take X=grad V or X to be the Hamiltonian flow generated by V. I can't think of any others and probably it's possible to prove these are the only such ones. (Assuming it's elliptic, the symbol of the operator must be an SO(3)-equivariant isomorphism from TS2 to TS2 and there can't be too many choices! Using the metric leads to grad and using the area form leads to the Hamiltonian flow.) Then you just have to decide which one to use. For the case of a general embedded surface $M$, you can ask "is it possible to obtain all deformations of conformal structure by deforming the embedding into R3?" To answer this we can again think of a normal vector field as a function V on the surface. There is a second-order linear differential operator $$D\colon C^\infty(M) \to \Omega^{0,1}(T)$$ which sends a normal vector field to the corresponding infinitesimal change of conformal structure. This operator will factor through the hessian with a homomorphism from `$T^ \otimes T^$` to `$T^{0,1}\otimes T^{1,0}$`. The operator $D$ will not be onto, but what we want to know is whether every cohomology class in $H^{0,1}(T)$ has a representative in the image of $D$. At least, this is how I would try and approach the question; I'm sure there are other methods. - You should have a look into the paper "Constrained Willmore Surfaces" by Bohle, Pinkall, Peters. There they give the formula (easy to derive) that the infinitesimal change of the complex structure $\cdot J$ by an infinitesimal normal variation $\cdot f=u N$ (N being the unit normal of the surface) is given by $2uA^\circ J,$ where $A^\circ$ is the trace free part of the Weingarten operator. Especially, in the case of the round sphere, you see that an infinitesimal normal variation does not change the complex structure. -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 37, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9325879812240601, "perplexity_flag": "head"}
http://mathhelpforum.com/trigonometry/71601-more-trig-pre-cal-identity-proofs.html	# Thread: 1. ## More Trig/Pre-Cal Identity Proofs Sorry to ask again, but I need help with these problems: (secx+1)/tanx = sinx/(1-cosx) I got (1+cosx)/sinx which is the opposite of what I need. tan^4x + tan^2x = sec^4 - sec^2x tan^2x - sin^2x = tan^2x * sin^2x sinx/(1+cosx) + (1+cosx)/sinx = 2cscx I could only figure out about half of my homework and I would ask the teacher, but she isn't at school in the morning for tutoring. 2. Originally Posted by Purpledog100 Sorry to ask again, but I need help with these problems: (secx+1)/tanx = sinx/(1-cosx) I got (1+cosx)/sinx which is the opposite of what I need. (This is good) tan^4x + tan^2x = sec^4 - sec^2x tan^2x - sin^2x = tan^2x * sin^2x sinx/(1+cosx) + (1+cosx)/sinx = 2cscx I could only figure out about half of my homework and I would ask the teacher, but she isn't at school in the morning for tutoring. Now multiply top and bottom by $1- \cos x$ $\frac{1 + \cos x}{\sin x} \cdot \frac{1 - \cos x}{1 - \cos x}$ expand the numerator, use some identites and simplify. 3. (secx+1)/tanx = sinx/(1-cosx) I got (1+cosx)/sinx which is the opposite of what I need. go from there ... $\frac{1+\cos{x}}{\sin{x}} \cdot \frac{1 - \cos{x}}{1 - \cos{x}} =$ $\frac{1 - \cos^2{x}}{\sin{x}(1 - \cos{x})} =$ $\frac{\sin^2{x}}{\sin{x}(1 - \cos{x})} =$ $\frac{\sin{x}}{1 - \cos{x}}$ 4. For the third one: $\tan^2 x -\sin^2 x$ $= \frac{\sin^2 x}{\cos^2 x} - \tan^2 x \cos^2 x$ $= \frac{\sin^2 x}{\cos^2 x} - \frac{\sin^2 x}{\cos^2 x} \cos^2x$ $= \frac{\sin^2 x - \sin^2 x \cos^2 x} {\cos^2 x}$ $= \frac {\sin^2 x (1- \cos^2 x)}{\cos^2 x}$ $= \frac {\sin^2 x}{\cos^2 x} (1- \cos^2 x)$ $= \tan^2 x \sin^2 x$ 5. Originally Posted by danny arrigo Now multiply top and bottom by $1- \cos x$ $\frac{1 + \cos x}{\sin x} \cdot \frac{1 - \cos x}{1 - \cos x}$ expand the numerator, use some identites and simplify. For the second, factor a $\tan^2 x$ from the LHS, i.e. $\tan^2x \left( 1 + \tan^2x \right)$ then use the identity $1 + \tan^2x = \sec^2x$ to eliminate the tan For the forth one, multiply the first term by $\frac{1 - \cos x}{1 - \cos x}$ then follow the idea what skeeter did above. 6. Originally Posted by Amanda H For the third one: $\tan^2 x -\sin^2 x$ $= \frac{\sin^2 x}{\cos^2 x} - \tan^2 x \cos^2 x$ $= \frac{\sin^2 x}{\cos^2 x} - \frac{\sin^2 x}{\cos^2 x} \cos^2x$ $= \frac{\sin^2 x - \sin^2 x \cos^2 x} {\cos^2 x}$ $= \frac {\sin^2 x (1- \cos^2 x)}{\cos^2 x}$ $= \frac {\sin^2 x}{\cos^2 x} (1- \cos^2 x)$ $= \tan^2 x \sin^2 x$ Why did you change $-sin^2x$to $-tan^2x*cos^2x$ For the forth one, multiply the first term by then follow the idea what skeeter did above. I'm sorry, but I still don't understand how to do it. I have: ${sinx(1-cosx)}/sin^2x + (1+cosx)/sinx$ Do I multiply the second fraction by sinx to have equal denominators? 7. It comes from $\frac {\sin x}{\cos x} = \tan x$ rearranged it becomes $\sin x = \tan x \cos x$ 8. Originally Posted by Amanda H It comes from $\frac {\sin x}{\cos x} = \tan x$ rearranged it becomes $\sin x = \tan x \cos x$ this works, but it's overkill. you could go directly from $\frac {\sin^2 x}{\cos^2 x} - \sin^2 x \to \frac {\sin^2 x - \sin^2 x \cos^2 x}{\cos^2 x}$. just combining fractions, as we know $\frac ab \pm \frac cd = \frac {ad \pm bc}{bd}$. from there, it's nice we could also do $\frac {\sin^2 x}{\cos^2 x} - \sin^2 x$ $= \sin^2 x \left( \frac 1{\cos^2 x} - 1 \right)$ $= \sin^2 x \cdot \frac {1 - \cos^2 x}{\cos^2 x}$ $= \sin^2 x \tan^2 x$ @ OP: the nice thing about trig identities is that there are usually several ways to solve them. it is good practice to use your imagination to come up with alternative methods of proving the identities than the ones posted here. just take note of the techniques used, like changing everything to sines and cosines. usually a good way to go. 9. Originally Posted by Amanda H It comes from $\frac {\sin x}{\cos x} = \tan x$ rearranged it becomes $\sin x = \tan x \cos x$ Oh, right. Duh. I had that in my notes and still couldn't figure it out. @ OP: the nice thing about trig identities is that there are usually several ways to solve them. it is good practice to use your imagination to come up with alternative methods of proving the identities than the ones posted here. just take note of the techniques used, like changing everything to sines and cosines. usually a good way to go. Yeah, but I usually pick the wrong ways to solve them. Thanks to everyone for all of their help!! 10. Originally Posted by Purpledog100 Yeah, but I usually pick the wrong ways to solve them. that's the point. as long as your moves are mathematically sound, and you can get from one side to the other, there is no wrong way. a problem here can be done several valid ways. the difference is, one way might be easier or quicker than another. if you do enough practice, this won't matter though, as you will either be able to move so quickly that the extra steps hardly slow you down, or you will naturally be able to spot the easier ways first	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 41, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.952759325504303, "perplexity_flag": "middle"}
http://www.physicsforums.com/showthread.php?t=219310	Physics Forums The de Broglie wavelength The mass of an electron is 9.1110^-31 kg. If the de Broglie wavelength for an electron in an hydrogen atom is 3.3110^-10 m, how fast is the electron moving relative to the speed of light? The speed of light is 3.0010^8 m/s. here's what I did: i solved for velocity=6.62610^-34J/(9.1110^-31kg)(3.3110^-10) v=2.197410^-74 and i tried to gain the percent by dividing the speed of light by velocity. where did i go wrong? PhysOrg.com science news on PhysOrg.com >> Front-row seats to climate change>> Attacking MRSA with metals from antibacterial clays>> New formula invented for microscope viewing, substitutes for federally controlled drug Mentor Your calculation is wrong. Don't just blindly do calculations. Think...does the number your calculator has spewed out actually make any sense? When it's something ridiculous like 10^-74 m/s, the answer is emphatically NO. Kind of slow for a particle, don't you think? I get v = (0.00732)c Quote by plstevens The mass of an electron is 9.1110^-31 kg. If the de Broglie wavelength for an electron in an hydrogen atom is 3.3110^-10 m, how fast is the electron moving relative to the speed of light? The speed of light is 3.0010^8 m/s. here's what I did: i solved for velocity=6.62610^-34J/(9.1110^-31kg)(3.3110^-10) v=2.197410^-74 and i tried to gain the percent by dividing the speed of light by velocity. where did i go wrong? how did u calculate.... watch the exponents first...... the magnitude is 10^7 m/s... $$\frac{6}{9\times3}\times\frac{10^{-34}}{10^{-31}\times10^{-10}}\approx\frac{2}{9}10^7 m/s$$ this suggest us that it is better to treat the electron relativistically if we want to penetrate deep in its properties... regards marco The de Broglie wavelength thanx Dirac :) so hows do i get the percentage here's what i'm doing: 3.0010^8 m/s /100 = 0.00732/x. x=2.410^8, but i know this isn't right so, what shall i do? Mentor I'm not sure what percentage you are talking about, since it's not mentioned in the original post. For the velocity of the particle, I get: $$v = 2.197 \times 10^6 \ \ \ \frac{\textrm{m}}{\textrm{s}}$$ The question asks how fast the particle is moving relative to the speed of light. Well, their ratio is $$\frac{v}{c} = \frac{2.197 \times 10^6 \ \ \ \textrm{m/s}}{3.00 \times 10^8 \ \ \ \textrm{m/s}} = 0.00732$$ So, expressed in units of the speed of light, the velocity is $$v = 0.00732c$$ The particle is moving at 0.00732 times the speed of light. Obviously, as a percentage, that's 0.732%. So I guess if you wanted to, you could say that the particle is moving at 0.732% of the speed of light. It's a completely equivalent statement though. It doesn't add any extra meaning. Thread Tools \| \| \| \| \|------------------------------------------------\|-------------------------------\|---------\| \| Similar Threads for: The de Broglie wavelength \| \| \| \| Thread \| Forum \| Replies \| \| \| Introductory Physics Homework \| 11 \| \| \| Quantum Physics \| 2 \| \| \| Introductory Physics Homework \| 2 \| \| \| Quantum Physics \| 7 \| \| \| Quantum Physics \| 4 \|	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 4, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.8955897688865662, "perplexity_flag": "middle"}
http://mathhelpforum.com/geometry/159571-area-circles-equilateral-triangle.html	# Thread: 1. ## Area - circles and equilateral triangle. 1. Find the area of the shaded region. this triangle is equilateral all sides are 2. Thanks for helping! 2. You have an equilateral triangle with sides of length 2. If you drop a perpendicular from one vertex of the triangle to the opposite side, you divide the triangle into two right triangles with hypotenuse 2 and one leg 1. You can use the Pythagorean theorem to determine that $x^2+ 1^2= 2^2$, $x^2= 4- 1= 3$, $x= \sqrt{3}$: the other leg has length $\sqrt{3}$, the altitude of the triangle. From that you can calculate the area of the triangle. Each of the three circles has radius 1 and so area $\pi$. Each angle of the triangle is 60 degrees which is 60/360= 1/6 of the complete circle. That is, each of the sectors of a circle has area $\frac{\pi}{6}$. The shaded area is the area of triangle minus the three sector areas.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 6, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.8916518092155457, "perplexity_flag": "head"}
http://mathhelpforum.com/math-challenge-problems/155005-units-polynomial-ring.html	# Thread: 1. ## Units in a polynomial ring Challenge question: Let $R$ be a commutative ring and $f(x)=a_0+...+a_nx^n \in R[x]$ then $f$ is a unit if and only if $a_0$ is a unit and $a_1,...a_n$ are nilpotent. I have a solution using more involved ring theory, but I'm really interested to see more elementary solutions (or even more involved or indirect, why not). Moderator approved CB 2. see my post in here.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 5, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.928043782711029, "perplexity_flag": "head"}
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I just want to know generally, what are the most common methods of model calibration(such as Black-Scholes Model, Stochastic ...	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 19, "mathjax_display_tex": 2, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.929500937461853, "perplexity_flag": "middle"}
http://math.stackexchange.com/questions/178054/what-does-f-is-a-function-on-s-mean/178076	# What does “$f$ is a function on $S$” mean? If somebody says "$f$ is a function on $S$", what do they mean? Does it mean that $S$ is the domain of the function, the codomain, or both? - 4 $S$ is a domain. The codomain is any superset of the image of $S$ under $f$. – Jayesh Badwaik Aug 2 '12 at 14:49 4 In what context? If I saw that alone I'd think it meant $f:S \to \mathbb{R}$, but the codomain might be something else depending on context. – Clive Newstead Aug 2 '12 at 14:50 I am guessing the exact expression of $f$ will provide image. – Jayesh Badwaik Aug 2 '12 at 14:53 2 I'd think that a function "on" $S$ is $f: S \to S,$ (no necessarily total or surjective) but I will leave it up to the more experienced users. – user2468 Aug 2 '12 at 14:55 @J.D.: What comes to my mind, probably because it's the only context in which I've seen this terminology used, is measure theory, where "let $f$ be a function on the measure space $E$" would mean "let $f : E \to \mathbb{R}$ be a measurable function". Of course, there are no doubt other contexts in which the meaning is different. – Clive Newstead Aug 2 '12 at 15:23 show 1 more comment ## 2 Answers When dealing with functions, one usually specifies "$f$ is a function from $X$ to $Y$" or "a function on $X$ into $Y$", or $f$ is a function $f:X\to Y$. In particular, a function $f$ is a subset of the cartesian product $X\times Y$ , that is, a relation from $X$ into $Y$, such that for every $x\in X$ there is a $y\in Y$ for which $(x,y)\in f$, and whenever $(x,y)\in f$ and $(x,y')\in f$ it follows that $y=y'$. That is, to every $x\in X$ there corresponds a unique $y\in Y$ such that $(x,y)\in f$. Note it can happen that for two different $x,x'\in X$, the pairs $(x',y)$ and $(x,y)$ are in $f$. We usually call $X$ the domain and $Y$ codomain, and we call the image of $A\subset X$ under $f$ to the subset $f(A)\subset Y$ whose elements are the points $y=f(x)$ with $x\in X$, or more precisely, to the subset $f(A)$ of points in $Y$ such that $(x,y)\in f$ for $x\in A$. As an example, let $f$ be a function $f:[-4,4]\to \Bbb R$ such that $(x,y)\in f$ whenever $y=x^2$. Then $f([-1,1])=[0,1]$ and $f([-4,0]) =[0,16]$ Now, when it happens that "$f$ is a function from (on) $X$ to (into) $X$" or $f:X\to X$, we might say simply that "$f$ is a function on $X$" (recall how a relation $R$ is said to be a relation on $X$ when boths its range and domain are the same set). This means $f$ is a subset of the product $X\times X$ with the aforementioned conditions. In this case the both the domain and codomain are $X$, and the image of $X$ under $f$ will be a subset of $X$ itself. As J.D. noted, this constraint can produce some problems and should be handled carefully: consider the correspondence $f$ such that $(x,y)\in f$ whenever $y=x^2$, on the set $[0,10]$. It is immediate that whenever we choose $x>\sqrt 10$ we wont be able to find an $y$ to map $x$ to. This misbehaviour is given a name, namely, we call $f$ a partial function from $X$ to $Y$ if $f$ is a function $f:X'\to Y$ such that ${\rm dom }\, f=X'\subset X$, that is, not every element of $X$ is mapped into $Y$. Note that $f:X'\to Y$ is a "good" function in its own right, and it is the worded specification of $X$ that is problematic. As Nate has commented, it might also happen that we say $f$ is a function on $X$ when the codomain $Y$ is understood, such as $\mathbb C$ or $\Bbb R$, and we need only to specify what the domain is. The word mapping is used as a synonym with function in nonspecific contexts, but might be given a special definition in other areas, as he notes. - 3 I wouldn't agree that this "usually" means a function from $X$ into $X$. For instance, in real analysis, "a function on $X$" usually means a function from $X$ into $\mathbb{R}$ (or perhaps $\mathbb{C}$). Often $X$ is some abstract space; in this case functions from $X$ to itself are much less commonly considered, and often called "mappings". – Nate Eldredge Aug 2 '12 at 15:40 @NateEldredge I'll try to rewrite this. Now I have to go. That is what I learned it means. I have that mapping is a synonym with function, also. – Peter Tamaroff Aug 2 '12 at 15:43 @NateEldredge Tell me what you think. – Peter Tamaroff Aug 2 '12 at 16:23 I like it! ${}$ – Nate Eldredge Aug 2 '12 at 16:42 I have never heard "function on X" to mean "function from X to itself", but always something like a function whose domain is X and whose values are something like numerical, for example, I heard it use for real valued functions, complex valued functions, and sections of structure sheaves in algebraic or analytic geometry. – Omar Aug 2 '12 at 17:01 show 1 more comment $S$ is the domain. This sentence doesn't specify the codomain of $f$; it would have to be determined from context. -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 103, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9451043605804443, "perplexity_flag": "head"}
http://nrich.maths.org/773/note	### Cubic Spin Prove that the graph of f(x) = x^3 - 6x^2 +9x +1 has rotational symmetry. Do graphs of all cubics have rotational symmetry? ### Sine Problem In this 'mesh' of sine graphs, one of the graphs is the graph of the sine function. Find the equations of the other graphs to reproduce the pattern. ### More Parabolic Patterns The illustration shows the graphs of twelve functions. Three of them have equations y=x^2, x=y^2 and x=-y^2+2. Find the equations of all the other graphs. # Parabolic Patterns ### Why dothis problem? In this problem, instead of giving the equations of some functions and asking learners to sketch the graphs, this challenge gives the graphs and asks them to find their equations. This encourages learners to experiment by changing the equations systematically to discover the effect on the graphs. ### Possible approach Start by showing the picture and asking learners to identify the graphs of $y=x^2$ and $y=-(x-4)^2$. Encourage discussion about the similarities and differences between these two graphs, and their equations. Give learners time to experiment with graphical calculators or graphine software with the aim of idenitfying rules for modifying equations to transform graphs and create those on the image. Draw ideas together with the purpose of defining a set of rules that can be applied in other similar situations. A nice extension is for pairs to create a picture of their own, which they then challenge another pair to reproduce. Those learners who are familiar with the idea of completing the square may wish to write the equation of each parabola in this form and consider the coordinates of the vertex. You can read about one teacher's experience of using this task in the classroom. ### Key questions • You are being asked to sketch a family of graphs. What makes this a family? • What is the same and what is different about the equations $y=x^2$ and $y=-(x-4)^2$? • How might these similarities and differences relate to the way they look and their positions on the axes? • Can you convince us that the rules you have found will work with graphs of other functions. ### Possible extension More Parabolic Patterns and Parabolas again offer similar pictures to reproduce. Cubics uses graphs of cubic functions, and Ellipses gives the opportunity to investigate the equation of an ellipse. ### Possible Support Learners could begin by investigating translation of straight lines and look at how the equations change. The NRICH Project aims to enrich the mathematical experiences of all learners. To support this aim, members of the NRICH team work in a wide range of capacities, including providing professional development for teachers wishing to embed rich mathematical tasks into everyday classroom practice. More information on many of our other activities can be found here.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 4, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9205687642097473, "perplexity_flag": "middle"}
http://mathoverflow.net/questions/6033?sort=oldest	## Inverses in convolution algebras ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) Let $G$ be a locally compact totally disconnected group, and to make life easy let's suppose its Haar measure is bi-invariant. Let $C_c(G)$ be the space of locally constant complex functions on $G$ with compact support, which forms an algebra under convolution. Suppose $e \in C_c(G)$ is an idempotent, so that $H = eC_c(G)e$ is an algebra with identity. Is it now true that if $f \star g = e$ in $H$, then $g \star f = e$ as well? This may be too general to be true, so to be more specific: suppose $K < G$ is a compact subgroup such that $K\backslash G/K$ is countable, suppose $\phi : G \to \mathbf{C}^{\times}$ is a character, and suppose the idempotent $e$ is the function with support in $K$ such that $e(x) = \phi(x^{-1})/\mu(K)$ for $x \in K$. Now is it true that if $f \star g = e$ in $H$ then $g \star f = e$ as well? [I am reading something that claims some $f$ is a unit but then checks it by checking the existence of $g$ such that $f \star g = e$. So really, the question is whether it follows by some general business that $g \star f = e$, or whether one has to do another computation to check the other direction.] - Daft/ignorant question: does the countability hypothesis on the double coset space force $K$ to be open in $G$? – Yemon Choi Nov 19 2009 at 1:12 I certainly intended compact open subgroup, at any rate. – D. Savitt Nov 19 2009 at 1:59 1 Returning to this after a long gap, just in case you have any residual interest: it turns out that on any unimodular $G$, if you have a and b in C_c(G) such that $ab=a+b$, then $ba=a+b$. So now take $a=e-f$ and $b=e-g$, with $e,f,g$ as in your question, then this seems to answer your question in the positive. See Theorem 3.6 of arxiv.org/abs/1205.4354 – Yemon Choi Jul 1 at 22:26 ## 3 Answers I don't have a solution, but here are some thoughts which might be of use or interest. You may have seen this already, but if your group is discrete then its group von Neumann algebra $VN(G)$ is "directly finite" - that is, every left invertible element is invertible. I think this property is inherited by the algebra obtained when one compresses by an idempotent in $C_c(G)$. The earliest reference I know of is somewhere in Kaplansky's Fields and Rings; a proof of something slightly weaker, which can in fact be boosted to prove the original result, was given in Montgomery, M. Susan. Left and right inverses in group algebras. Bull. Amer. Math. Soc. 75 1969 539--540. MR0238967 (39 #327) (The proof uses the existence of a faithful tracial state on $VN(G)$, plus the fact that every idempotent in a $C^$-algebra is similar in the algebra to a self-adjoint idempotent -- something which was not all that obvious to me the first time I saw this result.) I don't know what the state of play is for algebras of the form $H$, as described in your question. I think enough is known about $C^$-algebras of some totally disconnected groups (work of Plymen et al.) that one might have similar results, but the arguments have to be different from the discrete case because one no longer has the faithful positive trace that is used by Kaplansky and Montgomery's arguments. Of course in the more special case you have at hand, we might have enough structure to force left-invertibles to be right-invertible; but off the top of my head nothing comes to mind. - 1 I checked with the author of what I was reading, who confirms that it was a mistake -- they should have been checking invertibility in both directions. (Well, in fact it was only the left-invertibility of $f$ that was actually being used subsequently.) So although Yemon's comment doesn't answer the question, it may be as helpful as we're going to get. – D. Savitt Nov 19 2009 at 17:07 For a final answer, see Yemon's second comment to the question. – D. Savitt Jul 3 at 23:56 ### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you. I think the general problem with infinite matrices (that are row-column-finite) arises from the fact that the shift matrix ei -> ei+1 (for i ranging over positive integers) only has a left inverse. - Duh. Good call. Editing. – D. Savitt Nov 18 2009 at 23:43 Hey DLS. The finiteness assumptions you put on the matrices mean that they can be interpreted as linear maps from an infinite direct sum of copies of k to itself. Call this spae X. So now your question seems to be "if f,g:X-->X and fg=id, is gf=id?" and now it's easy to see counterexamples: e.g. imagine one of the matrices sends e_i to e_{i+1} and the other sends e_i to e_{i-1} and kills e_1. - Hi! Yes, let's move on to the non-trivially-false part of the question. – D. Savitt Nov 18 2009 at 23:47 To clarify: this was an answer to warm-up part (0) of the question, which has since been deleted. I don't know the answer to the question in its current form. – Kevin Buzzard Nov 19 2009 at 21:53	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 37, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.953510046005249, "perplexity_flag": "head"}
http://mathhelpforum.com/differential-geometry/146991-prove-mathematical-induction.html	# Thread: 1. ## Prove By Mathematical Induction I'm stuck on this question: State the principle of mathematical induction, and use it to prove that for $n \geq 4$ $4n! > 2^{n+2}$ So far I've done this: $P(n)$ is a statement for each natural number $n$, where $n \geq 4$. $P(n)$ is true if: • P(n) is true for n = 4. • If it is true for arbitrary n, it is also true for n + 1. For $n = 4$, $4(4!) > 2^{6}$, $96 > 64$ hence true. For $n + 1$: $4(n + 1)! > 2^{n + 3}$ $(n + 1)! > 2^{n + 1}$ $n!(n+1) > 2^{n}(n+1)$ (Using hypothesis) And that's where I get stuck. RHS should be $2^{n+1}$, but that only occurs when n = 1. 2. Originally Posted by splbooth I'm stuck on this question: State the principle of mathematical induction, and use it to prove that for $n \geq 4$ $4n! > 2^{n+2}$ So far I've done this: $P(n)$ is a statement for each natural number $n$, where $n \geq 4$. $P(n)$ is true if: • P(n) is true for n = 4. • If it is true for arbitrary n, it is also true for n + 1. For $n = 4$, $4(4!) > 2^{6}$, $96 > 64$ hence true. For $n + 1$: $4(n + 1)! > 2^{n + 3}$ $(n + 1)! > 2^{n + 1}$ $n!(n+1) > 2^{n}(n+1)$ (Using hypothesis) And that's where I get stuck. RHS should be $2^{n+1}$, but that only occurs when n = 1. you can try to prove whether or not $4(n+1)!>2^{(n+1)+2}$ if $4n!>2^{n+2}$ P(k) $4k!>2^{k+2}$ P(k+1) $4(k+1)!>2^{(k+1)+2}$ Proof $4(k+1)!=4(k+1)k!=(k+1)4k!$ If $k\ge4$ then $(k+1)4k!\ge(5)4k!$ $2^{k+3}=(2)2^{k+2}$ Then if $4k!>2^{k+2}$ $4(k+1)!>2^{k+3}$ certainly since $(5)4k!>(2)2^{k+2}$ 3. That's great, thanks	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 39, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.8784508109092712, "perplexity_flag": "head"}
http://sagemath.org/doc/tutorial/tour_linalg.html	# Linear Algebra¶ Sage provides standard constructions from linear algebra, e.g., the characteristic polynomial, echelon form, trace, decomposition, etc., of a matrix. Creation of matrices and matrix multiplication is easy and natural: ```sage: A = Matrix([[1,2,3],[3,2,1],[1,1,1]]) sage: w = vector([1,1,-4]) sage: wA (0, 0, 0) sage: Aw (-9, 1, -2) sage: kernel(A) Free module of degree 3 and rank 1 over Integer Ring Echelon basis matrix: [ 1 1 -4] ``` Note that in Sage, the kernel of a matrix $$A$$ is the “left kernel”, i.e. the space of vectors $$w$$ such that $$wA=0$$. Solving matrix equations is easy, using the method solve_right. Evaluating A.solve_right(Y) returns a matrix (or vector) $$X$$ so that $$AX=Y$$: ```sage: Y = vector([0, -4, -1]) sage: X = A.solve_right(Y) sage: X (-2, 1, 0) sage: A * X # checking our answer... (0, -4, -1) ``` A backslash \ can be used in the place of solve_right; use A \ Y instead of A.solve_right(Y). ```sage: A \ Y (-2, 1, 0) ``` If there is no solution, Sage returns an error: ```sage: A.solve_right(w) Traceback (most recent call last): ... ValueError: matrix equation has no solutions ``` Similarly, use A.solve_left(Y) to solve for $$X$$ in $$XA=Y$$. Sage can also compute eigenvalues and eigenvectors: ```sage: A = matrix([[0, 4], [-1, 0]]) sage: A.eigenvalues () [-2I, 2I] sage: B = matrix([[1, 3], [3, 1]]) sage: B.eigenvectors_left() [(4, [ (1, 1) ], 1), (-2, [ (1, -1) ], 1)] ``` (The syntax for the output of eigenvectors_left is a list of triples: (eigenvalue, eigenvector, multiplicity).) Eigenvalues and eigenvectors over QQ or RR can also be computed using Maxima (see Maxima below). As noted in Basic Rings, the ring over which a matrix is defined affects some of its properties. In the following, the first argument to the matrix command tells Sage to view the matrix as a matrix of integers (the ZZ case), a matrix of rational numbers (QQ), or a matrix of reals (RR): ```sage: AZ = matrix(ZZ, [[2,0], [0,1]]) sage: AQ = matrix(QQ, [[2,0], [0,1]]) sage: AR = matrix(RR, [[2,0], [0,1]]) sage: AZ.echelon_form() [2 0] [0 1] sage: AQ.echelon_form() [1 0] [0 1] sage: AR.echelon_form() [ 1.00000000000000 0.000000000000000] [0.000000000000000 1.00000000000000] ``` For computing eigenvalues and eigenvectors of matrices over floating point real or complex numbers, the matrix should be defined over RDF (Real Double Field) or CDF (Complex Double Field), respectively. If no ring is specified and floating point real or complex numbers are used then by default the matrix is defined over the RR or CC fields, respectively, which do not support these computations for all the cases: ```sage: ARDF = matrix(RDF, [[1.2, 2], [2, 3]]) sage: ARDF.eigenvalues() [-0.0931712199461, 4.29317121995] sage: ACDF = matrix(CDF, [[1.2, I], [2, 3]]) sage: ACDF.eigenvectors_right() [(0.881845698329 - 0.820914065343I, [(0.750560818381, -0.616145932705 + 0.238794153033I)], 1), (3.31815430167 + 0.820914065343I, [(0.145594698293 + 0.37566908585I, 0.915245825866)], 1)] ``` ## Matrix spaces¶ We create the space $$\text{Mat}_{3\times 3}(\QQ)$$ of $$3 \times 3$$ matrices with rational entries: ```sage: M = MatrixSpace(QQ,3) sage: M Full MatrixSpace of 3 by 3 dense matrices over Rational Field ``` (To specify the space of 3 by 4 matrices, you would use MatrixSpace(QQ,3,4). If the number of columns is omitted, it defaults to the number of rows, so MatrixSpace(QQ,3) is a synonym for MatrixSpace(QQ,3,3).) The space of matrices has a basis which Sage stores as a list: ```sage: B = M.basis() sage: len(B) 9 sage: B[1] [0 1 0] [0 0 0] [0 0 0] ``` We create a matrix as an element of M. ```sage: A = M(range(9)); A [0 1 2] [3 4 5] [6 7 8] ``` Next we compute its reduced row echelon form and kernel. ```sage: A.echelon_form() [ 1 0 -1] [ 0 1 2] [ 0 0 0] sage: A.kernel() Vector space of degree 3 and dimension 1 over Rational Field Basis matrix: [ 1 -2 1] ``` Next we illustrate computation of matrices defined over finite fields: ```sage: M = MatrixSpace(GF(2),4,8) sage: A = M([1,1,0,0, 1,1,1,1, 0,1,0,0, 1,0,1,1, ... 0,0,1,0, 1,1,0,1, 0,0,1,1, 1,1,1,0]) sage: A [1 1 0 0 1 1 1 1] [0 1 0 0 1 0 1 1] [0 0 1 0 1 1 0 1] [0 0 1 1 1 1 1 0] sage: rows = A.rows() sage: A.columns() [(1, 0, 0, 0), (1, 1, 0, 0), (0, 0, 1, 1), (0, 0, 0, 1), (1, 1, 1, 1), (1, 0, 1, 1), (1, 1, 0, 1), (1, 1, 1, 0)] sage: rows [(1, 1, 0, 0, 1, 1, 1, 1), (0, 1, 0, 0, 1, 0, 1, 1), (0, 0, 1, 0, 1, 1, 0, 1), (0, 0, 1, 1, 1, 1, 1, 0)] ``` We make the subspace over $$\GF{2}$$ spanned by the above rows. ```sage: V = VectorSpace(GF(2),8) sage: S = V.subspace(rows) sage: S Vector space of degree 8 and dimension 4 over Finite Field of size 2 Basis matrix: [1 0 0 0 0 1 0 0] [0 1 0 0 1 0 1 1] [0 0 1 0 1 1 0 1] [0 0 0 1 0 0 1 1] sage: A.echelon_form() [1 0 0 0 0 1 0 0] [0 1 0 0 1 0 1 1] [0 0 1 0 1 1 0 1] [0 0 0 1 0 0 1 1] ``` The basis of $$S$$ used by Sage is obtained from the non-zero rows of the reduced row echelon form of the matrix of generators of $$S$$. ## Sparse Linear Algebra¶ Sage has support for sparse linear algebra over PIDs. ```sage: M = MatrixSpace(QQ, 100, sparse=True) sage: A = M.random_element(density = 0.05) sage: E = A.echelon_form() ``` The multi-modular algorithm in Sage is good for square matrices (but not so good for non-square matrices): ```sage: M = MatrixSpace(QQ, 50, 100, sparse=True) sage: A = M.random_element(density = 0.05) sage: E = A.echelon_form() sage: M = MatrixSpace(GF(2), 20, 40, sparse=True) sage: A = M.random_element() sage: E = A.echelon_form() ``` Note that Python is case sensitive: ```sage: M = MatrixSpace(QQ, 10,10, Sparse=True) Traceback (most recent call last): ... TypeError: __classcall__() got an unexpected keyword argument 'Sparse' ``` Basic Rings Polynomials ### Quick search Enter search terms or a module, class or function name.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 12, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.8314211368560791, "perplexity_flag": "middle"}
http://mathhelpforum.com/geometry/130740-find-area-shaded-region.html	# Thread: 1. ## Find the area of the shaded region... In the given fig., AB and AC are two equal chords of length 7cm and ∟BAC=60 degrees. Find the area of the shaded region A,C,B in clockwise direction. Attached Thumbnails 2. The problem would be easy if you knew the radius of the circle. To find the radius, draw a radius from the center to A and to B. Clearly this is a 30, 30, 120 triangle. Since you know AB=7, the law of cosines can get you the radius. Once you have r, you have 2 triangles and a section of the circle of degree measure 120 to add up. 3. Originally Posted by snigdha In the given fig., AB and AC are two equal chords of length 7cm and ∟BAC=60 degrees. Find the area of the shaded region A,C,B in clockwise direction. Hi snighda, Since AB and AC are equal, the geometry is symmetrical with respect to the centre of the circle. if you join all 3 vertices A, B, C to the circle centre O, you will have two isosceles triangles ABO and ACO, with opposite angles of 60/2 = 30 degrees and obtuse angle 180-2(30) = 120 degrees and a segment, with a 60(2) = 120 degree angle at O. If you use the Sine Rule for one of the isosceles triangles, then you can find the radius, easily solving for segment and triangle areas. The sum of all the areas of all three regions is the shaded area. $\frac{R}{Sin30^o}=\frac{7}{Sin120^o}$ $R=\frac{7Sin30^o}{Sin120^o}$	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 2, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.8828344941139221, "perplexity_flag": "middle"}
http://math.stackexchange.com/questions/169693/a-mapsto-log-left-lvert-f-lvert-1-a-right-is-a-convex-map	# $a\mapsto \log\left(\lVert f\lVert_{1/a}\right)$ is a convex map Proof for $a\in (0,1)$ that $a\to F(a)= \log\left(\lVert f\lVert_{1/a}\right)$ is a convex map, if $f\in L^p(X)$ for all $p\geq1$ in some measure space $(X,\mu)$. I'd like to prove that for all $a_1,a_2\in(0,1)$ and $t\in[0,1]$ the following holds: \begin{align} F(t a_1+(1-t)a_2)\leq t F(a_1)+(1-t)F(a_2)\end{align} using the properties of the logarithm yields $F(a)=a \log\left(\int_X \lvert f\lvert^{1/a}\right)$. Plugging this in the above inequality reads \begin{align} (t a_1+(1-t)a_2)\log\left( \int_X \lvert f\lvert^{1/(t a_1+(1-t)a_2)}\right)\leq\\ t a_1 \log\left( \int_X \lvert f\lvert^{1/a_1}\right)+(1-t)a_2 \log\left( \int_X \lvert f\lvert^{1/a_2}\right)\end{align} Is the $\log$ on the LHS dominated by those on the RHS? I don't know why? How can I continue? Apply Hölder? - ## 1 Answer Let $a_1, a_2\in (0,1)$ and $t\in(0,1)$, then by Hölder inequality with $$p=\frac{ta_1+(1-t)a_2}{ta_1}\quad\text{ and }\quad q=\frac{ta_1+(1-t)a_2}{(1-t)a_2}$$ we get $$\int\limits_X \|f\|^{\frac{1}{ta_1+(1-t)a_2}}d\mu= \int\limits_X \|f\|^{\frac{t}{ta_1+(1-t)a_2}}\|f\|^{\frac{1-t}{ta_1+(1-t)a_2}}d\mu \leq \left(\int\limits_X \left(\|f\|^{\frac{t}{ta_1+(1-t)a_2}}\right)^{\frac{ta_1+(1-t)a_2}{ta_1}}d\mu\right)^{\frac{ta_1}{ta_1+(1-t)a_2}}\left(\int\limits_X \left(\|f\|^{\frac{1-t}{ta_1+(1-t)a_2}}\right)^{\frac{ta_1+(1-t)a_2}{(1-t)a_2}}d\mu\right)^{\frac{(1-t)a_2}{ta_1+(1-t)a_2}}= \left(\int\limits_X \|f\|^{\frac{1}{a_1}}d\mu\right)^{\frac{ta_1}{ta_1+(1-t)a_2}}\left(\int\limits_X \|f\|^{\frac{1}{a_2}}d\mu\right)^{\frac{(1-t)a_2}{ta_1+(1-t)a_2}}$$ Hence $$\left(\int\limits_X \|f\|^{\frac{1}{ta_1+(1-t)a_2}}d\mu\right)^{ta_1+(1-t)a_2}\leq \left(\int\limits_X \|f\|^{\frac{1}{a_1}}d\mu\right)^{ta_1}\left(\int\limits_X \|f\|^{\frac{1}{a_2}}d\mu\right)^{(1-t)a_2}$$ After taking lorarithms we see that the this inequality is equivalent to $$F(t a_1+(1-t)a_2)\leq tF(a_1)+(1-t)F(a_2)$$ Hence $F$ is convex. -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 13, "mathjax_display_tex": 4, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.8236624598503113, "perplexity_flag": "head"}
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It is about how to implement simple codes in Mathematica efficiently exploiting Mathematica's programming ...	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 11, "mathjax_display_tex": 2, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.8876099586486816, "perplexity_flag": "middle"}
http://mathoverflow.net/questions/70451/g-bundles-in-classical-mechanics/70458	## G-bundles in classical mechanics ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) The paper Geometry of the Prytz Planimeter described a mechanical instrument whose configuration space is an $S^1$-bundle with an $SU(1,1)$ action. That paper goes on to study the holonomies of various paths in the base space. Motion is described by a connection on this bundle. I am interested because it's an example of a $G$-bundle appearing in classical mechanics. Are there other explicit classical mechanical systems that engineers study that exhibit the basic concepts of differential geometry so lucidly? - ## 5 Answers Another possible example is the falling cat problem which has been studied among the others by Richard Montgomery. Cf.Gauge Theory of the falling cat - 1 A wonderful quote from the article. "The main point of these earlier works is that a dictionary can be developed between the gauge theory of the physicist's and mathematicians, and the problems occuring in the orientation control of deformable bodies. Briefly, in this dictionary the space of shapes of the body plays the role of the base space, or space-time in the physicist's gauge theory. Its tangent space is the space of controls. The state space, or configuration space of the body, is principal bundle of the theory. ... – Tom LaGatta Dec 29 at 8:11 ... The gauge group is the group of rigid reorientations of the body. The gauge field summarizes the condition that the angular momentum be zero." – Tom LaGatta Dec 29 at 8:11 ### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you. Another couple of examples are the sphere rolling on the plane (or any surface, for that matter) without twisting or slipping, which is described by a connection on a principal SO(3)-bundle over the surface, and the motion of a trailer (or series of hitched trailers) being pulled by a tractor, which is described by a PSL(2,R)-connection (or a product of these) over the space of positions of the tractor. (This latter example is probably closely related to the planimeter problem described in Foote's article.) - Take the planar three-body problem. Or, said a bit differently, take that 'cat' to consist of three point masses moving about in the plane -- a triangle! Fix the center of the mass at the origin by the usual trick. Take G = SO(2). Now take the quotient and you get the cone over the usual Hopf fibration. The points of the sphere in the base spacerepresent oriented similarity classes of triangles. This geometry is at the heart of much modern understanding of the planar three body problem. You can find references in my 2000 paper with Chenciner `A remarkable periodic solution of the three-body problem in the case of equal masses` and the geometry explained in some detail in the 1st few pages and in the appendix to my 1996 paper `The geometric phase of the three-body problem`. You can download these from http://count.ucsc.edu/~rmont/papers/list.html - Take the planar three-body problem. Or, said a bit differently, take that 'cat' to consist of three point masses moving about in the plane -- a triangle! Fix the center of the mass at the origin by the usual trick. Take G = SO(2). Now take the quotient and you get the cone over the usual Hopf fibration. The points of the sphere in the base spacerepresent oriented similarity classes of triangles. This geometry is at the heart of much modern understanding of the planar three body problem. You can find references in my 2000 paper with Chenciner ```A remarkable periodic solution of the three-body problem in the case of equal masses' and the geometry explained in some detail in the 1st few pages and in the appendix to my 1996 paper ```The geometric phase of the three-body problem'. You can download these from http://count.ucsc.edu/~rmont/papers/list.html - Just to record what I heard from another source, Geometric Control Theory has some promise. Many examples from classical mechanics examined from the point of view of connections on their phase space. The text is Geometric Control of Mechanical Systems by Francesco Bullo and Andrew D. Lewis. -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 3, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9246286749839783, "perplexity_flag": "middle"}
http://sciencehouse.wordpress.com/2010/08/06/bayesian-model-comparison/	Scientific Clearing House Carson C. Chow Bayesian model comparison This is the follow up post to my earlier one on the Markov Chain Monte Carlo (MCMC) method for fitting models to data. I really should have covered Bayesian parameter estimation before this post but as an inadvertent demonstration of the simplicity of the Bayesian approach I’ll present these ideas in random order. Although it won’t be immediately apparent in this post, the MCMC is the only computational method you ever really need to learn to do all Bayesian inference. It often comes up in biology and other fields that you have several possible models that could explain some data and you want to know which model is best. The first thing you could do is to see which model fits the data best. However, if a model has more parameters then certainly it will fit the model better. You could always have a model with as many parameters as data points and that would fit the data perfectly. So, it is necessary to balance how good you fit the data with the complexity of the model. There have been several proposals to address this issue. For example, the Akaike Information Criterion (AIC) or Bayes Information Criterion (BIC). However, as I will show here, these criteria are just approximations to Bayesian model comparison. So suppose you have a set of data, which I will denote by D. This could be any set of measurements over any domain. For example it could be a time series of points such as the amount of glucose in the blood stream after a meal measured at each minute. We then have some proposed models that could fit the data and we want to know which is the best model. The model could be a polynomial, a differential equation or anything. Let’s say we have two models M1 and M2. The premise behind Bayesian inference is that anything can be assigned a probability so what we want to assess is the probability of a model given the data D. The model with the highest probability is then the most likely model. So we want to evaluate P(M1 \|D) and compare it to P(M2\|D). A frequentist could not do this because only random variables can be assigned probabilities and in fact there is no systematic way to evaluate models using frequentist statistics. Now, how do you valuate P(M\|D)? This is where Bayes rule comes in $P(M\|D) = \frac{ P(D\|M) P(M)}{P(D)}$ () where P(D\|M) is the likelihood function, P(M) is the prior probability for the model and P(D) is the evidence. This is also generally where people’s eyes start to glaze over. They can’t really see why simply rewriting probability can be of any use. The whole point is that while it is not clear how to estimate the probability of a model given the data, it is just a matter of computation to obtain the probability that the data would arise given a model i.e. P(D\|M). However, there is one problem with equation () and that is it is not clear how to compute P(D), which is the probability of data marginalized over all models! The finesse around this problem is to look at odds ratios. So the odds ratio of M1 compared to M2 is given by $\frac{P(M1\|D)}{P(M2\|\|D)} = \frac{P(D\|M1)}{P(D\|M2)}\frac{P(M1)}{P(M2)}$. The ratio $\frac{P(D\|M1)}{P(D\|M2)}$ is called the Bayes factor and this is the quantity we want to compute. If the prior probabilities of the models are equal then the Bayes factor is all you need to determine the best model. The AIC and BIC are approximations of the Bayes factor although some may quibble that the AIC is something different. Generally, we take logs of the Bayes factor and thus whichever model has the highest model log likelihood (i.e. $\ln P(M\|D)$) is the best model. This approach also generalizes to comparing many models. You just compare them pairwise and if you use log likelihoods then the model with the highest log likelihood of all the models wins. The next move is what fuels a massive debate between Bayesians and frequentists. Suppose each model depends on a set of parameters denoted by $\theta$. Whenever we compare the model to the data, we need to pick an instantiation of the parameters. For example, if you have some differential equation model for a time series, then you pick some parameters to specify the model, simulate the differential equation and compare the results to the data. You can then try other parameters and compare the results to the data. The Bayesian argument is that to compute P(D\|M), you must marginalize over the parameters to obtain $P(D\|M)= \int P(D\|M,\theta)P(\theta\|D) d\theta$ () So what does this mean? Well it says that the probability that the data came from a model, is the “average” performance of a model “weighted” by the prior probability of the parameters. So, now you can see why this step is so controversial. It suggests that a model is only as good as the prior probability for the parameters of the model. The Bayesian viewpoint is that even if some model has a set of parameters that makes the model fit exactly to the data, if you have no idea where those parameters are then that model is not very good. But there is even more to this. How good a model is also depends on how “broad” the prior distribution is compared to the likelihood function because the “overlap” between the two functions in the integrand of the integral in () determines how large the probability is. As a simple way to see this suppose there is one parameter and the prior for the parameter has the form $P(\theta \| M)=\frac{1}{\sigma}, 0 < \theta <\sigma$ Then the likelihood for the model is given by $P(D\|M) =\frac{1}{\sigma}\int_0^\sigma P(D\|M,\theta) d\theta$ Now suppose that the likelihood is a single peaked function with a maximum likelihood value of L. Then we can rewrite the model likelihood as $P(D\|M) =L(\delta/\sigma)$, where $\delta$ is just a constant that gives a “measure” of the “width” of the distribution. We thus see that the probability of a model is given by the maximum likelihood, weighted by the ratio of the width of the data likelihood and the prior $\delta/\sigma\le 1$. This ratio is called the Occam’s factor. When generalized to k parameters Occam’s factor has the form $(\delta/\sigma)^k$. Thus, we see that as the number of parameters increases, Occam’s factor decreases and thus models with more parameters are always penalized. If we take the log of the likelihood we get $\ln P(D\|M) \simeq \ln L - k \ln\sigma/\delta$, which can be compared to the BIC and AIC. So to compare two models we just compute the Bayesian log likelihood of the model and the model with the highest value is more likely. If you have more than one model you just compare all the models to each other pairwise and the model with the highest Bayesian log likelihood is the best. The beautiful thing about Bayesian inference is that the maximum likelihood value for the parameters, the posterior distribution and the Bayes factor can all be estimated in a single MCMC computation. I’ll cover this in the next post. Like this: This entry was posted on August 6, 2010 at 10:49 and is filed under Bayes, Pedagogy, Probablity, Statistics. You can follow any responses to this entry through the RSS 2.0 feed. You can leave a response, or trackback from your own site. 7 Responses to “Bayesian model comparison” 1. Bayesian parameter estimation « Scientific Clearing House Says: November 11, 2010 at 10:41 [...] is the third post on Bayesian inference. The other two are here and here. This probably should be the first one to read if you are completely unfamiliar with the [...] 2. Moore’s law and science « Scientific Clearing House Says: February 21, 2011 at 16:58 [...] I need to manipulate very large matrices. It is true that we can now do Bayesian inference and model comparison on larger models. However, the curse of dimensionality strongly works against us here. If you [...] 3. monobhu Says: May 2, 2011 at 10:13 Excellent article. Do you still plan to write a post about estimating posterior distribution by MCMC? 4. Carson Chow Says: May 2, 2011 at 10:55 Hi, I probably should write a post making the connection between the posterior and the MCMC explicit. The pieces are all there in the three Bayesian posts I have written so far but perhaps a fourth is warranted. Thanks for the suggestion. 5. monobhu Says: May 3, 2011 at 09:16 Hi, Thanks for reply. Noticed your blog days before and found lots of interesting posts here. Excellent job. 6. Approaches to theoretical biology « Scientific Clearing House Says: October 15, 2011 at 11:53 [...] data, what we did was to explore classes of models that could explain a given set of data and use Bayesian model comparison to decide which was better. This approach was used in the work on quantifying insulin’s [...] 7. Bayesian model comparison part 2 \| Scientific Clearing House Says: May 11, 2013 at 16:14 […] a previous post, I summarized the Bayesian approach to model comparison, which requires the calculation of the […] %d bloggers like this:	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 13, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9227283000946045, "perplexity_flag": "head"}
http://mathoverflow.net/questions/84594?sort=newest	## Find the maximum set whose subset sum is unique for every of its subset. ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) We are given a set of $n$ positive integers. We assume all of them are bounded by a polynomial of $n$. We would like to find a subset $S$ of numbers such that for any $T_1,T_2\subseteq S$, the sum of numbers in $T_1$ is not equal to that of $T_2$. We want the size of $S$ is maximized. Clearly, the problem is in NP and can be solved in $n^{O(\log n)}$ time (Since $\|S\|\leq O(\log n)$). Is this problem polynomial time solvable? Is this problem studied before? Any help is appreciated. - ## 2 Answers In the modern language, sets with all subset sums pairwise distinct are often called dissociated. Maximal dissociated subsets of a given set are important in Additive Combinatorics and Fourier analysis, although I cannot recall anybody ever addressing the algorithmic aspect. Yet another reference you may find useful: http://math.haifa.ac.il/~seva/Papers/DisBases.pdf. Added: taking into account the interpretation of a maximal dissociated subset of a given set $A$ as a "basis of $A$ over the set ${-1,0,1}$", it may be reasonable to try and adjust one of the standard algorithms of finding a maximal independent subset of a given set in a vector space. - ### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you. This has been studied, see http://www.mathnet.or.kr/mathnet/kms_tex/978590.pdf and the somewhat more interesting: NEWMAN POLYNOMIALS WITH PRESCRIBED VANISHING AND INTEGER SETS WITH DISTINCT SUBSET SUMS PETER BORWEIN AND MICHAEL J. MOSSINGHOFF (available online) See also the references in: http://garden.irmacs.sfu.ca/?q=op/sets_with_distinct_subset_sums None of these answers your question, but I think this is the state of the art... -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 12, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9152297973632812, "perplexity_flag": "head"}
http://stats.stackexchange.com/questions/31108/regression-with-categorical-factor-variable-and-the-correlation-among-the-variab	# Regression with categorical factor variable and the correlation among the variables ````lm(y~x1 + x2 -1) ```` where `x1` is a continuous numerical variable and `x2` is a categorical factor variable with 4 levels. Is there a way to measure the "correlation" between the `x1` variable and each level of the factor variable `x2`? By putting correlation into double quotes, I admit that I don't really know what is a good definition for the associatedness between a continuous variable and a specific level of the factor variable. Hopefully readers get my intuition. I mean that some levels of `x2` may associated with `x1` more actively than other levels of `x2`. Not knowing how to measure it, I am thinking of the following procedure: 1. run `lm(y~x2 +1)` 2. run `lm(y~x2 + x1 -1)` i.e. replace the intercept in "Step 1" by the continous variable `x1` in "Step 2" and then see which $\beta$ (of associated factor level) changed most. My questions are: 1. Does my approach make sense? 2. How do I measure if a beta (of a specific associated factor level) changed and by how much? Is there a way to make fair comparison and draw some meaningful conclusions? Could anybody please shed some lights on me? Thanks a lot! - For the purposes of this question, are you only interested in the association between `x1` and `x2`? – Macro Jun 25 '12 at 23:50 Luna, I am struggling to understand what role $y$ plays in this question about association between $x_1$ and $x_2$. Could you enlighten us about this? – whuber♦ Jun 26 '12 at 0:06 @whuber, it could be sensible to only be interested in the association between $x_1, x_2$ insofar as it affects the regression coefficient estimates when they are both entered into the same regression model. This seems like an ad hoc method of diagnosing confounding. Although, if $y$ doesn't play any role at all, then of course the problem becomes much simpler. – Macro Jun 26 '12 at 0:10 Thanks, @Macro. I'm still confused, because I understand "association" quite generally to refer to a relationship between two things whereas "affects the regression coefficients" appears to refer to something else altogether. To circumvent guessing and possible misinterpretation of what is being asked, I am hoping that Luna can make this all clear with a suitable edit. – whuber♦ Jun 26 '12 at 0:13 Hi Macro and whuber, thanks a lot for your help. The more complete picture is: initially we did the regression lm(y~x1 + x2 -1)... since our end goal was to fit the data y~(x1, x2) and it seems that whenver factor is involved, we should do y~x1+x2-1, so did we. But then we wanted the further study the relationship between the two variables x1 and x2. And what impact do they have on the regression, partially and jointly... and do they have a lot of collinearity between them and how does that impact the regression, etc. These are typical data exploring steps... – Luna Jun 26 '12 at 2:26 show 1 more comment ## 2 Answers The two variables' degree of association (put another way, the extent to which x1 means differ by level of x2) can be tested with an anova, among other ways. And ordering up an eta-squared statistic as part of an anova procedure will tell you the percent of variance in x1 that can be explained by level of x2. Also, in places it seems that you are seeking to learn about this relationship between x1 and x2, but in other places you say things such as [...] some levels of x2 may associated with x1 more actively than other levels of x2. This seems to indicate a misunderstanding. If x2 is fixed at a single level, it cannot be meaningful to talk about its association or correlation with anything. It is only when x2 is free to vary that it can covary with something else. - If the covariate `x1` is relevant to explaining variation in `y`, it should be included in your model. Adding it in and then taking it out does not do you much good...if should definitely always be in your model. Theory will help guide you in this decision. Once you've decided to include or exclude the regressor `x1`, then you can move on to 'diagnostics' such as exploring multicollinearity. I highly doubt that multicollinearity will be a problem, since your categorical variable can only take on four values. For multicollinearity diagnostics, google "variance inflation factor" which for you, using R's 'HH' package, would be `vif(lm(y ~ x1 + x2))`. 1) Does my approach make sense? No. The first regression `lm(y~x2)` merely will report the mean of `y` in each category of `x2`. The second regression will report the intercepts of each category, and an additional slope parameter on `x1` for the entire sample. Do you understand how these are different? So, all the parameters will likely change, because they represent different things. 2) How do I measure if a beta (of a specific associated factor level) changed and by how much? Is there a way to make fair comparison and draw some meaningful conclusions? Could anybody please shed some lights on me? See my answer to your first question. You do not want to approach the problem in this way. Also, it would help clarify things if you (1) stated what your variables represent (e.g., apples, temperature, ??) and (2) what research question you are attempting to answer. - 1 (-1) Model is selection involves too many considerations for your "definitely always" statement in paragraph 1 to hold true. Also, it doesn't seem right to conclude that just because x2 only takes 4 values it cannot be closely associated with x1. For all we know, the mean of x1 might depend in large part on the level of x2, which could be group membership, region, industry, type of apple, etc. – rolando2 Sep 24 '12 at 13:31	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 6, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9515940546989441, "perplexity_flag": "middle"}
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In mechanical engineering, the torque due to a couple is given by $\tau = P\times d$, where $\tau$ is the resulting couple, $P~$ is one of the force vectors in the couple and $d$ is the arm of the ...	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 12, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9419737458229065, "perplexity_flag": "middle"}
http://physics.stackexchange.com/questions/tagged/coordinate-systems?page=2&sort=active&pagesize=15	# Tagged Questions The coordinate-systems tag has no wiki summary. 1answer 175 views ### Inverting Generalized Coordinates In Corben's classical mechanics on pg. 9, it says that given generalized coordinates $q_m = q_m(x_1, ..., x_n,t)$, then if the Jacobian is non-zero everywhere, you may express \$x_i = ... 3answers 253 views ### Is the equivalence principle strictly fulfilled by general relativity? The equivalence principle states The outcome of any local experiment in a freely falling laboratory is independent of the velocity of the laboratory and its location in spacetime. Any real local ... 3answers 270 views ### Need some basic help with notation and the Christoffel symbols Apologies in advance if some of the questions below seem overly simple. In an introductory GR book, I find the following expression for the autoparallel of the affine connection (the upper bound of ... 1answer 221 views ### Is spherical coordinates enough to calculate the exact position of a point with rotations included? With "Geographic coordinate system" we only get the location, no rotations or anything. Translated into latitude and longitude. Where with "Spherical coordinate system", we should be able to get the ... 1answer 257 views ### Understanding P-, CP-, CPT-violation etc. in field theory and in relation to the principle of relativity I can never get my head around the violations of $P-$, $CP-$, $CPT-$ violations and their friends. Since the single term "symmetry" is so overused in physics and one has for example to watch out and ... 1answer 166 views ### Is General Relativity applicable for all coordinate systems? My understanding was that relativistic physics can be expressed in any inertial coordinate system, but not arbitrary systems. That is, no experiment can determine if we are "still" or "moving" at a ... 3answers 4k views ### Find radius of curvature, given a velocity vector and acceleration magnitude? The particle P moves along a space curve. At one instant it has velocity $v = (4i-2j-k)$ $m/s$. The magnitude of the acceleration is 8 $m/s^2$. The angle between the acceleration and the velocity ... 4answers 464 views ### Why do Engineers manipulate coordinate systems? [closed] When tackling a physics problem, An Engineer will manipulate the axes/coordinate system where a Mathematicians and/or Physicists will use the original coordinate system and math. Why do Engineers ... 2answers 306 views ### What is a Kustaanheimo-Stiefel transformation? What is a Kustaanheimo-Stiefel transformation? Which applications has it in physics? Can you point me to a reference, where this transformation is explained? 1answer 313 views ### When are we now? Time coordinate system [closed] body is missing please delete it 1answer 365 views ### Does change of coordinate system require acceleration? This question came about from a side discussion that arose on this: Does GR provide a maximum electric field limit? Can we change our choice of coordinate system completely independent of physical ... 1answer 483 views ### converting force from spherical to Cartesian coords So I am working on my assignment, and have a question about converting coordinates. I dont know whether I should ask here or the math SE, so lets give it a try here. The force in question is ... 2answers 380 views ### Is there any situation in Physics where the Right Hand Rule is not arbitrary? We use Right Hand Rule in calculating Torque not because that's the direction torque is pointing in the real, physical world, but because it's a convenient way to indicate the "sign" of the rotation ...	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 7, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.8753266334533691, "perplexity_flag": "middle"}
http://math.stackexchange.com/questions/85243/small-generating-set-of-third-degree-polynomials-in-r-mathbbz-2x-1-dots-x?answertab=oldest	# Small generating set of third degree polynomials in $R=\mathbb{Z}_2[x_1,\dots,x_n]/\langle x_1^2-x_1,\dots,x_n^2-x\rangle$ Let $R=\mathbb{Z}_2[x_1,\dots,x_n]/\langle x_1^2-x_1,\dots,x_n^2-x\rangle$, i.e. we can think of $R$ as the ring of multivariate polynomials with the additional property that one can "linearize" higher powers (e.g. $x^2+y^3$ is the same as $x+y$). Note that I sometimes use "$y$","$z$" etc. instead of $x_2,x_3$. I wish to generate the set of all third degree monomials, e.g. monomials of the form $xyz$. That is, I wish to find a set $p_1,\dots,p_t\in R$ such that every monomial can be written as sums and products of $p_1,\dots,p_t$. One can see that we have two trivial cases: If we allow products of any number of elements, then $xyz$ can always be written as a product of three elements $x\cdot y\cdot z$ and so $x_1,\dots,x_n$ is a generating set which seems to me to be minimal (although I have no proof), so in this case the size of the generating set is probably $\Theta(n)$. Also, if we disallow products, i.e. only allow linear combinations, than the generating set must be of size $\Theta(n^3)$ since this is the dimension of the set of monomials $xyz$ seen as a vector space. The interesting case is when we allow products of pairs of the generating set, i.e. we look at combinations of the form $\sum p_ip_j$. Trivial arguments can only show a lower bound of $\Omega(n^{1.5})$ but I have no idea how to search for a generating set of such size and it seems to be that $\Theta(n^2)$ is the correct size in this case (it's very easy to see it's an upper bound: take all monomials $xy$ to be in the generating set and now $xyz=xy\cdot xz$). So my question is what is the correct bound, and even more - what is the "correct" way to attack a problem like that. I highly wish to exploit the underlying structure of $R$, which seems to be a classical algebraic-geometry object, but so far I haven't got a clue. - I expected you would know better to use `\langle` rather than `<`. – Asaf Karagila Nov 24 '11 at 15:14 I embarrass myself so much in this question that this is the least of my problems. But I'll fix it. – Gadi A Nov 24 '11 at 15:20 The ring $R$ is isomorphic to the ring of functions from $\mathbb{F}_2^n$ to $\mathbb{F}_2$, with $x_j$ being projection onto the $j$th coordinate, and with addition and multiplication being pointwise. But this doesn't make the answer any clearer to me. – David Speyer Nov 24 '11 at 17:16	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 28, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9602919220924377, "perplexity_flag": "head"}
http://mathoverflow.net/questions/71687?sort=oldest	## Metrizable dual space ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) I've got the following questions concerning the theory of locally convex spaces : Let $X$ be a locally convex metrizable space, what is the necessary and sufficient condition to have its dual $X^$ metrizable? Is it possible that $X^$ is the F-space when $X$ is a locally convex non-complete metrizable space which is not a normed space? Thank you in advance for the answer. - See also mathoverflow.net/questions/63383/… – Andrey Rekalo Jul 30 2011 at 23:03 ## 1 Answer The nLab cites a theorem that the dual of a Fréchet space $X$ is Fréchet if and only if $X$ is a Banach space. (Reference: paragraph 29.1 (7) in Gottfried Koethe, Topological Vector Spaces I.) Even if $X$ is non-complete, the dual of $X$ is isomorphic to the dual of its completion, so $X^\ast$ cannot be Fréchet if $X$ is a non-normable locally convex metrizable TVS. - Thank you very much. – Romanov Jul 30 2011 at 22:43	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 10, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.8470095992088318, "perplexity_flag": "middle"}
http://math.stackexchange.com/questions/130697/show-x2-in-the-interval-0-1-3-has-no-fixed-points	# Show $x^2$ in the interval $(0,1/3]$ has no fixed points. Show $x^2$ in the interval $(0,1/3]$ has no fixed points. I understand that the range of that domain is always lower than $y=x$, but what is a proper way of showing this? $$\left(0,\frac13\right] \to \left(0,\frac19\right]$$ - 1 Funny looking intervals! – lhf Apr 12 '12 at 2:24 @lhf: I now have a pain in the neck, straining to tilt my head to the right... – Aryabhata Apr 12 '12 at 2:25 ## 2 Answers Let $f(x)=x^2$. You’re being asked to show that there is no $x\in(0,1/3]$ such that $f(x)=x$. Set up the equation $f(x)=x$, solve it, and discover that its only solutions are outside of $(0,1/3]$. - 1 @lhf: Perfectly true, but that wasn’t the question. – Brian M. Scott Apr 12 '12 at 1:09 Don't you mean that x > x^2 for (0,1) ? – Mary Apr 12 '12 at 1:19 The quadratic equation $x^2=x$ can be written as $x^2-x=0$, and then as $x(x-1)=0$. That has two solutions. Neither of them is in the interval $(0,1/3]$. -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 14, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9517936706542969, "perplexity_flag": "middle"}
http://stats.stackexchange.com/questions/47916/bayesian-batting-average-prior	# Bayesian Batting Average Prior I wanted to ask a question inspired by an excellent answer to the query about the intuition for the beta distribution. I wanted to get a better understanding of the derivation for the prior distribution for the batting average. It looks like David is backing out the parameters from the mean and the range. Under the assumption that the mean is $0.27$ and the standard deviation is $0.18$, can you back out $\alpha$ and $\beta$ by solving these two equations: \begin{equation} \frac{\alpha}{\alpha+\beta}=0.27 \\ \frac{\alpha\cdot\beta}{(\alpha+\beta)^2\cdot(\alpha+\beta+1)}=0.18^2 \end{equation} - Honestly, I just kept graphing values in R until it looked right. – David Robinson Jan 17 at 3:09 Incidentally, that's the formula for the variance, not the standard deviation- which is `.18`? – David Robinson Jan 17 at 3:18 I added the missing ^2 above. – Dimitriy V. Masterov Jan 17 at 18:11 ## 1 Answer Notice that: \begin{equation} \frac{\alpha\cdot\beta}{(\alpha+\beta)^2}=(\frac{\alpha}{\alpha+\beta})\cdot(1-\frac{\alpha}{\alpha+\beta}) \end{equation} This means the variance can therefore be expressed in terms of the mean as \begin{equation} \sigma^2=\frac{\mu*(1-\mu)}{\alpha+\beta+1} \\ \end{equation} If you want a mean of $.27$ and a standard deviation of $.18$ (variance $.0324$), just calculate: \begin{equation} \alpha+\beta=\frac{\mu(1-\mu)}{\sigma^2}-1=\frac{.27\cdot(1-.27)}{.0324}-1=5.083333 \\ \end{equation} Now that you know the total, $\alpha$ and $\beta$ are easy: \begin{equation} \alpha=\mu(\alpha+\beta)=.27 \cdot 5.083333=1.372499 \\ \beta=(1-\mu)(\alpha+\beta)=(1-.27) \cdot 5.083333=3.710831 \end{equation} You can check this answer in R: ````> mean(rbeta(10000000, 1.372499, 3.710831)) [1] 0.2700334 > var(rbeta(10000000, 1.372499, 3.710831)) [1] 0.03241907 ```` - David, do you happen to follow any baseball research? There are several competing techniques out there for finding the proper $\alpha$ and $\beta$, so I was wondering if you had any opinion on the matter if you were doing something besides just trying to find a graph that looked reasonable. – Michael McGowan Jan 17 at 14:53 I don't particularly follow sabermetrics- in the other answer it just happened to provide a very convenient example of estimating p from a binomial with a prior. I don't even know if this is how it's done in sabermetrics, and if it is, I know there are many components I left out (players having different priors, stadium adjustments, weighting recent hits over old ones...) – David Robinson Jan 17 at 15:02 I am impressed that your eyeballing was this accurate. – Dimitriy V. Masterov Jan 19 at 1:03	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 11, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.8462719321250916, "perplexity_flag": "middle"}
http://mathoverflow.net/revisions/62683/list	2 edited tags 1 # fgf = f, gfg = g, fg not necessarily identity, what was that called? A very simple question, I just totally forgot how it was called, and google is not helping. There's a pair of functions $f:X\to Y$, $g:Y\to X$. $fgf = f$, $gfg = g$, but $fg$ and $gf$ don't need to be identities (and usually are not in interesting cases). A simple example would be $f(a,b,c)=(a,b)$, $g(a,b)=(a,b,0)$ What were $f$ and $g$ called?	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 10, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9794632196426392, "perplexity_flag": "middle"}
http://physics.stackexchange.com/questions/tagged/photons+quantum-mechanics	# Tagged Questions 1answer 27 views ### How do particles become entangled? A person asked me this and I'm just a lowly physical chemist. I used a classical analogy (how good or bad is this and how to fix?) Basically, light has a net angular momentum of zero, insofar as ... 0answers 19 views ### Does quantum mechanics depend solely on electromagnetic waves? [duplicate] I am beginning to learn quantum mechanics. Since determining the position of an object involves probing by electromagnetic waves and since i have read a simple derivation of Heisenberg's uncertainty ... 1answer 45 views ### Is the de Broglie wavelength of a photon equal to the EM wavelength of the radiation? Is the de Broglie (matter) wavelength $\lambda=\frac{h}{p}$ of a photon equal to the electromagnetic wavelength of the radiation? 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http://mathoverflow.net/questions/55162/how-can-i-write-down-polynomial-relations-that-define-when-a-polynomial-is-a-squa/55164	## How can I write down polynomial relations that define when a polynomial is a square? ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) It's easy to tell when a polynomial is squarefree (or not): that's just the question of the vanishing of the discriminant, which can be dealt with as the resultant of $f$ and $f'$. However, given a polynomial of degree $2n$ $f$, when is it of the form $g^2$ for $g$ a polynomial of degree $n$? I've been trying to work out the relations on the coefficients that will guarantee this for a specific degree ($n=6$ is my case) but whenever I take the obvious equations in the coefficients of $g$ and of $f$ and try to use Groebner bases to eliminate the coefficients of $g$, I run out of memory and my software crashes. Is there a way to understand the locus of polynomials which are squares concretely without having to do a (seemingly unrealistically) big computation? Or perhaps a clever trick that will give these polynomial identities in a more computable way? - Probably not the answer you are looking for, but for a square polynomial $gcd(f,f')$ has all roots of odd multiplicity, and if $2k_i-1$ is the multiplicity of the $x_i$th root then $\sum_i k_i=n$. The converse is also true, the condition $\sum_i k_i=n$ guarantees that $f$ doesn't have any simple roots. – Nick S Feb 11 2011 at 20:12 Not directly related to your question: what did you use as a Groebner solver? Last I checked (admittedly a while ago), Mathematica or Maple were vastly outperformed by more specialized software. This might be an interesting avenue to explore, regardless of a more theoretical solution. – Thierry Zell Feb 11 2011 at 22:54 1 Trivial observation: You must require the field to be algebraically closed, or the polynomial to be monic or something stronger. Else, $aX^2$ will reduce the undecidable is-$a$-a-square to your is-my-polynomial-a-square. – darij grinberg Feb 11 2011 at 23:06 Also, the answers require the field to have characteristic $0$. This seems to have a good reason: If the field has characteristic $2$, then you can reduce the question is-a-field-element-$a$-a-square to your is-my-polynomial-a-square (apply to the polynomial $X^2-a$). Of course, there is no polynomial equation that tells you when an element of a field of characteristic $2$ is a square. – darij grinberg Feb 11 2011 at 23:11 Note on darij's remark: Let $k$ have characteristic other than $2$. Let $f$ be a polynomial in $k[x]$. Then $f$ is square in $k[x]$ if and only if $f$ is square in $k^{\mathrm{alg}}[x]$ and the leading term of $f$ is square in $k$. So the issues about algebraic closure are comparatively mild. (This is evident from looking at Greg's explicit solution.) – David Speyer Feb 12 2011 at 17:34 show 1 more comment ## 4 Answers Say, for simplicity, you are working over $\mathbb{C}$ or in characteristic zero in general. Then you can guess one of the two values of $g(0)$ (say) and then compute the Taylor series of $\sqrt{f}$. The approach is similar to Hensel lifting: The equation for the first coefficient is non-linear; the equations for the others are all locally linear (so that you get explicit formulas for the coefficients of $g$ in terms of existing data). I first misread Charles' question, but now that I have it right (I think), here is why I think that the above is still a solution. If you read the coefficients of a polynomial of degree $n$ as projective coordinates, then over $\mathbb{C}$ the set of squares of degree $2n$ is some projective variety $S$ in $\mathbb{C}P^{2n}$. Charles is interested in projective equations for this variety $S$. For simplicity let's rescale the polynomial $f(x)$ so that $f(0) = 1$. (And I guess we're working the affine chart in which $f(0) \ne 0$ before the rescaling. It shouldn't change things much or at all.) Then you can assume that $g = \sqrt{f}$ also satisfies $g(0) = 1$, and you can make explicit expressions for its Taylor series. Then $g$ is a polynomial of degree $n$ if and only if its Taylor series vanishes in degree $n < k \le 2n$. I think that this gives you the desired equations. - Maybe I'm not understanding properly, but wouldn't the conditions I want then be that for all $k>n$, the $k$th coefficient of $sqrt(f)$'s Taylor series vanishes? Doesn't that give me infinitely many conditions (a priori) and no obvious way to know which finite set of them suffice? – Charles Siegel Feb 11 2011 at 20:23 2 Of course once you reach $k=n$, you can square the result and see if it works. – Greg Kuperberg Feb 11 2011 at 20:26 Ok...perhaps you could be more explicit about what you're doing, because when I Taylor expand $\sqrt{f}$, I get that the degree of the coefficient of $x^i$ is $i$ in the coefficients of $f$ (well, with a $\sqrt{a_0}$ to some power in the denominator). I'm not sure what you're doing, or what the output should be, but I'm looking for a finite set of polynomials in the coefficients of $f$ whose vanishing determines that $f$ is a square. – Charles Siegel Feb 11 2011 at 20:38 It's true that I misinterpreted the problem, but I think that, at the expense of a $\sqrt{a_0}$, the solution still works. The coefficients of the Taylor series of $g$ vanish for $n < k \le 2n$ if and only if $g$ is a polynomial of degree $n$. – Greg Kuperberg Feb 11 2011 at 21:07 ### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you. So unless I am misunderstanding the question, temporarily normalize so that the coefficient of $x^6$ in $f$ is 1. One is left with three degrees of freedom, coming from the quadradic, linear and constant terms of the degree 3 polynomial square root. For concreteness, let $f(x) = x^6 + c_5x^5 + c_4x^4 + c_3x^3 + c_2x^2 + c_1x + c_0$ Then I work out that necessary relations on the coefficients are: $c_2 = 2(\frac{1}{2}c_5)(\frac{1}{2}c_3-\frac{1}{4}c_4c_5+\frac{1}{16}c_5^3)+(\frac{1}{2}c_4-\frac{1}{8}c_5^2)^2$ $c_1 = 2(\frac{1}{2}c_4-\frac{1}{8}c_5^2)(\frac{1}{2}c_3 - \frac{1}{4}c_4c_5 + \frac{1}{16}c_5^3)$ $c_0 = (\frac{1}{2}c_3 - \frac{1}{4}c_4c_5 + \frac{1}{16}c_5^3)^2$ These are also sufficient since if they hold, then $f$ is the square of $x^3+(\frac{1}{2}c_5)x^2+(\frac{1}{2}c_4-\frac{1}{8}c_5^2)x + (\frac{1}{2}c_3-\frac{1}{4}c_4c_5+\frac{1}{16}c_5^3)$ Unless I did something wrong, it doesn't seem like these computations should be crashing the system. What are you using to run the Groebner calculations? - 1 This is equivalent to the solution that I describe, except of course working from the other end. The explicit calculation is nice; it shows quite clearly that it works. – Greg Kuperberg Feb 12 2011 at 2:33 @Greg: It is indeed essentially the same as what your answer says; I was interested in carrying through the solution concretely. – ARupinski Feb 12 2011 at 3:22 When I said "$n=6$", I meant the case where $\deg f=12$, which makes this much, much worse for a CAS to work out. – Charles Siegel Feb 12 2011 at 19:44 http://en.wikipedia.org/wiki/Square-free_polynomial gives a method for finding a square-free factorization of a polynomial (over characteristic zero field), ie $f=a_1\cdot a_2^2\cdots a_n^n$ where each $a_i$ is a square-free polynomial. Then $f$ is a perfect square iff $a_{2i+1}=1$ for each $i$. - It seems to me that the OP almost contains the answer: the gcd of $f$ and $f^\prime$ (let's assume characteristic zero) contains all the irreducible factors of $f$ which appear with exponent greater than $1.$ This is surely enough to figure out if the polynomial is a square. EDIT to answer the revised version of the question: Write down $$\sum_{i=0}^n a_i x^i = (\sum_{j=0}^{n/2} b_j x^j)^2.$$ This gives a collection of $n+1$ quadratic equations in $3n/2 + 2$ variables. You now eliminate the $b_j$ to get the variety of perfect squares. Needless to say, this is not algorithmically very pleasant (the degree is going to be exponential in $n$), but you can use successive resultants or Grobner bases to do it for small degrees, and you might see a pattern. Another Edit If you have Mathematica, you can perform the above-mentioned experiments with the program below: genpoly[deg_, name_, var_] := Sum[name[i] var^i, {i, 0, deg}] quadraticeq[deg_, name1_, name2_, var_]:= Eliminate[MapThread[Equal, {CoefficientList[genpoly[2deg, name1, var], var],CoefficientList[Expand[genpoly[deg, name2, var]^2], var]}], Table[name2[i], {i, 0, deg}]] (for example, to see what the variety is describing quadratic polynomials which are squares, you do: quadraticeq[1, a, b, x] a and b are dummy variables, a[0], ..., a[2 deg] are the variables you care about. For quadratic polynomials you get (no surprise): 4 a[0] a[2]==a[1]^2 While for quartic polynomials you get: a[0] a[3]^2==a[1]^2 a[4]&&-4 a[0] a[1] a[2]+8 a[0]^2 a[3]==-a[1]^3&&8 a[0] a[3] a[4]==a[1] (-a[3]^2+4 a[2] a[4])&&16 a[0] a[4]^2==-a[2] a[3]^2+4 a[2]^2 a[4]-2 a[1] a[3] a[4]&&8 a[1] a[4]^2==a[3] (-a[3]^2+4 a[2] a[4])&&a[0] (-4 a[2]^2+2 a[1] a[3])+16 a[0]^2 a[4]==-a[1]^2 a[2]&&a[0] (-4 a[2] a[3]+8 a[1] a[4])==-a[1]^2 a[3] which is a little more painful. - 1 That would be fine for determining if a given polynomial is a square, but I'm looking for the equations that define the locus of square polynomials, and I don't see how to do that in this way. – Charles Siegel Feb 11 2011 at 20:47 1 @Charles: everyone who has answered the question has misread it so far, so I've taken the liberty of making the title more precise. Hope I have captured your meaning. – Qiaochu Yuan Feb 11 2011 at 22:37 @Igor: This is essentially what I was doing in Macaulay2, but the elimination step was too bad by the time I got to 12th degree polynomials – Charles Siegel Feb 12 2011 at 19:46 @Charles: yes, mathematica seems to choke on this as well (not surprisingly -- macaulay 2 should be more efficient). But what are you trying to do with these equations? – Igor Rivin Feb 12 2011 at 22:33	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 91, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9310258030891418, "perplexity_flag": "head"}
http://math.stackexchange.com/questions/148351/finding-the-domain-and-distribution-of-y-x3-6-for-x-sim-u-2-2?answertab=active	# Finding the domain and distribution of $Y = X^3 + 6$ for $X \sim U[-2,2]$ A small probability problem that I am struggling with... Let $X \sim U[-2 , 2]$. Find the distribution of $Y = X^3 + 6$. My main problem is the domain of $Y$. I found that the domain of $Y$ is $-2 \leq Y \leq 14$, but I believe that the correct one is $6 \leq Y \leq 14$. Also I am a bit confused now, is that $6 \leq Y \leq 14$ actually the domain of $y$ and not $Y$? Should I rather say that: $Y$'s domain is $[-2,14]$ So $F_Y(y) = P(Y \leq y) = P(x^3+6 \leq y) = P(x \leq \sqrt[3]{y-6})$ So actually $y$'s domain is $[6,14]$? And then, what integrals should I take? Using what domain? Thanks a lot! Thanks all, i think it is sufficiently answered the question. - This looks like homework, and if so, please add the `homework` tag. Also, it would help if you used LaTex to clean up your equations. Now to the problem: Why do you think the domain of $Y$ is $[6,14]$? You found it correctly as $[-2,14]$. For each real number $y \in [-2,14]$, $$P\{Y \leq y\} = P\{X^3+6 \leq y\} = P\{X\leq (y-6)^{1/3}\}$$ and since $X$ is uniformly distributed, you should be able to compute this probability without needing to integrate. – Dilip Sarwate May 22 '12 at 19:15 Thanks for your answer. Its not that i need to find a probability but i need to find the distribution of Y. So i have to integrate , dont i ? – Peter Mk_dir May 22 '12 at 19:26 For any given real number $y$, the value of the distribution function $F_Y(y)$ at $y$ is a probability: $F_Y(3) = P\{Y \leq 3\}$ is a probability. You will find life a lot easier if you forget all about fancy words and remember instead that for continuous random variables, probabilities are areas under the density curve. $P\{X \leq \sqrt[3]{y-6}\}$ is the area under the density $f_X(x)$ to the left of $\sqrt[3]{y-6}$. Draw a sketch of $f_X(x)$ and see if you can figure out the desired area without writing the symbol $\displaystyle \int$. – Dilip Sarwate May 22 '12 at 19:34 ## 2 Answers What you have written is correct, but why do you say that "So actually $y$'s domain is $[6,14]$?". Note that you are dealing with cube-root (and not square-root) of $(y-6)$ which is defined even when $y-6$ is negative. Just to finish it off, from what you have written we get that $$F_Y(y) = \mathbb{P}(X \leq \sqrt[3]{y-6}) = \begin{cases} 0 & \text{if $\sqrt[3]{y-6} \leq -2$}\\ \frac{\sqrt[3]{y-6} + 2}{4} & \text{if $\sqrt[3]{y-6} \in [-2,2]$}\\ 1 & \text{if $\sqrt[3]{y-6} \geq 2$} \end{cases}$$ Getting rid of the cube-roots in the domain of definition of $F_Y(y)$, we get that $$F_Y(y) = \mathbb{P}(X \leq \sqrt[3]{y-6}) = \begin{cases} 0 & \text{if $y \leq -2$}\\ \frac{\sqrt[3]{y-6} + 2}{4} & \text{if $y \in [-2,14]$}\\ 1 & \text{if $y \geq 14$} \end{cases}$$ - and thanks!!! finally i got it ! – Peter Mk_dir May 22 '12 at 20:00 @PeterMk_dir Kindly accept one of the answers if you have understood and obtained what you want. – user17762 May 22 '12 at 20:17 Let's not worry about words, let's solve the problem. We will find the cumulative distribution function $F_Y(y)$ of the random variable $Y$. This function is, as usual, defined for all reals. It is clear that if $y<-2$, then $P(Y \le y)=0$. It is also clear that if $y>14$, then $P(Y\le y)=1$. Finally, we deal with $y$ in the interval $[-2,14]$. We have $Y\le y$ iff $X^3+6\le y$ iff $X\le (y-6)^{1/3}$. There is no problem below $y=6$, since $w^{1/3}$ can be thought of as defined for all $w$, even negative $w$. For $y$ between $2$ and $14$, $$P(X\le (y-6)^{1/3})=\frac{1}{4}\left((y-6)^{1/3} -(-2)\right).$$ This is directly obtainable from the geometry. However, to deal with more general situations, we observe that $X$ has density function $\frac{1}{4}$ on the interval $[-2,2]$, so the probability is $$\int_{-2}^{(y-6)^{1/3}}\frac{dx}{4}.$$ The conclusion is that $F_Y(y)=0$ if $y<2$, $F_Y(y)= \frac{1}{4}\left((y-6)^{1/3} -(-2)\right)$ if $-2\le y\le 14$, and $F_Y(y)=1$ if $y>2$. For the density function, differentiate. There is a point of non-differentiability at $y=6$, which one should not worry overly about. - Very nice explanation, thank you :) – Peter Mk_dir May 22 '12 at 21:49	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 63, "mathjax_display_tex": 5, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9268813133239746, "perplexity_flag": "head"}
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I think I can prove this ... 1answer 2k views ### Divisor — line bundle correspondence in algebraic geometry I know a little bit of the theory of compact Riemann surfaces, wherein there is a very nice divisor -- line bundle correspondence. But when I take up the book of Hartshorne, the notion of Cartier ... 2answers 178 views ### Usefulness of completion in commutative algebra After studying about the completion of a module $M$ over a ring $A$ (e.g. $I$-adic completion), I am left with the following questions: (i) What is the usefulness of the concept of completion in ... 2answers 283 views ### Variety vs. Manifold In the ambit of differential geometry the aim is to study smooth manifolds. Why the objects studied in algebraic geometry are called algebraic varieties and not for example algebraic manifolds? I am ... 2answers 118 views ### Is a divisor in the hyperplane class necessarily a hyperplane divisor? Let $V$ be a smooth irreducible projective curve over an algebraically closed field $k$, embedded in some projective space $\mathbb{P}^n$, and let $[H]$ be the induced hyperplane divisor class on $V$. ... 2answers 387 views ### Global sections of $\mathcal{O}(-1)$ and $\mathcal{O}(1)$, understanding structure sheaves and twisting. In chapter 2 section 7 (pg 151) of Hartshorne's algebraic geometry there is an example given that talks about automorphisms of $\mathbb{P}_k^n$. In that example Hartshorne states that ... 2answers 263 views ### Good books/expository papers in moduli theory I have been studying mathematics for 4 years and I know schemes (I studied chapters II, III and IV of Hartshorne). I would like to learn some moduli theory, especially moduli of curves. I began ... 6answers 478 views ### Visualizations of some of the abstractions of algebraic geometry Where, or do there exist, good visualizations of sheaves, stalks, stacks, and/or schemes? I'm a better visual thinker than I am a symbolic thinker, and it would be easier for me to follow some of the ... 2answers 128 views ### Can an algebraic variety be described as a category, in the same way as a group? Can an algebraic variety be described as a category, in the same way as a group? A group can be considered a category with one object, with elements of the group the morphisms on the object. 1answer 422 views ### What is an intuitive meaning of genus? I read from the Finnish version of the book "Fermat's last theorem, Unlocking the Secret of an Ancient Mathematical Problem", written by Amir D. Aczel, that genus describes how many handles there are ... 2answers 201 views ### Good problems in Algebraic Geometry I am now using Fulton's book Algebraic Curves to learn algebraic geometry from and have just finished chapter 2. However I feel that the problems are not very inspiring (at the moment at least) and ... 1answer 329 views ### Geometric meaning of primary decomposition In the book "Commutative Algebra with a view toward Algebraic Geometry of David Eisenbud, he wrote about the Geometric interpretation of primary decomposition. I summary as follows : Let ... 2answers 128 views ### Elliptic curves over Spec Z I want to show that there are only finitely many elliptic curves over Spec $\mathbf Z$ without appealing to Siegel's theorem or Shafarevich' theorem. Firstly, I think (but I am not sure) that such an ... 1answer 178 views ### intrinsic proof that the grassmannian is a manifold I was trying to prove that the grassmannian is a manifold without picking bases, is that possible? Here's what I've got, let's start from projective space. Take $V$ a vector space of dimension n, and ... 2answers 133 views ### What does $Tor_{R}^n(M,N)$ represent? Let $R$ be a commutative ring and $M$ and $N$ be $R$-modules (I am not sure if one really needs commutativity in the following). It is well-known that $Ext_{R}^n(M,N)$ for $n>1$ parametrizes ... 1answer 424 views ### My first course in algebraic geometry: two simple questions I'm attending my first course in algebraic geometry, and my professor has chosen an approach which is a middle-way between the basic algebraic geometry done in $\mathbb A^n_k$ and the approach with ... 1answer 216 views ### Why study schemes? Why study schemes instead of only affine/projective varieties, given by zeros of polynomials in the affine/projective space? I mean, what is gained by introducing the concept of schemes? Thank you! 1answer 105 views ### Curious about Hilbert-Zariski theorem involving homogeneous variety and set of zeroes. I got myself in a confusing situation the other week while trying to read a bit of algebraic geometry. I'm hoping someone can pull me out. Suppose $k$ is a field, and $V$ a homogeneous variety with ... 1answer 80 views ### Ext between two coherent sheaves Let $X$ be a smooth projective variety over a field $k = \overline k$. From Hartshorne we know, that $\textrm{dim} \, H^i (X,F)<\infty$ for any coherent sheaf $F$. How to show, that all \$Ext^i ... 2answers 220 views ### Can there be a point on a Riemann surface such that every rational function is ramified at this point? Let $X$ be a compact connected Riemann surface, and let $S\subset X$ be a finite subset. Does there exist a morphism $f:X\to \mathbf{P}^1(\mathbf{C})$ which is unramified at the points of $S$? I'm ... 1answer 311 views ### Stacks in arithmetic geometry [closed] Stacks, of varying kinds, appear in algebraic geometry whenever we have moduli problems, most famously the stacks of (marked) curves. But these seem to be to be very geometric in motivation, so I was ... 1answer 288 views ### vector bundles on affine schemes Serre's theorem (one of them) states that for a quasi-coherent sheaf $\mathscr F$ on an affine noetherian scheme $H^i(X,\mathscr{F})$ vanish for $i >0$. I used to think that this would imply that ...	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 111, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9242402911186218, "perplexity_flag": "head"}
http://www.all-science-fair-projects.com/science_fair_projects_encyclopedia/Linear_algebra/Linearly_independent_vectors	# All Science Fair Projects ## Science Fair Project Encyclopedia for Schools! Search Browse Forum Coach Links Editor Help Tell-a-Friend Encyclopedia Dictionary # Science Fair Project Encyclopedia For information on any area of science that interests you, enter a keyword (eg. scientific method, molecule, cloud, carbohydrate etc.). Or else, you can start by choosing any of the categories below. # Linear independence (Redirected from Linear algebra/Linearly independent vectors) In linear algebra, a set of elements of a vector space is linearly independent if none of the vectors in the set can be written as a linear combination of finitely many other vectors in the set. For instance, in three-dimensional Euclidean space R3, the three vectors (1, 0, 0), (0, 1, 0) and (0, 0, 1) are linearly independent, while (2, −1, 1), (1, 0, 1) and (3, −1, 2) are not (since the third vector is the sum of the first two). Vectors which are not linearly independent are called linearly dependent. Contents ## Definition Let V be a vector space over a field K. If v1, v2, ..., vn are elements of V, we say that they are linearly dependent over K if there exist elements a1, a2, ..., an in K not all equal to zero such that: $a_1 \mathbf{v}_1 + a_2 \mathbf{v}_2 + \cdots + a_n \mathbf{v}_n = \mathbf{0}$ or, more concisely: $\sum_{i=1}^n a_i \mathbf{v}_i = \mathbf{0} \,$ (Note that the zero on the right is the zero element in V, not the zero element in K.) If there do not exist such field elements, then we say that v1, v2, ..., vn are linearly independent. An infinite subset of V is said to be linearly independent if all its finite subsets are linearly independent. To focus the definition on linear independence, we can say that the vectors v1, v2, ..., vn are linearly independent, if and only if the following condition is satisfied: Whenever a1, a2, ..., an are elements of K such that: a1v1 + a2v2 + ... + anvn = 0 then ai = 0 for i = 1, 2, ..., n. The concept of linear independence is important because a set of vectors which is linearly independent and spans some vector space, forms a basis for that vector space. ## The projective space of linear dependences A linear dependence among vectors v1, ..., vn is a vector (a1, ..., an) with n scalar components, not all zero, such that $a_1 v_1 + \cdots + a_n v_n=0. \,$ If such a linear dependence exists, then the n vectors are linearly dependent. It makes sense to identify two linear dependences if one arises as a non-zero multiple of the other, because in this case the two describe the same linear relationship among the vectors. Under this identification, the set of all linear dependences among v1, ...., vn is a projective space. ## Example I The vectors (1, 1) and (−3, 2) in R2 are linearly independent. Proof: Let a, b be two real numbers such that: $a(1, 1) + b(-3, 2) = (0, 0) \,$ Then: $( a - 3 b , a + 2 b ) = (0, 0) \,$ and $a - 3b = 0 \,$ and $a + 2b = 0. \,$ Solving for a and b, we find that a = 0 and b = 0. ## Example II Let V=Rn and consider the following elements in V: $e_1 = (1,0,0,\ldots,0) \,$ $e_2 = (0,1,0,\ldots,0) \,$ $\cdots \,$ $e_n = (0,0,0,\ldots,1) \,$ Then e1,e2,...,en are linearly independent. Proof: Suppose that a1, a2,...,an are elements of Rn such that $a_1 e_1 + a_2 e_2 + \cdots + a_n e_n = 0 \,$ Since $a_1 e_1 + a_2 e_2 + \cdots + a_n e_n = (a_1 ,a_2 ,\ldots, a_n) \,$ then ai = 0 for all i in {1, .., n}. ## Example III: (calculus required) Let V be the vector space of all functions of a real variable t. Then the functions et and e2t in V are linearly independent. Proof: Suppose a and b are two real numbers such that $a e^t + b e^{2 t} = 0 \,$ (1) for all values of t. We need to show that a = 0 and b = 0. In order to do this, we differentiate both sides of (1) to get $a e^t + 2 b e^{2 t} = 0 \,$ (2) which also holds for all values of t. Subtracting the first relation from the second relation, we obtain: $b e^{2 t} = 0 \,$ and, by plugging in t = 0, we get b = 0. From the first relation we then get: $a e^t = 0 \,$ and again for t = 0 we find a = 0. ## See also 03-10-2013 05:06:04 The contents of this article is licensed from www.wikipedia.org under the GNU Free Documentation License. Click here to see the transparent copy and copyright details Science kits, science lessons, science toys, maths toys, hobby kits, science games and books - these are some of many products that can help give your kid an edge in their science fair projects, and develop a tremendous interest in the study of science. When shopping for a science kit or other supplies, make sure that you carefully review the features and quality of the products. Compare prices by going to several online stores. Read product reviews online or refer to magazines. Start by looking for your science kit review or science toy review. Compare prices but remember, Price \$ is not everything. Quality does matter. Science Fair Coach What do science fair judges look out for? ScienceHound Science Fair Projects for students of all ages All Science Fair Projects.com Site All Science Fair Projects Homepage Search \| Browse \| Links \| From-our-Editor \| Books \| Help \| Contact \| Privacy \| Disclaimer \| Copyright Notice	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 17, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.8591156005859375, "perplexity_flag": "middle"}
http://math.stackexchange.com/questions/101837/how-to-prove-this-relation-between-ramsey-numbers-rs-t-rs-t-1-rs-1	# How to prove this relation between Ramsey Numbers: $R(s, t) ≤ R(s, t-1) + R(s-1, t)$ for $s,t>2$ I am trying to prove that $$R(s, t) ≤ R(s, t-1) + R(s-1, t)$$ for $s,t>2$, where $R(s,t)$ is the Ramsey number of $(s,t)$, and I'd be really grateful for a hint that gets me started. - Hint: pick a vertex $v$ in the graph, and split the other vertices in the graph into two sets: those connected to $v$ by red edges and those connected to $v$ by blue edges. – David Moews Jan 24 '12 at 0:17 ## 1 Answer The usual approach is very nice: Suppose you are given the complete graph on $n=R(s,t-1)+R(s-1,t)$ vertices, and a coloring of its edges with colors red and blue. Pick one of the vertices, call it $v$. Divide the remaining $n-1$ into two sets $A$ and $B$, according to whether they are joined to $v$ by a red or a blue edge, respectively. Let $a=\|A\|$ and $b=\|B\|$. Then $a+b=n-1$, so either $a\ge R(s,t-1)$ or $b\ge R(s-1,t)$. This is because otherwise, $a+b\le n-2$. It should be easy to see how to continue from here. Just for fun, let's compute some upper bounds using this inequality, knowing that $R(2,t)=R(t,2)=t$. We have $R(3,3)\le R(2,3)+R(3,2)=3+3=6$. In fact, we have equality in this case. Moreover, the usual "party argument" one sees sometimes showing $R(3,3)\le 6$ is precisely the argument above. $R(4,3)\le R(3,3)+R(4,2)=6+4=10$. In fact, one can extend the argument above a bit to show that if both $R(s,t-1)$ and $R(s-1,t)$ are even, then we have strict inequality. This means that $R(4,3)\le 9$, and again this is sharp. $R(4,4)\le R(3,4)+R(4,3)=18$. Again, equality holds. $R(5,3)\le R(5,2)+R(4,3)=5+9=14$ and, again, equality holds. $R(5,4)\le R(5,3)+R(4,4)=14+18=32$. (So $R(5,4)\le31$ as both $14$ and $18$ are even.) In fact, $R(5,4)=24$, and the bound cease being optimal around here. Improving this bound turns out to be remarkably difficult and the subject of much work. For a nice up to date list of the known values and bounds for Ramsey numbers, together with references, see the dynamic survey on "Small Ramsey numbers" by Stanisław Radziszowski, last updated Aug. 22, 2011, in the Electronic Journal of Combinatorics. (I see I had suggested the same paper as an answer to this other question.) -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 34, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9560412764549255, "perplexity_flag": "head"}
http://www.reference.com/browse/pi	Definitions Nearby Words # pi [pahy] /paɪ/ pi, in mathematics, the ratio of the circumference of a circle to its diameter. The symbol for pi is π. The ratio is the same for all circles and is approximately 3.1416. It is of great importance in mathematics not only in the measurement of the circle but also in more advanced mathematics in connection with such topics as continued fractions, logarithms of imaginary numbers, and periodic functions. Throughout the ages progressively more accurate values have been found for π; an early value was the Greek approximation 31/7, found by considering the circle as the limit of a series of regular polygons with an increasing number of sides inscribed in the circle. About the mid-19th cent. its value was figured to 707 decimal places and by the mid-20th cent. an electronic computer had calculated it to 100,000 digits. Although it has now been calculated to some 2.6 trillion digits, the exact value of π cannot be computed. It was shown by the German mathematician Johann Lambert in 1770 that π is irrational and by Ferdinand Lindemann in 1882 that π is transcendental; i.e., cannot be the root of any algebraic equation with rational coefficients. The important connection between π and e, the base of natural logarithms, was found by Leonhard Euler in the famous formula eiπ=-1, where i=-1. List of numbers – Irrational numbers ζ(3) – √2 – √3 – √5 – φ – α – e – π – δ Binary 11.00100100001111110110… Decimal 3.14159265358979323846… Hexadecimal 3.243F6A8885A308D31319… Continued fraction $3 + cfrac\left\{1\right\}\left\{7 + cfrac\left\{1\right\}\left\{15 + cfrac\left\{1\right\}\left\{1 + cfrac\left\{1\right\}\left\{292 + ddots\right\}\right\}\right\}\right\}$Note that this continued fraction is not periodic. Pi or π is a mathematical constant which represents the ratio of any circle's circumference to its diameter in Euclidean geometry, which is the same as the ratio of a circle's area to the square of its radius. It is approximately equal to 3.14159. Pi is one of the most important mathematical constants: many formulae from mathematics, science, and engineering involve π. Pi is an irrational number, which means that it cannot be expressed as a fraction m/n, where m and n are integers. Consequently its decimal representation never ends or repeats. Beyond being irrational, it is a transcendental number, which means that no finite sequence of algebraic operations on integers (powers, roots, sums, etc.) could ever produce it. Throughout the history of mathematics, much effort has been made to determine π more accurately and understand its nature; fascination with the number has even carried over into culture at large. The Greek letter π, often spelled out pi in text, was adopted for the number from the Greek word for perimeter "περίμετρος", probably by William Jones in 1706, and popularized by Leonhard Euler some years later. The constant is occasionally also referred to as the circular constant, (not to be confused with an Archimedes number), or . ## Fundamentals ### The letter π The name of the Greek letter π is pi, and this spelling is used in typographical contexts where the Greek letter is not available or where its usage could be problematic. When referring to this constant, the symbol π is always pronounced like "pie" in English, the conventional English pronunciation of the letter. In Greek, the name of this letter is IPA /pi/. The constant is named "π" because "π" is the first letter of the Greek words περιφέρεια (periphery) and περίμετρος (perimeter), probably referring to its use in the formula to find the circumference, or perimeter, of a circle. π is Unicode character U+03C0 ("Greek small letter pi"). ### Definition In Euclidean plane geometry, π is defined as the ratio of a circle's circumference to its diameter: $pi = frac\left\{c\right\}\left\{d\right\}.$ Note that the ratio c/d does not depend on the size of the circle. For example, if a circle has twice the diameter d of another circle it will also have twice the circumference c, preserving the ratio c/d. This fact is a consequence of the similarity of all circles. Alternatively π can be also defined as the ratio of a circle's area (A) to the area of a square whose side is equal to the radius: $pi = frac\left\{A\right\}\left\{r^2\right\}.$ The constant π may be defined in other ways that avoid the concepts of arc length and area, for example, as twice the smallest positive x for which cos(x) = 0. The formulas below illustrate other (equivalent) definitions. ### Irrationality and transcendence The constant π is an irrational number; that is, it cannot be written as the ratio of two integers. This was proven in 1761 by Johann Heinrich Lambert. In the 20th century, proofs were found that require no prerequisite knowledge beyond integral calculus. One of those, due to Ivan Niven, is widely known. A somewhat earlier similar proof is by Mary Cartwright. Furthermore, π is also transcendental, as was proven by Ferdinand von Lindemann in 1882. This means that there is no polynomial with rational coefficients of which π is a root. An important consequence of the transcendence of π is the fact that it is not constructible. Because the coordinates of all points that can be constructed with compass and straightedge are constructible numbers, it is impossible to square the circle: that is, it is impossible to construct, using compass and straightedge alone, a square whose area is equal to the area of a given circle. ### Numerical value The numerical value of π truncated to 50 decimal places is: 3.14159 26535 89793 23846 26433 83279 50288 41971 69399 37510 See the links below and those at sequence A000796 in OEIS for more digits. While the value of pi has been computed to more than a trillion (1012) digits, elementary applications, such as calculating the circumference of a circle, will rarely require more than a dozen decimal places. For example, a value truncated to 11 decimal places is accurate enough to calculate the circumference of the earth with a precision of a millimeter, and one truncated to 39 decimal places is sufficient to compute the circumference of any circle that fits in the observable universe to a precision comparable to the size of a hydrogen atom. Because π is an irrational number, its decimal expansion never ends and does not repeat. This infinite sequence of digits has fascinated mathematicians and laymen alike, and much effort over the last few centuries has been put into computing more digits and investigating the number's properties. Despite much analytical work, and supercomputer calculations that have determined over 1 trillion digits of π, no simple pattern in the digits has ever been found. Digits of π are available on many web pages, and there is software for calculating π to billions of digits on any personal computer. ### Calculating π π can be empirically estimated by drawing a large circle, then measuring its diameter and circumference and dividing the circumference by the diameter. Another geometry-based approach, due to Archimedes, is to calculate the perimeter, Pn , of a regular polygon with n sides circumscribed around a circle with diameter d. Then $pi = lim_\left\{n to infty\right\}frac\left\{P_\left\{n\right\}\right\}\left\{d\right\}$ That is, the more sides the polygon has, the closer the approximation approaches π. Archimedes determined the accuracy of this approach by comparing the perimeter of the circumscribed polygon with the perimeter of a regular polygon with the same number of sides inscribed inside the circle. Using a polygon with 96 sides, he computed the fractional range: $begin\left\{smallmatrix\right\}3frac\left\{10\right\}\left\{71\right\} < pi < 3frac\left\{1\right\}\left\{7\right\}end\left\{smallmatrix\right\}$. π can also be calculated using purely mathematical methods. Most formulas used for calculating the value of π have desirable mathematical properties, but are difficult to understand without a background in trigonometry and calculus. However, some are quite simple, such as this form of the Gregory-Leibniz series: $pi = frac\left\{4\right\}\left\{1\right\}-frac\left\{4\right\}\left\{3\right\}+frac\left\{4\right\}\left\{5\right\}-frac\left\{4\right\}\left\{7\right\}+frac\left\{4\right\}\left\{9\right\}-frac\left\{4\right\}\left\{11\right\}cdots!$. While that series is easy to write and calculate, it is not immediately obvious why it yields π. In addition, this series converges so slowly that 300 terms are not sufficient to calculate π correctly to 2 decimal places. However, by computing this series in a somewhat more clever way by taking the midpoints of partial sums, it can be made to converge much faster. Let $pi_\left\{0,1\right\} = frac\left\{4\right\}\left\{1\right\}, pi_\left\{0,2\right\} =frac\left\{4\right\}\left\{1\right\}-frac\left\{4\right\}\left\{3\right\}, pi_\left\{0,3\right\} =frac\left\{4\right\}\left\{1\right\}-frac\left\{4\right\}\left\{3\right\}+frac\left\{4\right\}\left\{5\right\}, pi_\left\{0,4\right\} =frac\left\{4\right\}\left\{1\right\}-frac\left\{4\right\}\left\{3\right\}+frac\left\{4\right\}\left\{5\right\}-frac\left\{4\right\}\left\{7\right\}, cdots!$ and then define $pi_\left\{i,j\right\} = frac\left\{pi_\left\{i-1,j\right\}+pi_\left\{i-1,j+1\right\}\right\}\left\{2\right\}$ for all $i,jge 1$ then computing $pi_\left\{10,10\right\}$ will take similar computation time to computing 150 terms of the original series in a brute force manner, and $pi_\left\{10,10\right\}=3.141592653cdots$, correct to 9 decimal places. This computation is an example of the Van Wijngaarden transformation. ## History The history of π parallels the development of mathematics as a whole. Some authors divide progress into three periods: the ancient period during which π was studied geometrically, the classical era following the development of calculus in Europe around the 17th century, and the age of digital computers. ### Geometrical period That the ratio of the circumference to the diameter of a circle is the same for all circles, and that it is slightly more than 3, was known to ancient Egyptian, Babylonian, Indian and Greek geometers. The earliest known approximations date from around 1900 BC; they are 25/8 (Babylonia) and 256/81 (Egypt), both within 1% of the true value. The Indian text gives π as 339/108 ≈ 3.139. The Tanakh appears to suggest, in the Book of Kings, that π = 3, which is notably worse than other estimates available at the time of writing (600 BC). The interpretation of the passage is disputed, as some believe the ratio of 3:1 is of an exterior circumference to an interior diameter of a thinly walled basin, which could indeed be an accurate ratio, depending on the thickness of the walls (See: Biblical value of π). Archimedes (287-212 BC) was the first to estimate π rigorously. He realized that its magnitude can be bounded from below and above by inscribing circles in regular polygons and calculating the outer and inner polygons' respective perimeters: By using the equivalent of 96-sided polygons, he proved that 223/71 < π < 22/7. Taking the average of these values yields 3.1419. In the following centuries further development took place in India and China. Around 265, the Wei Kingdom mathematician Liu Hui provided a simple and rigorous iterative algorithm to calculate π to any degree of accuracy. He himself carried through the calculation to 3072-gon and obtained an approximate value for π of 3.1416. $$ begin{align} pi approx A_{3072} & {} = 768 sqrt{2 - sqrt{2 + sqrt{2 + sqrt{2 + sqrt{2 + sqrt{2 + sqrt{2 + sqrt{2 + sqrt{2+1}}}}}}}}} & {} approx 3.14159. end{align} Later, Liu Hui invented a quick method of calculating π and obtained an approximate value of 3.1416 with only a 96-gon, by taking advantage of the fact that the difference in area of successive polygons forms a geometric series with a factor of 4. Around 480, the Chinese mathematician Zu Chongzhi demonstrated that π ≈ 355/113, and showed that 3.1415926 < π < 3.1415927 using Liu Hui's algorithm applied to a 12288-gon. This value would stand as the most accurate approximation of π over the next 900 years. ### Classical period Until the second millennium, π was known to fewer than 10 decimal digits. The next major advancement in the study of π came with the development of calculus, and in particular the discovery of infinite series which in principle permit calculating π to any desired accuracy by adding sufficiently many terms. Around 1400, Madhava of Sangamagrama found the first known such series: $frac\left\{pi\right\}\left\{4\right\} = 1 - frac\left\{1\right\}\left\{3\right\} + frac\left\{1\right\}\left\{5\right\} - frac\left\{1\right\}\left\{7\right\} + cdots!$ This is now known as the Madhava-Leibniz series or Gregory-Leibniz series since it was rediscovered by James Gregory and Gottfried Leibniz in the 17th century. Unfortunately, the rate of convergence is too slow to calculate many digits in practice; about 4,000 terms must be summed to improve upon Archimedes' estimate. However, by transforming the series into $pi = sqrt\left\{12\right\} , left\left(1-frac\left\{1\right\}\left\{3 cdot 3\right\} + frac\left\{1\right\}\left\{5 cdot 3^2\right\} - frac\left\{1\right\}\left\{7 cdot 3^3\right\} + cdotsright\right)!$ Madhava was able to calculate π as 3.14159265359, correct to 11 decimal places. The record was beaten in 1424 by the Persian mathematician, Jamshīd al-Kāshī, who determined 16 decimals of π. The first major European contribution since Archimedes was made by the German mathematician Ludolph van Ceulen (1540–1610), who used a geometrical method to compute 35 decimals of π. He was so proud of the calculation, which required the greater part of his life, that he had the digits engraved into his tombstone. Around the same time, the methods of calculus and determination of infinite series and products for geometrical quantities began to emerge in Europe. The first such representation was the Viète's formula, $frac2pi = frac\left\{sqrt2\right\}2 cdot frac\left\{sqrt\left\{2+sqrt2\right\}\right\}2 cdot frac\left\{sqrt\left\{2+sqrt\left\{2+sqrt2\right\}\right\}\right\}2 cdot cdots!$ found by François Viète in 1593. Another famous result is Wallis' product, $frac\left\{pi\right\}\left\{2\right\} = frac\left\{2\right\}\left\{1\right\} cdot frac\left\{2\right\}\left\{3\right\} cdot frac\left\{4\right\}\left\{3\right\} cdot frac\left\{4\right\}\left\{5\right\} cdot frac\left\{6\right\}\left\{5\right\} cdot frac\left\{6\right\}\left\{7\right\} cdot frac\left\{8\right\}\left\{7\right\} cdot frac\left\{8\right\}\left\{9\right\} cdots!$ written down by John Wallis in 1655. Isaac Newton himself derived a series for π and calculated 15 digits, although he later confessed: "I am ashamed to tell you to how many figures I carried these computations, having no other business at the time." In 1706 John Machin was the first to compute 100 decimals of π, using the formula $frac\left\{pi\right\}\left\{4\right\} = 4 , arctan frac\left\{1\right\}\left\{5\right\} - arctan frac\left\{1\right\}\left\{239\right\}!$ with $arctan , x = x - frac\left\{x^3\right\}\left\{3\right\} + frac\left\{x^5\right\}\left\{5\right\} - frac\left\{x^7\right\}\left\{7\right\} + cdots!$ Formulas of this type, now known as Machin-like formulas, were used to set several successive records and remained the best known method for calculating π well into the age of computers. A remarkable record was set by the calculating prodigy Zacharias Dase, who in 1844 employed a Machin-like formula to calculate 200 decimals of π in his head. The best value at the end of the 19th century was due to William Shanks, who took 15 years to calculate π with 707 digits, although due to a mistake only the first 527 were correct. (To avoid such errors, modern record calculations of any kind are often performed twice, with two different formulas. If the results are the same, they are likely to be correct.) Theoretical advances in the 18th century led to insights about π's nature that could not be achieved through numerical calculation alone. Johann Heinrich Lambert proved the irrationality of π in 1761, and Adrien-Marie Legendre proved in 1794 that also π2 is irrational. When Leonhard Euler in 1735 solved the famous Basel problem – finding the exact value of $frac\left\{1\right\}\left\{1^2\right\} + frac\left\{1\right\}\left\{2^2\right\} + frac\left\{1\right\}\left\{3^2\right\} + frac\left\{1\right\}\left\{4^2\right\} + cdots!$ which is π2/6, he established a deep connection between π and the prime numbers. Both Legendre and Leonhard Euler speculated that π might be transcendental, a fact that was proved in 1882 by Ferdinand von Lindemann. William Jones' book A New Introduction to Mathematics from 1706 is cited as the first text where the Greek letter π was used for this constant, but this notation became particularly popular after Leonhard Euler adopted it in 1737. He wrote: There are various other ways of finding the Lengths or Areas of particular Curve Lines, or Planes, which may very much facilitate the Practice; as for instance, in the Circle, the Diameter is to the Circumference as 1 to (16/5 - 4/239) - 1/3(16/5^3 - 4/239^3) + ... = 3.14159... = π ### Computation in the computer age The advent of digital computers in the 20th century led to an increased rate of new π calculation records. John von Neumann used ENIAC to compute 2037 digits of π in 1949, a calculation that took 70 hours. Additional thousands of decimal places were obtained in the following decades, with the million-digit milestone passed in 1973. Progress was not only due to faster hardware, but also new algorithms. One of the most significant developments was the discovery of the fast Fourier transform (FFT) in the 1960s, which allows computers to perform arithmetic on extremely large numbers quickly. In the beginning of the 20th century, the Indian mathematician Srinivasa Ramanujan found many new formulas for π, some remarkable for their elegance and mathematical depth. Two of his most famous formulas are the series $frac\left\{1\right\}\left\{pi\right\} = frac\left\{2 sqrt 2\right\}\left\{9801\right\} sum_\left\{k=0\right\}^infty frac\left\{\left(4k\right)!\left(1103+26390k\right)\right\}\left\{\left(k!\right)^4 396^\left\{4k\right\}\right\}!$ and $frac\left\{426880 sqrt\left\{10005\right\}\right\}\left\{pi\right\} = sum_\left\{k=0\right\}^infty frac\left\{\left(6k\right)! \left(13591409 + 545140134k\right)\right\}\left\{\left(3k\right)!\left(k!\right)^3 \left(-640320\right)^\left\{3k\right\}\right\}!$ which deliver 14 digits per term. The Chudnovsky brothers used this formula to set several π computing records in the end of the 1980s, including the first calculation of over one billion (1,011,196,691) decimals in 1989. It remains the formula of choice for π calculating software that runs on personal computers, as opposed to the supercomputers used to set modern records. Whereas series typically increase the accuracy with a fixed amount for each added term, there exist iterative algorithms that multiply the number of correct digits at each step, with the downside that each step generally requires an expensive calculation. A breakthrough was made in 1975, when Richard Brent and Eugene Salamin independently discovered the Brent–Salamin algorithm, which uses only arithmetic to double the number of correct digits at each step. The algorithm consists of setting $a_0 = 1 quad quad quad b_0 = frac\left\{1\right\}\left\{sqrt 2\right\} quad quad quad t_0 = frac\left\{1\right\}\left\{4\right\} quad quad quad p_0 = 1!$ and iterating $a_\left\{n+1\right\} = frac\left\{a_n+b_n\right\}\left\{2\right\} quad quad quad b_\left\{n+1\right\} = sqrt\left\{a_n b_n\right\}!$ $t_\left\{n+1\right\} = t_n - p_n \left(a_n-a_\left\{n+1\right\}\right)^2 quad quad quad p_\left\{n+1\right\} = 2 p_n!$ until an and bn are close enough. Then the estimate for π is given by $pi approx frac\left\{\left(a_n + b_n\right)^2\right\}\left\{4 t_n\right\}!$. Using this scheme, 25 iterations suffice to reach 45 million correct decimals. A similar algorithm that quadruples the accuracy in each step has been found by Jonathan and Peter Borwein. The methods have been used by Yasumasa Kanada and team to set most of the π calculation records since 1980, up to a calculation of 206,158,430,000 decimals of π in 1999. The current record is 1,241,100,000,000 decimals, set by Kanada and team in 2002. Although most of Kanada's previous records were set using the Brent-Salamin algorithm, the 2002 calculation made use of two Machin-like formulas that were slower but crucially reduced memory consumption. The calculation was performed on a 64-node Hitachi supercomputer with 1 terabyte of main memory, capable of carrying out 2 trillion operations per second. An important recent development was the Bailey–Borwein–Plouffe formula (BBP formula), discovered by Simon Plouffe and named after the authors of the paper in which the formula was first published, David H. Bailey, Peter Borwein, and Plouffe. The formula, $pi = sum_\left\{k=0\right\}^infty frac\left\{1\right\}\left\{16^k\right\} left\left(frac\left\{4\right\}\left\{8k + 1\right\} - frac\left\{2\right\}\left\{8k + 4\right\} - frac\left\{1\right\}\left\{8k + 5\right\} - frac\left\{1\right\}\left\{8k + 6\right\}right\right),$ is remarkable because it allows extracting any individual hexadecimal or binary digit of π without calculating all the preceding ones. Between 1998 and 2000, the distributed computing project PiHex used a modification of the BBP formula due to Fabrice Bellard to compute the quadrillionth (1,000,000,000,000,000:th) bit of π, which turned out to be 0. ### Memorizing digits Even long before computers have calculated π, memorizing a record number of digits became an obsession for some people. In 2006, Akira Haraguchi, a retired Japanese engineer, claimed to have recited 100,000 decimal places. This, however, has yet to be verified by Guinness World Records. The Guinness-recognized record for remembered digits of π is 67,890 digits, held by Lu Chao, a 24-year-old graduate student from China. It took him 24 hours and 4 minutes to recite to the 67,890th decimal place of π without an error. There are many ways to memorize π, including the use of "piems", which are poems that represent π in a way such that the length of each word (in letters) represents a digit. Here is an example of a piem: How I need a drink, alcoholic in nature (or: of course), after the heavy lectures involving quantum mechanics. Notice how the first word has 3 letters, the second word has 1, the third has 4, the fourth has 1, the fifth has 5, and so on. The contains the first 3834 digits of π in this manner. Piems are related to the entire field of humorous yet serious study that involves the use of mnemonic techniques to remember the digits of π, known as piphilology. See English mathematics mnemonics#Pi for examples. In other languages there are similar methods of memorization. However, this method proves inefficient for large memorizations of pi. Other methods include remembering patterns in the numbers. ## Advanced properties ### Numerical approximations Due to the transcendental nature of π, there are no closed form expressions for the number in terms of algebraic numbers and functions. Formulas for calculating π using elementary arithmetic typically include series or summation notation (such as "..."), which indicates that the formula is really a formula for an infinite sequence of approximations to π. The more terms included in a calculation, the closer to π the result will get. Consequently, numerical calculations must use approximations of π. For many purposes, 3.14 or 22/7 is close enough, although engineers often use 3.1416 (5 significant figures) or 3.14159 (6 significant figures) for more precision. The approximations 22/7 and 355/113, with 3 and 7 significant figures respectively, are obtained from the simple continued fraction expansion of π. The approximation 355⁄113 (3.1415929…) is the best one that may be expressed with a three-digit or four-digit numerator and denominator. The earliest numerical approximation of π is almost certainly the value 3. In cases where little precision is required, it may be an acceptable substitute. That 3 is an underestimate follows from the fact that it is the ratio of the perimeter of an inscribed regular hexagon to the diameter of the circle. ### Open questions The most pressing open question about π is whether it is a normal number — whether any digit block occurs in the expansion of π just as often as one would statistically expect if the digits had been produced completely "randomly", and that this is true in every base, not just base 10. Current knowledge on this point is very weak; e.g., it is not even known which of the digits 0,…,9 occur infinitely often in the decimal expansion of π. Bailey and Crandall showed in 2000 that the existence of the above mentioned Bailey-Borwein-Plouffe formula and similar formulas imply that the normality in base 2 of π and various other constants can be reduced to a plausible conjecture of chaos theory. It is also unknown whether π and e are algebraically independent, although Yuri Nesterenko proved the algebraic independence of {π, eπ, Γ(1/4)} in 1996. However it is known that at least one of πe and π + e is transcendental (see Lindemann–Weierstrass theorem). ## Use in mathematics and science π is ubiquitous in mathematics, appearing even in places that lack an obvious connection to the circles of Euclidean geometry. ### Geometry and trigonometry For any circle with radius r and diameter d = 2r, the circumference is πd and the area is πr2. Further, π appears in formulas for areas and volumes of many other geometrical shapes based on circles, such as ellipses, spheres, cones, and tori. Accordingly, π appears in definite integrals that describe circumference, area or volume of shapes generated by circles. In the basic case, half the area of the unit disk is given by: $int_\left\{-1\right\}^1 sqrt\left\{1-x^2\right\},dx = frac\left\{pi\right\}\left\{2\right\}$ and $int_\left\{-1\right\}^1frac\left\{1\right\}\left\{sqrt\left\{1-x^2\right\}\right\},dx = pi$ gives half the circumference of the unit circle. More complicated shapes can be integrated as solids of revolution. From the unit-circle definition of the trigonometric functions also follows that the sine and cosine have period 2π. That is, for all x and integers n, sin(x) = sin(x + 2πn) and cos(x) = cos(x + 2πn). Because sin(0) = 0, sin(2πn) = 0 for all integers n. Also, the angle measure of 180° is equal to π radians. In other words, 1° = (π/180) radians. In modern mathematics, π is often defined using trigonometric functions, for example as the smallest positive x for which sin x = 0, to avoid unnecessary dependence on the subtleties of Euclidean geometry and integration. Equivalently, π can be defined using the inverse trigonometric functions, for example as π = 2 arccos(0) or π = 4 arctan(1). Expanding inverse trigonometric functions as power series is the easiest way to derive infinite series for π. ### Higher analysis and number theory The frequent appearance of π in complex analysis can be related to the behavior of the exponential function of a complex variable, described by Euler's formula $e^\left\{ivarphi\right\} = cos varphi + isin varphi !$ where i is the imaginary unit satisfying i2 = −1 and e ≈ 2.71828 is Euler's number. This formula implies that imaginary powers of e describe rotations on the unit circle in the complex plane; these rotations have a period of 360° = 2π. In particular, the 180° rotation φ = π results in the remarkable Euler's identity $e^\left\{i pi\right\} = -1.!$ There are n different n-th roots of unity $e^\left\{2 pi i k/n\right\} qquad \left(k = 0, 1, 2, dots, n - 1\right).$ The Gaussian integral $int_\left\{-infty\right\}^\left\{infty\right\}e^\left\{-x^2\right\}dx=sqrt\left\{pi\right\}.$ A consequence is that the gamma function of a half-integer is a rational multiple of √π. ### Physics Although not a physical constant, π appears routinely in equations describing fundamental principles of the Universe, due in no small part to its relationship to the nature of the circle and, correspondingly, spherical coordinate systems. Using units such as Planck units can sometimes eliminate π from formulae. $Lambda = \left\{\left\{8pi G\right\} over \left\{3c^2\right\}\right\} rho$ • Heisenberg's uncertainty principle, which shows that the uncertainty in the measurement of a particle's position (Δx) and momentum (Δp) can not both be arbitrarily small at the same time: $Delta x, Delta p ge frac\left\{h\right\}\left\{4pi\right\}$ $R_\left\{ik\right\} - \left\{g_\left\{ik\right\} R over 2\right\} + Lambda g_\left\{ik\right\} = \left\{8 pi G over c^4\right\} T_\left\{ik\right\}$ • Coulomb's law for the electric force, describing the force between two electric charges (q1 and q2) separated by distance r: $F = frac\left\{left\|q_1q_2right\left\{4 pi varepsilon_0 r^2\right\}$ $mu_0 = 4 pi cdot 10^\left\{-7\right\},mathrm\left\{N/A^2\right\},$ • Kepler's third law constant, relating the orbital period (P) and the semimajor axis (a) to the masses (M and m) of two co-orbiting bodies: $frac\left\{P^2\right\}\left\{a^3\right\}=\left\{\left(2pi\right)^2 over G \left(M+m\right)\right\}$ ### Probability and statistics In probability and statistics, there are many distributions whose formulas contain π, including: • the probability density function for the normal distribution with mean μ and standard deviation σ, due to the Gaussian integral: $f\left(x\right) = \left\{1 over sigmasqrt\left\{2pi\right\} \right\},e^\left\{-\left(x-mu \right)^2/\left(2sigma^2\right)\right\}$ • the probability density function for the (standard) Cauchy distribution: $f\left(x\right) = frac\left\{1\right\}\left\{pi \left(1 + x^2\right)\right\}.$ Note that since $int_\left\{-infty\right\}^\left\{infty\right\} f\left(x\right),dx = 1$ for any probability density function f(x), the above formulas can be used to produce other integral formulas for π. Buffon's needle problem is sometimes quoted as a empirical approximation of π in "popular mathematics" works. Consider dropping a needle of length L repeatedly on a surface containing parallel lines drawn S units apart (with S > L). If the needle is dropped n times and x of those times it comes to rest crossing a line (x > 0), then one may approximate π using the Monte Carlo method: $pi approx frac\left\{2nL\right\}\left\{xS\right\}.$ Though this result is mathematically impeccable, it cannot be used to determine more than very few digits of π by experiment. Reliably getting just three digits (including the initial "3") right requires millions of throws, and the number of throws grows exponentially with the number of digits desired. Furthermore, any error in the measurement of the lengths L and S will transfer directly to an error in the approximated π. For example, a difference of a single atom in the length of a 10-centimeter needle would show up around the 9th digit of the result. In practice, uncertainties in determining whether the needle actually crosses a line when it appears to exactly touch it will limit the attainable accuracy to much less than 9 digits. ## See also • List of topics related to π • Proof that 22/7 exceeds π • Feynman point – comprising the 762nd through 767th decimal places of π, consisting of the digit 9 repeated six times. • Indiana Pi Bill. • Pi Day. • Software for calculating π on personal computers. • Mathematical constants: e and φ • SOCR resource hands-on activity for estimation of π using needle-dropping simulation	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 42, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9341980814933777, "perplexity_flag": "middle"}
http://mathoverflow.net/questions/16104?sort=votes	## Which Fréchet manifolds have a smooth partition of unity? ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) A classical theorem is saying that every smooth, finite-dimensional manifold has a smooth partition of unity. My question is: 1. Which Fréchet manifolds have a smooth partition of unity? 2. How is the existence of smooth partitions of unity on Fréchet manifolds related to paracompactness of the underlying topology? From some remarks in some literature, I got the impression that not all Fréchet manifolds have smooth partitions of unity, but some have, e.g. the loop space $LM$ of a finite-dimensional smooth manifold $M$. For $LM$, the proof seems to be that $LM$ is Lindelöf, hence paracompact. Is this true for all mapping spaces of the form $C^\infty (K,M)$ for $K$ compact? - ## 2 Answers Use the source, Luke. Specifically, chapters 14 (Smooth Bump Functions) to 16 (Smooth Partitions of Unity and Smooth Normality). You may be particularly interested in: Theorem 16.10 If $X$ is Lindelof and $\mathcal{S}$-regular, then $X$ is $\mathcal{S}$-paracompact. In particular, nuclear Frechet spaces are $C^\infty$-paracompact. For loop spaces (and other mapping spaces with compact source), the simplest argument for Lindelof/paracompactness that I know of goes as follows: 1. Embed $M$ as a submanifold of $\mathbb{R}^n$. 2. So the loop space $LM$ embeds as a submanifold of $L\mathbb{R}^n$. 3. $L\mathbb{R}^n$ is metrisable. 4. So $L M$ is metrisable. 5. Hence $L M$ is paracompact. (Paracompactness isn't inheritable by all subsets. Of course, if you can embed your manifold as a closed subspace then you can inherit the paracompactness directly.) I use this argument in my paper on Constructing smooth manifolds of loop spaces to show that most "nice" properties devolve from the model space to the loop space for "nice" model spaces (smooth, continuous, and others). See corollary C in the introduction of the published version. (I should note that the full statement of Theorem 16.10 (which I did not quote above) is not quite correct (at least in the book version, it may have been corrected online) in that the proof of the claim for strict inductive sequences is not complete. I needed a specific instance of this in my preprint The Smooth Structure of Piecewise-Smooth Loops (see section 5.4.2) which wasn't covered by 16.10 but fortunately I could hack together bits of 16.6 with 16.10 to get it to work. This, however, is outside the remit of this question as it deals with spaces more general than Frechet spaces.) On the opposite side of the equation, we have the following after 16.10: open problem ... Is every paracompact $\mathcal{S}$-regular space $\mathcal{S}$-paracompact? So the general case is not (at time of publishing) known. But for manifolds, the case is somewhat better: Ch 27 If a smooth manifold (which is smoothly Hausdorff) is Lindelof, and if all modelling vector spaces are smoothly regular, then it is smoothly paracompact. If a smooth manifold is metrisable and smoothly normal then it is smoothly paracompact. Since Banach spaces are Frechet spaces, any Banach space that is not $C^\infty$-paracompact provides a counterexample for Frechet spaces as well. The comment after 14.9 provides the examples of $\ell^1$ and $C([0,1])$. So, putting it all together: nuclear Frechet spaces are good, so Lindelof manifolds modelled on them are smoothly paracompact. Smooth mapping spaces (with compact source) are Lindelof manifolds with nuclear model spaces, hence smoothly paracompact. (Recall that smooth mapping spaces without compact source aren't even close to being manifolds. I know that Konrad knows this, I merely put this here so that others will know it too.) - Hi Andrew, I totally forgot about the source. Many thanks for your answer! When you are saying "smoothly paracompact" you mean exactly "has a smooth partition of unity", right? – Konrad Waldorf Feb 23 2010 at 17:55 Yes. I mean that every open cover has a smooth partition of unity subordinate to it (the closures of the supports are a locally finite cover refining the initial one). – Andrew Stacey Feb 23 2010 at 18:37 ### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you. One also has to assume that M is compact for LM to be paracompact, as well as finite dimensional. A question that should have a slightly easier answer is this: which Frechet spaces have partitions of unity? (Dang, I just registered and lost all my rep from when I was an unregistered user. I would leave a comment otherwise) - If you email the mods or leave a message on meta they can combine the accounts for you. – jc Feb 23 2010 at 1:45 can you explain why M needs to be compact? or give a reference or something? Thanks! (I know the moderators would be happy to combine your accounts). – Chris Schommer-Pries Feb 23 2010 at 4:10 You should now own all your posts and have all your rep. Let me know if you have any other account problems. – Anton Geraschenko♦ Feb 23 2010 at 4:52 Hi David, M does not have to be compact to make LM paracompact. See Brylinski's book, Prop. 3.1.2 (citing Pressley-Segal). [I should have written Lindelöf instead of metrizable above] – Konrad Waldorf Feb 23 2010 at 6:15 Whoops! Probably getting confused with needing compact source. I considered referring to my esteemed colleague above, but I'm not sure about the propriety of saying 'so-and-so knows this', since it is a cheap way of answering without answering. – David Roberts Feb 24 2010 at 22:56	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 23, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9180847406387329, "perplexity_flag": "middle"}
http://mathhelpforum.com/algebra/66898-solving-equation-contains-x-sqrtx.html	# Thread: 1. ## solving an equation which contains x and sqrtx Techniques of Differentiation When $3x+4\sqrt{x}-1=0$ then why is $\sqrt{x}=\frac{-2+\sqrt{7}}{3}?$ It wouldn't be a problem for me if the equation would be in the normal quadratic form. Multiplying all terms with $x$ wouldn't do any good since I would get $x\sqrt{x}$. Same if I would quadratize the equation by itself. I think that they used the quadratic formula somehow but I don't see how. Obviously the problem is the combination of 2. Originally Posted by Sebastian de Vries Techniques of Differentiation When $3x+4\sqrt{x}-1=0$ then why is $\sqrt{x}=\frac{-2+\sqrt{7}}{3}?$ It wouldn't be a problem for me if the equation would be in the normal quadratic form. Multiplying all terms with $x$ wouldn't do any good since I would get $x\sqrt{x}$. Same if I would quadratize the equation by itself. I think that they used the quadratic formula somehow but I don't see how. Obviously the problem is the combination of Put $y=\sqrt{x}$ to get the quadratic: $3y^2+4y-1=0$ Now use the quadratic formula and you get two solutions, one of which is negative and so is a spurious solution since $y=\sqrt{x}\ge0$, the remaining positive root gives the solution. . 3. Hmm, when I do this I get: $\sqrt{x}=\frac{-4\pm\sqrt{28}}{6}$ Not sure if this is something I've missed or what? My above answer was done using the quadratic forumula by the way 4. Originally Posted by craig Hmm, when I do this I get: $\sqrt{x}=\frac{-4\pm\sqrt{28}}{6}$ Not sure if this is something I've missed or what? $\sqrt{28} = 2 \sqrt{7}$ .... 5. how did I miss that one lol. 6. Thanks.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 14, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9701365232467651, "perplexity_flag": "middle"}
http://mathoverflow.net/questions/29232/the-unification-of-mathematics-via-topos-theory/29553	## The unification of Mathematics via Topos Theory ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) When the paper The unification of Mathematics via Topos Theory by Olivia Caramello, says "one can generate a huge number of new results in any mathematical field without any creative effort." is this an exaggeration, and if not is this a new idea or has it always been thought that topos theory could enable automatic generation of theorems. - 17 I'd love to see the first theorem on the Navier-Stokes equations proved by means of the topos theory... – Andrey Rekalo Jun 23 2010 at 13:37 26 Being able to create a huge number of new dishes doesn't make one a chef. – Charles Matthews Jun 23 2010 at 13:58 14 To those voting to close: honestly, people! Whether or not the question sounds polemical to you, there are already some interesting answers below and perhaps more are coming. That's what matters. – algori Jun 25 2010 at 23:27 15 One doesn't need to know anything about topos theory to know that the claim "One can generate a huge number of new results in any mathematical field without any creative effort" is trivially true if the new results are not required to be of interest to anyone and trivially false otherwise: almost by definition "creative effort" is that which produces interesting, new theorems. I found the paper (including its title) to be rather over the top. Probably others here feel similarly. But it shouldn't reflect negatively on the OP -- he just asked for our opinion on this curious statement. – Pete L. Clark Jun 26 2010 at 6:02 13 Why are there suddenly so many straw men being thrown around? Yes, this statement is trivially true if taken to cover any new theorems, and almost certainly false if very specific well-known theorems are asked for. But the original intent is clearly somewhere between: how many interesting non-trivial theorems are there (none? some? or many, as Caramello claims?) that can be read straight off via topos-theoretic dictionaries from theorems in other areas? And this is surely an interesting and reasonably precise question! – Peter LeFanu Lumsdaine Jun 27 2010 at 19:46 show 3 more comments ## 5 Answers Topos theory provides a dictionary between (certain areas of) logic and (certain areas of) geometry. As such, it provides all the benefits that mathematical dictionaries do: It lets you translate between two languages whose natural evolutions proceeded independently. An insight that is obvious in one domain may not be so obvious when translated to the other domain. Dictionaries cannot perform magic. In particular, it is usually too optimistic to think that a dictionary will allow you to prove significant new theorems with no effort. True, sometimes we do get lucky in this way. When Richard Stanley first discovered the dictionary between toric varieties and convex polytopes, he almost immediately reaped the reward of proving an important combinatorial conjecture with very little effort, because geometers had already put in a lot of work to solve exactly the problem he needed. But more commonly, the payoff of a dictionary is that it allows you to formulate good questions with very little effort. That is, you now have a new way to think about old problems, so you may be able to find your way more easily to a solution by borrowing concepts from both domains. You will still need to do hard work to solve hard problems, but your toolbox is now bigger. - I believe some things which were impossible to prove in the past are now possible to be proven due to new mathematical techniques. Is there a book on such history? – Fábio Torres Apr 20 at 7:06 ### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you. I can't really answer the question, but I attended Olivia's talk today at CT2010 in Genova where she was presenting this paper, and I asked her some questions afterwards. Any errors or misrepresentations in my following comments are entirely mine. When the informal statement is made technical, there appear to be some qualifications. First, it applies only to mathematical theories that can be formalized in geometric logic. This is a kind of logic with a restriction on how quantifiers can be used in axioms. According to Olivia, this is not much of a restriction, because it includes for example finitary logic, and more. Second, to be able to transport results from one theory to another, the theories must be Morita equivalent. Apparently this is more common than one might think, and once one has Morita equivalence, one can transport all kinds of results back and forth. So perhaps it would be fair to say that "some creative effort" goes into finding appropriate Morita equivalences in the first place. Olivia refered to her Ph.D. thesis for a number of examples where this had been done. I have not really seen the examples. The areas of mathematics that she specifically mentioned in her talk were algebra, topology, two different branches of logic (proof theory and model theory), and a 5th one which I forgot. Definitely not PDE's or Riemannian geometry. Two more remarks: a topos is a kind of set theory, so it is indeed plausible that all kinds of areas of mathematics can be formalized within a topos. Sometimes this is done by "real" mathematicians (i.e., non-logicians). For example, Tom Hales's ongoing formalization of the proof of the Kepler conjecture is done in HOL Light, which uses a topos logic as far as I remember. (The proof is not constructive, as excluded middle and the axiom of choice have been added as additional axioms). Tom has already given a complete machine-checkable proof of the Jordan curve theorem. The point of doing this in HOL Light, at this point, is not so much the topos aspect, but the fact that the proofs can be verified by machine, eliminating the referee's residual uncertainty. This is impractical with "ordinary" set theory. Last, when one claims to be able to generate "new theorems", this does not necessarily mean "interesting new theorems". I guess that in some cases, they can be interesting, but perhaps more so to logicians than to other mathematicians. Then again, it's happened before that disjoint areas of mathematics were related, and when this happens, it is always useful. Seems nice to have another avenue. - Welcome to MathOverflow, and thanks for the comment! – Neel Krishnaswami Jun 25 2010 at 16:32 This statement is true, but there's substantially less than meets the eye to it. Topoi are gadgets which are both models of both a fairly large fragment of logic (typed higher-order logic), and are generalizations of sheaves on a topological space. (The reason this connection is possible is morally that the open sets of a topological space form a Heyting algebra, which is a model of propositional intuitionistic logic.) As a result of their high logical strength, you can take many practical constructions and encode them in the internal logic of a topos. Then you can take an external view and hit these constructions with topologist technical wizardry. Since topologists and logicians talk to each other rather less frequently than we ought to, this method is a very fertile source of theorems. "Without creative effort" is just a rhetorical flourish intended to encourage people to learn about topoi -- there's no way (presently) that a computer could generate theorems via topos theory. But it is an effective tool for human mathematicians to port ideas from one area to another, which has always been a productive endeavor. The book to read on this subject is MacLane and Mordeijk's Sheaves in Geometry and Logic, which, as you can guess from the title, emphasizes the moral that taking a synthesizing perspective on mathematics is often valuable. - 8 Please give an example of an interesting theorem generated via the procedure of the second paragraph. – Boyarsky Jun 23 2010 at 15:54 8 Fiore and Simpson's "Lambda-Definability with Sums via Grothendieck Logical Relations" showed how to adapt the idea of a cover algebra or Grothendieck topology to the setting of structural proof theory, which Balat et al subsequently used to give a normalization algorithm for the lambda calculus with disjoint sum types. The insight was that the case-distinction in the disjunction elimination rule could be split apart and the pieces could be taken as a kind of "open cover" of the whole proof. – Neel Krishnaswami Jun 23 2010 at 16:34 7 Dear Neel: Unfortunately I don't understand what your example is saying (even the insight which you say hadn't been noticed before). I should have said "interesting theorem in a field of mathematics different from logic", since logicians appear to already be sold on topoi. Since you say the quote in the question is true, in the spirit of "any mathematical field" let's be specific and take 3 big fields: PDE, Riemannian geometry, and representation theory. – Boyarsky Jun 23 2010 at 17:23 2 Dear Neel: Good luck with the representation theory; it's a beautiful subject. (It boggles my mind how it could have any applications of the sort you suggest, but perhaps that is due to my own ignorance of logic.) – Boyarsky Jun 24 2010 at 12:54 3 I'm a fan of category theory in general and think toposes are just nifty (for geometry at least), but this answer made me cringe. – Harry Gindi Jun 25 2010 at 23:17 show 2 more comments It would be presumptuous on my part to attempt to answer this question, but I want to share with other MOers this recent paper http://www.ihes.fr/~lafforgue/math/TheorieCaramello.pdf of Laurent Lafforgue and this video of one of his recent lectures, [dont] le but [] est de poser cette question (inspirée par la théorie de Caramello) : l'indépendance de $l$ de la cohomologie $l$-adique et la correspondance de Langlands sont-elles des équivalences de Morita entre topos classifiants ? Here is a quote from the paper : La théorie de Caramello... offre déjà un très grand nombre d'exemples d'équivalences de Morita et de leurs applications. Ces exemples sont étonnament divers et ils apparaissent presque toujours comme surprenants. Beaucoup d'énoncés auraient été très difficiles à démontrer, et plus encore à imaginer, sans les topos et sans les méthodes de calcul que la théorie des topos classifiants et des équivalences de Morita rend possibles et naturelles. Quand on songe que la correspondance de Langlands ressemble beaucoup à une equivalence de Morita et qu'elle en est peut-être une, on se dit que le champ ouvert à cette théorie est immense. - 4 Thanks a lot for sharing this. So fields medalist Lafforgue seems to be convinced of the power and the future impact of Caramello's theory. – Martin Brandenburg Mar 18 at 17:07 3 and he's not isolated at the IHES with his opinion... ;) – BY Mar 18 at 18:48 When a phrase is taken from a context, it is easy to misread it. If one looks at the paper, "one can generate a huge number of new results in any mathematical field without any creative effort" makes sense and is justified. The idea is that most Mathematics can be formalized as a geometric theory, i.e, a logical theory with infinitary disjunction and finitary conjunction with the natural rules of inference. This is true since most Mathematics can be formalized in first-order logic which can be translated in geometric logic, and the translation preserves set-theoretic models. The model theory of geometric logic is naturally categorical; in fact, it is sound and complete (in a strong sense) when one considers the categorical models living inside Grothendieck toposes. The completeness theorem is proved by constructing a universal model which lives in a special topos, the classifying topos. Olivia's result shows that very different theories share the same classifying topos (or better, they have different but equivalent classifying toposes). When this fact happens, and it is not uncommon - Olivia provides very general techniques to achieve this result - the classifying topos allows to move results from one theory to another. Examples have shown that very simple results in one theory translate into very deep results in other theories. But the translation technique is not-dictionary oriented but rather "modulo an abstract invariant" which allows for an extremly high degree of freedom. The interesting thing in this machinery, which is not automatic, is exactly that a huge number of new mathematical results in any field of Mathematics can be generated without any significant creative effort. Of course, not every result can be generated in this way, as far as we know now, and not every generated result is significant. The mathematical intuition of an educated mathematician is needed to choose the right theories and the correct invariant to achieve a significant result. Olivia's examples show that very simple invariants applied to very classical theories provide explanations to deep results and a couple of new interpretation of classical results have been obtained. As far as I'm able to understand Olivia's results, I believe her conclusions because her articles prove them. Of course, my understanding is not perfect and the explanation above are approximations of the real results. And yes, she is doing a big claim - but I became convinced that it is a true one. - 6 @Marco: It will help to clarify matters if you can provide one example of a deep result in some part of mathematics apart from logic or set theory which is deduced from a simple result in another theory by means of topoi. (Note: results which are routine applications of Zorn's Lemma and/or easily equivalent to the Axiom of Choice do not count as deep.) – BCnrd Jun 29 2010 at 3:40 2 @BCnrd: a partial answer to your question can be find in arxiv.org/abs/0808.1972 which shows a non-trivial result obtained in Algebra, specifically in the theory of fields. On the other side, one should consider that Olivia's theory has been conceived in 2008. It has reached its full expression at the end of 2009, and in the last months it has been refined and extended, as one can check in the Math Archive. It is not strange that there is a limited number of examples outside logic: If one has to develop a new theory, s/he would first focus on the field where it naturally lives. – Marco Benini Jun 29 2010 at 14:31 8 Dear Marco: what is interesting (let alone deep) from the viewpoint of the theory of fields in the link you give? (I don't think I have ever encountered a situation where the DeMorgan issue is relevant, and I know the theory of fields extremely well and have used it in many many ways.) In particular I still don't see what justifies the claim that her works allows one to translate something simple in one part of math into something deep in another. As far as I can tell, the grand sweep of what Olivia claims is not appropriate at the present time. – BCnrd Jun 30 2010 at 4:34 9 @Marco: Thanks very much for the follow-up! Now that I see the definitions of terms I didn't know, the assertion with finite fields is very well-known, easy to prove directly (not deep), and has been used for a long time. If someone told you it's hard to see with standard methods, they're wrong. It supports my original "concern": after unraveling the logic terminology, the "transferred" result in the other area of math will be obvious if no creative effort was required. It's up to Olivia to justify her claim of interesting/deep applications to other areas of math; I'll leave it to her. – BCnrd Jul 12 2010 at 12:36 3 Dear BCnrd, I have to say that I'm quite surprised by your answer. I see now that the example we discussed is "easy" in the theory of fields, but the point I wanted to make is that, once you have a description of the classifying topos, this result as well as many others (e.g., the ones I previously reported) flow naturally and immediately from it (and I really mean by a one-line yet entirely rigorous proof) - note that to prove this result by standard methods, a basic knowledge of the theory of field extensions is needed, while the topos-theoretic proof follwos a completely different path. – Marco Benini Jul 16 2010 at 21:40 show 14 more comments	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.933496356010437, "perplexity_flag": "middle"}
http://quant.stackexchange.com/questions/7243/what-are-d-1-and-d-2-for-laplace/7256	What are $d_1$ and $d_2$ for Laplace? What are the formulae for `d1` & `d2` using a Laplace distribution? - 1 Answer Your question is interesting because I thought that the only chance with Lévy-processes is to use Fourier-transform approaches (see e.g. Cont,Tankov). But in the paper Option Pricing for Log-Symmetric Distributions of Returns by Fima C. Klebaner· Zinoviy Landsman they consider models, where the log of the price has a symmetric distribution. In Corollary 3.2 they propose an approximate formula if the log of the price follows the Laplace distribution where $$d_{1,2} = \frac{\ln(S_0/K) + (r \pm log(1-\sigma^2/2))T}{\sigma \sqrt{T}}. $$ They write $\log$ but I guess this is just an $\ln$ as before. But please try yourself and/or read the first pages of the paper to be sure. - Good to hear that. I have to play a bit with this distribution myself. Your other questions in this context look interesting too... – Richard Feb 9 at 23:00	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 4, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.918633759021759, "perplexity_flag": "middle"}
http://www.all-science-fair-projects.com/science_fair_projects_encyclopedia/Probability_distribution	# All Science Fair Projects ## Science Fair Project Encyclopedia for Schools! Search Browse Forum Coach Links Editor Help Tell-a-Friend Encyclopedia Dictionary # Science Fair Project Encyclopedia For information on any area of science that interests you, enter a keyword (eg. scientific method, molecule, cloud, carbohydrate etc.). Or else, you can start by choosing any of the categories below. # Probability distribution In mathematics, a probability distribution assigns to every interval of the real numbers a probability, so that the probability axioms are satisfied. In technical terms, a probability distribution is a probability measure whose domain is the Borel algebra on the reals. A probability distribution is a special case of the more general notion of a probability measure, which is a function that assigns probabilities satisfying the Kolmogorov axioms to the measurable sets of a measurable space. Every random variable gives rise to a probability distribution, and this distribution contains most of the important information about the variable. If X is a random variable, the corresponding probability distribution assigns to the interval [a, b] the probability Pr[a ≤ X ≤ b], i.e. the probability that the variable X will take a value in the interval [a, b]. The probability distribution of the variable X can be uniquely described by its cumulative distribution function F(x), which is defined by $F(x) = \Pr\left[ X \le x \right]$ for any x in R. A distribution is called discrete if its cumulative distribution function consists of a sequence of finite jumps, which means that it belongs to a discrete random variable X: a variable which can only attain values from a certain finite or countable set. A distribution is called continuous if its cumulative distribution function is continuous, which means that it belongs to a random variable X for which Pr[ X = x ] = 0 for all x in R. The so-called absolutely continuous distributions can be expressed by a probability density function: a non-negative Lebesgue integrable function f defined on the reals such that $\Pr \left[ a \le X \le b \right] = \int_a^b f(x)\,dx$ for all a and b. That discrete distributions do not admit such a density is unsurprising, but there are continuous distributions like the devil's staircase that also do not admit a density. The support of a distribution is the smallest closed set whose complement has probability zero. Contents ## List of important probability distributions Several probability distributions are so important in theory or applications that they have been given specific names: ### Discrete distributions #### With finite support • The Bernoulli distribution, which takes value 1 with probability p and value 0 with probability q = 1 − p. • The Rademacher distribution, which takes value 1 with probability 1/2 and value −1 with probability 1/2. • The binomial distribution, which describes the number of successes in a series of independent Yes/No experiments. • The degenerate distribution at x0, where X is certain to take the value x0. This does not look random, but it satisfies the definition of random variable. This is useful because it puts deterministic variables and random variables in the same formalism. • The discrete uniform distribution, where all elements of a finite set are equally likely. This is supposed to be the distribution of a balanced coin, an unbiased die, a casino roulette or a well-shuffled deck. Also, one can use measurements of quantum states to generate uniform random variables. All these are "physical" or "mechanical" devices, subject to design flaws or perturbations, so the uniform distribution is only an approximation of their behaviour. In digital computers, pseudo-random number generators are used to produced a statistically random discrete uniform distribution. • The Ewens sampling formula is a probability distribution on the set of all partitions of an integer n, arising in population genetics. • The hypergeometric distribution, which describes the number of successes in the first m of a series of n independent Yes/No experiments, if the total number of successes is known. • Zipf's law or the Zipf distribution. A discrete power-law distribution, the most famous example of which is the description of the frequency of words in the English language. • The Zipf-Mandelbrot law is a discrete power law distribution which is a generalization of the Zipf distribution. #### With infinite support • The Boltzmann distribution, a discrete distribution important in statistical physics which describes the probabilities of the various discrete energy levels of a system in thermal equilibrium. It has a continuous analogue. Special cases include: • The Gibbs distribution • The Maxwell-Boltzmann distribution • The Bose-Einstein distribution • The Fermi-Dirac distribution • The geometric distribution, a discrete distribution which describes the number of attempts needed to get the first success in a series of independent Yes/No experiments. • The logarithmic (series) distribution • The negative binomial distribution, a generalization of the geometric distribution to the nth success. • The parabolic fractal distribution • The Poisson distribution, which describes a very large number of individually unlikely events that happen in a certain time interval. • The Skellam distribution, the distribution of the difference between two independent Poisson-distributed random variables. • The Yule-Simon distribution • The zeta distribution has uses in applied statistics and statistical mechanics, and perhaps may be of interest to number theorists. It is the Zipf distribution for an infinite number of elements. ### Continuous distributions #### Supported on a bounded interval • The Beta distribution on [0,1], of which the uniform distribution is a special case, and which is useful in estimating success probabilities. • The continuous uniform distribution on [a,b], where all points in a finite interval are equally likely. • The rectangular distribution is a uniform distribution on [-1/2,1/2]. • The Dirac delta function although not strictly a function, is a limiting form of many continuous probability functions. It represents a discrete probability distribution concentrated at 0 — a degenerate distribution — but the notation treats it as if it were a continuous distribution. • The logarithmic distribution (continuous) • The triangular distribution on [a, b], a special case of which is the distribution of the sum of two uniformly distributed random variables (the convolution of two uniform distributions). • The von Mises distribution • The Wigner semicircle distribution is important in the theory of random matrices. #### Supported on semi-infinite intervals, usually [0,∞) • The chi distribution • The chi-square distribution, which is the sum of the squares of n independent Gaussian random variables. It is a special case of the Gamma distribution, and it is used in goodness-of-fit tests in statistics. • The inverse-chi-square distribution • The noncentral chi-square distribution • The scale-inverse-chi-square distribution • The exponential distribution, which describes the time between consecutive rare random events in a process with no memory. • The F-distribution, which is the distribution of the ratio of two normally distributed random variables, used in the analysis of variance. • The Gamma distribution, which describes the time until n consecutive rare random events occur in a process with no memory. • The Erlang distribution, which is a special case of the gamma distribution with integral shape parameter, developed to predict waiting times in queuing systems . • The inverse-gamma distribution • Fisher's z-distribution • The half-normal distribution • The log-logistic distribution • The log-normal distribution, describing variables which can be modelled as the product of many small independent positive variables. • The Pareto distribution, or "power law" distribution, used in the analysis of financial data and critical behavior. • The Rice distribution • The type-2 Gumbel distribution • The Wald distribution • The Weibull distribution, of which the exponential distribution is a special case, is used to model the lifetime of technical devices. #### Supported on the whole real line • The Beta prime distribution • The Cauchy distribution, an example of a distribution which does not have an expected value or a variance. In physics it is usually called a Lorentzian profile, and is associated with many processes, including resonance energy distribution, impact and natural spectral line broadening and quadratic stark line broadening. • The Fisher-Tippett, extreme value, or log-Weibull distribution • The Gumbel distribution, a special case of the Fisher-Tippett distribution • The Landau distribution • The Laplace distribution • The Levy stable distribution is often used to characterize financial data and critical behavior. • The map-Airy distribution • The normal distribution, also called the Gaussian or the bell curve. It is ubiquitous in nature and statistics due to the central limit theorem: every variable that can be modelled as a sum of many small independent variables is approximately normal. • Rayleigh distribution • Student's t-distribution, useful for estimating unknown means of Gaussian populations. • The noncentral t-distribution • The type-1 Gumbel distribution • The Voigt distribution, or Voigt profile, is the convolution of a normal distribution and a Cauchy distribution. It is found in spectroscopy when spectral line profiles are broadened by a mixture of Lorentzian and Doppler broadening mechanisms. ### Joint distributions #### Two or more random variables on the same sample space • Dirichlet distribution, a generalization of the beta distribution. • multinomial distribution, a generalization of the binomial distribution. • multivariate normal distribution, a generalization of the normal distribution. #### Matrix-valued distributions • Wishart distribution • matrix normal distribution • matrix t-distribution ## See also 03-10-2013 05:06:04	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 2, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.8585456013679504, "perplexity_flag": "head"}
http://mathhelpforum.com/discrete-math/54859-need-help-doing-proof.html	# Thread: 1. ## Need help doing a proof Prove: If a positive integer n is not prime, then 2^(n)-1 is not prime. If n is not prime, then it factors into n = ab. Use the identity x^(b) = (x-1)(x^(b-1) + x^(b-2) + ...+ x + 1) and substitute x = 2^(a). 2. Originally Posted by noles2188 Prove: If a positive integer n is not prime, then 2^(n)-1 is not prime. If n is not prime, then it factors into n = ab. Use the identity x^(b) = (x-1)(x^(b-1) + x^(b-2) + ...+ x + 1) and substitute x = 2^(a). There's a "-1" missing in that identity. It should be $x^b {\color{red}{} - 1} = (x-1)(x^{b-1} + x^{b-2} + \ldots + x + 1)$. Substitute $x=2^a$ in that and you should have no trouble getting the result.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 2, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.8316499590873718, "perplexity_flag": "middle"}
http://unapologetic.wordpress.com/2008/11/06/bialgebras/?like=1&source=post_flair&_wpnonce=c327cf8c6a	# The Unapologetic Mathematician ## Bialgebras In yesterday’s post I used the group algebra $\mathbb{F}[G]$ of a group $G$ as an example of a coalgebra. In fact, more is true. A bialgebra is a vector space $A$ equipped with both the structure of an algebra and the structure of a coalgebra, and that these two structures are “compatible” in a certain sense. The traditional definitions usually consist in laying out the algebra maps and relations, then the coalgebra maps and relations. Then they state that the algebra structure preserves the coalgebra structure, and that the coalgebra structure preserves the algebra structure, and they note that really you only need to require one of these last two conditions because they turn out to be equivalent. In fact, our perspective allows this equivalence to come to the fore. The algebra structure makes the bialgebra a monoid object in the category of vector space over $\mathbb{F}$. Then a compatible coalgebra structure makes it a comonoid object in the category of algebras over $\mathbb{F}$. Or in the other order, we have a monoid object in the category of comonoid objects in the category of vector spaces over $\mathbb{F}$. And these describe essentially the same things because internalizations commute! Okay, let’s be explicit about what we mean by “compatibility”. This just means that — on the one side — the coalgebra maps are not just linear maps between the underlying vector spaces, but actually are algebra homomorphisms. On the other side, it means that the algebra maps are actually coalgebra homomorphisms. Multiplication and comultiplication being compatible actually mean the same thing. Take two algebra elements and multiply them, then comultiply the result. Alternatively, comultiply each of them, and the multiply corresponding factors of the result. We should get the same answer whether we multiply or comultiply first. That is: $\Delta\circ\mu=(\mu\otimes\mu)\circ(1_A\otimes\tau_{A,A}\otimes1_A)\circ(\Delta\otimes\Delta)$, where $\tau$ is the twist map, exchanging two factors. Let’s check this condition for the group algebra $\mathbb{F}[G]$: $\begin{aligned}\left[\mu\otimes\mu\right]\left(\left[1_A\otimes\tau_{A,A}\otimes1_A\right]\left(\left[\Delta\otimes\Delta\right](e_g\otimes e_h)\right)\right)=\\\left[\mu\otimes\mu\right]\left(\left[1_A\otimes\tau_{A,A}\otimes1_A\right](e_g\otimes e_g\otimes e_h\otimes e_h)\right)=\\\left[\mu\otimes\mu\right](e_g\otimes e_h\otimes e_g\otimes e_h)=e_{gh}\otimes e_{gh}=\\\Delta(e_{gh})=\Delta\left(\mu(e_g\otimes e_h)\right)\end{aligned}$ Similarly, if we multiply two algebra elements and then take the counit, it should be the same as the product (in $\mathbb{F}$) of the counits of the elements. Dually, the product of two copies of the algebra unit should be the algebra unit again, and the counit of the algebra unit should be the unit in $\mathbb{F}$. It’s straightforward to verify that these hold for $\mathbb{F}[G]$. ### Like this: Posted by John Armstrong \| Algebra, Category theory ## 5 Comments » 1. [...] One more piece of structure we need. We take a bialgebra , and we add an “antipode”, which behaves sort of like an inverse operation. Then what [...] Pingback by \| November 7, 2008 \| Reply 2. [...] of Bialgebras What’s so great about bialgebras? Their categories of representations are [...] Pingback by \| November 11, 2008 \| Reply 3. [...] we said that a bialgebra is a comonoid object in the category of algebras over . But let’s consider this category [...] Pingback by \| November 18, 2008 \| Reply 4. [...] is cocommutative if we can swap the outputs from the comultiplication. That is, if . Similarly, bialgebras and Hopf algebras can be [...] Pingback by \| November 19, 2008 \| Reply 5. [...] actually seen this before, a long while back. The group algebra is an example of a bialgebra, with the map serving as a “comultiplication”. If this seems complicated, don’t [...] Pingback by \| November 5, 2010 \| Reply « Previous \| Next » ## About this weblog This is mainly an expository blath, with occasional high-level excursions, humorous observations, rants, and musings. The main-line exposition should be accessible to the “Generally Interested Lay Audience”, as long as you trace the links back towards the basics. Check the sidebar for specific topics (under “Categories”). I’m in the process of tweaking some aspects of the site to make it easier to refer back to older topics, so try to make the best of it for now.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 13, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.8984378576278687, "perplexity_flag": "head"}
http://inperc.com/wiki/index.php?title=Real_analysis:_final_2	This site contains: mathematics courses and book; covers: image analysis, data analysis, and discrete modelling; provides: image analysis software. Created and run by Peter Saveliev. # Real analysis: final 2 ### From Intelligent Perception This is the final exam for Real analysis: course. 1. Give two non-Euclidean metrics on $\mathbf{R}^{2}.$ Prove. 2. Prove that an open ball in a metric space is an open set. 3. Prove that a compact set in a metric space is bounded and closed. 4. Suppose $S,T$ are metric spaces and $f,g:S\rightarrow T$ are continuous functions. Prove that the set $A=\{x\in S:f(x)=g(x)\}$ is closed in $S.$ What can you say about $B=\{x\in S:f(x)\neq g(x)\}?$ 5. State and prove the fundamental lemma of differentiation for $f:\mathbf{R}^{2}\rightarrow\mathbf{R}$. 6. State the definition of a differentiable function $f:\mathbf{R}^{N}\rightarrow\mathbf{R.}$ Give an example of a function $f:\mathbf{R}^{2}\rightarrow\mathbf{R}$ such that both partial derivatives of $f$ exist at $x=a,$ but $f$ is not differentiable. 7. State the extension of the Mean Value Theorem to functions $f:\mathbf{R}^{n}\rightarrow\mathbf{R}.$ 8. Give example of such a function $f:\mathbf{R}^{2}\rightarrow\mathbf{R}$ that $f$ is not continuous at $(0,0)$ but both partial derivatives exist at $(0,0).$ 9. State and prove the Contraction Principle. Give examples of functions for which the theorem does or does not apply. 10. Describe Newton's method. Give an example of a function for which the method does not apply. 11. Let $S$ be a complete metric space. Then every subset $A$ of $S$ is also a metric space. Whan is and when is not $A$ a complete metric space? 12. Give examples of functions $f:\mathbf{R}\rightarrow\mathbf{R}$ that satisfy and don't satisfy the Lipschitz condition. 13. Find an parametric equation of an ascending spiral in space. Define the arc-length of a parametric curve and provide its basic properties. Provide the integral formula. 14. Define the curvature of a curve. Find the curvature of the curve $<t^{2},t,5>$ as a function of $t>0.$ Under what circumstances is the acceleration perpendicular to the velocity? 1. State and prove the Schwarz inequality. 2. Suppose $(S_{1},d_{1})$ and $(S_{2},d_{2})$ are metric spaces. Prove that $(T,D)$ is a metric space, where $T=$ $S_{1}\times S_{2}$ and $D((x_{1},x_{2}),(y_{1},y_{2}))=\max\{d_{1}(x_{1},y_{1}),d_{2}(x_{2},y_{2})).$ 3. Suppose both $(S_{1},d_{1})$ and $(S_{2},d_{2})$ in Problem 2 are Euclidean, $S_{1}=S_{2}=\mathbf{R.}$ Describe the open balls in $(T,D),$ convergent sequences, completeness, compactness, and connectedness. 4. Suppose $(S_{1},d_{1})$ and $(S_{2},d_{2})$ are metric spaces and function $f:S_{1}\rightarrow S_{2}$ is function. Provide three definitions of continuity of $f$: (a) in terms of sequences, (b) in terms of $\varepsilon -\delta$ (c) in terms of open or closed sets. Prove that if $A\subset S_{1}$ is connected then so is $f(A).$ 5. State the fundamental lemma of differentiation. State and prove the chain rule for the composition of functions of two variables. 6. State the Contraction Principle. State and prove the existence and uniqueness theorem for the initial value problem. 7. Define the arc-length of a parametric curve and provide its basic properties. Provide the integral formula. Use it to find the arc-length of a circle. 8. Define the curvature of a parametric curve. Find the curvature of the curve $(t^{2},t,5)$ as a function of $t>0.$ Under what circumstances is the acceleration perpendicular to the velocity? 9. Suggest parametric equations for (a) circle in the plane, (b) an ascending spiral in space. Compare their curvatures based on the definition.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 43, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.8947984576225281, "perplexity_flag": "head"}
http://mathoverflow.net/questions/95353/is-this-sequences-of-complexes-of-sheaves-exact	## Is this Sequences of Complexes of Sheaves Exact? ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) So in another question of mine there is a sequence of complexes of sheaves which the author asserts is exact. Let $K^{\bullet} = \underline{\mathbb{C}}^* \ \underrightarrow{d\ log} \ \underline{A}^1_{M, \mathbb{C}}$ and so we have an exact sequence of complexes of sheaves: $$0 \rightarrow {\mathbb{C}}^* \rightarrow K^{\bullet} \rightarrow \underline{A^2}{M, cl}[-1] \rightarrow 0$$ Where that nastily noted $\underline{A^2}{M, cl}[-1]$ means the two term complex with 0 in the first slot and closed 2 forms on $M$ in the second slot. The fact that this sequence is exact in itself seems to rely on the fact that the sheafification of the image of the contant $\mathbb{C}^$ sheaf is isomorphic to the sheaf of smooth functions $\underline{\mathbb{C}^ }$ right? Well that part is bothersome to me - ## 1 Answer This is not related to sheafification. The sheaf $\mathbb{C}^{}$ of locally constant functions on $M$ is already a sheaf, so sheafification will not change it. This sequence is not an exact sequence of complexes but it is an exact triangle of complexes. That is - it is an exact sequence of complexes, up to quasi-isomorphism. The obvious short exact sequence of complexes is ```\[ 0 \to \mathbb{C}^{} \to \left[\begin{array}{c} \underline{\mathbb{C}}^{*} \\ \downarrow \\ A^{1}_{M} \end{array} \right] \to \left[ \begin{array}{c} A^{1}_{M,cl} \\ \downarrow \\ A^{1}_{M} \end{array}\right] \to 0 . \]``` Now note that the last complex has an obvious surjective map ```\[ \left[ \begin{array}{c} A^{1}_{M,cl} \\ \downarrow \\ A^{1}_{M} \end{array}\right] \to \left[ \begin{array}{c} 0 \\ \downarrow \\ A^{2}_{M,cl} \end{array} \right] \]``` and that this surjective map is a quasi-isomorphism. So up to a quasi-isomorphism, you can replace the last term in the short exact sequence with the complex you wanted. -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 7, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9497370719909668, "perplexity_flag": "head"}
http://math.stackexchange.com/questions/200846/selection-through-identical-balls	# Selection through identical balls I was trying the following puzzle from this link : You randomly withdraw 3 balls out of a lot containing identical red, identical green and identical blue balls. What is the probability that you get 1 red, 1 green and 1 blue ball given the number of a) Red balls = 3, Green balls = 3, Blue balls = 3 b) Red balls = p, Green balls = q, Blue balls = r such that p,q,r>3 c) Red balls = infinity, Green balls = infinity, Blue balls = infinity My solution for the puzzle is as follows: Number of ways of selecting ‘r’ things from ‘n’ identical things is ‘1’. Therefore, first answer would be 1.1.1/(number of solutions of x+y+z=3, where x,y,z>=0) i.e. 1/10 For second part : 1/10, since p,q,r > 3 For third part : 1/10. Same logic. But if I apply the same logic to solve : "What is the probability to select a red ball out of a lot of 1 red ball and infinite blue balls?", my answer is 0.5 which intuitively is wrong(It should be 0). Please tell me where am I wrong. - Unless you are doing quantum mechanics, "identical" things don't exist. – celtschk Sep 22 '12 at 20:26 You case c) is undefined, since it does not tell you the probability that selecting one ball it will be red respectively green or blue. Contrary to the other cases you cannot assume a uniform probability: such a distribution does not exist; so some balls are necessarily more likely to be picked than others! – Marc van Leeuwen Sep 25 '12 at 8:00 ## 4 Answers Other answers have explained how to work the problem correctly, but not what’s wrong with your solution. You correctly calculated that there are $10$ distinguishable combinations of colors, all of which are possible in every version of the problem, and observed that only one of them contains one ball of each color. If the $10$ different combinations were equally likely to be drawn, your answer would be correct. However, they are not equally likely to be drawn. This is probably easiest to see in the first version, in which we have specific numbers of balls of each color. Imagine that the nine balls have invisible labels: the red balls are labelled $1,2$, and $3$, the green balls are labelled $4,5$, and $6$, and the blue balls are labelled $7,8$, and $9$. You reach into the box and without looking grab three balls. If the numbers were suddenly made visible, you’d find that you’d chosen a $3$-element subset of $\{1,2,3,4,5,6,7,8,9\}$. Each of these subsets is equally likely to be chosen. Thus, the probability of getting one of your $10$ color combinations can be found by dividing the number of subsets representing that combination by the total number of subsets, which is $\binom93=84$. The all-red combination, for instance, is obtained only when you draw the set $\{1,2,3\}$, so the probablity of getting this color combination is only $\frac1{84}$. If all $10$ of the possible combinations were equally likely, the probabilities of getting them would add up to only $10\cdot\frac1{84}=\frac5{42}$, which is absurd. Thus, they cannot be equally likely, and this fact immediately invalidates your reasoning. You can of course go on to use this approach to calculate the correct probability. In order to get one ball of each color, you must get a set containing exactly one member of $\{1,2,3\}$, one member of $\{4,5,6\}$, and one member of $\{7,8,9\}$. There are $3^3=27$ such sets, so the probability of getting balls of all three colors is $\frac{27}{84}=\frac9{28}$. Similarly, you can calculate that the probability of getting two red balls and one green ball is $\frac9{84}=\frac3{42}$, since there are $\binom32\binom31=9$ $3$-elements subsets of $\{1,2,3,4,5,6,7,8,9\}$ that result in that color combination. Each of the other two-color combinations of course has the same probability. This is enough to show that you can’t safely assume that the different combinations of colors are equally likely; we don’t have to demonstrate it anew for each of the other versions of the problem. - Ross Millikan's elegantly simple answer works whenever the number of each ball is the same. The same approach to version A of your questions yields: $1\cdot \tfrac{6}{8}\tfrac{3}{7}$, or $\tfrac{9}{28}$. A couple of less elegant approaches work in more cases. METHOD 1, PRODUCT OF PROBABILITIES: DETERMINE THE PROBABILITY OF A REPRESENTATIVE FAVORABLE OUTCOME. This works when every favorable outcome is equally likely. Let's determine the probability that you will draw first red, then green, then blue. The probability that you will draw red on your first draw is $$\frac{p}{p+q+r}.$$ If you draw red first, the probability that you will next draw blue is $$\frac{q}{p+q+r-1}.$$ If you draw red and blue on your first two draws, the probability that you will next draw green is $$\frac{r}{p+q+r-2}.$$ The probability that you will draw first red, then green, then blue is the product of these probabilities: $$\frac{pqr}{(p+q+r)(p+q+r-1)(p+q+r-2)}.$$ You can also express this as $$\frac{pqr}{(p+q+r)!/(p+q+r-3)!}.$$ MULTIPLY THIS PRODUCT BY THE NUMBER OF EQUALLY LIKELY FAVORABLE OUTCOMES. In this case there are $3!$ or $6$ equally likely favorable outcomes; bgr, brg, gbr, grb, rbg, rgb. So the probability is $$\frac{6pqr}{(p+q+r)(p+q+r-1)(p+q+r-2)}.$$ In this case, that's $\tfrac{6\cdot 3\cdot 3\cdot 3}{9\cdot 8\cdot 7}$ or $\tfrac{9}{28}$. METHOD 2, COUNTING This method requires basic knowledge of combinatorics, which I'm going to assume. Count the number of ways to get one of each color: $p$ ways to get red, $q$ ways to get blue, $r$ ways to get green gives you $pqr$ ways to get one of each color. Count the number of ways to choose $x$ different balls out of $p+q+r$ balls irrespective of order. That's $$\frac{(p+q+r)!}{(p+q+r-x)!x!}.$$ Divide the number of favorable outcomes by the total number of outcomes, $$\frac{pqr}{ (p+q+r)!/[(p+q+r-x)!x!]}.$$ In this case that means $\tfrac{3\cdot 3\cdot3}{9! / (6!3!)}=\tfrac{9}{28}$. - Is there a difference between picking all 3 balls simultaneously and picking them one by one ? – damned Sep 22 '12 at 21:02 No difference in the result, damned. Sometimes it's easier to think about it as a series of decisions, though. – Michael Schwartz Sep 22 '12 at 21:08 You are trying to grab $3$ balls. Does it make any difference whether your hand touches the three balls simultaneously, or with a delay of a few microseconds? – André Nicolas Sep 22 '12 at 21:10 It can not be 1/10, because if then the answer will be 1/10 even if there are 3R 10B and 20W balls. If you think, it's not possible that the probability remains same, for increasing numbers of balls of certain colours only. Hence 3C13C13C1/9C3=9/28 is correct. AND probability of getting same combination in three turns without replacement will be different=3/93/83/7=3/56 with replacement in three turns it shall be=3/93/93/9=1/27 - In this case, for infinite balls it seems you should consider the chance of any given color to be 1/3, and not to change as balls are drawn. The chance of one of each would be $1$(you can draw any one first)$\cdot \frac23$ (can't match the first)$\cdot \frac 13=\frac 29$ - I'd rate this answer below 52.8k. – Christian Blatter Sep 22 '12 at 21:13	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 42, "mathjax_display_tex": 8, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9397342801094055, "perplexity_flag": "head"}
http://mathoverflow.net/questions/47998/conjugate-groups-of-quasi-fuchsian-groups	## Conjugate Groups of (quasi) Fuchsian Groups ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) I apologize in advance if this question is so trivial or too low level. Let $\Gamma$ be a Fuchsian group. Let $\mathcal{F}$ be the set of pairs $(\mu,f)$, where $\mu \in L^\infty(\mathbb{C})$ such that $\mu(\overline{z})=\overline{\mu(z)}$, $\|\|\mu\|\| < 1$, and $f$ is a quasiconformal mapping of the plane satisfying beltrami differential equation with beltrami coefficient $\mu$: \begin{equation} \mu f_z = f_{\overline{z}} \end{equation} The solution exists up to a Mobius transformation. Let $\mathcal{F}(\Gamma)$ pairs $(\mu,f) \in \mathcal{F}$ be such that $\mu \circ \gamma \frac{\overline{\gamma'}}{\gamma'} = \mu$ for all $\gamma \in \Gamma$. In this case, $\Gamma_f = f\circ \Gamma \circ f^{-1}$ is a Fuchsian group. I'm wondering when $\Gamma = f\circ \Gamma \circ f^{-1}$ would hold. (Here, by "=" I mean equal as subsets of $PSL(2,\mathbb{R})$. What I do know are some special (trivial) cases. 1) $\Gamma = 1$. No conditions on $f$ needs to be imposed. 2) $\Gamma$ is generated by a single parabolic element $g$. Then, the following conditions are sufficient: a) $f$ fixes the fixed point of $g$. b) $f$ fixes $b$ and $g(b)$ for some other point $b$. 3) $\Gamma$ is generated by a single hyperbolic element $g$. Then, the following conditions are sufficient: a) $f$ fixes the fixed points of $g$. b) $f$ has some other fixed point. But outside of these cases, I don't have the slightest clue. - ## 1 Answer When $\Gamma$ is nonelementary, I think the condition is simply that the map is the identity on the limit set of $\Gamma$. This should be in Gardiner's "Teichmuller Theory and Quadratic Differentials" somewhere in the chapter on the Teichmuller space of a fuchsian group. - Aren't "Teichmuller trivial" differentials those solutions whose restriction to a real line is that of an element in $PSL(2,\mathbb{R})$? – BrainDead Dec 3 2010 at 13:34 Yes, but the normalized solutions for these are the identity on $\mathbb{R}$. Maybe I don't have the right terminology for the differentials in the infinite covolume case, but that should be the correct condition. – Richard Kent Dec 3 2010 at 13:39 Sorry, by "these" I meant the maps whose boundary values agree with a Mobius transformation. – Richard Kent Dec 3 2010 at 13:41 Sorry, perhaps my question was a bit round-about. What I'm really interested in getting out of this question is whether there are appropriate normalizations on the solutions so that $T(\Gamma)$ becomes an honest group from composition of the solutions. I imagine this is not going to be possible in general. – BrainDead Dec 3 2010 at 14:21 Sorry about all of my silly comments this morning, you can ignore all my previous comments (I'm a little doped up on cold medicine). You were right about the definition of Teichmuller trivial. I do think the answer that I have up now should be the correct one, and you should be able to derive it from the cases you mentioned. I do know that $\mathrm{Teich}(1)$ is a group as you say, but you have to be careful about the order of composition, or else the group law isn't continuous. I'm not sure about the more general situation. I can try and dig up a reference if you like. – Richard Kent Dec 3 2010 at 14:44 show 1 more comment	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 36, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9297336935997009, "perplexity_flag": "head"}
http://mathoverflow.net/questions/84797/can-the-difference-of-two-distinct-fibonacci-numbers-be-a-square-infinitely-often/84882	## Can the difference of two distinct Fibonacci numbers be a square infinitely often? ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) Can the difference of two distinct Fibonacci numbers be a square infinitely often? There are few solutions with indices $<10^{4}$ the largest two being $F_{14}-F_{13}=12^2$ and $F_{13}-F_{11}=12^2$ Probably this means there are no identities between near neighbours. Since Fibonacci numbers are the only integral points on some genus 0 curves the problem is equivalent to finding integral points on one of few varieties. Fixing $F_j$ leads to finding integral points on a quartic model of an elliptic curve. Are there other solutions besides the small ones? [Added later] Here is a link to elliptic curves per François Brunault's comment. According to DIOPHANTINE EQUATIONS, FIBONACCI HYPERBOLAS,. AND QUADRATIC FORMS. Keith Brandt and John Koelzer. Fibonacci numbers with consecutive odd indices are the only solutions to $$x^2-3xy+y^2 = -1 \qquad (1)$$ Fibonacci numbers with consecutive even indices are the only solutions to $$x^2-3xy+y^2 = 1 \qquad (2)$$ Given $F_n$ and $F_{n+2}$ one can compute $F_{n+k}$ using the linear Fibonacci recurrence and $F_{n+k}$ will be a linear combination $l(x,y)$ of $x,y$. Adding $l(x,y)-x=z^2$ to (1) or (2) gives a genus 1 curve. (Or just solve $l(x,y)-x=z^2$ and substitute in (1) or (2) to get a genus 1 quartic). The closed form of $l(x,y)$ might be of interest, can't find the identity at the moment. Probably a genus 0 curve with integral points $F_{2n},F_{2n+1}$ will be better. Added much later Does some generalization of the $abc$ conjecture predict something? For 3 Fibonacci numbers identities are much easier: $$F_{4n+1}+F_{4n+3}-F_3 = L_{2n+1}^2$$ - 2 Another note: for differences of consecutive Fibonacci's, or gaps of $2$ (i.e. F_{n+2}-F_n) the answer is that the only square values are $1$ and $144$. This follows from a result of J. Cohn, On square Fibonacci numbers, J. London Math. Soc. 39 1964. – Alan Haynes Jan 3 2012 at 12:01 4 Using Cohn's characterization of when $F_n = 2x^2$ (it's true only when n = 0, 3, or 6), and simplifying $F_{n+3} - F_n = 2F_{n+1}$, the only examples separated by 3 are $F_5 - F_2 = 4$, and $F_8 - F_5 = 16$. Similarly, to find all pairs separated by 4, simplify $F_{n+2} - F_{n-2} = F_{n+1} + F_{n-1} = L_n$ to get that the Lucas number $L_n$ must be a square, so n = 1 or 3, giving $F_5 - F_1 = 4$ as the only solution. Both characterizations come from J. Cohn's paper "Square Fibonacci Numbers, Etc." Fibonacci Quarterly 2 1964, pp. 109-113. – Zack Wolske Jan 3 2012 at 13:48 3 In a similar fashion, you can show that there is a solution (and exactly one) separated by 6 terms: $F_{15} - F_9 = 576 = 24^2$. But this example relies on the Lucas number $L_3$ being a square, so there's no hope that it produces an infinite family. – Zack Wolske Jan 3 2012 at 14:54 3 Shorey and Stewart proved that in any non-trivial second-order linear recurrence sequence, there are only finitely many perfect powers. (See Bugeaud-Mignotte-Siksek www-irma.u-strasbg.fr/~bugeaud/travaux/fibo.pdf) So after fixing the gap, there are only finitely many squares. I don't know whether this can be made effective. – François Brunault Jan 3 2012 at 19:35 3 The result of Shorey and Stewart is effective (though rather non-explicit!). For fixed $j$ with $F_k-F_j=x^2$, the link to elliptic curves comes from the fact that Fibonacci numbers are the $y$-values satisfying $x^2-5 y^2 = \pm 4$. I don't see how to prove finiteness for the general problem, ineffectively or otherwise (but 10 hours as department head has undeniably left my brain enfeebled). – Mike Bennett Jan 4 2012 at 3:28 show 16 more comments ## 4 Answers All of the solutions for $F_{n + 4m} - F_n$ are the ones listed, barring the cases $m=2$ and $m=12$ where I still have some kinks to work out. Following Will Sawin's suggestion, we write $F_{n + m} - F_n$ as a reccurence, with some initial terms. Taking $n=0$ tells us that $F_m$ is one term, and $F_{m+1} - 1$ is the next. Extrapolating backwards, we have the general term $F_{m-i} + (-1)^i F_i$. So we see that if $m=4k+2$, then the term $i = 2k+1$ gives us $0$, and the next term gives us $F_{2k} + F_{2k+2} = L_{2k+1}$, so we have a copy of the Fibonacci numbers, multiplied by $L_{2k+1}$. That is, $F_{n + 4k+2} - F_n = L_{2k+1}F_{n + 2k+1}$. If $m = 4k$, then the term $i=2k$ gives us $2F_{2k}$, and the next term gives us $F_{2k+1} - F_{2k-1}=F_{2k}$, so we have a copy of the Lucas numbers, multiplied by $F_{2k}$. That is, $F_{n + 4k} - F_n = F_{2k}L_{n + 2k}$. This is useful, because there are certain primes which never divide Lucas numbers (see http://oeis.org/A053028), so if $F_{2k}$ contains one of these primes (exactly one is easiest, but any odd multiple will do), then the product cannot be a square. Agol gave a link to a paper in the comments (http://www.fq.math.ca/Scanned/7-1/ferns.pdf) which also contains this result. Let's find all squares where $m = 2^k$. The cases $k=0, 1$ were given by AH in the comments, and the case $k=2$ is also included there. We need two facts about Lucas numbers, both of which are easy to show. First, $3 \| L_m$ iff $m \equiv 2 \pmod{4}$ and second, $7 \| L_m$ iff $m \equiv 4 \pmod{8}$. Since $4 \| 2^k$, we can write $F_{n + 2^k} - F_n = F_{2^{k-1}}L_{n + 2^{k-1}}$. I have not resolved the case $k = 3$ (this gap has been fixed, see 2nd edit below), it is equivalent to finding Lucas numbers $x$ in the Diophantine equation $x^2 + 2 = 3y^2$. For now assume $k > 3$. By the facts above, we know that $3$ and $7$ cannot both divide a Lucas number, so we want to show that each of them divide $F_{2^{k-1}}$, and that neither $3^2$ nor $7^2$ do so. That both 3 and 7 divide follows from $F_8=21$ and $F_n \| F_{2n}$. That both 9 and 49 don't divide follows by induction: the base case is $F_8$; and we have $F_{2^{k+1}} = F_{2^k}L_{2^k}$, and neither 3 nor 7 can divide $L_{2^k}$ when $k>2$. Then by induction the squares do not divide $F_{2^k}$ for $k \geq 3$, and neither $3$ nor $7$ divide $L_{2^k}$, since $2^k \equiv 0 \pmod{8}$. Hence there are no squares when $m = 2^k$ and $k > 3$. For the general case of $m=4k$, we write $F_{n + 4k} - F_n = F_{2k}L_{n + 2k} = F_{k}L_{k}L_{n + 2k}$ and invoke Carmichael's theorem: for each $n>3$, there is at least one prime $p \| F_n$ which divides no previous Fibonacci number. Such a prime is called a primitive. Further - this is not part of the theorem - $p^2 \nmid F_n$ (after trying to work out a proof of this, I went to the literature and found that this is a conjecture in P. Ribenboim "Square classes of Fibonacci and Lucas numbers" Port. Math 46 (1989), 159-175. I'm not sure if it's been proven or falsified since then). Taking $n$ to be odd, we exploit the fact that $p$ does not divide any Lucas number, since its Fibonacci entry point is odd (see C. Ballot and M. Elia, "Rank and period of primes in the Fibonacci sequence; a trichotomy," Fib. Quart., 45 (No. 1, 2007), 56-63). If the odd part of $k$ is greater than $3$, we are done, since we can continue splitting $F_{k} = F_{k/2}L_{k/2}$ until we have an odd indexed Fibonacci number, and use a primitive for it. So we now have only $k = 3\cdot2^i$ left to consider. $i = 0$ and $i=1$ are easy to deal with: $F_{n + 12} - F_n = F_{6}L_{n + 6} = 8L_{n+6}$ and we know $L_{n+6} = 2x^2$ only if $n=0$ or negative (from J. Cohn's paper "Square Fibonacci Numbers, Etc." Fibonacci Quarterly 2 1964, pp. 109-113). $F_{n + 24} - F_n = F_{12}L_{n + 12} = 12^2L_{n+6}$ and we know $L_{n+12} = x^2$ has no solutions (again barring negative Fibonacci numbers - in these cases no solutions are distinct from the positive ones). $i=2$ causes me some trouble, and led to the negative solution $F_{36} - F_{-12} = 3864^2$. I also leave this unresolved. For $i > 2$, we proceed as in the argument for powers of $2$. Both $7$ and $47$ divide $F_{3\cdot2^{i+1}}$ exactly once, with the base case being $F_{48}$, and they cannot both divide $L_{n + 3\cdot2^{i+1}}$, since $47 \| L_m$ iff $m \equiv 8 \pmod{16}$. The other even differences should fall the same way, although the formula for those fixes a Lucas number and varies the Fibonacci numbers, we can still find a primitive for the odd part of the Fibonacci index, but I'll have to patch up some pieces where the odd part is 3 or 1. I've had some success with the odd differences using J. Cohn's trick: $L_m \| (F_{n+2m} + F_n)$ when $3 \nmid m$ and $2\|m$, and the fact that $L_m \equiv 3 \pmod{4}$ for such $m$, but no arguments covering infinitely many differences. EDIT: The other even numbers are easier. Writing $F_{n + 4k+2} - F_n = L_{2k+1}F_{n + 2k+1}$ and considering $F_{n + 2k+1}$, the primitive argument removes all indices $n+2k+1$ with an odd part greater than $3$. The same argument as above works for all powers of $2$, since $3$ never divides $L_{2k+1}$. Finally, when the odd part is $3$, it's easier than before, since $7$ divides $F_{3\cdot2^i}$ for $i > 2$, $7^2$ never does, and $7$ and cannot divide the Lucas numbers since they have odd index. $i= 0, 1$ and $2$ are solved by finding all Lucas numbers which are squares, or $2$ times a square. EDIT 2: For $m=2^3$, we want to know if $F_4L_{n+4} = 3L_{n+4}$ is a square. This is solved in M. Goldman. "On Lucas Numbers of the Form $px^2$ where $p=3,7,47,2207$". Math. Reports Canada Acad. Sci. (June 1988). The only example is n+4 = 2, which either does or doesn't happen according to your taste. - There is a gap here where my argument assumes that no higher powers of a primitive divides its Fibonacci number. As far as I know, it's not resolved. – Zack Wolske Jan 5 2012 at 11:49 Something similar to yours is: warwick.ac.uk/~maseap/papers/fnpm.pdf FIBONACCI NUMBERS AT MOST ONE AWAY FROM A PERFECT POWER YANN BUGEAUD, FLORIAN LUCA, MAURICE MIGNOTTE, SAMIR SIKSEK – joro Jan 7 2012 at 7:27 @joro: It looks like their decomposition works because 1 appears with both even and odd index in the sequence, so they can apply the argument for even differences (or sums) twice. That's why the same argument won't work for $F_n + 2 = y^p$. – Zack Wolske Jan 7 2012 at 15:02 The gap can be resolved as follows: If $p$ is an odd prime dividing a Fibonacci number with odd index (it need not be a primitive factor), then it does not divide any Lucas number. This follows from $\gcd(F_a, F_b) = F_{\gcd(a,b)}$. We can always find such a prime with odd exponent in the factorization of $F_q$ for q >3 and odd, since otherwise we get $F_q = x^2$ or $F_q = 2x^2$. – Zack Wolske Jan 7 2012 at 15:08 @Zack With "the squares do not divide $F_{2^k}$ " you mean $F_{2^k}$ is squarefree? – joro Jan 4 at 9:40 show 1 more comment ### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you. Reporting on some computations: the only solutions to $F_k - F_j = x^2$ with $0\leq j < k$ and $1 \leq x \leq 10^6$ are these: {0,1,1}, {0,2,1}, {1,3,1}, {2,3,1}, {3,4,1}, {1,5,2}, {2,5,2}, {5,8,4}, {6,11,9}, {0,12,12}, {11,13,12}, {13,14,12}, {6,13,15}, {9,15,24} where the format is {j,k,x}. - 7 It's easy enough to search well past $x = 10^6$. The following gp code takes little only ~10 minutes here to try $F_k-F_j$ for all $j < k \leq 10^4$, and thus all $x$ up to $\sqrt{F_{9998}} > 10^{1000}$. Not too surprisingly, though, it finds no further solutions other than the spurious $(j,k,x)=(1,2,0)$. for(j=0,10^4,for(k=j+1,10^4,if(issquare(fibonacci(k)-fibonacci(j),&x),print([j,k,x])))) – Noam D. Elkies Jan 4 2012 at 3:32 and I find no new ones up to $10^{200}$... – Mike Bennett Jan 4 2012 at 3:40 3 oops. sure wish I could take that back.... – Mike Bennett Jan 4 2012 at 3:41 2 @Noam Wouldn't your code be significantly faster if you precompute fibonacci(i) in f[i] and then check issquare(f[k]-f[i])? You need not compute fibonacci(i) many times. – joro Jan 4 2012 at 7:22 3 @joro Good question. I didn't do it that way because (i) A table vector(N+1,i,fibonacci(i-1)) takes about $N^2$ space, (ii) I thought the issquare test would take longer in any case (that turns out not to be the case, though $F_j$ takes only $O(\log(j))$ operations to compute), and (iii) Precomputing the table would make the code longer than one command ;-) – Noam D. Elkies Jan 4 2012 at 14:56 show 2 more comments This is a slight elaboration on joro's comment; I was hoping that someone else would write a better version of this. The integer points on $x^2-xy-y^2 = 1$ are precisely the pairs $(F_{2n+1}, F_{2n})$. So looking for solutions of the form $F_{2n+1} - F_{2m+1} = z^2$ is looking for integer points on $$x_1^2 - x_1 y_1 - y_1^2 = x_2^2 - x_2 y_2 - y_2^2 = 1,\ x_1 -x_2 = z^2.$$ The other three possibilities give similar equations. Each of these is a $K3$ surface. Here is where a better answer would review the major results on integer points on $K3$ surfaces. But I don't know them, so I'm going to stop here and hope someone else fills it in. - Thank you. Is it possible $x^2-x y-y^2 = -1$ to be a typo?. I get $x^2-x y-y^2 = +1$ – joro Jan 6 2012 at 10:34 Right, thank you. – David Speyer Jan 6 2012 at 13:12 The package "desing" de-singularized the surface. If I have done it right I see no curves on it (and the remaining solutions must be quite large). – joro Jan 6 2012 at 13:22 1 AFAIK, there are no proven major results on integer points on affine pieces of K3 surfaces. However, Vojta's conjecture predicts that the set of such points lies on a finite union of curves. So assuming Vojta's conjecture, one might be able to make further progress. – Joe Silverman Jan 4 at 23:21 A finite union of genus zero curves (plus finitely many counterexamples), due to Siegel's theorem. I think that what joro was saying above was that there are no rational curves on this surface, but I may have misunderstood him. – David Speyer Jan 5 at 1:24 I get numerical support for the link with integral points on genus 1 curves. For $x,y=F_n,F_{n+2}$ experimentally the closed form is $F_{n+k}=l(x,y)= F_{k} y -F_{k-2} x$ (probably provable by induction). For a fixed gap $k$ solutions correspond to (positive?) integral points on: $$z^{4} + (2 {F_{k-2}} - 3 {F_{k}} + 2) x z^{2} + ({F_{k-2}}^{2} - 3 {F_{k-2}} {F_{k}} + {F_{k}}^{2} + 2 {F_{k-2}} - 3 {F_{k}} + 1) x^{2} + {F_{k}}^{2}=0$$ or $$z^{4} + (2 {F_{k-2}} - 3 {F_{k}} + 2) x z^{2} + ({F_{k-2}}^{2} - 3 {F_{k-2}} {F_{k}} + {F_{k}}^{2} + 2 {F_{k-2}} - 3 {F_{k}} + 1) x^{2} - {F_{k}}^{2}=0$$ Just noticed that if one accepts negative Fibonacci numbers (as described on wikipedia) $F_{-n}=(-1)^{n+1} F_n$ there are some more small solutions. Here are curves where $x$ corresponds to $F_{2n+1}$ on the first curve and $F_{2n}$ on the second curve. 1 (z^4 + xz^2 - x^2 + 1, z^4 + xz^2 - x^2 - 1) 2 (z^4 - xz^2 - x^2 + 1, z^4 - xz^2 - x^2 - 1) 3 (z^4 - 2xz^2 - 4x^2 + 4, z^4 - 2xz^2 - 4x^2 - 4) 4 (z^4 - 5xz^2 - 5x^2 + 9, z^4 - 5xz^2 - 5x^2 - 9) 5 (z^4 - 9xz^2 - 11x^2 + 25, z^4 - 9xz^2 - 11x^2 - 25) 6 (z^4 - 16xz^2 - 16x^2 + 64, z^4 - 16xz^2 - 16x^2 - 64) 7 (z^4 - 27xz^2 - 29x^2 + 169, z^4 - 27xz^2 - 29x^2 - 169) 8 (z^4 - 45xz^2 - 45x^2 + 441, z^4 - 45xz^2 - 45x^2 - 441) 9 (z^4 - 74xz^2 - 76x^2 + 1156, z^4 - 74xz^2 - 76x^2 - 1156) 10 (z^4 - 121xz^2 - 121x^2 + 3025, z^4 - 121xz^2 - 121x^2 - 3025) 11 (z^4 - 197xz^2 - 199x^2 + 7921, z^4 - 197xz^2 - 199x^2 - 7921) 12 (z^4 - 320xz^2 - 320x^2 + 20736, z^4 - 320xz^2 - 320x^2 - 20736) 13 (z^4 - 519xz^2 - 521x^2 + 54289, z^4 - 519xz^2 - 521x^2 - 54289) 14 (z^4 - 841xz^2 - 841x^2 + 142129, z^4 - 841xz^2 - 841x^2 - 142129) 15 (z^4 - 1362xz^2 - 1364x^2 + 372100, z^4 - 1362xz^2 - 1364x^2 - 372100) 16 (z^4 - 2205xz^2 - 2205x^2 + 974169, z^4 - 2205xz^2 - 2205x^2 - 974169) 17 (z^4 - 3569xz^2 - 3571x^2 + 2550409, z^4 - 3569xz^2 - 3571x^2 - 2550409) 18 (z^4 - 5776xz^2 - 5776x^2 + 6677056, z^4 - 5776xz^2 - 5776x^2 - 6677056) 19 (z^4 - 9347xz^2 - 9349x^2 + 17480761, z^4 - 9347xz^2 - 9349x^2 - 17480761) 20 (z^4 - 15125xz^2 - 15125x^2 + 45765225, z^4 - 15125xz^2 - 15125x^2 - 45765225) 21 (z^4 - 24474xz^2 - 24476x^2 + 119814916, z^4 - 24474xz^2 - 24476x^2 - 119814916) 22 (z^4 - 39601xz^2 - 39601x^2 + 313679521, z^4 - 39601xz^2 - 39601x^2 - 313679521) 23 (z^4 - 64077xz^2 - 64079x^2 + 821223649, z^4 - 64077xz^2 - 64079x^2 - 821223649) 24 (z^4 - 103680xz^2 - 103680x^2 + 2149991424, z^4 - 103680xz^2 - 103680x^2 - 2149991424) Lack of integral points on both curves would mean no solution for the given gap. - You can definitely handle this kind of curves using Magma : magma.maths.usyd.edu.au/magma/handbook/text/… If you don't have access to Magma you can do limited computation at magma.maths.usyd.edu.au/calc – François Brunault Jan 4 2012 at 16:32 Treating $F_k,F_{k-2}$ as variables and using the curves described in the question appears to lead to a surface. – joro Jan 5 2012 at 13:37	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 174, "mathjax_display_tex": 6, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.8794791102409363, "perplexity_flag": "middle"}
http://mathoverflow.net/questions/37423/is-there-an-easy-way-to-describe-the-sheaf-of-smooth-functions-on-a-product-manif/37485	## Is there an easy way to describe the sheaf of smooth functions on a product manifold? ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) A smooth structure on a manifold $M$ can be given in the form of a sheaf of functions $\mathcal{F}$ such that there is an open cover $\mathcal{U}$ of $M$ with every $U\in \mathcal{U}$ isomorphic (along with $\mathcal{F}\|_U$) to an open subset $V$ of $\mathbb{R}^n$ (along with $\mathcal{O}\|_V$, where $\mathcal{O}$ is the sheaf of smooth functions on $\mathbb{R}^n$). I think we might also need to say that this satisfies a smooth-coordinate-change axiom, although maybe that's already tied up in the definition of a sheaf. In any case, here is my question: Given two smooth manifolds $(M,\mathcal{F})$ and $(N,\mathcal{G})$, is there an easy way to write the sheaf of functions on $M\times N$ without reference to coordinate neighborhoods? I'm wondering this because in one of my classes we defined smooth manifolds in this way (and we defined analytic and holomorphic manifolds similarly). It seems like some people are very fond of this alternative definition because it doesn't refer to an atlas, which at first seems like it's an inherent part of the structure of the manifold. So okay fine, everyone loves a canonical definition. However, this is only going to be useful as long as we can tell our whole story in this canonical language. In class, the only way the professor was able to give the sheaf on the product was by breaking down and using coordinates. (Admittedly, he was on the spot and presumably unprepared for the question.) This also suggests the broader, more open-ended question: Are there longer-run advantages to the above definition (compared to the usual definition involving an atlas and perhaps a maximal atlas)? - This way of defining manifolds is the same one as the one which consists of giving an atlas, up to very minor differences: if you are given a sheaf satisfying the conditions you mention in your first paragraph, the atlas can be immediately be found by looking at all subsets $U$ involved there together with the isomorphisms to open subsets of $\mathbb R^n$; going the other way requires as little effort, too. – Mariano Suárez-Alvarez Sep 1 2010 at 21:00 I think over open sets which are of the form V= $U_1$ x $U_2$ where U_i is an open set in M_i you can define F(V) to be the "projective tensor product" of $F_i(U_i)$.Then this should determine the sheaf of smooth functions on the product by the concept of "sheaves on a base". I don't know if this "sheafy" approach to smooth manifolds buys you much... but a different theorem that the functor which assigns M---> $C^{\infty}(M)$, the algebra of global sections is fully faithful is really cool and might have some applications. – Daniel Pomerleano Sep 1 2010 at 23:33 3 Yes, $C^{\infty}(M \times N)$ is some completed tensor product $C^{\infty}(M) \widehat{\otimes} C^{\infty}(N)$. – Martin Brandenburg Sep 2 2010 at 0:19 This is subject to one's taste, but a lot of things (e.g. tangent sheaf, cotangent sheaf) are "easier" if you use the sheaf definition. It also generalizes better -- c.f. Hartshorne Chapter II. – Kevin Lin Sep 2 2010 at 21:11 ## 4 Answers Smooth manifolds are affine, thus the sheaf of smooth functions is determined by its global sections. Now C^∞(M×N)=C^∞(M)⊗C^∞(N). The tensor product here is the projective tensor product of complete locally convex Hausdorff topological algebras. - Projective space is a non-affine smooth manifold... – Daniel Loughran Sep 2 2010 at 10:59 @Daniel: Projective space is an affine smooth manifold. – Dmitri Pavlov Sep 2 2010 at 11:04 1 Ahh I see sorry and thanks for the clarification. I have only come across the word "affine" in algebraic geometry before, but it does indeed make sense in this context. – Daniel Loughran Sep 2 2010 at 11:21 1 I see. This is probably the answer I'm looking for, but I don't think I know enough to be sure one way or the other. Could you either explain what this means in other words, or say a little bit about "projective tensor product" and "complete locally convex Hausdorff topological algebras"? Thanks! – Aaron Mazel-Gee Sep 3 2010 at 6:01 1 Dimitri: I think it goes back to Grothendieck's book Produits tensoriels topologiques et espaces nucléaires but I don't have that in front of me at the moment so I'll quote Schaefer's Topological Vector Spaces instead, chapter III section 6. He doesn't go in to much detail on the other topologies on tensor products, but there are plenty. At the other end is the inductive topology where one asks simply for the bilinear maps to be separately continuous. Nuclear spaces, such as smooth functions, can be characterised in terms of their behaviour with tensor products. – Andrew Stacey Sep 3 2010 at 13:15 show 7 more comments ### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you. Let $Mfd$ be the category of smooth manifolds (over $\mathbb{R},\mathbb{C}$) and $LRS$ be the category of locally ringed spaces (over $\mathbb{R},\mathbb{C}$). Then the functor $Mfd \to LRS$ is full and faithful and indeed, you may define $Mfd$ to be the full subcategory of $LRS$, which consists of those locally ringed spaces, which are locally isomorphic to $\mathbb{R}^n$ together with the sheaf of smooth functions. Unfortunately, the functor $Mfd \to LRS$ does not preserve products; see below. It should be remarked that products in $LRS$ do exist (even infinite ones); take the obvious product in $RS$ and "make it local" by introducing new points, namely prime ideals in the stalks, and take the localizations of the stalks as the new stalks. Now if we take the product of two manifolds $M,N$ in $LRS$, we get as a topological space the usual product $M \times N$; however, the structure sheaf consists only of those functions, which are locally of the form $(u,v) \mapsto f(u) * g(v)$ for smooth functions $f,g$ defined locally on $M,N$, or sums and also quotients of these functions. These functions are also smooth with respect to the usual smooth structure on $M \times N$, but the reverse is not true: For example it seems to be true that $\mathbb{R}^2 \to \mathbb{R}, (u,v) \mapsto exp(uv)$ is not such a functions (however, I don't know how to prove this). However, I think that Stone-Weierstraß implies that these simple functions are dense within all smooth functions. Thus, we may regard $M \times N$ in $Mfd$ as the "completion" of $M \times N$ in $LRS$. - If I may be forgiven for climbing on my soap box ... This kind of thing highlights one of the problems with these "one sided" approaches to the theory of smooth spaces. If one takes smooth functions out, then the resulting functor, Manifolds → Functions, does not preserve limits (as Martin says). But if one takes smooth functions in then the resulting functor does not preserve colimits. The resolution is to take both smooth functions in and out, whereupon one does get a functor that preserves both colimits and limits. The resulting category is Hausdorff Frölicher spaces. Taking functions both in and out also follows in the footsteps of the study of manifolds. I asked a bunch of topologists not long ago whether a chart was a map to a manifold or from it. It was about a 50-50 split! At the least, this shows that the direction of charts is ambiguous. At best, it shows that it's useful to have access to both directions. So the answer to the question is as follows: the smooth functions $M \times N \to \mathbb{R}$ are those functions $f \colon M \times N \to \mathbb{R}$ with the property that $f \circ \gamma \in C^\infty(\mathbb{R},\mathbb{R})$ whenever $\gamma \colon \mathbb{R} \to M \times N$ is such that $(g,h) \circ \gamma \in C^\infty(\mathbb{R},\mathbb{R}^2)$ for all $g \in C^\infty(M)$ and $h \in C^\infty(N)$. This can probably be "sheafified" but I'm not a farmer. Obligatory nlab link: for more on $C^\infty$-rings and the like: take a look at smooth algebra. - 3 LOL. I'm not much of a farmer either. – Aaron Mazel-Gee Sep 2 2010 at 6:53 If one denotes $\pi_1:M\times N\to M$ and $\pi_2:M\times N\to N$ the two projections, then I think that the Sheaf of functions on $M\times N$ is simply $\pi_1^\mathcal{F}\otimes\pi_2^\mathcal G$. - 3 I don't think this is right. Taking $M=N=\mathbb{R}$, this would say that the smooth functions on $\mathbb{R}^2$ are exactly those that are sums of products of smooth functions of $x$ and $y$. This doesn't include e.g. $\sin(xy)$... – Aaron Mazel-Gee Sep 2 2010 at 6:57 1 @Aaron, you have to use the correct tensor product and then it works. – Mariano Suárez-Alvarez Sep 2 2010 at 12:22 1 @Aaron: The statement is true, because the tensor product used here is the projective tensor product, not the algebraic tensor product. See also my answer below. – Dmitri Pavlov Sep 2 2010 at 12:24 I prefer to take the tensor product of sheaves. First consider the naive tensor product (without completion). It's only a preasheaf... then sheafify it. – DamienC Apr 25 2011 at 21:25	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 62, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9453251361846924, "perplexity_flag": "head"}
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http://mathoverflow.net/questions/21852?sort=votes	Tensor product of sheaves and modules Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) Hello to all, I have been looking quite recently at the following theorem: Let $X$ be a projective variety and $T$ a tilting object for $X$. If $A:=End(T)$ is the associated endomorphism algebra, then the functor $RHom(T -): D^b(X) \rightarrow D^b(A)$ is in fact an equivalence. Now, this is proven (as in the claasical Bondal paper) by showing that the functor is fully faithful and essentially surjective. But in I have noticed another version where one defines a functor $-\otimes^L_A T: D^b(A) \longrightarrow D^b(X)$. My question is could someone maybe give a definition of this functor (I of course know what all types of tensor-products are, but I'm not really sure how to "tensor" a module over a noncommutative ring $A$ together with a sheaf to obtain another sheaf. - The functor should be $(\mathord-)\otimes^L_XT$, as $(\mathord-)$ does not have any possible action of $A$ on it... – Mariano Suárez-Alvarez Apr 19 2010 at 16:31 (Unless you swapped the domain and codomain of the functor!) – Mariano Suárez-Alvarez Apr 19 2010 at 16:33 I'm just curious, what is $D^a$? – Martin Brandenburg Apr 19 2010 at 17:55 1 Answer What I've seen is a construction of a quasi-inverse for RHom(T,-), defined as $-\otimes_A^L T$, as you wrote. This last symbol should be interpreted as follows. Given a left A-module M, we define a presheaf which with each U associates $M\otimes^L_A T(U)$. Finally $M\otimes^L_A T$ is defined as the sheafification of the former. I should have a reference for this, let me check. Reference Added: A. A. Beilinson. Coherent sheaves on $\mathbb{P}^n$ and problems in linear algebra. Funktsional.Anal. i Prilozhen., 12(3):68–69, 1978. (it's this one, if I remember correctly, but I might not...) - great, thanks, that was the only paper I couldn't get my hands on, I'll see if I can find it – louis de Thanhoffer de Völcsey Apr 19 2010 at 16:51 I had a hard time finding it as well! I had to ask my supervisor to ask the department for it. Unfortunately I don not have it with me at the moment, otherwise I could scan for you the interesting bit. – babubba Apr 19 2010 at 21:31 I've also found useful this paper by Alastair Craw while learning this stuff (and I'm still learning it) xxx.lanl.gov/abs/0807.2191 – babubba Apr 19 2010 at 21:34 @angeleirovero: I wonder how can you scan "the interesting bit" without essentially scanning the whole thing. It is one of those devilish short papers! – Mariano Suárez-Alvarez Apr 19 2010 at 22:14 @mariano: Well... I guess you can't. – babubba Apr 20 2010 at 7:46	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 15, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9249022603034973, "perplexity_flag": "middle"}
http://mathoverflow.net/questions/89801/a-limit-involving-a-regularizing-kernel	## A limit involving a regularizing kernel ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) I am studying the following article by Benoit Perthame: http://www.mendeley.com/research/uniqueness-error-estimates-first-order-quasilinear-conservation-laws-via-kinetic-entropy-defect-measure/# Somewhere in the middle of it, I'm stuck at proving a certain limit equality. Maybe it's obvious and I can't get it. $$\int_{(\Bbb{R})} \left(\chi(\xi,u)\star \varphi_\varepsilon \right)^2d \xi \to \|u\| \text{ in } {L}^1_{loc}$$ where $\varphi_\varepsilon(t,x)$ is a regularizing kernel, $u$ satisfies $$\partial_t u +\text{div}A(u)=0 \text{ and }\text{ in }\mathcal{D}^\prime((o,\infty)\times \Bbb{R}^d)$$ and $$\chi(\xi,u)=\begin{cases} 1 & {0\leq \xi\leq u} \newline -1 & u \leq \xi \leq 0 \newline 0 & \text{otherwise} \end{cases}$$ Thank you. Sorry. I forgot to mention that $u \in L^1_{loc}$. - 2 (a) I think if the equality were to hold, it will hold regardless of the equation. For continuous functions it is easy to check that the limit holds even pointwise, since $\|u\| = \int_{\mathbb{R}} \chi(\xi,u)^2 d\xi$ and it is just pushing integrals around and justifying a few Fubini's. (b) If $u\in D'$, you don't necessarily have that $\|u\|$ is some well-defined object in $L^1_{loc}$. For example, what is $\|\delta'\|$? So perhaps what the equation is doing is giving $u$ a priori some regularity so that the absolute value of $u$ is well-defined. – Willie Wong Feb 29 2012 at 14:16 2 Ah, your desired expression appear in the "Proof for (2.2)" step. In (2.2), which is part of the statement of Theorem 2.1. So by hypothesis $u$ is $L^1_{loc}$. So you can probably use the result for continuous functions and Luzin's theorem to get convergence. – Willie Wong Feb 29 2012 at 14:29 @Willie Wong: Thank you for your comment. I'm starting to understand now. – Beni Bogosel Feb 29 2012 at 16:38	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 12, "mathjax_display_tex": 3, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9431934356689453, "perplexity_flag": "middle"}
http://mathhelpforum.com/discrete-math/183849-help-understanding-infinity-axiom-zf-2.html	# Thread: 1. ## Re: Help with understanding the infinity axiom in ZF Originally Posted by MoeBlee in this sense of subsets, there is a least successor inductive set. No doubt the von Neumann construction is the typical approach to constructing the natural numbers as a least inductive successor set, but recognize how you previously stated it was unique and now with an indefinite article in the above quote. I was simply agreeing with Deveno's point that there is not a unique (definite article) least inductive successor set we call the natural numbers. There are, in fact, multiple such constructions that are separate representations. Once we fixate on such a construction, it is no doubt uniquely defined. We should remember that $\omega$ is defined as a concept, in our context, as the least inductive successor set. That concept, though, can be met by a number of actual constructions within a given theory of sets. 2. ## Re: Help with understanding the infinity axiom in ZF Originally Posted by bryangoodrich No doubt the von Neumann construction is the typical approach to constructing the natural numbers as a least inductive successor set, but recognize how you previously stated it was unique No, I did not say that there is only one way to construct the natural numbers; I did not say that the von Neumann construction is the only way to construct the natural numbers nor that the von Neumann construction is the only way to construct a Peano system. Rather, I said that there is a unique inductive successor set. Obviously, that is to be taken in the context that we are given some specific defintion of 'successor inductive'. And in this context, I used 'successor inductive' to correspond to the xu{x} method given in the axiom of infinity in its ordinary formulation and as formulated in the first post of this thread. GIVEN the definitions: y is successor inductive <-> (0 in y & Ax(x in y -> xu{x} in y)) y is a least successor inductive set <-> y is successor inductive & Ax(x is successor inductive -> y subset of x) then there is a unique least successor inductive set. And it would be seen that that definition of 'successor inductive' would be at play here since it corresponds exactly to the axiom of infinity as is ordinary and as stated in the first post in this thread. If it was not clear by context to you that I meant some specific definition of 'successor inductive' then fine, I made it explicit after you mentioned it. I thought that, since this is standard textbook stuff, and since I'm in context of the axiom of infinity as in the first post, which is the ordinary formulation, such a context would have been understood. If I had said, "there is a unique Peano system" or "there is a unique way to construct a system of natural numbers" or "there is a unique such system and no others are isomorphic with it", then, yes, I would have been quite in error. But what I said is that there is unique successor inductive set, and when I use a mathematical term like 'successor inductive' I mean it as having some specific definition. 3. ## Re: Help with understanding the infinity axiom in ZF P.S. As to the article 'a', my use is correct. There is a unique least successor inductive set. Of course, there is not a unique Peano system. Nor do I claim that one can't give a different definition of 'successor inductive' that provides a different set from the von Neumann naturals but still provides for a Peano system. For that matter, one can take ANY denumerable set and make a Peano system out of it. 4. ## Re: Help with understanding the infinity axiom in ZF Originally Posted by MoeBlee Successor inductive. In this context, "leastness" refers to the subset relation. There is a unique least successor inductive set in the sense that there is a unique set w such that w is successor inductive and for all x, if x is successor inductive then w is a subset of x. That set is not successor inductive. Recall that X is successor inductive if and only if 0 is in X and for all y, if y is in X then yu{y} is in X. Of course, we could define 'successor' or 'successor inductive' in a different way, indeed so that the Zermelo natural numbers you just mentioned suit such a definition. But in context, we refer to the more ordinary definitions I'm using. i see the distinction you draw between "inductive" and "successor inductive" (meaning a very specific function X--->XU{X}). as you have stated it, you are correct (i had to think about this for quite some time), obviously there are other set injections that might serve as a "successor function" (such as X-->{X}). the thing is, we really want to regard these in some way "as the same". the "usual" defintion of the natural numbers borders on the contradictory: the natural numbers are the unique set possessing all the properties of the natural numbers (i am being deliberately vague here, so as to avoid a lengthy discussion of what those properties actually ARE). in fact, i see this as a fundamental flaw in set theory itself: we want to characterize a set by the properties its elements have. of course, we can't just go ahead and say that will do, or we run into various antimonies (such as the Russell set). so we restrict comprehension, by agreeing that if we have some other (previously defined) set, we can describe a subset by listing the properties this subset will have. of course, this pre-supposes we can find a big enough set to be sure the set we wish to describe is actually a subset of our "big set". making THAT precise, gets a bit away from the basic axioms of ZF. The axiom schema of separation: For all P and x, if P is a formula and x is a variable not free in P, then all closures of the following are axioms: ExAy(y in x <-> (y in z & P)) So you see that the only technical matter about variables is that x is not free in the defining formula P. That is hardly "convoluted". Indeed it's straightforward and common sense. It's common sense that you wouldn't define a set, calling it 'x' with a condition that itself mentions x. Indeed convolultion (and contradiction) would result by allowing x to be free in P. As to what predicates are allowable. This is rigorous and straightforward. Any formula P in which x does not occur free. The axiom schema of separation expresses a straightforward notion: Given a set z, we can form the set x that is a subset of z and x has as members all and only those objects that are in z and that have property P. That is even common everyday reasoning. Given the set of all umbrellas we can also form the set of all umbrellas that are red. That is given the set z of all umbrellas, we have the set x that is a subset of z and such that the members of x are all and only those objects in z that are red. can we actually, just from the ZF axioms, show that the set of umbrellas exists? or, even more basically, can we establish that {umbrella} (one umbrella) forms a set? how does one define "conceptual objects" (the abstract idea of a single umbrella) in a certain well-defined manner? naively, of course, i am perfectly willing to talk about "a set of umbrellas". but when it comes to formalizing how to talk logically about umbrellas, it gets trouble-some. does a partially destroyed umbrella count? what defines "umbrella-ness?" if, at some point, we are reduced to defining logic in purely linguistic terms, how can we ever say that logical reasoning applies to the "real world" we wish it to? i again humbly submit that there are well-formed logical predicates that are "meaningless" in natural language. when it comes to questions of foundation, we are faced with an (i believe) unavoidable dilemma: undefinition, or circularity. Yes, as does any formal first order theory. What formal first order theory that axiomatizes mathematics for the sciences would one point to as being less complicated than ZC set theories? Of course, one doesn't have to concern oneself with formal theories for mathematics, but if one is interested in a formal axiomatization, then I don't see that ZFC is espeically convoluted or complicated. and how does one define a first-order logic in a set-free way (after all, it is rather unfair to use sets to define a first-order logic, and then use first-order logic to define sets)? this is what i mean by "it begs the question" of which predicates are allowable. But the machinery is not JUST to be able to say "1 is a natural number". of course not, i was being facetious. the idea is to generate all of the usual sets of mathematics we are accustomed to dealing with: N, Z, Q, R, C, various function sets involving the previous, etc. my point being, expressing even the simplest statements, like: one thing and one thing make two things (something self-evident to a small child), have a very dense expression in terms of ZF. Just to be clear, I'm not an "adherent" of ZFC. I find that it has certain virtues and also that one may have certain philosophical objections to it. I enjoy studying it, but I don't take it as some theory that I must "adhere" to. i see it as a stop-gap measure until we get something better. i think topoi are a step in the right direction. i think math should take a cue from natural language, and focus more on verbs as the pivotal elements, rather than nouns, that is to say: who cares what things ARE, what do they DO? it's quite evident to me you enjoy ZF set theory as an object of study. i'm not saying you're some sort of zealot, lol. 5. ## Re: Help with understanding the infinity axiom in ZF MoeBlee, I'm not really disagreeing with you on that point. As Deveno pointed out, "successor" can be defined differently. The context that is important is, as you pointed out, which definition of inductive set we are using. I understand the context from the original post, but I was merely emphasizing that the von Neumann ordinals is just one approach as Deveno provided an alternative. Of course, once we fix the definition, then by the axiom of extension the natural numbers just is the inductive set with the minimality property we seek, and they are uniquely defined. The axiom of infinity defines an inductive set, and $\omega$ just is the inductive set common to all inductive sets. This is, of course, what you stated initially. 6. ## Re: Help with understanding the infinity axiom in ZF Originally Posted by Deveno we really want to regard these in some way "as the same" That is done as we define the notion of a Peano system and show that any two Peano systems are isomorphic. Originally Posted by Deveno the "usual" defintion of the natural numbers borders on the contradictory I don't know what it means for something to "border" on contradiction; a theory is inconsistent or it's not. Also, definitions (properly formed) do not introduce contradiction. Originally Posted by Deveno the natural numbers are the unique set possessing all the properties of the natural numbers (i am being deliberately vague here, so as to avoid a lengthy discussion of what those properties actually ARE). Any X is the unique thing having the properities of X. That's true. But, just to be clear, it's not a definition, since it's true but circular. Originally Posted by Deveno i see this as a fundamental flaw in set theory itself: we want to characterize a set by the properties its elements have. of course, we can't just go ahead and say that will do, or we run into various antimonies (such as the Russell set). so we restrict comprehension, by agreeing that if we have some other (previously defined) set, we can describe a subset by listing the properties this subset will have. of course, this pre-supposes we can find a big enough set to be sure the set we wish to describe is actually a subset of our "big set". making THAT precise, gets a bit away from the basic axioms of ZF. If you look at the actual applications of the axioms, from theorem to theorem, you won't find any "getting away from the basic axioms" that you imagine. Originally Posted by Deveno can we actually, just from the ZF axioms, show that the set of umbrellas exists? or, even more basically, can we establish that {umbrella} (one umbrella) forms a set? ZF doesn't have a predicate "is an umbrella". That's aside the point here. I used umbrellas merely for concrete example of everyday reasoning. Originally Posted by Deveno how does one define "conceptual objects" Mathematics has a method of definitions. First is to prove in the theory E!xP (for some formula P), then define some constant c by Ax(x=c <-> P). Then c is the unique x such that P. Originally Posted by Deveno (does a partially destroyed umbrella count? what defines "umbrella-ness?" Those kinds of questions are not addressed by set theory. Rather, those are matters in the study of outdoor weather gear. Originally Posted by Deveno (if, at some point, we are reduced to defining logic in purely linguistic terms, how can we ever say that logical reasoning applies to the "real world" we wish it to? i again humbly submit that there are well-formed logical predicates that are "meaningless" in natural language. Those are philosophical questions that you may wish to entertain yourself with, but they hardly vitiate (or make "borderline contradictory") the simple mathematical content I've presented. Originally Posted by Deveno when it comes to questions of foundation, we are faced with an (i believe) unavoidable dilemma: undefinition, or circularity. Whatever the merits of your belief, it does not prevent us from forming mathematical definitions. Originally Posted by Deveno and how does one define a first-order logic in a set-free way (after all, it is rather unfair to use sets to define a first-order logic, and then use first-order logic to define sets)? this is what i mean by "it begs the question" of which predicates are allowable. Yes, we recognize that we use some of the same notions in our meta discussion about our formal theory that are themselves formalized in the formal theory. Aside from the notion of sets, we use such notions as modus ponens reasoning, identity, etc. to formalize modus ponens reasoning, identity theory, etc. I don't know how one would propose to escape the use of certain basic notions such as those. In any case though, your criticism in that regard is across the board as to formal theories; the matter of definining a least successor inductive set then is merely one out of EVERY mathematical definition you could have such objections to. Originally Posted by Deveno expressing even the simplest statements, like: one thing and one thing make two things (something self-evident to a small child), have a very dense expression in terms of ZF. Yes, and that's typical of axiomatic development. We define such things as natural numbers, addition on natural numbers, etc. because logically we don't NEED to take them as primitive. That's part of the point of what we're doing when we work from axioms. Originally Posted by Deveno i see it as a stop-gap measure until we get something better. That's fine. Except that we might find that some systems are better in certain aspects and other systems better in other aspects. Originally Posted by Deveno i think topoi are a step in the right direction. I would be very suprised if it turns out that using topoi in whatever manner you have in mind avoids the kind of philosophical difficulties you mentioned above. Originally Posted by Deveno i think math should take a cue from natural language, and focus more on verbs as the pivotal elements, rather than nouns, that is to say: who cares what things ARE, what do they DO? You're welcome to point out whatever theory or developments along those lines that you endorse. 7. ## Re: Help with understanding the infinity axiom in ZF Originally Posted by bryangoodrich once we fix the definition, then by the axiom of extension[ality] the natural numbers just is the inductive set with the minimality property we seek, and they are uniquely defined. The axiom of infinity defines an inductive set, and $\omega$ just is the inductive set common to all inductive sets. I would state it this way: Given the definition of 'successor inductive' I used, the axiom of infinity is that there exists a set that is successor inductive. Then an instance of the separation schema gives us the existence of a successor inductive set that is a subset of all successor inductive sets. Then the axiom of extensionality gives us that there is a unique set that is succesor inductive and a subset of all successor inductive sets. Then we define w (read 'omega') as that unique such set. 8. ## Re: Help with understanding the infinity axiom in ZF i want to point out (and obviously i'm doing it wrong), that depending on how the axiom of infinity is formulated, we get differing notions of what a minimal inductive set IS (in terms of the actual elements). as long as you call that set ω, and make no claim to it actually "being" the natural numbers, that's defensible. in fact, even referring to a set as "successor inductive" is misleading: if what ones actually means by "successor inductive" is that it can serve as a model for the Peano axioms. there is a tendency to view the "succession" function X--> X U {X} as the most "natural" one, and to go even further with identifying this with the set of natural numbers. this, i believe, is a mistake. for example, the wikipedia article Axiom of infinity - Wikipedia, the free encyclopedia says: "The infinite set I is a superset of the natural numbers. To show that the natural numbers themselves constitute a set, the axiom schema of specification can be applied to remove unwanted elements, leaving the set N of all natural numbers. This set is unique by the axiom of extensionality." this is just wrong. what is left is "a" set which can be identified with the natural numbers. and yet many respected authors/texts/websites will make the mistake of saying ω is N. to go a bit further, i was dead serious when talking about umbrellas. people use "sets" of everyday objects to illustrate the basics of set theory, such as: {Bob, Alice, Ted} ∩ {Bob, Barry} = {Bob}. if we don't have a predicate "is an umbrella", than using umbrellas to illustrate set theory isn't very kosher, is it? in fact, proving the existence of anything BUT certain sets guaranteed by the ZF axioms is problemmatic. for example, take the free semigroup on one letter (let's use A. it's a fine letter). we cannot prove (or disprove, for that matter) from the ZF axioms that {A} is a set. even more worrisome, we cannot show that there is a set consisting of any "string", although it seems to me, that we would like very much to be able to consider these "strings" as sets, so that we can use what we know about relations on sets to manipulate them. it is somewhat of a victory that we CAN define (a version of) the natural numbers flowing just from the axioms (as one can prove that the trivial group with trivial operation is a group just from the axioms), but it is also dismaying that ZF isn't "big enough" for what we want it for. you're probably right....any system powerful enough to express enough mathematics as to be considered a "base" is likely to have a "hole" in it....something mysterious, a kind of we-don't-really-know-what-it-is-ness. i wish i was clever enough to think of a way around this dilemma. i wll point out that the consistency of ZF is, to my knowledge, an open question. we hope it's consistent, so far, so good.... i hope you can recognize, that there is a certain problem in the very concept of "definition". mathematicians, in particular, seem to feel very satisfied with they come up with a "suitable" definition (this goes back to Euclid at least, who presented his "postulates" as self-evident). one of the main insights of the 20th century, is the realization that you can't talk about a theory wholly inside the theory (except for some theories of limited scope), and at some point you have to "step outside the theory", either in the form of a "meta-theory" or in some non-mathematical way (such as appealing to common experience). but, we go rather far afield (a field, it's a math joke, get it?), waaay past the concerns the original poster had. and i am rather poor at sharpening my point, which is this: i think the "content" of the ZF axioms is rather deep, and our discussion of the "interpretation" of it here is somewhat superficial (relating it to non-mathematical ideas about induction and a "hereditary property"). there are subtleties here, buried in the terseness of the first-order logical statement of the axiom. the original poster couldn't "relate", but to you it's "straight-forward". why do you suppose that might be? 9. ## Re: Help with understanding the infinity axiom in ZF Originally Posted by Deveno i want to point out (and obviously i'm doing it wrong), that depending on how the axiom of infinity is formulated, we get differing notions of what a minimal inductive set IS (in terms of the actual elements). as long as you call that set ω, and make no claim to it actually "being" the natural numbers, that's defensible. For my own part, I don't opine as to what the natural numbers are in terms of "actual being" aside from a given formal definition in a formal theory such as Z set theory. I have various notions about such things, but I don't assert that any particular notion along those lines must be taken as definitive or preeminent Originally Posted by Deveno in fact, even referring to a set as "successor inductive" is misleading: if what ones actually means by "successor inductive" is that it can serve as a model for the Peano axioms. Yes, that would be using a different definition from the one I was using. Originally Posted by Deveno there is a tendency to view the "succession" function X--> X U {X} as the most "natural" one, and to go even further with identifying this with the set of natural numbers. Just to be clear about my own view, I don't assert anything about such naturalness. The xu{x} method is convenient, it is the most common, and it has certain virtues. But I don't opine as to whether it is the most natural. Originally Posted by Deveno Wikipedia, the free encyclopedia says: "The infinite set I is a superset of the natural numbers. To show that the natural numbers themselves constitute a set, the axiom schema of specification can be applied to remove unwanted elements, leaving the set N of all natural numbers. This set is unique by the axiom of extensionality." this is just wrong. No, it's not wrong, given the definitions that are in use in that context. Originally Posted by Deveno what is left is "a" set which can be identified with the natural numbers. and yet many respected authors/texts/websites will make the mistake of saying ω is N. It seems to me that you're confusing two different matters. When we say that the set of natural numbers is defined as just mentioned, we are giving a particular formal definition for the purpose of working in certain theories. We are not necessarily saying that there are not other ways one could form definitions. We are stating a convention; we are not disallowing that one may have a different convention in play in another context, nor are we necessarily saying that the natural numbers "actually are" (whatever that means) such and such a formally defined set. Originally Posted by Deveno if we don't have a predicate "is an umbrella", than using umbrellas to illustrate set theory isn't very kosher, is it? There's nothing in the least that would make a rabbi blush. When I mentioned umbrellas, I was just giving an informal explanation of an idea. Of course, I don't use the notion of umbrellas in actual formal proofs in set theory. But to use such common items in certain general explanations is harmless. I merely illustrated the notion of the schema of separation. There was nothing ESSENTIAL in choosing umbrellas. I could have mentioned "schumbrellas" or "crumbellas" or "x's" or "y's" or whatever. However, of course, whether Orthodox Judaism permits the use of umbrellas on the Sabbath, on that question we must defer to the Talmudic scholars... Originally Posted by Deveno in fact, proving the existence of anything BUT certain sets guaranteed by the ZF axioms is problemmatic. Of course, yes, ZF only proves the existence statements that ZF proves. Originally Posted by Deveno for example, take the free semigroup on one letter (let's use A. it's a fine letter). we cannot prove (or disprove, for that matter) from the ZF axioms that {A} is a set. I don't recall the details about what 'free' means regarding semigroups, but generally speaking, in ZF we can define such predicates as 'is a semigroup', etc. (and I guess we could define 'is a free semigroup'?). Then also, given any A, of course in set theory we have that there exists a unique set {A} such that members of {A} are all and only A. Originally Posted by Deveno even more worrisome, we cannot show that there is a set consisting of any "string", although it seems to me, that we would like very much to be able to consider these "strings" as sets, so that we can use what we know about relations on sets to manipulate them. Strings are taken as sets commonly. We can take strings to be finite sequences, or we can take strings to be n-tuples. This is discussed in various texts in mathematical logic and in computability. Originally Posted by Deveno it is also dismaying that ZF isn't "big enough" for what we want it for. Depends on what we want it for. For example, if you want full category theory, my understanding (I'm not expert on this) is that adding to ZFC the axiom that there exists a Grothendieck universe gives you category theory. Originally Posted by Deveno i hope you can recognize, that there is a certain problem in the very concept of "definition". mathematicians, in particular, seem to feel very satisfied with they come up with a "suitable" definition (this goes back to Euclid at least, who presented his "postulates" as self-evident). one of the main insights of the 20th century, is the realization that you can't talk about a theory wholly inside the theory (except for some theories of limited scope), and at some point you have to "step outside the theory", either in the form of a "meta-theory" or in some non-mathematical way (such as appealing to common experience). Of course we use meta-theories. But I don't know what specific problem you find with the ordinary method of mathematical definitions. Originally Posted by Deveno there are subtleties here, buried in the terseness of the first-order logical statement of the axiom. the original poster couldn't "relate", but to you it's "straight-forward". why do you suppose that might be? The original poster had a question on how to prove something. I don't know whether he was having a problem relating to anything. As to why I find the schema of separation straightforward, I think it's because I studied such basic set theory in a clear, step by step manner. I have no special talent for mathematics or set theory, but most of this basic stuff presents to me as pretty clear when I approach it calmly and systematically.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 3, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9581771492958069, "perplexity_flag": "middle"}
http://math.stackexchange.com/questions/197576/transcendental-extension?answertab=votes	# Transcendental extension Is there any easy nice example to say that this element of $k((t))$ is transcendental over $k(t)$ (We can use the cardinality argument for the existence of transcendental element.But, I am looking example). Thank you - 2 $k((t))$ denotes formal power series? What about $\phi=\prod_n \frac1{1-t^n}$? It has infinitely many "poles" in $\bar k$, hence in any polynomial with coefficients in $k(t)$ there is some pole $z$ not occuring as pole or zero of any coefficient. Therefore $z$ is also a pole of order $\deg p$ of $p(\phi)$. – Hagen von Eitzen Sep 16 '12 at 15:17	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 10, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.8556038737297058, "perplexity_flag": "head"}
http://cs.stackexchange.com/questions/6732/is-there-any-nontrivial-problem-in-the-theory-of-serial-algorithms-with-a-nontri	# Is there any nontrivial problem in the theory of serial algorithms with a nontrivial polynomial lower bound of $\Omega(n^2)$? In the theory of distributed algorithms, there are problems with lower bounds, as $\Omega(n^2)$, that are "big" (I mean, bigger than $\Omega(n\log n)$), and nontrivial. I wonder if are there problems with similar bound in the theory of serial algorithm, I mean of order much greater than $\Omega(n\log n)$. EDIT: with trivial I mean "obtained just considering that we must read the whole input" and similarly. - Are you asking for lower bounds for problems or for lower bounds for specific algorithms? – A.Schulz Nov 18 '12 at 11:21 @A.Schulz I'm asking for lower bounds for problems. – natema Nov 18 '12 at 11:36 – natema Nov 18 '12 at 11:53 quite interestingly, we don't find much attention given to such classes - as it was done with the sorting problem. Matrix multiplication is $\Omega(n^2)$. The all-shortest path problem is $\Omega(n^2)$ - which you probably already know. But is there a class for these algorithms? I really dont know. The similarity between these problems is that every input (among the $n$) will have to do an action with every other input. Such action is at least $\Omega(1)$. – AJed Nov 18 '12 at 13:32 @RealzSlaw: I agree with you. I lacked some details in my answer. But you understand what I want to say. – AJed Nov 18 '12 at 16:29 show 3 more comments ## 2 Answers There are such problems by time hierarchy theorem. Just take any problem that is complete for a large complexity class. For example, take a problem that is complete for $\mathsf{ExpTime}$. Such a problem will be $\Omega(n^c)$ for all $c\in\mathbb{N}$. However also note that for problems in $\mathsf{NP}$, there are not strong time lower-bounds in multiple-tape TM model and existence of linear time algorithms for SAT is consistent with current state of knowledge. (In single-tape TM model it is not difficult to show that many problems like palindromes require $\Omega(n^2)$ time but such lower-bounds depend essentially on the particulars of the single-tape TM model.) - Some simple problems that have lower bounds greater than the size of their inputs, are algorithms that have output sizes greater than their input sizes. Some examples: • The problem of listing all solutions to 3-SAT, or similarly, the problem of listing all Hamiltonian cycles. These problems both have exponential number of solutions in the worst case. Thus they have a lower bound of $\Omega (c^n),c>1$. Interestingly however, the 3-SAT problem itself has no known super-linear (greater than $\Omega (n)$) bounds! This means we do not know if it is harder than linear! • You can even make up new algorithms like this: "completing a graph" that is, given $G = V,E$, where $E=\varnothing$, and $n=\|V\|$, the algorithm will output a graph $G' = V,E'$, where $E'=\left\{u,v\|u\neq v\space\wedge\space u,v\in V \right\}$. Furthermore, you might be able to compose a problem that has $\Omega (n^2)$-sized outputs, with a problem that takes $\Omega (n^2)$ as input, and outputs $\Omega (n)$ or even $\Omega (1)$-sized outputs (for example, something that counts the number outputs) to obtain a problem that takes $\Omega(n)$-sized input, and outputs $\Omega (n)$-sized output, and yet has a running time greater than $\Omega (n)$. However it might be very difficult to prove (that there is no shortcut to obtain the answer in less time). Another way some problems have known lower bounds, is to restrict the model of computation. Though comparison sort's lower bound does not exceed $\Omega (n\log n)$, I think its worth discussing. Comparison sort is also a problem that has a greater lower bound than its input size, but it's lower bound does not exceed $\Omega (n\log n)$, and in . However, as I was researching this, I found this question on mathoverflow: Super-linear time complexity lower bounds for any natural problem in NP. Further examples listed in the answer there are far below $\Omega (n\log n)$. I think the gist of it is, if you restrict the model of computation, you can get lower bounds for problems for which we otherwise do not have them. And if you do not restrict the model of computation, it is very difficult to prove lower bounds on problems. - There is something I dont understand. Long multiplication can be done with less than O(n^2) according to wikipedia ? Therefore, $\Omega(n^2)$ is nto a lower bound. – AJed Nov 18 '12 at 13:20 Matrix multiplication can be done with $O(n^2.8)$ and this bound has been improved. However, a natural lower bound is $\Omega(n^2)$. – AJed Nov 18 '12 at 13:24 1 @AJed its not a lower bound on the problem, but its a lower bound on the algorithm. – Realz Slaw Nov 18 '12 at 14:33 And now he edited his question to address "problem" instead of algorithm. – Realz Slaw Nov 18 '12 at 14:34 1 @RealzSlaw I apologize for being not enough accurate in the text of my question at the beginning. – natema Nov 18 '12 at 17:39 show 1 more comment	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 33, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.953951895236969, "perplexity_flag": "head"}
http://mathhelpforum.com/trigonometry/170944-how-can-i-approximately-solve-trigonometric-equations-simultaneously.html	# Thread: 1. ## How can I approximately solve trigonometric equations simultaneously? if I have p/k=sin(3x)/sin(2x), how could I solve for x in terms of p/k? I understand that these are not linear functions so they cannot be solved simultaneously but is it possible to approximately solve this problem using taylor expansions? If there is another way I could solve this, I would like to hear what you think. Thank you 2. You need to use Trigonometric Identities... $\displaystyle \sin{3x} \equiv \sin{x}(3 - 4\sin^2{x}), \sin{2x} \equiv 2\sin{x}\cos{x}$ and $\displaystyle \sin^2{x} + \cos^2{x} = 1$. So if $\displaystyle \frac{\sin{3x}}{\sin{2x}} = \frac{p}{k}$ $\displaystyle \frac{\sin{x}(3-4\sin^2{x})}{2\sin{x}\cos{x}} = \frac{p}{k}$ $\displaystyle \frac{3 - 4\sin^2{x}}{2\cos{x}} = \frac{p}{k}$ $\displaystyle \frac{3 - 4(1 - \cos^2{x})}{\cos{x}} = \frac{p}{k}$ $\displaystyle \frac{3 - 4 + 4\cos^2{x}}{\cos{x}} = \frac{p}{k}$ $\displaystyle \frac{4\cos^2{x} - 1}{\cos{x}} = \frac{p}{k}$ $\displaystyle 4\cos^2{x} - 1 = \frac{p}{k}\cos{x}$ $\displaystyle 4\cos^2{x} - \frac{p}{k}\cos{x} - 1 = 0$. Now let $\displaystyle X = \cos{x}$ and the equation becomes $\displaystyle 4X^2 - \frac{p}{k}X - 1 = 0$, which is a quadratic equation that you can solve for $\displaystyle X$, the solution of which you can use to solve for $\displaystyle x$. 3. thank you very much	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 14, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9664930701255798, "perplexity_flag": "head"}
http://math.stackexchange.com/questions/tagged/power-series+algebra-precalculus	# Tagged Questions 3answers 174 views ### the sum of a series I am stuck on the computation of the following sum: $$\sum_{k=0}^{\infty} {\Big( {\frac{q}{k+1}} \Big)}^k ,$$ where $k$ is a natural number, and $0<q<1$. 1answer 226 views ### Solving a formal power series equation I want to find a function $f(x,y)$ which can satisfy the following equation, \prod _{n=1} ^{\infty} \frac{1+x^n}{(1-x^{n/2}y^{n/2})(1-x^{n/2}y^{-n/2})} = \exp \left[ \sum _{n=1} ^\infty ... 1answer 56 views ### Any dominance between these two functions? Let $f\left(x\right):=e^{x}+e^{-x}+2$ and $g_{\beta}\left(x\right):=4e^{\beta x^{2}}$. Do there exist $a>0$ and $\beta>0$, such that $f\left(x\right)\le g\left(x\right)$ for all $x$, \$0\le x\le ... 4answers 150 views ### Formula for calculating $\sum_{n=0}^{m}nr^n$ I want to know the general formula for $\sum_{n=0}^{m}nr^n$ for some constant r and how it is derived. For example, when r = 2, the formula is given by: $\sum_{n=0}^{m}n2^n = 2(m2^m - 2^m +1)$ ... 1answer 61 views ### Manipulating Indices on Series I have a series: $$\sum_{n_1-l_1=0}^{\infty}\sum_{n_2-l_2=0}^{\infty}\sum_{n_3-l_3=0}^{\infty}a_{n_1-l_1,n_2-l_2,n_3-l_3}r^{n_1-l_1}s^{n_2-l_2}t^{n_3-l_3}$$ Which is equal to another series: ... 1answer 151 views ### Sum of the polynomial roots raised to a power. How to prove? Problem: If we have a polynomial $f$ with a derivative $f\,'$ and quotient $q$ function defined as: $$q(x)=\sum_{i=1}^{\infty}a_ix^{-i}=\frac{f\,'(x)}{f(x)},$$ and the roots of $f$ are ... 1answer 156 views ### How to find the sum? Based on logarithm function expansion The problem: How to find the sum? $$-\sum_{i=1}^{\infty}\frac{(-x)^{i\; \bmod(k-1)}}{i}$$ Details: I tried find this sum using Mathematica ...	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 18, "mathjax_display_tex": 4, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.8861567378044128, "perplexity_flag": "head"}
http://en.wikipedia.org/wiki/Mellin_transform	# Mellin transform In mathematics, the Mellin transform is an integral transform that may be regarded as the multiplicative version of the two-sided Laplace transform. This integral transform is closely connected to the theory of Dirichlet series, and is often used in number theory and the theory of asymptotic expansions; it is closely related to the Laplace transform and the Fourier transform, and the theory of the gamma function and allied special functions. The Mellin transform of a function f is $\left\{\mathcal{M}f\right\}(s) = \varphi(s)=\int_0^{\infty} x^{s-1} f(x)dx.$ The inverse transform is $\left\{\mathcal{M}^{-1}\varphi\right\}(x) = f(x)=\frac{1}{2 \pi i} \int_{c-i \infty}^{c+i \infty} x^{-s} \varphi(s)\, ds.$ The notation implies this is a line integral taken over a vertical line in the complex plane. Conditions under which this inversion is valid are given in the Mellin inversion theorem. The transform is named after the Finnish mathematician Hjalmar Mellin. ## Relationship to other transforms The two-sided Laplace transform may be defined in terms of the Mellin transform by $\left\{\mathcal{B} f\right\}(s) = \left\{\mathcal{M} f(-\ln x) \right\}(s)$ and conversely we can get the Mellin transform from the two-sided Laplace transform by $\left\{\mathcal{M} f\right\}(s) = \left\{\mathcal{B} f(e^{-x})\right\}(s).$ The Mellin transform may be thought of as integrating using a kernel xs with respect to the multiplicative Haar measure, $\frac{dx}{x}$, which is invariant under dilation $x \mapsto ax$, so that $\frac{d(ax)}{ax} = \frac{dx}{x}$; the two-sided Laplace transform integrates with respect to the additive Haar measure $dx$, which is translation invariant, so that $d(x+a) = dx$. We also may define the Fourier transform in terms of the Mellin transform and vice-versa; if we define the two-sided Laplace transform as above, then $\left\{\mathcal{F} f\right\}(-s) = \left\{\mathcal{B} f\right\}(-is) = \left\{\mathcal{M} f(-\ln x)\right\}(-is).$ We may also reverse the process and obtain $\left\{\mathcal{M} f\right\}(s) = \left\{\mathcal{B} f(e^{-x})\right\}(s) = \left\{\mathcal{F} f(e^{-x})\right\}(-is).$ The Mellin transform also connects the Newton series or binomial transform together with the Poisson generating function, by means of the Poisson–Mellin–Newton cycle. ## Examples ### Cahen–Mellin integral For $c>0$, $\Re(y)>0$ and $y^{-s}$ on the principal branch, one has $e^{-y}= \frac{1}{2\pi i} \int_{c-i\infty}^{c+i\infty} \Gamma(s) y^{-s}\;ds$ where $\Gamma(s)$ is the gamma function. This integral is known as the Cahen-Mellin integral.[1] ### Number theory An important application in number theory includes the simple function $f(x)=\begin{cases} 0 & x < 1, \\ x^{a} & x > 1, \end{cases},$ for which $\mathcal M f (s)= - \frac 1 {s+a}.$ ## As a unitary operator on L2 In the study of Hilbert spaces, the Mellin transform is often posed in a slightly different way. For functions in $L^2(0,\infty)$ (see Lp space) the fundamental strip always includes $\tfrac{1}{2}+i\mathbb{R}$, so we may define a linear operator $\tilde{\mathcal{M}}$ as $\tilde{\mathcal{M}}\colon L^2(0,\infty)\to L^2(-\infty,\infty), \{\tilde{\mathcal{M}}f\}(t) := \frac{1}{\sqrt{2\pi}}\int_0^{\infty} x^{\frac{1}{2}+it} f(x)\,\frac{dx}{x}.$ In other words we have set $\{\tilde{\mathcal{M}}f\}(t):=\tfrac{1}{\sqrt{2\pi}}\{\mathcal{M}f\}(\tfrac{1}{2}+it).$ This operator is usually denoted by just plain $\mathcal{M}$ and called the "Mellin transform", but $\tilde{\mathcal{M}}$ is used here to distinguish from the definition used elsewhere in this article. The Mellin inversion theorem then shows that $\tilde{\mathcal{M}}$ is invertible with inverse $\tilde{\mathcal{M}}^{-1}\colon L^2(-\infty,\infty) \to L^2(0,\infty), \{\tilde{\mathcal{M}}^{-1}\varphi\}(x) = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty} x^{-(\frac{1}{2}+it)} \varphi(t)\,dt.$ Furthermore this operator is an isometry, that is to say $\\|\tilde{\mathcal{M}} f\\|_{L^2(-\infty,\infty)}=\\|f\\|_{L^2(0,\infty)}$ for all $f\in L^2(0,\infty)$ (this explains why the factor of $1/\sqrt{2\pi}$ was used). Thus $\tilde{\mathcal{M}}$ is a unitary operator. ## In probability theory In probability theory Mellin transform is an essential tool in studying the distributions of products of random variables.[2] If X is a random variable, and X+ = max{X,0} denotes its positive part, while X − = max{−X,0} is its negative part, then the Mellin transform of X is defined as [3] $\mathcal{M}_X(s) = \int_0^\infty x^s dF_{X^+}(x) + \gamma\int_0^\infty x^s dF_{X^-}(x),$ where γ is a formal indeterminate with γ2 = 1. This transform exists for all s in some complex strip D = {s: a ≤ Re(s) ≤ b}, where a ≤ 0 ≤ b.[3] The Mellin transform $\scriptstyle\mathcal{M}_X(it)$ of a random variable X uniquely determines its distribution function FX.[3] The importance of the Mellin transform in probability theory lies in the fact that if X and Y are two independent random variables, then the Mellin transform of their products is equal to the product of the Mellin transforms of X and Y:[4] $\mathcal{M}_{XY}(s) = \mathcal{M}_X(s)\mathcal{M}_Y(s)$ ## Applications The Mellin Transform is widely used in computer science for the analysis of algorithms because of its scale invariance property. The magnitude of the Mellin Transform of a scaled function is identical to the magnitude of the original function. This scale invariance property is analogous to the Fourier Transform's shift invariance property. The magnitude of a Fourier transform of a time-shifted function is identical to the original function. This property is useful in image recognition. An image of an object is easily scaled when the object is moved towards or away from the camera. ## Examples • Perron's formula describes the inverse Mellin transform applied to a Dirichlet series. • The Mellin transform is used in analysis of the prime-counting function and occurs in discussions of the Riemann zeta function. • Inverse Mellin transforms commonly occur in Riesz means. • The Mellin transform can be used in Audio timescale-pitch modification (needs substantive reference). ## Notes 1. Hardy, G. H.; Littlewood, J. E. (1916). "Contributions to the Theory of the Riemann Zeta-Function and the Theory of the Distribution of Primes". 41 (1): 119–196. doi:10.1007/BF02422942. (See notes therein for further references to Cahen's and Mellin's work, including Cahen's thesis.) 2. ^ a b c ## References • Galambos, Janos; Simonelli, Italo (2004). Products of random variables: applications to problems of physics and to arithmetical functions. Marcel Dekker, Inc. ISBN 0-8247-5402-6. • Paris, R. B.; Kaminski, D. (2001). Asymptotics and Mellin-Barnes Integrals. Cambridge University Press. • Polyanin, A. D.; Manzhirov, A. V. (1998). Handbook of Integral Equations. Boca Raton: CRC Press. ISBN 0-8493-2876-4. • Flajolet, P.; Gourdon, X.; Dumas, P. (1995). "Mellin transforms and asymptotics: Harmonic sums". Theoretical Computer Science 144 (1-2): 3–58. • Tables of Integral Transforms at EqWorld: The World of Mathematical Equations. • Hazewinkel, Michiel, ed. (2001), "Mellin transform", , Springer, ISBN 978-1-55608-010-4	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 34, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.854218065738678, "perplexity_flag": "head"}
http://crypto.stackexchange.com/questions/2149/secure-remote-protocol-are-these-2-equations-equivalent/2150	# Secure Remote Protocol - Are these 2 equations equivalent? I'm implementing the Secure Remote Protocol that is specified to have this equation for the parameter `S`, the premaster secret. ```` S = (B - (k * g^x)) ^ (a + (u * x)) % N ```` But the implementation I'm referring to, has done it this way. Is this equivalent to the author's description? Its too complex for me to understand. Can you please check it and/or explain? ```` S = ((B + (N - ((k * (g^x % N)) % N))) ^ (a + u * x) % N); ```` Some questions: • Why the 3 `mod`'s instead of just 1 at the end? • Why `B+` and `N-` instead of `B-`? • Can it be optimized in any way? - ## 1 Answer Well, the bottom line is that it is equivalent. The intermediate mod's do not modify the value of the expression; this is because all the arithmetic done by the expression (except for the calculation of the exponent (a + u * x)) is done modulo N; it is always safe to do intermediate reductions. And, in this case, it's actually necessary; g^x is quite a huge number, and you really have to compute it modulo N, the actual value is just too big. As for why B+ and N- instead of B-, well, it is likely that the modular exponentiation primitive that the programmer used wasn't able to handle negative bases. By reducing the value of k * g^x modulo N (which is safe, as above), that value is always less than N; and so N - (k * g^x % N) is always positive, and so is B + (N - (k * g^x % N)). Also note that this modification is also safe, because for all $A$, $B$, $N$, we have: $A - B = A + N - B = A + (N - B) \mod N$ I would note that there are some further reductions that could have been done: • The exponent could have been computed modulo (N-1), as in (a + ux) % (N-1). This is a safe reduction because N is prime (and Fermat's Little Theorem); this would speed up the second modular exponentiation operation by giving it a smaller exponent. Note that the modular exponentiation primitive cannot do it (it couldn't assume N is prime). • When computing (B + (N - ((k (g^x % N)) % N)), they could have done a final % N to it; this might speed up the second modular exponentiation operation (by giving it a smaller base). However, unlike the above operation, the code that computes the modular exponentiation might do that already (because that is always safe, whether or not N is prime). -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 4, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9227540493011475, "perplexity_flag": "middle"}
http://math.stackexchange.com/questions/151991/how-does-one-give-a-mathematical-talk/152138	# How does one give a mathematical talk? Sometime tomorrow morning I will be presenting a mathematics talk on something related to commutative algebra. The people present there will probably be two mathematicians (an algebraic geometer and a complex analyst) and some friends of mine. Now this is the first time in my life giving a mathematics talk; it will not be a "general" talk aimed at the public but will be more technical. It will involve technical terms (if that's what one calls it) like quotient rings and localisation. As this is my first time I am obviously a little worried! I have read Halmos' advice here but I feel it is more for a talk aimed at the general public. Several things bug me, one of them being how much detail in the proofs does one put in a talk? I am thinking obviously one does not check that maps are well - defined on the board but just says "one can check so and so is well-defined". Furthermore, what about the speed that one writes? I am comfortable writing on the board and some people tell me I speak and write too fast. Obviously that is a problem and I need to slow down, but also I don't want to be talking too slow to bore the audience and seem to be out of passion. What is a good indicator of how "fast" or "slow" should one give a talk? Besides, is there a way that one should "act when up on stage"? By that I mean so called "socially acceptable" do's and don'ts. I think any advice given from those who have "been there done that" would be useful for future wannabe mathematicians like me. Thanks Edit: Since many people have said it is difficult to give advice not knowing the audience, for the moment the audience will be a complex analyst, an algebraic geometer, one person who has just completed honours in orbifold theory, another friend in third year taking courses in measure theory,galois theory and differential geometry, and lastly a PhD student in operator/ $C^{\ast}$ - algebras - 5 Pause often enough to give people a chance to absorb what you’ve said. Move around enough so that everyone can see what you’ve written. (Ideally, avoid standing in front of what you’re writing even while you write it; this is quite possible, though not everyone has the knack. I always alternated between writing ahead of me and writing behind me, to open sight-lines from opposite sides of the room.) – Brian M. Scott May 31 '12 at 12:03 13 Prepare a talk for the time you have, then cut one-third of it. It really is amazing how much time is needed to say anything. – Gunnar Magnusson May 31 '12 at 12:05 5 @BenjaminLim If there are only two mathematicians in the audience, I don't see why you would want to get much more technical than a "general talk". Pick the most important proof (or two) and when the machinery comes in, do your best to give them a feel for how it works (without demanding the details). Never overestimate your audience's ability to keep up. Also, do not forget pants, if they are appropriate in your region. – rschwieb May 31 '12 at 12:06 4 @BenjaminLim To your question about "talk not rigorous enough": NO. You may sacrifice lots of rigor in the interests of getting the good ideas across. (But do not teach them wrong stuff.) If you will be using the board, then the best advice has already been given: avoid talking to the board, avoid sluggishly speaking what you are currently writing. If I'm going to write something important on the board, I usually first say what I'm going to write, write a little, pause, review the idea aloud, finish writing, then go over it one final time aloud. – rschwieb May 31 '12 at 12:18 3 @rschwieb "Also, do not forget pants, if they are appropriate in your region." I have a friend who forgot his pants on the day of his thesis defense. He'd bicycled to campus, and successfully remembered to pack all the rest of his suit.... – RBerteig May 31 '12 at 18:19 show 17 more comments ## 5 Answers Here are some miscellaneous tips. They are not of the "don't look at your shoes, speak clearly, etc." variety, but rather they assume that you want to give a really great talk, rather than just a bearable one. 1. Do not overrun. 2. Do not overrun! 3. Practice your talk in front of an empty room, beginning to end. Every "umm" will echo back at you so loud, that you will really want to weed those out. You will likely need to practice at least twice, taking care of tips 1. and 2. When you practice, you need to pretend that you really have people sitting there and taking notes. In particular, you will sometimes need to just pause and stare at the empty room, waiting for your imaginary audience to finish writing. 4. If you want to give a really interesting talk, it is not enough to know your stuff. You should try to tell your audience a story. So before you write the talk, decide what your story is about, then structure your talk around this. This involves deeper understanding of the topics than just understanding every step in every proof that you will be presenting. Once you know what your story is, you will likely also know, which proofs to leave out or to only sketch, where to give examples, etc. 5. Involve your audience, but ask only the most basic questions. Do not create uncomfortable silence (so don't wait for the answer for too long), but do not answer your questions too quickly either (you don't want to beat one of your audience members to it by half a second - it feels pretty bad to them). 6. Do not overrun!! 7. Since this is your first talk, it is less important to make it perfect, than to learn from it. I have written up some of my strategies to get useful feedback after my talks on another SE site, you might find it useful. 8. Enjoy the experience. Enjoy interacting with the audience. Don't forget that you are talking about something that you find really fun! But also don't forget that your audience members might be less enthusiastic than you, or slower on the take. - A lot of people have advised me (on this page here) not to drone with "umm" and stuff. Why is this the case? Also, the motivation for what I may be talking about comes from projective varieties and things like $\textrm{Proj} R$. I have not studied algebraic geometry, so I will have to really try to provide a lot of motivation from algebra. Already I can think of 2 or 3 reasons as to why we want to localise rings, so perhaps motivation from just the commutative algebra is enough .... – BenjaLim May 31 '12 at 14:33 4 To me, "umm" just sounds ugly. If you are serious about giving awesome talks, then you should see yourself as a performer. People should enjoy watching you "on stage". As for motivation, this is kind of hard to answer without knowing the exact background of your audience. Remember that motivation is something that motivates your audience to continue listening to you. From that point of view, it is always good to have motivation from as many points of view as possible. If you can give geometric or number theoretic examples, go for it. Otherwise make do with what you know. – Alex B. May 31 '12 at 14:46 Thanks! +1 By the way for your answer. – BenjaLim May 31 '12 at 14:50 1 (+1) But, I think there's one point that could stand to be emphasized: Do not overrun!! – cardinal May 31 '12 at 18:26 2 @user02138: Going over time. – Alex B. Jun 4 '12 at 2:10 show 5 more comments ## Did you find this question interesting? Try our newsletter Sign up for our newsletter and get our top new questions delivered to your inbox (see an example). email address I remember the first time I gave such talk, it was my last requirement as an undergrad (a part of the course I took in representation theory). I was stuttering the whole time and "ummm... uhhh..." was dense in my lecture. Since then I began teaching and gave several seminar lectures. Let me give you some general advices which may be applicable here. 1. Know what you are talking about in details. The better you understand, the better you can explain. In the future you may find yourself giving a talk without preparing at all, and it may end up being pretty good - this cannot happen if you only have a vague idea about the topic. The more you "grok" it, the better you will present it to others. 2. Be interesting. This, at least for me, means that you need to try and keep active interaction with the crowd (which is relatively easy if the crowd is small). Try and speak in a non-monotonic voice, a constant tone and speed is a good way to put people to sleep and despair. Try to insert a joke or two when appropriate. 3. It's fine not to prove everything. I learned this from a seminar lecture given by Stefan Geschke a year ago. He proved some of the theorems but he skipped some of the proofs, instead talking about the idea behind the proof or what can be done with the theorem. I remember coming out of the lecture wide awake and with a good sense of understanding the general idea. In contrast, some talks I attend to are filled with proofs of every single detail and lemma, and I get lost. This advice is particularly useful if the crowd already knows some of the things you are going to say. You can cite them and keep forward. 4. Don't overdo 3. You still have to prove something, otherwise people feel that you only gave some vague framework - not a talk about a mathematical concept. If you aim to prove some big theorem, write it on one side of the board along with two or three lemmas you will need, then work on the other side. This will allow people who zone-out to tune back into the talk. 5. Relax and try to have fun. Regardless to all of the above, I doubt most people give good talks the first few times they give talks. So just remember to take it easy, it gets better with time. - This is great advice! – Eugene May 31 '12 at 12:29 1 +1 Asaf great words of advice! – BenjaLim May 31 '12 at 12:30 @BenjaminLim What is the talk going to be about, if I might ask? – rschwieb May 31 '12 at 12:33 @BenjaminLim Just one more thing to add to this. Be ready to answer questions. Some audience members will ask for clarification on certain issues and showing that you know your stuff down pat is important. – Eugene May 31 '12 at 12:36 4 Related to what Eugene said, don't be afraid to not know the answer to a question - it's going to happen sooner or later. One thing I've learned is that it's really easy to ask an incredibly difficult question! – Jason DeVito May 31 '12 at 13:02 show 7 more comments A lot of good advice has already been given. However, there is a key question which I'm surprised that no one has yet asked: WHY are you giving this talk? Every mathematical talk has a purpose, and there are many different purposes for a mathematical talk. For instance: 1) The job talk: you are being (explicitly or implicitly) screened for an academic job, and the goal of your talk is to make the listeners want to hire you. (As a good friend and colleague of mine likes to point out, in some sense every talk that you travel to a new place to deliver is a job talk in that if you do very well you will advance your career and if you do very badly your career may suffer.) 2) The research seminar talk: In this type of talk the purpose is to convey something about your own recent work to an audience of relative experts. Note that this is formally similar to a job talk (and see the above parenthetical comment), but if it is really not a job talk -- and, in particular, if it's a talk you're giving to your colleagues / peers / students -- it has a quite different purpose: to inform rather than impress. 3) The colloquium talk: This is somewhat similar to 2) but with a general mathematical audience. (Note that the phrase "general mathematical audience" actually carries relatively little meaning: if you are asked to give such a talk, you should find out more exactly what it means! Sometimes colloquia are attended primarily by research mathematicians with a particular interest in your field, and sometimes they are attended primarily by undergraduates who may or may not be math majors.) The purpose of a colloquium talk is somewhere between information and entertainment. Although you are expected to talk about your research, this can sometimes be in a very general way: for instance, you can give an excellent colloquium talk in which you do not explicitly state a theorem of your own (but it takes some guts and confidence in yourself to do so). 4) The learning seminar talk: This is a talk that you give as a participant in a seminar. Generally you have chosen, or been assigned, some specific paper or part of a paper and your job is to present as much of this as possible in the time allotted, starting from an overview but often including key technical details. 5) The student project talk: This is a talk that a student gives in the context of a particular course. Often, but not always, it is given "in class" and only attended by other members of the class and the instructor. Often there will be an accompanying written paper that you are summarizing. This is similar to 4) but probably at a lower level. 6) The expository talk: This is a talk with the goal of teaching the audience some material which is (usually) not due to you. Really it is like teaching a course except it all takes place in one sitting (or maybe a small series of sittings), which of course makes it challenging. Generally you have some specific reason to do this, e.g. it functions as background / prelude to some other activity. This is probably not a complete list, and if anyone wants to suggest any additions, please do so. So...what kind of talk will you be giving? Any of the above? Did someone ask you to give this talk? If so, why? - 4 To echo Pete's answer: indeed, there is a purpose to any talk, and that should determine most aspects. Sometimes people slip into thinking that there is an "abstract" notion of "talk" that is independent of context or purpose, but that leads to self-sabotage. The purpose, the audience, the expectations, the goals, ... must be considered and responded-to. These and the time constraint usually (nearly-) completely determine what one should do. – paul garrett May 31 '12 at 18:32 I have been giving long talks professionally for about twenty years, with great success, and the best advice I can give you or anyone is to give at least one practice version of the talk first. Find someone to give it to if you can, or if not then give it to your dog, or to an empty room. Actually say the words aloud. This will give you a better sense of how long your talk is than anything else you can do. But equally important, you will discover all sorts of things that seemed clear and straightforward in your head that somehow turn into garble when they come out of your mouth. The practice talk is a chance to fix this. One problem that many beginning (and experienced) speakers have is the droning problem that other posters in this thread have alluded to. This is not just the problem of pausing and saying "ummmm." (Or its British counterpart "Errrr"; in Norwegian it is "Øøøø".) It's that a lot of people who are perfectly good speakers around a café table or in a bar suddenly lose all their intonation, inflection, and pacing when they get up in front of an audience, and speak in a robotic monotone in which all the syllables are the same length. One strategy for avoiding this is to make a conscious effort to pretend that you are sitting at a bar with these people after a round of drinks, and try to speak as if that were the case. Another strategy is to pick one person in the audience who seems friendly and pretend that you are giving the talk just for them. I sympathize with your problem of talking too quickly. I do this too. If you can't talk more slowly, there are two things you can try that might be almost as good. You can insert a long pause after every sentence or two. (I once gave a talk in Taiwan that went very well, in part because I had to pause after every sentence or two to let the translator repeat my remark in Chinese.) Or you can repeat everything two or three times, which is not as awful as it sounds. - There is one problem with talking "to one person", you feel that you ignore the rest of the audience and it can bother you as a speaker. I know that sometimes when talking to a class full of people I may feel that I focus on one person and it makes me feel a bit uncomfortable. – Asaf Karagila May 31 '12 at 16:40 3 Yes, it's easy to go too far with that, particularly in a small room. But I think it might be better to give the talk to one person in the room than to give it to nobody in the room! – MJD May 31 '12 at 16:42 1 (+1) I've found that practicing once helps, but practicing 3-4 times is usually necessary to work most of the verbal kinks out. Stand in front of a mirror, point at your laptop screen as if it were the projector. The mirror will help catch any unusual mannerisms or other quirks that need polishing. – cardinal May 31 '12 at 18:29 1 Talking to one person is better than talking to nobody, but it usually better to look at everyone. Some people feel compelled to test the speaker with questions if they feel ignored. Also, accoustics often force you to speak differently than at the coffee table. Stage actors spend lots of time on their voice. – Michael Greinecker Jun 1 '12 at 8:01 When I did my first math talk, I read all the answers here. From those good advices, I made my slides and my script. Those two documents illustrate the message of the answers. Learning by example is helpfull, so take a look. For a 30 minutes talk, I settled for 12 slides, and the slide are not crowded. The script, to read in combination with the slides, side-by-side, consists of what I said for each slide. It shows how to keep the balance between details and birdviews. I received very good feedback for this talk, so I think it is good enough to be post as an example. -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9718419313430786, "perplexity_flag": "middle"}
http://motls.blogspot.com/2012/03/most-of-research-of-nonlocality-is.html?showComment=1332276259014&m=0	# The Reference Frame ## Tuesday, March 20, 2012 ... ///// ### Most of research of "nonlocality" is pseudoscience In June, Sabine Hossenfelder is organizing a workshop on nonlocality: Workshop on Nonlocality, June 27-29 (Backreaction) Current registrants (Workshop's website) So far, 10 people have shared their intent to attend. Some of them are just Mr or Ms. George Musser belongs to the best that Scientific American has to offer these days but it's still true that "Scientific" American has sucked for a decade or so. And there are a few people who have something to say but no one will listen to them, anyway. "Nonlocality" is a cool buzzword but the actual research that uses this buzzword is split to several categories that have nothing to do with each other and the research in most of these categories is pseudoscientific in character. An experimental physicist training telekinesis, a version of nonlocality What do I mean? Let us spend some additional time with locality and nonlocality. As this blog has discussed many times, locality is an inevitable consequence of the special theory of relativity. The Lorentz symmetry underlying this theoretical framework implies that an action that would be faster than light in the vacuum would be equivalent, by a Lorentz transformation (a change of the reference frame), to an action that influences your past. This can't happen because one would deal with the paradox of your prematurely castrated grandfather (you could do it to him before he had sex with your grandma). Consequently, locality has to hold. Signals are not only forbidden from propagating into the past; they're not allowed to propagate faster than light, either. One of the corollaries is that if you want to influence a distant place in the Universe, something will have to happen (or go through) any plane that sits in between you and the distant place. Voodoo, telepathy, telekinesis, and similar things are banned by the laws of physics. Now, there exist lots of confusion about this point. People believe in many kinds of "loopholes" that actually don't exist. The research that loves to use the term "nonlocality" may be divided to these basic groups: 1. Attempts to introduce nonlocality to "get rid of" dark matter, dark energy, or other things 2. Nonlocality as a consequence of otherwise unmotivated and unjustifiable "alternative" theories of physics such as the Hořava-Lifshitz gravity 3. Effects in theories that would love to be "somewhat relativistic" and "somewhat non-relativistic", especially the so-called Double (or Deformed) Special Relativity 4. Misconceptions about nonlocality in quantum mechanics, motivated by people's fundamental misunderstandings of the meaning of quantum mechanics in general and entanglement in particular 5. Appearance of weak nonlocal effects in the process of the preservation of the black hole information and related weak nonlocal effects in string theory (in limited environments) in general. Among these categories, all of them except for the last one are scientifically nonsensical or indefensible. Let's look at those things in some detail. Special relativity For thousands of years, people understood that we can't change the past – although totalitarian ideologies often try to rewrite the history (and e.g. HadCRUT4 was just released to make the past cooler than it was according to HadCRUT3, a self-evident main goal of raising the digit). We may only influence the future. In other words, if we use the usual convention for the sign of time $t$ that increases into the future, the coordinates of the cause and its effect have to obey\[ t_\text{cause} \lt t_\text{effect}. \] That's a very simple assertion ("the cause precedes its effect") that holds in Newtonian physics – and that was known long before Isaac Newton was born. We may call this inequality "causality" (and reserve the term "determinism" for the idea that everything is determined by the initial state). It's logically necessary because we know that we may affect the future. If it were possible for influences to propagate in the opposite direction as well, we could easily identify "closed loops" or generalized "closed time-like curves" that would lead to logical paradoxes. However, special relativity says that all inertial reference frames related by Lorentz transformations are equally good as frameworks to describe the physical phenomena. Imagining that the only other spacetime coordinate besides $t$ is $z$, the Lorentz transformation takes the form\[ t = \frac{t'-vz'/c^2}{\sqrt{1-v^2/c^2}}, \qquad z = \frac{z'-vt'}{\sqrt{1-v^2/c^2}} \] where $v$ is the relative velocity between the two reference frames. We may translate the inequality above to the reference frame that uses the primed coordinates as follows:\[ t'_\text{cause} - \frac{vz'_\text{cause}}{c^2} \lt t'_\text{effect} - \frac{vz'_\text{effect}}{c^2} \] The square root in the denominator could have been cancelled. It's important that the inequality above – which is just the original "causality" condition – has to hold in all reference frames, i.e. for all values of $v$ that are smaller than $c$. The velocity may approach $v\approx \pm c$ arbitrarily closely. Write the inequality above in these two limiting cases and you will see that the inequality also implies\[ t'_\text{effect} - t'_\text{cause} \geq \frac{1}{c} \abs{ z'_\text{effect} - z'_\text{cause} } \] If we dealt with all the spatial coordinates, the absolute value would simply become the distance between the places of where the "cause" and "effect" (two events in spacetime) occur. At any rate, the formula above says that the effect can't come earlier than the time that light needs to fly from one place to another. Light in the vacuum – or anything with the same speed (e.g. gravitons) – is the fastest "messenger" or a "tool to influence" that Nature allows. It doesn't matter whether your theory is classical, quantum, male, female, or produced out of sockpuppets. The derivation above didn't make any assumptions about these matters so it has to hold generally. And be sure that it does hold. Still, I've listed "five flavors" of nonlocality that people like to talk about. Let's look at them from the last one. Black hole information puzzle When Stephen Hawking made his calculation of the black hole radiation that is named after him today, he was able to compute the temperature and many other details about the radiation. But one property of the radiation was easy to see without the full-fledged formalism of quantum field theory on curved backgrounds: the Hawking radiation couldn't have depended on the initial state of matter that collapsed into the black hole. The initial deviations of the newborn black hole from its ultimate static shape are quickly, exponentially disappearing. This stabilization of the newborn black hole may be (at least approximately) described by the ringing or quasinormal modes. Very quickly, the black hole becomes nearly static. Whatever happens with the collapsed matter inside the black hole can't influence the Hawking radiation because of locality: the region where the Hawking radiation propagates is spacelike-separated (i.e. requiring superluminal signals) from the region where the collapsed matter approaches the black hole singularity. The nonlocality arguments sketched above, when generalized to the case of curved spacetimes, strictly prohibit such an influence. That's why Hawking deduced that the radiation had to be exactly thermal and independent of the initial state. In the mid 1990s, string theory made it clear that while this conclusion is correct for all practical purposes, the information is fundamentally not lost. Subtle nonlocal influences imprint the properties of the initial state into the Hawking radiation. The latter still looks almost exactly thermal but it's not quite thermal. One may say that the information may "tunnel" from the black hole interior to the exterior. It may violate the cosmic speed limit "for a while" and produce exponentially tiny imprints in the Hawking radiation that remember the initial state although the information can't be extracted in practice. In particular, Matrix theory and AdS/CFT make it totally indisputable that even though string/M-theory contains black holes and they behave exactly as they should in all the approximations we may consider, it's still a theory that fully conserves the information. A few years ago, Stephen Hawking surrendered in his black hole war against Lenny Susskind, gave up his bet against John Preskill, and declared himself as a co-discoverer of the preservation of the information in the black hole, too. ;-) Of course, he wasn't really a discoverer of that but he developed a psychological framework that allowed him to understand why the actual theory of quantum gravity may find a loophole and produce a different qualitative conclusion than one he had believed for 30 years. His semiclassical calculations are extremely accurate but they're expansions around a point that is "qualitatively misleading". It forces you to destroy the information to all orders. But the truth isn't destroying it because the black hole interior isn't really "sharply well-defined". A particle located in the black hole interior is similar to an alpha-particle inside a nucleus that may alpha-decay. There's no real "permanent state" of the alpha-particle in which it would be strictly isolated from the exterior of the nucleus. The particle may simply tunnel out and escape. The same is true for the matter or mass confined by the black hole. It may also get out. Just to be sure, if I talk about the tunneling, it doesn't allow you to send "small amounts of information" superluminally in any context. It only works in the presence of black holes. I don't want to go into details. Michele Arzano, the first participant of the Nonlocality Workshop, knows quite something about the interpretation of the Hawking radiation in terms of quantum tunneling but I am afraid that it will be above the other participants' heads. String/M-theory is based on elementary objects that are extended in space – e.g. strings. So in some sense, it allows "slightly nonlocal effects" to operate. However, this intuition is still misleading. Physical phenomena (including the splitting and merging of the stringy pieces) are still local on the string or the world sheet. And even if you describe string theory as a string field theory in which the whole string is represented by a field at one point, the string's center of mass, you still reproduce some conditions that you expect from a fully local theory. In particular, perturbative string theory seems to saturate some conditions for high-energy scattering that may be derived for local quantum field theories. One must be very careful about these things. There are "some effects" that might be called "nonlocal" that string theory allows but it still vehemently forbids many others. String theory isn't a garden-variety nonlocal "rebel" theory that violates everything. It is highly conservative and when you study it using the right language, it's actually as local as local quantum field theory. I don't want to focus on this topic here so let's switch to another category I mentioned above: Misconceptions about nonlocality in quantum mechanics This is the most frequently cited "nonlocality" in popular physics books. In fact, the popular physics books seem to present a near "consensus" that there's something nonlocal about the way how quantum mechanics operates. Despite this consensus and the fact that some otherwise pretty good physicists are participating in it, all these claims are complete rubbish. The "quantum nonlocality" claim revolves around the EPR phenomena and entanglement. Readers of popular and sometimes even not-so-popular books and articles are led to believe that an object in an entangled pair may remotely control its cousin. But nothing like that occurs. Entanglement is nothing else than the quantum variation of the concept of correlation. It either represents any correlation between two subsystems that is properly described and understood in the language of quantum mechanics; or it refers to those correlations that make the subsystems behave differently than anything in classical physics. Technically, an entangled state is an element of a tensor product Hilbert space that can't be written as a tensor product of two states:\[ \ket\psi\in \HH_\text{here}\otimes \HH_\text{there},\qquad \ket\psi\neq \ket\psi_\text{here} \otimes \ket\psi_\text{there} \] Those states represent immense psychological problems for people who can't reconcile themselves with the fact that the world doesn't obey the laws of classical physics. It doesn't even obey the "bare realist logic" of classical physics. Everyone who tries to "squeeze" quantum physics into a classical straitjacket ultimately starts to talk about nonlocality and similar nonsense. However, entangled states are nothing "extraordinary" that would force us to revise the rules of relativity, e.g. the proof that there can't be any nonlocality that was presented at the beginning of this blog entry. Entanglement states aren't miraculous or special or supernatural. They're generic states in the Hilbert spaces. Almost all states in a tensor product Hilbert space are entangled. If you don't learn how to properly predict situations involving entangled states, then you misunderstand at least 99.99999 percent of quantum mechanics. There isn't any quantum mechanics without entangled states. Quantum mechanics is a theoretical framework that makes probabilistic predictions of the outcomes of measurements. Everyone who tries to pretend that quantum mechanics is something else, e.g. a version of classical physics, inevitably drowns in the quantum nonlocal new-age crackpottery at some point. If the physical system is composed of two subsystems, various quantities such as $x_1$ and $y_2$ may be measured on these subsystems $1,2$. Quantum mechanics gives us some probabilistic predictions – the probability density $\rho(x_1,y_2)$ for any pair of values of the two quantities that may be measured, whatever they are – according to its totally well-defined rules. Quantum mechanics is a toolkit to deal with the actual observations. It surely isn't a propagandistic tool designed for you to confirm your medieval prejudices about the world as a classical place or something like that. Quite on the contrary, it says that these prejudices are wrong. It says many other things, too. In almost all cases (at least when the systems $1,2$ were in contact sometime in the past), the probability distribution for the two quantities will reveal some correlations:\[ \rho(x_1,y_2) \neq \rho_1(x_1) \rho_2(y_2) \] In the typical case, you will simply be unable to find the distributions for the individual quantities so that the probabilistic distribution for both would be a simple product. This simply says that the two systems have correlated properties. It's not shocking that two objects that met in the past exhibit some correlated properties. Bertlmann's socks do the same thing. Quantum mechanics allows these correlations to affect many more pairs of possibly measured quantities etc. than classical physics could. But that shouldn't be shocking, either. Quantum mechanics isn't classical physics, stupid. So it predicts different statements about things. But they're really the same things as the things in classical physics – observables, their values, and correlations between those values. Now, when quantum mechanics predicts the probability distribution $\rho(x_1,y_2)$, then Nature – and She knows quantum mechanics very well – will have no problem to determine what will happen. If it has to produce a result for $x_1$ that the first experimenter measures, it may simply start by calculating the probabilistic distribution for $x_1$ itself. If you were given the two-variable distribution, the one-variable distribution is simply\[ \rho_1(x_1) = \int \dd y_2~\rho(x_1,y_2) \] The funny thing is that maths of quantum mechanics guarantees that if the second experimenter decides to measure $z_2$ instead of $y_2$, whatever these variables are, the integral over $z_2$ will yield the same $\rho_1(x_1)$. This natural result isn't an independent dogma that is inserted to quantum mechanics; it is a result of equations. You may check that it works and the world is "well-behaved" in this sense. But it works for a different reason than the wrong reason that the world is classical and these facts are a priori incorporated into a classical world! The world isn't classical. Our quantum world shares some consistency conditions with classical physics but it isn't equivalent to any classical or realist theory. It's a quantum world, stupid. Some things you have believed to be true may be untrue in our quantum world; other things you have believed to be true may still be true but their proof may require some nontrivial linear algebra instead of "self-evident dogmas"; another class of things you have believed may be just "approximately true". The independence of the integral above on the choice of the variable $y_2$ or $z_2$ ultimately boils down to the completeness relation\[ {\bf 1} = \int \dd y_2~\ket{y_2}\bra{y_2} = \int \dd z_2~\ket{z_2}\bra{z_2} \] for two orthonormal bases (in this example continuous bases normalized to the Dirac delta-function). This simple piece of linear algebra with a crisp physical interpretation is a reason why the decision of the second experimenter what to measure – whether $y_2$ or $z_2$ – has no impact whatsoever on the probabilistic distribution for $x_1$. That's a part of the reason why there are no nonlocal influences running between the two subsystems. So Nature knows the distribution for $x_1$. It gives the subsystem some freedom but the resulting $x_1$ must look random, selected according to the calculable distribution. The same thing holds for $y_2$ or, if you decide to measure it, $z_2$. The role of $x_1$ and $y_2$ is obviously totally symmetric. The two measurements may be spatially separated and very distant and relativity doesn't allow you to say which of them was done first. The time ordering of these two events may depend on the reference frame. But Nature doesn't need to know it. Quantum mechanics only predicts a correlation between $x_1$ and $y_2$; it doesn't predict or require any causation. Correlation isn't causation. Whenever there's some correlation in the world – in our quantum world – it's a consequence of the two subsystems' interactions (or common origin) in the past. You may imagine that Nature first calculates the distribution for $x_1$, throws dice, collapses the wave function for $y_2$ so that the freshly measured value of $x_1$ is already assumed, and then throws dice again according to the distribution $\rho(x_1^\text{measured},y_2)$. But the "dice" and "collapses" is just a stupid fairy-tale for babies. No "visual mechanism" like that exists in Nature. What exists are the measurable results of experiments and they follow the same distributions whether you imagine that $x_1$ was decided before $y_2$ or vice versa. If you imagine that Nature is a hot babe who throws dice because it makes you more excited or pleased, be my guest. But even in principle, it's impossible to distinguish between the two orderings how the babe throws the dice so this ordering is totally unphysical. If you don't understand that things that are even in principle indistinguishable by any measurements are fundamentally indistinguishable in physics, and that all the crutches that distinguish them are unphysical superconstructions that mustn't ever be treated as physics, then you're simply not thinking about the world scientifically. You may be thinking about the world in some hardcore materialist (and implicitly classical) way but it isn't the same thing as the scientific approach to the natural phenomena! The dice touching Nature's breasts (or His penis, so that I can't be accused of sexism or homophobia) after She makes a guy called $x_1$ collapse after an intense intercourse are just your sexual fantasies, stupid. The actual physics doesn't contain anything of the sort. The paragraphs above were written to make you think twice or thrice or 1,000 times if you still fail to understand that there's no "nonlocal influence" needed to explain the results of measurements of entangled particles. If you still feel that there's some influence here, it's because you're making additional, invalid assumptions about the world. Quantum field theory Now, the moment of the measurement is just one part of the events that happen to the two subsystems. I've explained that there's no influence here: quantum mechanics only predicts correlations, not causation between the two measurements. The other part of the events according to quantum mechanics is the evolution in time. (That's how the Copenhagen interpretation of quantum mechanics divides "processes" in physics. This division only makes sense "psychologically": the measurement is the process in which we're learning some actual data. But the future properties of a system are unaffected by the question whether or not we consider some process a measurement; these properties are calculable independently of these colorful interpretations.) I haven't discussed it yet. In general, the evolution in time may include actions at a distance. In nonrelativistic physics, the Sun is instantly influencing the planets. If it quickly explodes, the planetary orbits are immediately affected. You know that according to relativity, such an influence would still take 8 minutes to get to the Earth. The simplest class of theories that obey the general postulates of quantum mechanics as well as the rules of special relativity is the class of (relativistic or local) quantum field theories. They may be "constructed" by adding hats above the observables in relativistic classical field theories such as electrodynamics with its Maxwell's equations. The hats don't change anything about the locality of these theories. In the operator approach to quantum mechanics, the evolution in time is governed by the Hamiltonian, the operator of energy. In quantum field theory, it's essentially an integral of the energy density over space:\[ H = \int \dd^3 x~\rho_{00}(\vec x) \] That's great because if we study the evolution of two subsystems in two regions, here and there, the relevant part of the Hamiltonian becomes a simple sum of the two Hamiltonians for the separate regions:\[ H = H_1+H_2 \] This additivity or extensiveness of energy is equivalent to locality of the evolution in time. The funny thing is that the Hamiltonians $H_1,H_2$ commute with the quantities describing the second and first particle, respectively, i.e.\[ [H_1,y_2]=0, \qquad [H_2,x_1] = 0 \] and so on. This follows from the fact that the Hamiltonian density is a function of the local fields and their derivatives and the fields commute (or anticommute) if they're spatially separated. But if you use the Heisenberg picture, the vanishing commutators above make it very clear that the Hamiltonian $H_2$ has no impact on the probabilistic distributions for $x_1$ or any other quantity describing the first subsystem or vice versa. So if you want to predict the probabilistic distributions such as $\rho(x_1)$ only, you only need to deal with $H_1$. The evolution of the second, distant subsystem and the measurements that could occur there (as I have already discussed) make absolutely no impact on the probabilities that the first system will experience something or something else. And vice versa. This doesn't mean that the results of the distant measurements won't be correlated. They will be correlated but the Hamiltonian argument makes it clear that such correlations between the two subsystems can't result from the separate evolution of these subsystems in time; they must always be a consequence of the subsystems' contact in the past. Indeed, if your two subsystems are unentangled i.e. uncorrelated in the quantum sense, their state vector has to be a tensor product\[ \ket\psi_{12} = \ket\psi_1 \otimes \ket\psi_2. \] But if that's the case, this factorized form of the state vector will be true forever. The state above will evolve to\[ U_{12}\ket\psi_{12} = U_1\ket\psi_1 \otimes U_2\ket\psi_2. \] where the evolution operators $U_1,U_2$ may be calculated by simply exponentiating multiples of the Hamiltonians $H_1,H_2$, respectively. The evolution of the composite systems according to the local laws of quantum field theory won't change the degree of their entanglement at all! It may only change the pairs of quantities in which the subsystems are highly correlated or less correlated but for any pair of strongly correlated initial quantities, there will be a pair of quantities describing the two subsystems at any later time that will be exactly equally correlated! There are just no nonlocal influences running in quantum field theory. All predictions respect the independent existence of spatially separated regions and objects. This assertion may be verified by explicit calculations in quantum mechanics and these calculations are indeed needed. It's wrong to assume that this thing has been obvious to you since you were in the kindergarten. The actual reasons why these statements holds are derived and can only be verified by a person who has learned quantum mechanics. You can't verify these features of the real world by some arguments based on your classical intuition – where these properties could be trivial – because our world isn't classical, stupid. Entanglement isn't any sign of a nonlocality. Bell's inequalities guaranteed that the experimentally observed correlations can't be explained by a local realist theory. But in a striking contrast with the popular scientific literature, the wrong assumption isn't locality; it's realism. Locality is just a property of relativistic and similar theories, whether they're quantum or classical. And indeed, it holds. The validity of locality was one of the key results of Einstein's special relativistic revolution of 1905, a revolution that can't be undone anymore. On the contrary, realism is an assumption behind all classical theories, whether they're relativistic or not. And it's been shown invalid in the 1920s because classical physics has been shown wrong. Only probabilities of actual measurements may be predicted by physics. This is what the quantum revolution of the mid 1920s is all about. The new picture of the world is "local, non-realist". Everyone who suggests that it's "nonlocal, realist" apparently misunderstands both major revolutions of the 20th century physics, quantum mechanics and relativity. DSR It isn't surprising I spent so much time with the foundations of quantum mechanics. It was my original goal. Let me say a few words on the status of the first three "kinds of nonlocality". Doubly special relativity isn't a viable framework to produce realistic theories in physics, at least not in spacetimes of dimension 4 and higher, such as ours. Sabine Hossenfelder actually became one of the people who understood this point so it's unfortunate that she's still organizing workshops for the people who remain in the dark about this elementary point. Doubly Special Relativity may be connected with a well-defined "quantum group", a quantum deformation of the Lorentz group. But if you actually try to construct any theory that has this symmetry, you will fail. You will either get theories that are fully equivalent to local, Lorentz-symmetric theories by a field redefinition; or you will end up with the class of totally nonrelativistic, and therefore nonlocal, theories that are totally ruled out experimentally because they prevent you from objectively saying whether a lethally poisonous snake is closed in a box. Such important questions become observer-dependent in any scheme of DSR that would be inequivalent to proper special relativity. Sabine Hossenfelder knows the simple proof of this assertion but it's probably still helpful for her to invite Jerzy Kowalski-Glikman who will surely offer his delusions about DSR's not being dead yet. Lifshitz and other random Lorentz-violating theories The second category is about random ideas how to build fundamentally Lorentz-violating theories. The Hořava-Lifshitz theory is an example from recent years. It's a theory one can write down. It's inconsistent with basic properties of black hole physics and many other things but the Lagrangians superficially have some nice properties that lead dozens of people to study such things. Fine. Those theories still disagree with the observations. The accuracy with which special relativity works in the real world around us is amazing. For some particular terms in the effective equations, people have been able to show that the coefficients of the Lorentz-violating terms are much smaller than expected from dimensional analysis, even at the Planck scale. So they almost certainly don't exist in Nature. To say the least, they're not responsible for any large effects. Also, their existence would create a new hierarchy problem (lots of them) because people would have to be explaining why many terms that can a priori be of order one are so incredibly tiny that we haven't seen them yet. I have no clue what motivates people to write papers about this stuff. In my opinion, it's the money. People get paid for that even though they know that the papers are pretty much worthless for physics. We're not learning any physics because those theories aren't correct descriptions of the Universe; we're not learning any maths because these theories don't have any deep, unifying, pretty, or profound maths in them. Replacing dark matter, dark energy etc. Various nonrelativistic and/or nonlocal theories were also proposed to get rid of a "problem" – a word that the proponents may use for the need for dark matter or dark energy. Well, I have nothing against such attempts to "simplify" cosmology a priori but the resulting proposed theories must still be fairly compared with the "alternative" – i.e. with general relativity including dark energy and dark matter. When one looks at the ability of both competitors to naturally explain the observed phenomena and the general lessons extracted from them, including the Lorentz symmetry, equivalence principle etc., I find it obvious that proper general relativity is, regardless of the "need" for dark matter and dark energy, more convincing – by many orders of magnitude. This is not a proof that e.g. a particular nonrelativistic MOND theory has to be wrong but it is surely a sufficient explanation why I consider it to be a huge waste of money and fossil fuels if people fly to conferences about this physically unlikely and mathematically unappealing alternative theories. So most of the nonlocality research is pure rubbish. I believe that the actual reason why this stuff hasn't gone extinct in the Academia yet is the support from the laymen who just find it cool if someone does something that looks like telekinesis or telepathy etc. That's why I really think that the picture at the top of this blog entry is a great visual representation of the nonlocality research. It's a P.R. game building on ordinary people's ignorance, ordinary people's thirst for magic and mystery, and ordinary people's misconceptions that these people consider particularly "cool". And that's the memo. Posted by Luboš Motl \| Email This BlogThis! Share to Twitter Share to Facebook Other texts on similar topics: philosophy of science, science and society #### snail feedback (4) : reader Shawn Halayka said... Wicked. George Musser on Twitter is awesome. reader Harlow said... I can't turn back the clock, but if I could I couldn't kill my grandfather because I wouldn't be there. reader Vladimir Kalitvianski said... I personally encountered strange objections from 't Hooft to my letter where he says that my "theory" in non relativistic and non local, so he is not interested in such. And non locality he meant was about smearing the electron charge due to its interaction with the vacuum electromagnetic field. "Non locality" is a regular thing even in Classical Mechanics when you describe the system of two particles in terms of the center of mass coordinate R and the relative motion coordinate r=r1-r2 and an external force acts only on one particle (on particle-1, for example). Then the CM equation for R contains the external force F_ext that depends on a "shifted" argument: \$M \ddot{R}=F_{ext} = F(R + \epsilon\cdot r)\$. This "shifted" argument is nothing but r_1 so no non locality is implied, but in terms of R it looks as non local potential or force. Hence objections. reader Luboš Motl said... Vladimir, Gerard 't Hooft is obviously excessively polite... 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Subscribe to: new TRF blog entries (RSS)	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 46, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9453970193862915, "perplexity_flag": "middle"}
http://mathhelpforum.com/discrete-math/35663-help-prove.html	# Thread: 1. ## help to prove prove if a A is uncountable and A is a subset of B, then B is uncountable 2. Originally Posted by dbhakta prove if a A is uncountable and A is a subset of B, then B is uncountable Since $A$ is uncountable it means $\|A\| > \aleph_0$. Since $A\subseteq B$ it means $\|A\|\leq \|B\|$ because the identity map on $A$ will be an injective function. Thus, $\aleph_0 < \|A\| \leq \|B\|$.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 6, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9327294230461121, "perplexity_flag": "head"}
http://dsp.stackexchange.com/questions/741/why-should-i-zero-pad-a-signal-before-taking-the-fourier-transform	# Why should I zero-pad a signal before taking the Fourier transform? In an answer to a previous question, it was stated that one should zero-pad the input signals (add zeros to the end so that at least half of the wave is "blank") What's the reason for this? - It depends on what you're doing. This could have been a comment on my answer. I added some explanation to it. – endolith Nov 30 '11 at 21:31 @endolith: I initially thought to put it as a comment, but I think the question may be of general interest, and that it would be a pity if a good answer to it was buried in comments somewhere. If you disagree, I'll delete this question. – Jonas Nov 30 '11 at 21:49 4 Well it's a very general question. You can zero pad to make something a power-of-2, you can zero pad to make circular transform behave like non-circular transform, you can do it to resample a signal, to change frequency resolution, etc. etc. – endolith Nov 30 '11 at 22:08 – finnw Jun 16 '12 at 20:21 ## 6 Answers Zero padding allows one to use a longer FFT, which will produce a longer FFT result vector. A longer FFT result has more frequency bins that are more closely spaced in frequency. But they will be essentially providing the same result as a high quality Sinc interpolation of a shorter non-zero-padded FFT of the original data. This might result in a smoother looking spectrum when plotted without further interpolation. Although this interpolation won't help with resolving or the resolution of and/or between adjacent or nearby frequencies, it might make it easier to visually resolve the peak of a single isolated frequency that does not have any significant adjacent signals or noise in the spectrum. Statistically, the higher density of FFT result bins will probably make it more likely that the peak magnitude bin is closer to the frequency of a random isolated input frequency sinusoid, and without further interpolation (parabolic, et.al.). But, essentially, zero padding before a DFT/FFT is a computationally efficient method of interpolating a large number of points. Zero-padding for cross correlation is used to not mix convolution results. The full result of a convolution is longer than either of the two input vectors. If you don't provide a place to put the end of this longer convolution result, FFT fast convolution will just mix it in with and cruft up your desired result. Zero-padding provides a bunch zeros into which to mix the longer result. And it's far far easier to un-mix something that has only been mixed/summed with a zero vector. - 3 The last paragraph is the key answer to the original question, although I think it could be stated more clearly. Zero-padding in the context of correlation or convolution can be done to ensure that implementing the process in the frequency domain yields linear instead of circular convolution/correlation. However, doing so is not a requirement if you're willing to do some bookkeeping work on the side, like in the overlap-save and overlap-add algorithms. – Jason R Dec 1 '11 at 2:47 2 @Jason R: Actually, they are both circular convolution. A normal (non-pruned) FFT does all the multiplies and adds for the wrap around part of the result. It's just that in the sufficiently zero-padded case, all those multiplies and adds are of the value zero, so nobody cares about the nothing that is computed and wrapped around the circle. – hotpaw2 Dec 1 '11 at 7:06 3 Indeed; multiplication of the DFTs of two signals does always implement circular convolution. I should have worded it differently: you stuff zeros at the end of one signal to ensure that the result obtained by circularly convolving them is the same as what you get if you linearly convolve them (assuming that linear convolution is what you want, which is usually the case). – Jason R Dec 1 '11 at 16:23 There are a few things to consider before you decide to zero pad your time-domain signal. You may not need to zero pad the signal at all! 1) Lengthen the time-domain data (not zero padding) to get better resolution in the frequency domain. 2) Increase the number of FFT points beyond your time-domain signal length (zero padding) if you would like to see better definition of the FFT bins, though it doesn't buy you any more true resolution. You can also pad to get to a power of 2 number of FFT points. 3) When fiddling with the FFT points (in the previous point), make sure your frequency points end up where you want them. The spacing of the points is $f_s/N$, where $f_s$ is the sampling frequency and $N$ is the number of FFT points. There are some nice figures illustrating these points at http://www.bitweenie.com/listings/fft-zero-padding/ One last thing to mention: If you zero pad the signal in the time domain and you want to use a windowing function, make sure you window the signal before you zero pad. If you apply the window function after zero padding, you won't accomplish what the window is supposed to accomplish. More specifically, you'll still have a sharp transition from the signal to zero instead of a smooth transition to zero. - In general zero-padding prior to DFT is equivalent to interpolation, or sampling more often, in the transformed domain. Here is a quick visualization. If you sample a bandlimited signal in time at higher rate, you get a more 'squashed' spectrum, i.e., spectrum with more zeros at both ends. In other words, you can obtain more samples in time by simply zero-padding in frequency after DFT and IDFTing the result. The same effect holds in reverse when zero-padding occurs in time. This is all because the perfect signal reconstruction is possible as long as a signal is bandlimited and sampled at least at the Nyquist rate. The term 'resolution' depends on how you define it. For me, it means how well the two adjacent points of observation in time or frequency can be reliably (statistically) discriminated. In this case the resolution actually depends on the DFT size due to spectral leakage. That is, smaller the window size, more blurry or smeared the transformed signal, and vice versa. It is different from how often you sample, or what I term 'definition'. For example, you can have a very blurry image sampled at high rate (high definition), yet you still cannot obtain more information than sampling at lower rate. So in summary, zero-padding does not improve resolution at all since you do not gain any more information than before. - If one has any interest in the spectrum of the windowing function used to isolate the time-domain sample, then zero-padding WILL increase the frequency resolution of the windowing function. If the time signal is $x(t)w(t)$, where $w(t)$ is the windowing function, then the overall spectrum is $X(f) * W(f)$, where $*$ inidicates convolution. If your windowing function is a simple rectangle (an extraction of some set of values from $x(t)$. Then $X(f)$ is the sync function. So, for example, if Nfft is the same as the width of your rectangle, and you had a sinusoid at precisely one of the bin frequencies, then the samples of the sync function that would appear centered on that bin happen to fall exactly at the off-peak zero crossings, and you don't see the shape of the sync in the spectrum at all. If you now zero pad your data going into the FFT, you'll see some samples at places other than the peak and the zero crossings, revealing the shape of the sync function in the resulting spectrum. So of what use is zero-padding? It is certainly of educational use in revealing the nature of the discrete transform of windowed signals, which is the usual case. In a practical sense, it could be useful in any case where you are interested in the spectral shape of an isolated envelope riding on a carrier wave. - There can be different reasons for this depending on any processes carried out before and after the Fourier transform. The most common reason is to achieve greater frequency resolution in any resulting transform. That is to say that, the larger the number of samples used in your transform, the narrower the binwidth in the resulting power spectrum. Remember: binwidth = sample_frequency/transform_size (often called window size). You can imagine from this, that as you increase your transform size, binwidth reduces (=better frequency resolution). Zero padding is a way of increasing the transform size without introducing new information to the signal. So why not just take a bigger transform without zero padding? Would that not achieve the same effect? Good question. In many cases you may want to analyze a stream of time domain data, for which you may be using a short time Fourier transform (stft). This involves taking a transform every N samples according to the time resolution you need in order to characterize changes in the frequency spectrum. Here in lies the problem. Too big a window and you will lose time resolution, too small a window and you will lose frequency resolution. The solution then is to take small time domain windows giving you good time resolution and then zero pad them to give you good frequency resolution. Hope this is useful for you Update I did not explain this well. I should have clarified it better. Referring to a windowed transform, indeed you get no 'actual' greater frequency resolution but for visualisation purposes (reading the power spectrum with the eye) it can provide clearer results. Using the critical sampling rate, each side lobe occupies a single bin, which depending on the graphing technique could be misleading. Zero padding provides an interpolated frequency spectrum which may be more revealing. Additionally, if you are using a simple peak picking method for frequency estimation, the spectral interpolation effect of zero padding will give you a spectral sample closer to the true peak of the main lobe. This link provides some useful diagrams: http://www.dsprelated.com/dspbooks/sasp/Practical_Zero_Padding.html - 6 This answer is not correct. Zero-padding does not improve frequency resolution at all; it merely interpolates between the outputs of the smaller transform. You can think of zero-padding as adding more frequency bins that have the same bandwidth as they do with the smaller transform; therefore, from a filter bank perspective, their passbands overlap. – Jason R Dec 1 '11 at 2:41 If it helps understanding: You can also do the opposite: take the FFT of a signal, then zero-pad the result, and inverse FFT. This will have the effect of interpolating the original signal. But of course the signal will still be the same signal, with the same Nyquist bandwidth. Interpolation won't give you more higher frequency information than was originally there. – endolith Dec 1 '11 at 19:18 1 @Jason R - You are right, my answer was misleading, I have attempted to clarify above on the original post. I should not have stated that zero padding increases frequency resolution. – Dan Barry Dec 7 '11 at 11:56 DFT of the zero padding signal is just a better approximation of the DTFT of the original signal. -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 9, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9275749325752258, "perplexity_flag": "middle"}
http://mathoverflow.net/questions/49243/comparing-diffusions/49246	## comparing diffusions ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) Consider a probability distribution $\pi$ on the real axis that has a density (w.r.t Lebesgue) proportional to $e^{-V(x)}$, where $V(\cdot)$ is a potential function. For any reasonable volatility function $\sigma:\mathbb{R} \to (0:+\infty)$ the diffusion $$dX^{\sigma}_t = [ -\frac{1}{2} \sigma(X_t^{\sigma})^2 V'(X_t^{\sigma}) + \sigma(X_t^{\sigma}) \sigma'(X_t^{\sigma}) ] dt + \sigma(X_t^{\sigma}) \, dW_t$$ has $\pi$ as unique invariant distribution. Question: Given two volatility functions $\sigma_1, \sigma_2$, are there tractable ways of comparing the speed of convergence to equilibrium of the two associated diffusions? For example, if $\sigma_2(x) = \alpha \cdot \sigma_1(x)$, the diffusion $X^{\sigma_2}$ is just $X^{\sigma_1}$ slowed down by a factor $\alpha$: any ways of comparing the two diffusions should say that if $\alpha > 1$ then $X^{\sigma_2}$ converges 'faster' than $X^{\sigma_1}$. Spectral Gaps work but are not very tractable when comparing two non-proportional diffusions. Is it hopeless ? Motivations: I consider several MCMC algorithms with target density $\pi$: each one of them, after some time-rescaling, looks like a diffusion $X^{\sigma}$. Which algorithm is the best $i.e.$ what diffusion $X^{\sigma}$ mixes the fastest ? - Hi Alekk, I was wondering if there exists theorems regarding speed of convergence to $\pi$ when the problem is not about diffusions but about Markov Chains. If there exist such results, maybe you can tackle the problem through Markov Chain discretisations of your diffusions and then try to obtain asymptotic results for the speed of convergence of the diffusions you want to compare. Best Regrads "et bon courage" – The Bridge Dec 16 2010 at 8:56 ## 2 Answers Depending on the boundary and regularity assumptions, the time evolution of the probability distribution is described by a Fokker-Planck equation (see Wikipedia). For a time-homogenous process with a unique stationary solution, the time evolution is described by an exponential decay of the initial distribution acoording to an eigenvalue expansion of the form $$p(x, t) = \sum_{k = 0}^{\infty} q_k(x) \exp{(- \lambda_k t)}$$ for eigenvalues $\lambda_k$ with $\lambda_0 = 0$ corresponding to the stationary distribution. The bigger the eigenvalues are, the faster the decay to the stationary distribution will be, so some kind of measure could be the smallest non-zero eigenvalue of the Fokker-Planck operator. For some concrete examples, have a look at the book Gardiner: "Handbook of Stochastic Methods", chapter 5.2.5 Eigenfunction Methods (Homogeneous Processes). I don't know though if it is possible to calculate or approximate the eigenvalues for the general equation you stated. - thanks Tim: yes, even the case where we are only interested in the most significant eigenvalue i.e. spectral gap, the computations become intractable. I was wondering if there were any other ways of comparing these diffusions, without spectral theory and the non tractable ODEs that are involved. – Alekk Dec 13 2010 at 13:07 Okay, at least I can help to make the question a little bit more precise :-) Do you need quantitative data or would the order itself be enough? – Tim van Beek Dec 13 2010 at 13:20 BTW, is the initial condition supposed to be fixed? – Tim van Beek Dec 13 2010 at 13:24 Tim, thanks for the interest: in fact, the initial distribution is not really what interests me - I would rather like to quantify how fast the diffusion visits the state space. I added more motivations in the question. – Alekk Dec 13 2010 at 15:51 ### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you. Hi Alekk You might take a look at this paper : Debussche,Faou - Weak Backward Error Analysis for SDEs Regards - Thanks TheBridge. You might want to check the link though: it is pointing to another paper. – Alekk May 4 2011 at 12:47 @ alekk : My mistake correction done – The Bridge May 4 2011 at 14:35	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 20, "mathjax_display_tex": 2, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.8900526762008667, "perplexity_flag": "head"}
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I want to draw a circle that cuts through the center of a sphere and has an inclination of 15 degrees with the ... 4answers 1k views ### How do I draw a triangle given the lengths of the sides? I know, of course, how to draw a triangle in the plane given the vertices: Graphics[Polygon[{{1, 0}, {0, Sqrt[3]}, {-1, 0}}]] But I'm not sure how to simply draw ... 2answers 239 views ### How to punch a hole in some 3D distribution of points Suppose we have a long list of 3D Cartesian coordinates, defining a distribution of random points in 3D space. How could we remove all the points inside a sphere of radius ... 2answers 194 views ### Triangle mapped on a sphere in $\mathbb R^3$? How can I map a triangle on an sphere? I want to visualize (plot or animate) it for my student in my Non Euclidean geometry. I have no restrictions on the triangle's kind or on the sphere in \$\mathbb ... 5answers 475 views ### How to plot rectangles aligned by their center? Supose I have a rectangle which area is $x^2$. In some cases I may not know what is the size of each side, for $x=12,$ we have several possibilites: ... 1answer 87 views ### Why is FindInstance failing when I relax a set of constraints? I'm attempting to use FindInstance to generate coordinate sets for plausible triangles with edge length distance constraints. E.g.: ... 0answers 88 views ### How to type function terms into metric? I submitted my question before on site:Anything wrong with Atlas 2 or Mathematica?. Due to my vagueness, my problems might not be clearly understood. I am trying to solve several functions based on ...	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 11, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9211270809173584, "perplexity_flag": "middle"}
http://mathoverflow.net/questions/88387/limits-of-binomial-distribution/88395	## Limits of binomial distribution ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) We know that as $n \to \infty$, the binomial distribution $B(n, p)$, with fixed $p$, after appropriate normalization, converges to a normal distribution. If $p = c/n$ for some constant $c$, then it converges to the Poisson distribution. What happens for intermediate cases, say when $p = c n^{-\alpha}$ for fixed $c$ and $0 < \alpha < 1$. After appropriate normalization, what is the limiting distribution of $B(n,p)$? - ## 5 Answers Assume the distribution of $X_n$ is binomial $(n,p)$. Then, for every real number $t$, $$\mathrm E(\mathrm e^{\mathrm it(X_n-np)})=\mathrm e^{\mathrm itnp}(1-p+p\mathrm e^{\mathrm it})^n.$$ Assuming that $p\to0$ when $n\to\infty$, standard limited expansions yield $$\mathrm E(\mathrm e^{\mathrm it(X_n-np)})=\exp\left(-\tfrac12npt^2(1-p)+O(npt^3)\right).$$ Let us assume that $np\to\infty$ when $n\to\infty$ and choose $t=s/\sqrt{np}$ for a given $s$. Then $npt^2(1-p)\to s^2$ and $npt^3=s^3/\sqrt{np}\to0$, hence $$\mathrm E(\mathrm e^{\mathrm is(X_n-np)/\sqrt{np}})\to\mathrm e^{-s^2/2}.$$ Thus, $(X_n-np)/\sqrt{np}$ converges in distribution to the standard gaussian distribution when $n\to\infty$, as soon as $p\to0$ and $np\to\infty$. - ### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you. I believe with a fairly standard scaling and shifting, we can recover a normal limit for any $0 < \alpha < 1$ by a standard application of the Lindeberg–Feller theorem. Define the triangular array of random variables $X_{n,m}$, $1 \leq m \leq n$ such that each row contains iid elements distributed as $\mathrm{Bernoulli}(cn^{-\alpha})$ random variables. Then $S_n = \sum_{m=1}^n X_{n,m}$ is $B(n,p)$ with $p = c n^{-\alpha}$. Now, shift and rescale by taking $Y_{n,m} = n^{-(1-\alpha)/2} (X_{n,m} - c n^{-\alpha})$. Note that $\mathbb E Y_{n,m} = 0$ and $\mathbb E Y_{n,m}^2 = n^{-1} c (1 - c n^{-\alpha})$. We need only check the Lindeberg–Feller conditions: 1. $\sum_{m=1}^n \mathbb E Y_{n,m}^2 = c (1-cn^{-\alpha}) \to c > 0$, and 2. For all $\epsilon > 0$, $\lim_{n\to\infty} \sum_{m=1}^n \mathbb E \|Y_{n,m}\|^2 1_{(\|Y_{n,m}\|^2 > \epsilon)} = 0$. The second one follows since $1_{(\|Y_{n,m}\|^2 > \epsilon)} = 0$ for all sufficiently large $n$ since $\|Y_{n,m}\|^2 \leq n^{-(1-\alpha)} (1+c)^2$ almost surely. Hence $n^{-(1-\alpha)/2} (S_n - c n^{1-\alpha}) \;\xrightarrow{\;d\;}\; \mathcal N(0,c)$. To see what fails in the case where $\alpha = 1$, note that the indicator function $1_{(\|Y_{n,m}\|^2 > \epsilon)}$ will no longer go to zero almost surely. Therefore the second condition will fail because there will be a contribution of $c n^{-1} (1-c/n)^2$ in expectation for each of the $n$ terms, which, when added up yields (obviously) a nonzero limit. - The total variation distance between $B(n,p)$ and Poisson($np$) goes to 0 as $n\to\infty$ whenever $p\to 0$. See http://journals.cambridge.org/action/displayAbstract?fromPage=online&aid=2086772 for example. This means that the Poisson and normal ranges overlap considerably and you don't need anything else. - 1 I disagree with the last part of your last sentence, which puts undue significance in the result recalled in the first sentence. For example, if $np\to0$, the distance between Poisson $(np)$ and a Dirac mass at $0$ ALSO goes to $0$ hence in this regime, the result simply says that $B(n,p)$ converges to a Dirac mass at $0$. Similarly, if $np\to\infty$, Poisson $(np)$ is itself (when shifted by $np$ and rescaled by $\sqrt{np}$) approximately standard gaussian hence in this regime, the result is simply an unusual way of stating that a CLT holds. – Didier Piau Feb 14 2012 at 10:30 I don't see your point. If $np\to 0$, the explicit bounds on the tdv given in the linked paper show more than just accumulation at 0. However, there are stronger results in that case to be sure (very easily obtained). For $np\to\infty$, you are right that scaled Poisson is close to normal, but that's what "the Poisson and normal ranges overlap considerably" means. The OP asked what happens between the ranges where binomial is like Poisson and where binomial is like normal, and the correct answer is that there is nothing between them. – Brendan McKay Feb 14 2012 at 19:10 My point is that in these two regimes, the Poisson distribution is a red herring, for a Dirac distribution or for a gaussian one. (I read your comment by chance, please use @ when commenting to someone else than the author of the post.) – Didier Piau Feb 15 2012 at 0:48 @Didier Piau: Depending on the way $p$ varies as a function of $n$ (except for $p=1-o(1/n)$), $B(n,p)$ is well approximated by either the Poisson or Normal distribution. The details depend on the distance measure, but for most measures (total variation distance being one) the Poisson approximation remains better than the normal well into the range where the binomial is asymptotically normal. For tiny $p$, the Poisson approximation is better than the Dirac approximation. Do you disagree with any of these statements? – Brendan McKay Feb 15 2012 at 9:08 $B(n,p)$ for $n$ very large and $p$ very small is approximately a Poisson distribution with $\lambda=pn$. Normalizing by a constant gives you approximately a constant times the Poisson distribution. You might as well take your constant to be $1/\lambda$. In this case, as $\lambda \to \infty$, the variance, $1/\lambda$, will go to $0$, and the random variable becomes constant. - 1. Let $X_1,X_2,...,X_n$ be independent and identically distributed random variables such that $E(X_i) = 0, Var(X_i) = 1$ and denote $W=n^{-1/2}\sum X_i$. By using Stein's method, we have that $$d_W(\mathcal{L}(W),N(0,1)) \leq \frac{5 E\|X_1\|^3}{n^{1/2}},$$ where $d_W$ is the Wasserstein metric. 2. Any binomial distribution $B(n, p)$ is the sum of $n$ independent Bernoulli trials $B(1, p)$, so you can consider $X_i$ as a normalized Bernoulli random variable and apply the above inequality. -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 82, "mathjax_display_tex": 4, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9231011271476746, "perplexity_flag": "head"}
http://nrich.maths.org/255/solution	### Pericut Two semicircle sit on the diameter of a semicircle centre O of twice their radius. Lines through O divide the perimeter into two parts. What can you say about the lengths of these two parts? ### Kissing Two perpendicular lines are tangential to two identical circles that touch. What is the largest circle that can be placed in between the two lines and the two circles and how would you construct it? ### Logosquares Ten squares form regular rings either with adjacent or opposite vertices touching. Calculate the inner and outer radii of the rings that surround the squares. # Baby Circle ##### Stage: 5 Challenge Level: Rosalind of Madras College sent in this solution, well done Rosalind. All three circles touch each other and have a common tangent $DFE$. The radius of the baby circle is $r$ and the radii of the other circles are $BE = 1$ unit and $AD = 2$ units. So $AC = 1$ unit, $AB = 3$ units, $AH = 2 - r$, $AR = 2 +r$, $BR = 1 + r$ and $BG = 1 - r$. Using Pythagoras Theorem: $$\eqalign{ \; RG &=& \sqrt{(1+r)^2 - (1-r)^2} = \sqrt{4r} = 2\sqrt{r} \\ \; HR &=& \sqrt{(2+r)^2 - (2-r)^2} = \sqrt{8r} = 2\sqrt{2r} \\ \; CB &=& 2\sqrt{2} = DE \\ RG + HR = CB &\Rightarrow& 2\sqrt{r} + 2\sqrt{2r} = 2\sqrt{2}.}$$ This gives $\sqrt{r}(1+\sqrt{2}) = \sqrt{2}$ and hence by squaring $$\eqalign{ r &=& \frac{2}{3+2\sqrt{2}} \\ \; &=& \frac{2(3 - 2\sqrt{2})}{(3+2\sqrt{2})(3-2\sqrt{2})} \\ \; &=& \frac{6 - 4\sqrt{2}}{9 - 4\times2} \\ \; &=& 6 - 4\sqrt{2}}$$ So the radius of the baby circle is $6 - 4\sqrt{2}$. The NRICH Project aims to enrich the mathematical experiences of all learners. To support this aim, members of the NRICH team work in a wide range of capacities, including providing professional development for teachers wishing to embed rich mathematical tasks into everyday classroom practice. More information on many of our other activities can be found here.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 12, "mathjax_display_tex": 2, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9144737720489502, "perplexity_flag": "middle"}
http://twistedone151.wordpress.com/2010/12/31/physics-friday-149/	# Twisted One 151's Weblog Just another WordPress.com weblog ## Physics Friday 149 For a block sliding down an inclined surface with angle of incline θ and with coefficient of kinetic friction μ. If we consider the forces, we see that the block can slide down the incline with constant velocity when the net forces are zero. Balancing forces perpendicular to the plane, the normal force is thus $N=mg\cos\theta$. The kinetic friction is thus $f=\mu{N}=\mu{mg}\cos\theta$, while the component of gravity parallel to the plane is thus $mg\sin\theta$. We see that these cancel when $\mu\cos\theta=\sin\theta$, or when $\mu=\tan\theta$. Now, suppose we have a large incline with $\mu=\tan\theta$, and we start our block sliding with a velocity v0 in a horizontal direction; that is to say, along the plane in a direction perpendicular to the direction of the slope. What, then, will be the speed a long time later? Let v be the total speed of the block, and let vy be the velocity component in the downslope direction. Now, we have two forces parallel to the plane; the friction force, with magnitude $f=\mu{N}=\mu{mg}\cos\theta=mg\sin\theta$ directed opposite to the motion of the block, and the component of gravity, also of magnitude $mg\sin\theta$, directed downslope. The former produces an acceleration of magnitude $g\sin\theta$ opposite the direction of motion, while the gravity gives the same acceleration $g\sin\theta$ in the downslope direction; thus $\frac{dv}{dt}=-g\sin\theta$, and $\frac{dv_y}{dt}=g\sin\theta$, so that $\frac{dv}{dt}+\frac{dv_y}{dt}=0$, and thus $v+v_y=C$, for C some constant. Now, at time t=0, our motion is purely horizontal, and so vy=0, v=v0, and so C=v0. Now, after a long time, the horizontal component of the velocity will be effectively zero, so that the velocity is entirely downslope. Then v=vy, and so $v+v_y=v_0$ becomes $2v=v_0$, and so the speed after a long time is $v=v_y=\frac12v_0$, half the initial speed; now in the downslope direction. ### Like this: Tags: Friday Physics, Gravity, Kinetic Friction, physics This entry was posted on December 31, 2010 at 1:01 am and is filed under Math/Science. You can follow any responses to this entry through the RSS 2.0 feed. You can leave a response, or trackback from your own site. ### 2 Responses to “Physics Friday 149” 1. Dr.D Says: December 31, 2010 at 8:04 am \| Reply In the definition of vo, what on earth do you mean by the phrase, “along the plane in a direction perpendicular to the direction of the slope?” That is about as muddy as any definition I have ever seen. I have no idea what you are saying vo is, so the rest of the problem is meaningless to me. 2. twistedone151 Says: January 3, 2011 at 3:47 am \| Reply When we usually do this problem, we usually only consider motion directly up or down the plane, such as with the ball rolling down the plane here. I mean motion in the direction on the surface of the plane which is horizontal, neither uphill nor downhill. In the case of a diagram like in the above problem, it would be motion directly into, or directly out of, the plane of the diagram. If you draw the plane from above, and indicate the direction of the slope of the plane (the direction straight uphill or downhill, as it were), then this direction is along the line in the plane perpendicular to this direction (line of constant elevation). %d bloggers like this:	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 17, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.904255211353302, "perplexity_flag": "middle"}
http://www.mashpedia.com/Program_refinement	# Refinement (computing) Language Data transformation/Source transformation Concepts Languages Techniques and transforms Applications Application fields Software languages macro preprocessing template Refinement is a generic term of computer science that encompasses various approaches for producing correct computer programs and simplifying existing programs to enable their formal verification. ## Program refinement In formal methods, program refinement is the verifiable transformation of an abstract (high-level) formal specification into a concrete (low-level) executable program.[citation needed] Stepwise refinement allows this process to be done in stages. Logically, refinement normally involves implication, but there can be additional complications. ## Data refinement Data refinement is used to convert an abstract data model (in terms of sets for example) into implementable data structures (such as arrays).[citation needed] Operation refinement converts a specification of an operation on a system into an implementable program (e.g., a procedure). The postcondition can be strengthened and/or the precondition weakened in this process. This reduces any nondeterminism in the specification, typically to a completely deterministic implementation. For example, x ∈ {1,2,3} (where x is the value of the variable x after an operation) could be refined to x ∈ {1,2}, then x ∈ {1}, and implemented as x := 1. Implementations of x := 2 and x := 3 would be equally acceptable in this case, using a different route for the refinement. However, we must be careful not to refine to x ∈ {} (equivalent to false) since this is unimplementable; it is impossible to select a member from the empty set. The term reification is also sometimes used (coined by Cliff Jones). Retrenchment is an alternative technique when formal refinement is not possible. The opposite of refinement is abstraction. ## Refinement calculus Refinement calculus is a formal system (inspired from Hoare logic) that promotes program refinement. The FermaT Transformation System is an industrial-strength implementation of refinement. The B-Method is also a formal method that extends refinement calculus with a component language: it has been used in industrial developments. ## Refinement types In type theory, a refinement type[1][2][3] is a type endowed with a predicate which is assumed to hold for any element of the refined type. Refinement types can express preconditions when used as function arguments or postconditions when used as return types: for instance, the type of a function which accepts natural numbers and returns natural numbers greater than 5 may be written as $f: \mathbb{N} \rarr \{n: \mathbb{N} \| n > 5\}$. Refinement types are thus related to behavioral subtyping. ## References 1. Freeman, T.; Pfenning, F. (1991). "Reﬁnement types for ML". Proceedings of the ACM Conference on Programming Language Design and Implementation. pp. 268–277. doi:10.1145/113445.113468. 2. Hayashi, S. (1993). "Logic of reﬁnement types". Proceedings of the Workshop on Types for Proofs and Programs. pp. 157–172. doi:10.1007/3-540-58085-9_74. 3. Denney, E. (1998). "Reﬁnement types for speciﬁcation". Proceedings of the IFIP International Conference on Programming Concepts and Methods. 125. Chapman & Hall. pp. 148–166. This software engineering-related article is a stub. You can help Wikipedia by expanding it.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 1, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.814817488193512, "perplexity_flag": "head"}
http://en.wikipedia.org/wiki/Definite_integral	# Integral (Redirected from Definite integral) A definite integral of a function can be represented as the signed area of the region bounded by its graph. Calculus Definitions Concepts Rules and identities Integral calculus Definitions Integration by Formalisms Definitions Specialized calculi Integration is an important concept in mathematics and, together with its inverse, differentiation, is one of the two main operations in calculus. Given a function f of a real variable x and an interval [a, b] of the real line, the definite integral $\int_a^b \! f(x)\,dx \,$ is defined informally to be the signed area of the region in the xy-plane bounded by the graph of f, the x-axis, and the vertical lines x = a and x = b, such that area above the x-axis adds to the total, and that below the x-axis subtracts from the total. The term integral may also refer to the notion of the antiderivative, a function F whose derivative is the given function f. In this case, it is called an indefinite integral and is written: $F = \int f(x)\,dx.$ The principles of integration were formulated independently by Isaac Newton and Gottfried Leibniz in the late 17th century. Through the fundamental theorem of calculus, which they independently developed, integration is connected with differentiation: if f is a continuous real-valued function defined on a closed interval [a, b], then, once an antiderivative F of f is known, the definite integral of f over that interval is given by $\int_a^b \! f(x)\,dx = F(b) - F(a)\,$ Integrals and derivatives became the basic tools of calculus, with numerous applications in science and engineering. The founders of the calculus thought of the integral as an infinite sum of rectangles of infinitesimal width. A rigorous mathematical definition of the integral was given by Bernhard Riemann. It is based on a limiting procedure which approximates the area of a curvilinear region by breaking the region into thin vertical slabs. Beginning in the nineteenth century, more sophisticated notions of integrals began to appear, where the type of the function as well as the domain over which the integration is performed has been generalised. A line integral is defined for functions of two or three variables, and the interval of integration [a, b] is replaced by a certain curve connecting two points on the plane or in the space. In a surface integral, the curve is replaced by a piece of a surface in the three-dimensional space. Integrals of differential forms play a fundamental role in modern differential geometry. These generalizations of integrals first arose from the needs of physics, and they play an important role in the formulation of many physical laws, notably those of electrodynamics. There are many modern concepts of integration, among these, the most common is based on the abstract mathematical theory known as Lebesgue integration, developed by Henri Lebesgue. ## History See also: History of calculus ### Pre-calculus integration The first documented systematic technique capable of determining integrals is the method of exhaustion of the ancient Greek astronomer Eudoxus (ca. 370 BC), which sought to find areas and volumes by breaking them up into an infinite number of shapes for which the area or volume was known. This method was further developed and employed by Archimedes in the 3rd century BC and used to calculate areas for parabolas and an approximation to the area of a circle. Similar methods were independently developed in China around the 3rd century AD by Liu Hui, who used it to find the area of the circle. This method was later used in the 5th century by Chinese father-and-son mathematicians Zu Chongzhi and Zu Geng to find the volume of a sphere (Shea 2007; Katz 2004, pp. 125–126). The next significant advances in integral calculus did not begin to appear until the 16th century. At this time the work of Cavalieri with his method of indivisibles, and work by Fermat, began to lay the foundations of modern calculus, with Cavalieri computing the integrals of xn up to degree n = 9 in Cavalieri's quadrature formula. Further steps were made in the early 17th century by Barrow and Torricelli, who provided the first hints of a connection between integration and differentiation. Barrow provided the first proof of the fundamental theorem of calculus. Wallis generalized Cavalieri's method, computing integrals of x to a general power, including negative powers and fractional powers. ### Newton and Leibniz The major advance in integration came in the 17th century with the independent discovery of the fundamental theorem of calculus by Newton and Leibniz. The theorem demonstrates a connection between integration and differentiation. This connection, combined with the comparative ease of differentiation, can be exploited to calculate integrals. In particular, the fundamental theorem of calculus allows one to solve a much broader class of problems. Equal in importance is the comprehensive mathematical framework that both Newton and Leibniz developed. Given the name infinitesimal calculus, it allowed for precise analysis of functions within continuous domains. This framework eventually became modern calculus, whose notation for integrals is drawn directly from the work of Leibniz. ### Formalizing integrals While Newton and Leibniz provided a systematic approach to integration, their work lacked a degree of rigour. Bishop Berkeley memorably attacked the vanishing increments used by Newton, calling them "ghosts of departed quantities". Calculus acquired a firmer footing with the development of limits. Integration was first rigorously formalized, using limits, by Riemann. Although all bounded piecewise continuous functions are Riemann integrable on a bounded interval, subsequently more general functions were considered – particularly in the context of Fourier analysis – to which Riemann's definition does not apply, and Lebesgue formulated a different definition of integral, founded in measure theory (a subfield of real analysis). Other definitions of integral, extending Riemann's and Lebesgue's approaches, were proposed. These approaches based on the real number system are the ones most common today, but alternative approaches exist, such as a definition of integral as the standard part of an infinite Riemann sum, based on the hyperreal number system. ### Historical notation Isaac Newton used a small vertical bar above a variable to indicate integration, or placed the variable inside a box. The vertical bar was easily confused with $\dot{x}$ or $x'\,\!$, which Newton used to indicate differentiation, and the box notation was difficult for printers to reproduce, so these notations were not widely adopted. The modern notation for the indefinite integral was introduced by Gottfried Leibniz in 1675 (Burton 1988, p. 359; Leibniz 1899, p. 154). He adapted the integral symbol, ∫, from the letter ſ (long s), standing for summa (written as ſumma; Latin for "sum" or "total"). The modern notation for the definite integral, with limits above and below the integral sign, was first used by Joseph Fourier in Mémoires of the French Academy around 1819–20, reprinted in his book of 1822 (Cajori 1929, pp. 249–250; Fourier 1822, §231). ## Terminology and notation The simplest case, the integral over x of a real-valued function f(x), is written as $\int f(x)\,dx .$ The integral sign ∫ represents integration. The dx indicates that we are integrating over x; x is called the variable of integration. In correct mathematical typography, the dx is separated from the integrand by a space (as shown). Some authors use an upright d (that is, dx instead of dx). Inside the ∫...dx is the expression to be integrated, called the integrand. In this case the integrand is the function f(x). Because there is no domain specified, the integral is called an indefinite integral. When integrating over a specified domain, we speak of a definite integral. Integrating over a domain D is written as $\int_D f(x)\,dx ,$ or $\int_a^b f(x)\,dx$ if the domain is an interval [a, b] of x; The domain D or the interval [a, b] is called the domain of integration. If a function has an integral, it is said to be integrable. In general, the integrand may be a function of more than one variable, and the domain of integration may be an area, volume, a higher dimensional region, or even an abstract space that does not have a geometric structure in any usual sense (such as a sample space in probability theory). In the modern Arabic mathematical notation, which aims at pre-university levels of education in the Arab world and is written from right to left, a reflected integral symbol is used (W3C 2006). The variable of integration dx has different interpretations depending on the theory being used. It can be seen as strictly a notation indicating that x is a dummy variable of integration; if the integral is seen as a Riemann sum, dx is a reflection of the weights or widths d of the intervals of x; in Lebesgue integration and its extensions, dx is a measure; in non-standard analysis, it is an infinitesimal; or it can be seen as an independent mathematical quantity, a differential form. More complicated cases may vary the notation slightly. In Leibniz's notation, dx is interpreted an infinitesimal change in x, but his interpretation lacks rigour in the end. Nonetheless Leibniz's notation is the most common one today; and as few people are in need of full rigour, even his interpretation is still used in many settings. ## Introduction Integrals appear in many practical situations. If a swimming pool is rectangular with a flat bottom, then from its length, width, and depth we can easily determine the volume of water it can contain (to fill it), the area of its surface (to cover it), and the length of its edge (to rope it). But if it is oval with a rounded bottom, all of these quantities call for integrals. Practical approximations may suffice for such trivial examples, but precision engineering (of any discipline) requires exact and rigorous values for these elements. Approximations to integral of √x from 0 to 1, with ■ 5 right samples (above) and ■ 12 left samples (below) To start off, consider the curve y = f(x) between x = 0 and x = 1 with f(x) = √x. We ask: What is the area under the function f, in the interval from 0 to 1? and call this (yet unknown) area the integral of f. The notation for this integral will be $\int_0^1 \sqrt x \, dx \,\!.$ As a first approximation, look at the unit square given by the sides x = 0 to x = 1 and y = f(0) = 0 and y = f(1) = 1. Its area is exactly 1. As it is, the true value of the integral must be somewhat less. Decreasing the width of the approximation rectangles shall give a better result; so cross the interval in five steps, using the approximation points 0, 1/5, 2/5, and so on to 1. Fit a box for each step using the right end height of each curve piece, thus √(1⁄5), √(2⁄5), and so on to √1 = 1. Summing the areas of these rectangles, we get a better approximation for the sought integral, namely $\textstyle \sqrt {\frac {1} {5}} \left ( \frac {1} {5} - 0 \right ) + \sqrt {\frac {2} {5}} \left ( \frac {2} {5} - \frac {1} {5} \right ) + \cdots + \sqrt {\frac {5} {5}} \left ( \frac {5} {5} - \frac {4} {5} \right ) \approx 0.7497.\,\!$ Notice that we are taking a sum of finitely many function values of f, multiplied with the differences of two subsequent approximation points. We can easily see that the approximation is still too large. Using more steps produces a closer approximation, but will never be exact: replacing the 5 subintervals by twelve as depicted, we will get an approximate value for the area of 0.6203, which is too small. The key idea is the transition from adding finitely many differences of approximation points multiplied by their respective function values to using infinitely many fine, or infinitesimal steps. As for the actual calculation of integrals, the fundamental theorem of calculus, due to Newton and Leibniz, is the fundamental link between the operations of differentiating and integrating. Applied to the square root curve, f(x) = x1/2, it says to look at the antiderivative F(x) = (2/3)x3/2, and simply take F(1) − F(0), where 0 and 1 are the boundaries of the interval [0,1]. So the exact value of the area under the curve is computed formally as $\int_0^1 \sqrt x \,dx = \int_0^1 x^{1/2} \,dx = F(1)- F(0) = 2/3.$ (This is a case of a general rule, that for f(x) = xq, with q ≠ −1, the related function, the so-called antiderivative is F(x) = xq + 1/(q + 1).) The notation $\int f(x) \, dx \,\!$ conceives the integral as a weighted sum, denoted by the elongated s, of function values, f(x), multiplied by infinitesimal step widths, the so-called differentials, denoted by dx. The multiplication sign is usually omitted. Historically, after the failure of early efforts to rigorously interpret infinitesimals, Riemann formally defined integrals as a limit of weighted sums, so that the dx suggested the limit of a difference (namely, the interval width). Shortcomings of Riemann's dependence on intervals and continuity motivated newer definitions, especially the Lebesgue integral, which is founded on an ability to extend the idea of "measure" in much more flexible ways. Thus the notation $\int_A f(x) \, d\mu \,\!$ refers to a weighted sum in which the function values are partitioned, with μ measuring the weight to be assigned to each value. Here A denotes the region of integration. Differential geometry, with its "calculus on manifolds", gives the familiar notation yet another interpretation. Now f(x) and dx become a differential form, ω = f(x) dx, a new differential operator d, known as the exterior derivative is introduced, and the fundamental theorem becomes the more general Stokes' theorem, $\int_{A} d\omega = \int_{\part A} \omega , \,\!$ from which Green's theorem, the divergence theorem, and the fundamental theorem of calculus follow. More recently, infinitesimals have reappeared with rigor, through modern innovations such as non-standard analysis. Not only do these methods vindicate the intuitions of the pioneers; they also lead to new mathematics. Although there are differences between these conceptions of integral, there is considerable overlap. Thus, the area of the surface of the oval swimming pool can be handled as a geometric ellipse, a sum of infinitesimals, a Riemann integral, a Lebesgue integral, or as a manifold with a differential form. The calculated result will be the same for all. ## Formal definitions There are many ways of formally defining an integral, not all of which are equivalent. The differences exist mostly to deal with differing special cases which may not be integrable under other definitions, but also occasionally for pedagogical reasons. The most commonly used definitions of integral are Riemann integrals and Lebesgue integrals. ### Riemann integral Main article: Riemann integral Integral approached as Riemann sum based on tagged partition, with irregular sampling positions and widths (max in red). True value is 3.76; estimate is 3.648. The Riemann integral is defined in terms of Riemann sums of functions with respect to tagged partitions of an interval. Let [a,b] be a closed interval of the real line; then a tagged partition of [a,b] is a finite sequence $a = x_0 \le t_1 \le x_1 \le t_2 \le x_2 \le \cdots \le x_{n-1} \le t_n \le x_n = b . \,\!$ Riemann sums converging as intervals halve, whether sampled at ■ right, ■ minimum, ■ maximum, or ■ left. This partitions the interval [a,b] into n sub-intervals [xi−1, xi] indexed by i, each of which is "tagged" with a distinguished point ti ∈ [xi−1, xi]. A Riemann sum of a function f with respect to such a tagged partition is defined as $\sum_{i=1}^{n} f(t_i) \Delta_i ;$ thus each term of the sum is the area of a rectangle with height equal to the function value at the distinguished point of the given sub-interval, and width the same as the sub-interval width. Let Δi = xi−xi−1 be the width of sub-interval i; then the mesh of such a tagged partition is the width of the largest sub-interval formed by the partition, maxi=1…n Δi. The Riemann integral of a function f over the interval [a,b] is equal to S if: For all ε > 0 there exists δ > 0 such that, for any tagged partition [a,b] with mesh less than δ, we have $\left\| S - \sum_{i=1}^{n} f(t_i)\Delta_i \right\| < \varepsilon.$ When the chosen tags give the maximum (respectively, minimum) value of each interval, the Riemann sum becomes an upper (respectively, lower) Darboux sum, suggesting the close connection between the Riemann integral and the Darboux integral. ### Lebesgue integral Main article: Lebesgue integration Riemann–Darboux's integration (blue) and Lebesgue integration (red). It is often of interest, both in theory and applications, to be able to pass to the limit under the integral. For instance, a sequence of functions can frequently be constructed that approximate, in a suitable sense, the solution to a problem. Then the integral of the solution function should be the limit of the integrals of the approximations. However, many functions that can be obtained as limits are not Riemann integrable, and so such limit theorems do not hold with the Riemann integral. Therefore it is of great importance to have a definition of the integral that allows a wider class of functions to be integrated (Rudin 1987). Such an integral is the Lebesgue integral, that exploits the following fact to enlarge the class of integrable functions: if the values of a function are rearranged over the domain, the integral of a function should remain the same. Thus Henri Lebesgue introduced the integral bearing his name, explaining this integral thus in a letter to Paul Montel: I have to pay a certain sum, which I have collected in my pocket. I take the bills and coins out of my pocket and give them to the creditor in the order I find them until I have reached the total sum. This is the Riemann integral. But I can proceed differently. After I have taken all the money out of my pocket I order the bills and coins according to identical values and then I pay the several heaps one after the other to the creditor. This is my integral. Source: (Siegmund-Schultze 2008) As Folland (1984, p. 56) puts it, "To compute the Riemann integral of f, one partitions the domain [a,b] into subintervals", while in the Lebesgue integral, "one is in effect partitioning the range of f". The definition of the Lebesgue integral thus begins with a measure, μ. In the simplest case, the Lebesgue measure μ(A) of an interval A = [a,b] is its width, b − a, so that the Lebesgue integral agrees with the (proper) Riemann integral when both exist. In more complicated cases, the sets being measured can be highly fragmented, with no continuity and no resemblance to intervals. Using the "partitioning the range of f" philosophy, the integral of a non-negative function f : R → R should be the sum over t of the areas between a thin horizontal strip between y = t and y = t + dt. This area is just μ{ x : f(x) > t} dt. Let f∗(t) = μ{ x : f(x) > t}. The Lebesgue integral of f is then defined by (Lieb & Loss 2001) $\int f = \int_0^\infty f^*(t)\,dt$ where the integral on the right is an ordinary improper Riemann integral (note that f∗ is a strictly decreasing positive function, and therefore has a well-defined improper Riemann integral). For a suitable class of functions (the measurable functions) this defines the Lebesgue integral. A general measurable function f is Lebesgue integrable if the area between the graph of f and the x-axis is finite: $\int_E \|f\|\,d\mu < + \infty.$ In that case, the integral is, as in the Riemannian case, the difference between the area above the x-axis and the area below the x-axis: $\int_E f \,d\mu = \int_E f^+ \,d\mu - \int_E f^- \,d\mu$ where $\begin{align} f^+(x)&=\max(\{f(x),0\}) &=&\begin{cases} f(x), & \text{if } f(x) > 0, \\ 0, & \text{otherwise,} \end{cases}\\ f^-(x) &=\max(\{-f(x),0\})&=& \begin{cases} -f(x), & \text{if } f(x) < 0, \\ 0, & \text{otherwise.} \end{cases} \end{align}$ ### Other integrals Although the Riemann and Lebesgue integrals are the most widely used definitions of the integral, a number of others exist, including: • The Darboux integral which is equivalent to a Riemann integral, meaning that a function is Darboux-integrable if and only if it is Riemann-integrable, and the values of the two integrals, if they exist, are equal. Darboux integrals have the advantage of being simpler to define than Riemann integrals. • The Riemann–Stieltjes integral, an extension of the Riemann integral. • The Lebesgue-Stieltjes integral, further developed by Johann Radon, which generalizes the Riemann–Stieltjes and Lebesgue integrals. • The Daniell integral, which subsumes the Lebesgue integral and Lebesgue-Stieltjes integral without the dependence on measures. • The Haar integral, used for integration on locally compact topological groups, introduced by Alfréd Haar in 1933. • The Henstock–Kurzweil integral, variously defined by Arnaud Denjoy, Oskar Perron, and (most elegantly, as the gauge integral) Jaroslav Kurzweil, and developed by Ralph Henstock. • The Itō integral and Stratonovich integral, which define integration with respect to semimartingales such as Brownian motion. • The Young integral, which is a kind of Riemann–Stieltjes integral with respect to certain functions of unbounded variation. • The rough path integral defined for functions equipped with some additional "rough path" structure, generalizing stochastic integration against both semimartingales and processes such as the fractional Brownian motion. ## Properties ### Linearity • The collection of Riemann integrable functions on a closed interval [a, b] forms a vector space under the operations of pointwise addition and multiplication by a scalar, and the operation of integration $f \mapsto \int_a^b f \; dx$ is a linear functional on this vector space. Thus, firstly, the collection of integrable functions is closed under taking linear combinations; and, secondly, the integral of a linear combination is the linear combination of the integrals, $\int_a^b (\alpha f + \beta g)(x) \, dx = \alpha \int_a^b f(x) \,dx + \beta \int_a^b g(x) \, dx. \,$ • Similarly, the set of real-valued Lebesgue integrable functions on a given measure space E with measure μ is closed under taking linear combinations and hence form a vector space, and the Lebesgue integral $f\mapsto \int_E f d\mu$ is a linear functional on this vector space, so that $\int_E (\alpha f + \beta g) \, d\mu = \alpha \int_E f \, d\mu + \beta \int_E g \, d\mu.$ • More generally, consider the vector space of all measurable functions on a measure space (E,μ), taking values in a locally compact complete topological vector space V over a locally compact topological field K, f : E → V. Then one may define an abstract integration map assigning to each function f an element of V or the symbol ∞, $f\mapsto\int_E f \,d\mu, \,$ that is compatible with linear combinations. In this situation the linearity holds for the subspace of functions whose integral is an element of V (i.e. "finite"). The most important special cases arise when K is R, C, or a finite extension of the field Qp of p-adic numbers, and V is a finite-dimensional vector space over K, and when K=C and V is a complex Hilbert space. Linearity, together with some natural continuity properties and normalisation for a certain class of "simple" functions, may be used to give an alternative definition of the integral. This is the approach of Daniell for the case of real-valued functions on a set X, generalized by Nicolas Bourbaki to functions with values in a locally compact topological vector space. See (Hildebrandt 1953) for an axiomatic characterisation of the integral. ### Inequalities for integrals A number of general inequalities hold for Riemann-integrable functions defined on a closed and bounded interval [a, b] and can be generalized to other notions of integral (Lebesgue and Daniell). • Upper and lower bounds. An integrable function f on [a, b], is necessarily bounded on that interval. Thus there are real numbers m and M so that m ≤ f (x) ≤ M for all x in [a, b]. Since the lower and upper sums of f over [a, b] are therefore bounded by, respectively, m(b − a) and M(b − a), it follows that $m(b - a) \leq \int_a^b f(x) \, dx \leq M(b - a).$ • Inequalities between functions. If f(x) ≤ g(x) for each x in [a, b] then each of the upper and lower sums of f is bounded above by the upper and lower sums, respectively, of g. Thus $\int_a^b f(x) \, dx \leq \int_a^b g(x) \, dx.$ This is a generalization of the above inequalities, as M(b − a) is the integral of the constant function with value M over [a, b]. In addition, if the inequality between functions is strict, then the inequality between integrals is also strict. That is, if f(x) < g(x) for each x in [a, b], then $\int_a^b f(x) \, dx < \int_a^b g(x) \, dx.$ • Subintervals. If [c, d] is a subinterval of [a, b] and f(x) is non-negative for all x, then $\int_c^d f(x) \, dx \leq \int_a^b f(x) \, dx.$ • Products and absolute values of functions. If f and g are two functions then we may consider their pointwise products and powers, and absolute values: $(fg)(x)= f(x) g(x), \; f^2 (x) = (f(x))^2, \; \|f\| (x) = \|f(x)\|.\,$ If f is Riemann-integrable on [a, b] then the same is true for \|f\|, and $\left\| \int_a^b f(x) \, dx \right\| \leq \int_a^b \| f(x) \| \, dx.$ Moreover, if f and g are both Riemann-integrable then f 2, g 2, and fg are also Riemann-integrable, and $\left( \int_a^b (fg)(x) \, dx \right)^2 \leq \left( \int_a^b f(x)^2 \, dx \right) \left( \int_a^b g(x)^2 \, dx \right).$ This inequality, known as the Cauchy–Schwarz inequality, plays a prominent role in Hilbert space theory, where the left hand side is interpreted as the inner product of two square-integrable functions f and g on the interval [a, b]. • Hölder's inequality. Suppose that p and q are two real numbers, 1 ≤ p, q ≤ ∞ with 1/p + 1/q = 1, and f and g are two Riemann-integrable functions. Then the functions \|f\|p and \|g\|q are also integrable and the following Hölder's inequality holds: $\left\|\int f(x)g(x)\,dx\right\| \leq \left(\int \left\|f(x)\right\|^p\,dx \right)^{1/p} \left(\int\left\|g(x)\right\|^q\,dx\right)^{1/q}.$ For p = q = 2, Hölder's inequality becomes the Cauchy–Schwarz inequality. • Minkowski inequality. Suppose that p ≥ 1 is a real number and f and g are Riemann-integrable functions. Then \|f\|p, \|g\|p and \|f + g\|p are also Riemann integrable and the following Minkowski inequality holds: $\left(\int \left\|f(x)+g(x)\right\|^p\,dx \right)^{1/p} \leq \left(\int \left\|f(x)\right\|^p\,dx \right)^{1/p} + \left(\int \left\|g(x)\right\|^p\,dx \right)^{1/p}.$ An analogue of this inequality for Lebesgue integral is used in construction of Lp spaces. ### Conventions In this section f is a real-valued Riemann-integrable function. The integral $\int_a^b f(x) \, dx$ over an interval [a, b] is defined if a < b. This means that the upper and lower sums of the function f are evaluated on a partition a = x0 ≤ x1 ≤ . . . ≤ xn = b whose values xi are increasing. Geometrically, this signifies that integration takes place "left to right", evaluating f within intervals [x i , x i +1] where an interval with a higher index lies to the right of one with a lower index. The values a and b, the end-points of the interval, are called the limits of integration of f. Integrals can also be defined if a > b: • Reversing limits of integration. If a > b then define $\int_a^b f(x) \, dx = - \int_b^a f(x) \, dx.$ This, with a = b, implies: • Integrals over intervals of length zero. If a is a real number then $\int_a^a f(x) \, dx = 0.$ The first convention is necessary in consideration of taking integrals over subintervals of [a, b]; the second says that an integral taken over a degenerate interval, or a point, should be zero. One reason for the first convention is that the integrability of f on an interval [a, b] implies that f is integrable on any subinterval [c, d], but in particular integrals have the property that: • Additivity of integration on intervals. If c is any element of [a, b], then $\int_a^b f(x) \, dx = \int_a^c f(x) \, dx + \int_c^b f(x) \, dx.$ With the first convention the resulting relation $\begin{align} \int_a^c f(x) \, dx &{}= \int_a^b f(x) \, dx - \int_c^b f(x) \, dx \\ &{} = \int_a^b f(x) \, dx + \int_b^c f(x) \, dx \end{align}$ is then well-defined for any cyclic permutation of a, b, and c. Instead of viewing the above as conventions, one can also adopt the point of view that integration is performed of differential forms on oriented manifolds only. If M is such an oriented m-dimensional manifold, and M' is the same manifold with opposed orientation and ω is an m-form, then one has: $\int_M \omega = - \int_{M'} \omega \,.$ These conventions correspond to interpreting the integrand as a differential form, integrated over a chain. In measure theory, by contrast, one interprets the integrand as a function f with respect to a measure $\mu,$ and integrates over a subset A, without any notion of orientation; one writes $\textstyle{\int_A f\,d\mu = \int_{[a,b]} f\,d\mu}$ to indicate integration over a subset A. This is a minor distinction in one dimension, but becomes subtler on higher dimensional manifolds; see Differential form: Relation with measures for details. ## Fundamental theorem of calculus Main article: Fundamental theorem of calculus The fundamental theorem of calculus is the statement that differentiation and integration are inverse operations: if a continuous function is first integrated and then differentiated, the original function is retrieved. An important consequence, sometimes called the second fundamental theorem of calculus, allows one to compute integrals by using an antiderivative of the function to be integrated. ### Statements of theorems • Fundamental theorem of calculus. Let f be a continuous real-valued function defined on a closed interval [a, b]. Let F be the function defined, for all x in [a, b], by $F(x) = \int_a^x f(t)\, dt\,.$ Then, F is continuous on [a, b], differentiable on the open interval (a, b), and $F'(x) = f(x)\,$ for all x in (a, b). • Second fundamental theorem of calculus. Let f be a real-valued function defined on a closed interval [a, b] that admits an antiderivative g on [a, b]. That is, f and g are functions such that for all x in [a, b], $f(x) = g'(x).\$ If f is integrable on [a, b] then $\int_a^b f(x)\,dx\, = g(b) - g(a).$ ## Extensions ### Improper integrals Main article: Improper integral The improper integral $\int_{0}^{\infty} \frac{dx}{(x+1)\sqrt{x}} = \pi$ has unbounded intervals for both domain and range. A "proper" Riemann integral assumes the integrand is defined and finite on a closed and bounded interval, bracketed by the limits of integration. An improper integral occurs when one or more of these conditions is not satisfied. In some cases such integrals may be defined by considering the limit of a sequence of proper Riemann integrals on progressively larger intervals. If the interval is unbounded, for instance at its upper end, then the improper integral is the limit as that endpoint goes to infinity. $\int_{a}^{\infty} f(x)dx = \lim_{b \to \infty} \int_{a}^{b} f(x)dx$ If the integrand is only defined or finite on a half-open interval, for instance (a,b], then again a limit may provide a finite result. $\int_{a}^{b} f(x)dx = \lim_{\epsilon \to 0} \int_{a+\epsilon}^{b} f(x)dx$ That is, the improper integral is the limit of proper integrals as one endpoint of the interval of integration approaches either a specified real number, or ∞, or −∞. In more complicated cases, limits are required at both endpoints, or at interior points. Consider, for example, the function $1/((x+1)\sqrt{x})$ integrated from 0 to ∞ (shown right). At the lower bound, as x goes to 0 the function goes to ∞, and the upper bound is itself ∞, though the function goes to 0. Thus this is a doubly improper integral. Integrated, say, from 1 to 3, an ordinary Riemann sum suffices to produce a result of π/6. To integrate from 1 to ∞, a Riemann sum is not possible. However, any finite upper bound, say t (with t > 1), gives a well-defined result, $2\arctan (\sqrt{t}) - \pi/2$. This has a finite limit as t goes to infinity, namely π/2. Similarly, the integral from 1/3 to 1 allows a Riemann sum as well, coincidentally again producing π/6. Replacing 1/3 by an arbitrary positive value s (with s < 1) is equally safe, giving $\pi/2 - 2\arctan (\sqrt{s})$. This, too, has a finite limit as s goes to zero, namely π/2. Combining the limits of the two fragments, the result of this improper integral is $\begin{align} \int_{0}^{\infty} \frac{dx}{(x+1)\sqrt{x}} &{} = \lim_{s \to 0} \int_{s}^{1} \frac{dx}{(x+1)\sqrt{x}} + \lim_{t \to \infty} \int_{1}^{t} \frac{dx}{(x+1)\sqrt{x}} \\ &{} = \lim_{s \to 0} \left(\frac{\pi}{2} - 2 \arctan{\sqrt{s}} \right) + \lim_{t \to \infty} \left(2 \arctan{\sqrt{t}} - \frac{\pi}{2} \right) \\ &{} = \frac{\pi}{2} + \left(\pi - \frac{\pi}{2} \right) \\ &{} = \frac{\pi}{2} + \frac{\pi}{2} \\ &{} = \pi . \end{align}$ This process does not guarantee success; a limit may fail to exist, or may be unbounded. For example, over the bounded interval 0 to 1 the integral of 1/x does not converge; and over the unbounded interval 1 to ∞ the integral of $1/\sqrt{x}$ does not converge. The improper integral $\int_{-1}^{1} \frac{dx}{\sqrt[3]{x^2}} = 6$ is unbounded internally, but both left and right limits exist. It may also happen that an integrand is unbounded at an interior point, in which case the integral must be split at that point, and the limit integrals on both sides must exist and must be bounded. Thus $\begin{align} \int_{-1}^{1} \frac{dx}{\sqrt[3]{x^2}} &{} = \lim_{s \to 0} \int_{-1}^{-s} \frac{dx}{\sqrt[3]{x^2}} + \lim_{t \to 0} \int_{t}^{1} \frac{dx}{\sqrt[3]{x^2}} \\ &{} = \lim_{s \to 0} 3(1-\sqrt[3]{s}) + \lim_{t \to 0} 3(1-\sqrt[3]{t}) \\ &{} = 3 + 3 \\ &{} = 6. \end{align}$ But the similar integral $\int_{-1}^{1} \frac{dx}{x} \,\!$ cannot be assigned a value in this way, as the integrals above and below zero do not independently converge. (However, see Cauchy principal value.) ### Multiple integration Main article: Multiple integral Double integral as volume under a surface. Integrals can be taken over regions other than intervals. In general, an integral over a set E of a function f is written: $\int_E f(x) \, dx.$ Here x need not be a real number, but can be another suitable quantity, for instance, a vector in R3. Fubini's theorem shows that such integrals can be rewritten as an iterated integral. In other words, the integral can be calculated by integrating one coordinate at a time. Just as the definite integral of a positive function of one variable represents the area of the region between the graph of the function and the x-axis, the double integral of a positive function of two variables represents the volume of the region between the surface defined by the function and the plane which contains its domain. (The same volume can be obtained via the triple integral — the integral of a function in three variables — of the constant function f(x, y, z) = 1 over the above mentioned region between the surface and the plane.) If the number of variables is higher, then the integral represents a hypervolume, a volume of a solid of more than three dimensions that cannot be graphed. For example, the volume of the cuboid of sides 4 × 6 × 5 may be obtained in two ways: • By the double integral $\iint_D 5 \ dx\, dy$ of the function f(x, y) = 5 calculated in the region D in the xy-plane which is the base of the cuboid. For example, if a rectangular base of such a cuboid is given via the xy inequalities 3 ≤ x ≤ 7, 4 ≤ y ≤ 10, our above double integral now reads $\int_4^{10}\left[ \int_3^7 \ 5 \ dx\right] dy$ From here, integration is conducted with respect to either x or y first; in this example, integration is first done with respect to x as the interval corresponding to x is the inner integral. Once the first integration is completed via the $F(b) - F(a)$ method or otherwise, the result is again integrated with respect to the other variable. The result will equate to the volume under the surface. • By the triple integral $\iiint_\mathrm{cuboid} 1 \, dx\, dy\, dz$ of the constant function 1 calculated on the cuboid itself. ### Line integrals Main article: Line integral A line integral sums together elements along a curve. The concept of an integral can be extended to more general domains of integration, such as curved lines and surfaces. Such integrals are known as line integrals and surface integrals respectively. These have important applications in physics, as when dealing with vector fields. A line integral (sometimes called a path integral) is an integral where the function to be integrated is evaluated along a curve. Various different line integrals are in use. In the case of a closed curve it is also called a contour integral. The function to be integrated may be a scalar field or a vector field. The value of the line integral is the sum of values of the field at all points on the curve, weighted by some scalar function on the curve (commonly arc length or, for a vector field, the scalar product of the vector field with a differential vector in the curve). This weighting distinguishes the line integral from simpler integrals defined on intervals. Many simple formulas in physics have natural continuous analogs in terms of line integrals; for example, the fact that work is equal to force, F, multiplied by displacement, s, may be expressed (in terms of vector quantities) as: $W=\vec F\cdot\vec s.$ For an object moving along a path in a vector field $\vec F$ such as an electric field or gravitational field, the total work done by the field on the object is obtained by summing up the differential work done in moving from $\vec s$ to $\vec s + d\vec s$. This gives the line integral $W=\int_C \vec F\cdot d\vec s.$ ### Surface integrals Main article: Surface integral The definition of surface integral relies on splitting the surface into small surface elements. A surface integral is a definite integral taken over a surface (which may be a curved set in space); it can be thought of as the double integral analog of the line integral. The function to be integrated may be a scalar field or a vector field. The value of the surface integral is the sum of the field at all points on the surface. This can be achieved by splitting the surface into surface elements, which provide the partitioning for Riemann sums. For an example of applications of surface integrals, consider a vector field v on a surface S; that is, for each point x in S, v(x) is a vector. Imagine that we have a fluid flowing through S, such that v(x) determines the velocity of the fluid at x. The flux is defined as the quantity of fluid flowing through S in unit amount of time. To find the flux, we need to take the dot product of v with the unit surface normal to S at each point, which will give us a scalar field, which we integrate over the surface: $\int_S {\mathbf v}\cdot \,d{\mathbf {S}}.$ The fluid flux in this example may be from a physical fluid such as water or air, or from electrical or magnetic flux. Thus surface integrals have applications in physics, particularly with the classical theory of electromagnetism. ### Integrals of differential forms Main article: differential form A differential form is a mathematical concept in the fields of multivariable calculus, differential topology and tensors. The modern notation for the differential form, as well as the idea of the differential forms as being the wedge products of exterior derivatives forming an exterior algebra, was introduced by Élie Cartan. We initially work in an open set in Rn. A 0-form is defined to be a smooth function f. When we integrate a function f over an m-dimensional subspace S of Rn, we write it as $\int_S f\,dx^1 \cdots dx^m.$ (The superscripts are indices, not exponents.) We can consider dx1 through dxn to be formal objects themselves, rather than tags appended to make integrals look like Riemann sums. Alternatively, we can view them as covectors, and thus a measure of "density" (hence integrable in a general sense). We call the dx1, …,dxn basic 1-forms. We define the wedge product, "∧", a bilinear "multiplication" operator on these elements, with the alternating property that $dx^a \wedge dx^a = 0 \,\!$ for all indices a. Note that alternation along with linearity and associativity implies dxb∧dxa = −dxa∧dxb. This also ensures that the result of the wedge product has an orientation. We define the set of all these products to be basic 2-forms, and similarly we define the set of products of the form dxa∧dxb∧dxc to be basic 3-forms. A general k-form is then a weighted sum of basic k-forms, where the weights are the smooth functions f. Together these form a vector space with basic k-forms as the basis vectors, and 0-forms (smooth functions) as the field of scalars. The wedge product then extends to k-forms in the natural way. Over Rn at most n covectors can be linearly independent, thus a k-form with k > n will always be zero, by the alternating property. In addition to the wedge product, there is also the exterior derivative operator d. This operator maps k-forms to (k+1)-forms. For a k-form ω = f dxa over Rn, we define the action of d by: $d\omega = \sum_{i=1}^n \frac{\partial f}{\partial x_i} dx^i \wedge dx^a.$ with extension to general k-forms occurring linearly. This more general approach allows for a more natural coordinate-free approach to integration on manifolds. It also allows for a natural generalisation of the fundamental theorem of calculus, called Stokes' theorem, which we may state as $\int_{\Omega} d\omega = \int_{\partial\Omega} \omega \,\!$ where ω is a general k-form, and ∂Ω denotes the boundary of the region Ω. Thus, in the case that ω is a 0-form and Ω is a closed interval of the real line, this reduces to the fundamental theorem of calculus. In the case that ω is a 1-form and Ω is a two-dimensional region in the plane, the theorem reduces to Green's theorem. Similarly, using 2-forms, and 3-forms and Hodge duality, we can arrive at Stokes' theorem and the divergence theorem. In this way we can see that differential forms provide a powerful unifying view of integration. ### Summations The discrete equivalent of integration is summation. Summations and integrals can be put on the same foundations using the theory of Lebesgue integrals or time scale calculus. ## Methods for computing integrals ### Analytical The most basic technique for computing definite integrals of one real variable is based on the fundamental theorem of calculus. Let f(x) be the function of x to be integrated over a given interval [a, b]. Then, find an antiderivative of f; that is, a function F such that F' = f on the interval. Provided the integrand and integral have no singularities on the path of integration, by the fundamental theorem of calculus, $\textstyle\int_a^b f(x)\,dx = F(b)-F(a).$ The integral is not actually the antiderivative, but the fundamental theorem provides a way to use antiderivatives to evaluate definite integrals. The most difficult step is usually to find the antiderivative of f. It is rarely possible to glance at a function and write down its antiderivative. More often, it is necessary to use one of the many techniques that have been developed to evaluate integrals. Most of these techniques rewrite one integral as a different one which is hopefully more tractable. Techniques include: Alternate methods exist to compute more complex integrals. Many nonelementary integrals can be expanded in a Taylor series and integrated term by term. Occasionally, the resulting infinite series can be summed analytically. The method of convolution using Meijer G-functions can also be used, assuming that the integrand can be written as a product of Meijer G-functions. There are also many less common ways of calculating definite integrals; for instance, Parseval's identity can be used to transform an integral over a rectangular region into an infinite sum. Occasionally, an integral can be evaluated by a trick; for an example of this, see Gaussian integral. Computations of volumes of solids of revolution can usually be done with disk integration or shell integration. Specific results which have been worked out by various techniques are collected in the list of integrals. ### Symbolic Main article: Symbolic integration Many problems in mathematics, physics, and engineering involve integration where an explicit formula for the integral is desired. Extensive tables of integrals have been compiled and published over the years for this purpose. With the spread of computers, many professionals, educators, and students have turned to computer algebra systems that are specifically designed to perform difficult or tedious tasks, including integration. Symbolic integration has been one of the motivations for the development of the first such systems, like Macsyma. A major mathematical difficulty in symbolic integration is that in many cases, a closed formula for the antiderivative of a rather simple-looking function does not exist. For instance, it is known that the antiderivatives of the functions exp(x2), xx and (sin x)/x cannot be expressed in the closed form involving only rational and exponential functions, logarithm, trigonometric and inverse trigonometric functions, and the operations of multiplication and composition; in other words, none of the three given functions is integrable in elementary functions, which are the functions which may be built from rational functions, roots of a polynomial, logarithm, and exponential functions. The Risch algorithm provides a general criterion to determine whether the antiderivative of an elementary function is elementary, and, if it is, to compute it. Unfortunately, it turns out that functions with closed expressions of antiderivatives are the exception rather than the rule. Consequently, computerized algebra systems have no hope of being able to find an antiderivative for a randomly constructed elementary function. On the positive side, if the 'building blocks' for antiderivatives are fixed in advance, it may be still be possible to decide whether the antiderivative of a given function can be expressed using these blocks and operations of multiplication and composition, and to find the symbolic answer whenever it exists. The Risch algorithm, implemented in Mathematica and other computer algebra systems, does just that for functions and antiderivatives built from rational functions, radicals, logarithm, and exponential functions. Some special integrands occur often enough to warrant special study. In particular, it may be useful to have, in the set of antiderivatives, the special functions of physics (like the Legendre functions, the hypergeometric function, the Gamma function, the Incomplete Gamma function and so on — see Symbolic integration for more details). Extending the Risch's algorithm to include such functions is possible but challenging and has been an active research subject. More recently a new approach has emerged, using D-finite function, which are the solutions of linear differential equations with polynomial coefficients. Most of the elementary and special functions are D-finite and the integral of a D-finite function is also a D-finite function. This provide an algorithm to express the antiderivative of a D-finite function as the solution of a differential equation. This theory allows also to compute a definite integrals of a D-function as the sum of a series given by the first coefficients and an algorithm to compute any coefficient.[1] ### Numerical Main article: Numerical integration The integrals encountered in a basic calculus course are deliberately chosen for simplicity; those found in real applications are not always so accommodating. Some integrals cannot be found exactly, some require special functions which themselves are a challenge to compute, and others are so complex that finding the exact answer is too slow. This motivates the study and application of numerical methods for approximating integrals, which today use floating-point arithmetic on digital electronic computers. Many of the ideas arose much earlier, for hand calculations; but the speed of general-purpose computers like the ENIAC created a need for improvements. The goals of numerical integration are accuracy, reliability, efficiency, and generality. Sophisticated methods can vastly outperform a naive method by all four measures (Dahlquist & Björck 2008; Kahaner, Moler & Nash 1989; Stoer & Bulirsch 2002). Consider, for example, the integral $\int_{-2}^{2} \tfrac{1}{5} \left( \tfrac{1}{100}(322 + 3 x (98 + x (37 + x))) - 24 \frac{x}{1+x^2} \right) dx ,$ which has the exact answer 94/25 = 3.76. (In ordinary practice the answer is not known in advance, so an important task — not explored here — is to decide when an approximation is good enough.) A “calculus book” approach divides the integration range into, say, 16 equal pieces, and computes function values. x f(x) x f(x) −2.00 −1.50 −1.00 −0.50 0.00 0.50 1.00 1.50 2.00 2.22800 2.45663 2.67200 2.32475 0.64400 −0.92575 −0.94000 −0.16963 0.83600 −1.75 −1.25 −0.75 −0.25 0.25 0.75 1.25 1.75 2.33041 2.58562 2.62934 1.64019 −0.32444 −1.09159 −0.60387 0.31734 Numerical quadrature methods: ■ Rectangle, ■ Trapezoid, ■ Romberg, ■ Gauss Using the left end of each piece, the rectangle method sums 16 function values and multiplies by the step width, h, here 0.25, to get an approximate value of 3.94325 for the integral. The accuracy is not impressive, but calculus formally uses pieces of infinitesimal width, so initially this may seem little cause for concern. Indeed, repeatedly doubling the number of steps eventually produces an approximation of 3.76001. However, 218 pieces are required, a great computational expense for such little accuracy; and a reach for greater accuracy can force steps so small that arithmetic precision becomes an obstacle. A better approach replaces the horizontal tops of the rectangles with slanted tops touching the function at the ends of each piece. This trapezium rule is almost as easy to calculate; it sums all 17 function values, but weights the first and last by one half, and again multiplies by the step width. This immediately improves the approximation to 3.76925, which is noticeably more accurate. Furthermore, only 210 pieces are needed to achieve 3.76000, substantially less computation than the rectangle method for comparable accuracy. Romberg's method builds on the trapezoid method to great effect. First, the step lengths are halved incrementally, giving trapezoid approximations denoted by T(h0), T(h1), and so on, where hk+1 is half of hk. For each new step size, only half the new function values need to be computed; the others carry over from the previous size (as shown in the table above). But the really powerful idea is to interpolate a polynomial through the approximations, and extrapolate to T(0). With this method a numerically exact answer here requires only four pieces (five function values)! The Lagrange polynomial interpolating {hk,T(hk)}k = 0…2 = {(4.00,6.128), (2.00,4.352), (1.00,3.908)} is 3.76 + 0.148h2, producing the extrapolated value 3.76 at h = 0. Gaussian quadrature often requires noticeably less work for superior accuracy. In this example, it can compute the function values at just two x positions, ±2⁄√3, then double each value and sum to get the numerically exact answer. The explanation for this dramatic success lies in error analysis, and a little luck. An n-point Gaussian method is exact for polynomials of degree up to 2n−1. The function in this example is a degree 3 polynomial, plus a term that cancels because the chosen endpoints are symmetric around zero. (Cancellation also benefits the Romberg method.) Shifting the range left a little, so the integral is from −2.25 to 1.75, removes the symmetry. Nevertheless, the trapezoid method is rather slow, the polynomial interpolation method of Romberg is acceptable, and the Gaussian method requires the least work — if the number of points is known in advance. As well, rational interpolation can use the same trapezoid evaluations as the Romberg method to greater effect. Method Points Rel. Err. Value Trapezoid Romberg Rational Gauss 1048577 257 129 36 −5.3×10−13 −6.3×10−15 8.8×10−15 3.1×10−15 $\textstyle \int_{-2.25}^{1.75} f(x)\,dx = 4.1639019006585897075\ldots$ In practice, each method must use extra evaluations to ensure an error bound on an unknown function; this tends to offset some of the advantage of the pure Gaussian method, and motivates the popular Gauss–Kronrod quadrature formulae. Symmetry can still be exploited by splitting this integral into two ranges, from −2.25 to −1.75 (no symmetry), and from −1.75 to 1.75 (symmetry). More broadly, adaptive quadrature partitions a range into pieces based on function properties, so that data points are concentrated where they are needed most. Simpson's rule, named for Thomas Simpson (1710–1761), uses a parabolic curve to approximate integrals. In many cases, it is more accurate than the trapezoidal rule and others. The rule states that $\int_a^b f(x) \, dx \approx \frac{b-a}{6}\left[f(a) + 4f\left(\frac{a+b}{2}\right)+f(b)\right],$ with an error of $\left\|-\frac{(b-a)^5}{2880} f^{(4)}(\xi)\right\|.$ The computation of higher-dimensional integrals (for example, volume calculations) makes important use of such alternatives as Monte Carlo integration. A calculus text is no substitute for numerical analysis, but the reverse is also true. Even the best adaptive numerical code sometimes requires a user to help with the more demanding integrals. For example, improper integrals may require a change of variable or methods that can avoid infinite function values, and known properties like symmetry and periodicity may provide critical leverage. ### Mechanical The area of an arbitrary two-dimensional shape can be determined using a measuring instrument called planimeter. The volume of irregular objects can be measured with precision by the fluid displaced as the object is submerged; see Archimedes's Eureka. ### Geometrical Main article: Quadrature (mathematics) Area can be found via geometrical compass-and-straightedge constructions of an equivalent square, e.g., squaring the circle. ## Some important definite integrals Mathematicians have used definite integrals as a tool to define identities. Among these identities is the definition of the Euler–Mascheroni constant: $\gamma = \int_1^ \infty\left({1\over\lfloor x\rfloor}-{1\over x}\right)\,dx \, ,$ the Gamma function: $\Gamma(z) = \int_0^\infty e^{-t} t^{z-1} dt \, ,$ and the Laplace transform which is widely used in engineering: $F(s) = \int_0^{\infty} e^{-st} f(t) \,dt \, .$ ## References • Apostol, Tom M. (1967), Calculus, Vol. 1: One-Variable Calculus with an Introduction to Linear Algebra (2nd ed.), Wiley, ISBN 978-0-471-00005-1 • Bourbaki, Nicolas (2004), Integration I, Springer Verlag, ISBN 3-540-41129-1 . In particular chapters III and IV. • Burton, David M. (2005), The History of Mathematics: An Introduction (6th ed.), McGraw-Hill, p. 359, ISBN 978-0-07-305189-5 • • Dahlquist, Germund; Björck, Åke (2008), "Chapter 5: Numerical Integration", Numerical Methods in Scientific Computing, Volume I, Philadelphia: SIAM • Folland, Gerald B. (1984), Real Analysis: Modern Techniques and Their Applications (1st ed.), John Wiley & Sons, ISBN 978-0-471-80958-6 • Fourier, Jean Baptiste Joseph (1822), Théorie analytique de la chaleur, Chez Firmin Didot, père et fils, p. §231 Available in translation as Fourier, Joseph (1878), The analytical theory of heat, Freeman, Alexander (trans.), Cambridge University Press, pp. 200–201 • (Originally published by Cambridge University Press, 1897, based on J. L. Heiberg's Greek version.) • Hildebrandt, T. H. (1953), "Integration in abstract spaces", 59 (2): 111–139, ISSN 0273-0979 • Kahaner, David; Moler, Cleve; Nash, Stephen (1989), "Chapter 5: Numerical Quadrature", Numerical Methods and Software, Prentice Hall, ISBN 978-0-13-627258-8 • Katz, Victor J. (2004), A History of Mathematics, Brief Version, Addison-Wesley, ISBN 978-0-321-16193-2 • Leibniz, Gottfried Wilhelm (1899), in Gerhardt, Karl Immanuel, Der Briefwechsel von Gottfried Wilhelm Leibniz mit Mathematikern. Erster Band, Berlin: Mayer & Müller • Lieb, Elliott; Loss, Michael (2001), Analysis (2 ed.), AMS Chelsea, ISBN 978-0821827833 • Miller, Jeff, Earliest Uses of Symbols of Calculus, retrieved 2009-11-22 • O’Connor, J. J.; Robertson, E. F. (1996), A history of the calculus, retrieved 2007-07-09 • Rudin, Walter (1987), "Chapter 1: Abstract Integration", Real and Complex Analysis (International ed.), McGraw-Hill, ISBN 978-0-07-100276-9 • Saks, Stanisław (1964), Theory of the integral (English translation by L. C. Young. With two additional notes by Stefan Banach. Second revised ed.), New York: Dover • Shea, Marilyn (May 2007), Biography of Zu Chongzhi, University of Maine, retrieved 9 January 2009 • Siegmund-Schultze, Reinhard (2008), "Henri Lebesgue", in Timothy Gowers, June Barrow-Green, Imre Leader, Princeton Companion to Mathematics, Princeton University Press . • Stoer, Josef; Bulirsch, Roland (2002), "Chapter 3: Topics in Integration", Introduction to Numerical Analysis (3rd ed.), Springer, ISBN 978-0-387-95452-3 . •	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 81, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.8944292664527893, "perplexity_flag": "head"}
http://physics.stackexchange.com/questions/tagged/work?sort=active&pagesize=15	Tagged Questions The work tag has no wiki summary. 1answer 69 views What is the difference of work $W$ and thermal energy $Q$ in thermodynamic Stirling-process for ideal gas? What is the difference of work $W$ and thermal energy $Q$ in thermodynamic Stirling-process (in simple form) for ideal gas? I think that you need work to preserve this process and you bring thermal ... 1answer 55 views Work done by Static friction Here $v1$ is relative to the block on which sphere is pure rolling but static friction isn't $0$ as of now . In the following diagram, is work done by static friction $0$ ?, since the point of ... 2answers 65 views Centripetal Force Acceleration Suppose you want to perform a uniform circular motion . Then a body performing uniform circular motion horizontally needs an acceleration $= \frac{v^2}{r}$ at each point on the circular path with ... 2answers 158 views First law of thermodynamics? 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To find the work exerted by the force on the object, compute the line ... 1answer 838 views Thermodynamics - Sign convention I use the sign convention: Heat absorbed by the system = $q+$ (positive) Heat evolved by the system = $q-$ (negative) Work done on the system = $w +$ (positive) Work done by the system = $w -$ ... 2answers 241 views Intuitively Understanding Work and Energy It is easy to understand the concepts of momentum and impulse. The formula $mv$ is simple, and easy to reason about. It has an obvious symmetry to it. The same cannot be said for kinetic energy, ... 5answers 265 views Is there a mathematical derivation of potential energy that is not rooted in the conservation of energy? For simplicity I'll consider only gravity, but in general this question only applies to conservative forces. As per my understanding, the way one gets to the equation for gravitational potential ... 3answers 95 views Mechanics Problem I'm trying to follow Feynman's lecture. Unfortunately I'm a bit stuck on a small piece, so if you could show me what I'm doing wrong then I'd greatly appreciate your help. I want to exactly know how ... 0answers 79 views How can I measure the calories consumed in daily activity? [closed] From a nice question here: How are the calories in food calculated? I can roughly understand how people provide the data of calories in certain food. However, I am still confused about 2 things. 1) ...	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 47, "mathjax_display_tex": 3, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9334142208099365, "perplexity_flag": "middle"}
http://mathoverflow.net/questions/26917/a-question-on-group-action-on-algebras/26921	## A question on group action on algebras ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) Let $G$ be a finite group acting on an associative $k$-algebra with unity ($k$ algebraically closed of characteristic zero). The action is called ergodic if $A^G = \{ a \in A \| g \cdot a = a, \forall g \in G \} = k$. Anyone know an example of a non-semisimple finite dimensional algebra with an ergodic action of a finite group? - ## 2 Answers Let $A = k[x,y]/(x^2,xy,y^2)$ and let the generator of the two-element group act by $x \mapsto -x$, $y \mapsto -y$. More generally, take any finite group acting on $R = k[x_1, \dots, x_n]$ with invariant subring $S = k[f_1, \dots, f_m]$, and set $A = R/(f_1, \dots, f_m)$. - ### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you. Another example: Let $V$ be a finite dimensional representation of G such that $V^G=0.$ Then you can take sem-direct product algebra $A=k[G]\oplus V,$ where $V^2=0$ and $g\cdot v=gv, g\in G, v\in V.$ - How do you want $G$ to act? – Mariano Suárez-Alvarez Jun 3 2010 at 19:24 You are right, my answer was incorrect. – Bedini Jun 3 2010 at 20:06	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 16, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.8955662846565247, "perplexity_flag": "head"}
http://math.stackexchange.com/questions/34028/simplification-of-the-erdos-gallai-theorem/34044	# Simplification of the Erdos-Gallai Theorem A week or two ago back I was pointed to the Erdos-Gallai Theorem in this question. I've been unable to locate a satisfactory proof of this theorem, since the reference on wikipedia is in Hungarian. I'm particularly curious to see the proof of just one direction. Suppose you are already given that a degree sequence $(d_1,\ldots,d_n)$ is graphic. How could you then show $$\sum_{i=1}^k d_i\leq k(k-1)+\sum_{i=k+1}^n\min(k,d_i)?$$ A suggestion I read said to think about any set $U$ of $k$ vertices, and then consider how many edges have at least one end vertex in that set $U$. My guess is $k$, one connected to each vertex in $U$, but I don't see how that is very insightful. Is there a nice way of showing the above implication. - 1 Shouldn't that $d_k$ on the left side of the display be $d_i$? – Gerry Myerson Apr 20 '11 at 13:05 ## 1 Answer Page 3-4 of Blitzstein and Diaconis's paper "A Sequential Importance Sampling Algorithm For Generating Random Graphs With Prescribed Degrees" has a brief discussion. Quoting from the paper: The necessity of these conditions is not hard to see: for any set $S$ of $k$ vertices in a realization of $d$, there are at most $\binom{k}{2}$ "internal" edges within $S$, and for each vertex $v \notin S$, there are at most $\min(k, \deg(v))$ edges from $v$ into $S$. The sufficiency of the Erdos-Gallai conditions is more difficult to show.... From their paper, they give a reference to: • Berge, C. (1976). Graphs and Hypergraphs, Elsevier, New York. • Choudum, S. A. (1986). A simple proof of the Erdos-Gallai theorem on graph sequences. Bulll. Austral. Math. Soc. 33, 1, 67-70 (pdf) • Tripathi, A. and Vijay, S. (2003). A note on a theorem of Erdos & Gallai. Discrete Math. 265, 1-3, 417-420 and of course the original paper that you found in hungarian: • Erdos, P. and Gallai, T. (1960). Graphen mit punkten vorgeshriebenen grades. Math. Lapok 11, 264-274. Searching around I also found a short paper by Tripathi, Venugopalan and West here. Hope that helps. - Thanks for the resources. – Hobbie Apr 22 '11 at 3:19	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 17, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9218740463256836, "perplexity_flag": "middle"}
http://en.wikipedia.org/wiki/NP-Complete?rdfrom=http%3A%2F%2Fwiki.apidesign.org%2Findex.php%3Ftitle%3DNP-Complete%26redirect%3Dno	# NP-complete (Redirected from NP-Complete) Euler diagram for P, NP, NP-complete, and NP-hard set of problems. The left side is valid under the assumption that P≠NP, while the right side is valid under the assumption that P=NP In computational complexity theory, the complexity class NP-complete (abbreviated NP-C or NPC) is a class of decision problems. A decision problem L is NP-complete if it is in the set of NP problems and also in the set of NP-hard problems. The abbreviation NP refers to "nondeterministic polynomial time." Although any given solution to such a problem can be verified quickly, there is no known efficient way to locate a solution in the first place; indeed, the most notable characteristic of NP-complete problems is that no fast solution to them is known. That is, the time required to solve the problem using any currently known algorithm increases very quickly as the size of the problem grows. This means that the time required to solve even moderately sized versions of many of these problems can easily reach into the billions or trillions of years, using any amount of computing power available today. As a consequence, determining whether or not it is possible to solve these problems quickly, called the P versus NP problem, is one of the principal unsolved problems in computer science today. While a method for computing the solutions to NP-complete problems using a reasonable amount of time remains undiscovered, computer scientists and programmers still frequently encounter NP-complete problems. NP-complete problems are often addressed by using approximation algorithms. ## Formal overview NP-complete is a subset of NP, the set of all decision problems whose solutions can be verified in polynomial time; NP may be equivalently defined as the set of decision problems that can be solved in polynomial time on a nondeterministic Turing machine. A problem p in NP is also NP-complete if every other problem in NP can be transformed into p in polynomial time. NP-complete can also be used as an adjective: problems in the class NP-complete are known as NP-complete problems. NP-complete problems are studied because the ability to quickly verify solutions to a problem (NP) seems to correlate with the ability to quickly solve that problem (P). It is not known whether every problem in NP can be quickly solved—this is called the P = NP problem. But if any single problem in NP-complete can be solved quickly, then every problem in NP can also be quickly solved, because the definition of an NP-complete problem states that every problem in NP must be quickly reducible to every problem in NP-complete (that is, it can be reduced in polynomial time). Because of this, it is often said that the NP-complete problems are harder or more difficult than NP problems in general. ## Formal definition of NP-completeness A decision problem $\scriptstyle C$ is NP-complete if: 1. $\scriptstyle C$ is in NP, and 2. Every problem in NP is reducible to $\scriptstyle C$ in polynomial time. $\scriptstyle C$ can be shown to be in NP by demonstrating that a candidate solution to $\scriptstyle C$ can be verified in polynomial time. Note that a problem satisfying condition 2 is said to be NP-hard, whether or not it satisfies condition 1. A consequence of this definition is that if we had a polynomial time algorithm (on a UTM, or any other Turing-equivalent abstract machine) for $\scriptstyle C$, we could solve all problems in NP in polynomial time. ## Background The concept of NP-completeness was introduced in 1971 by Stephen Cook in a paper entitled The complexity of theorem-proving procedures on pages 151–158 of the Proceedings of the 3rd Annual ACM Symposium on Theory of Computing, though the term NP-complete did not appear anywhere in his paper. At that computer science conference, there was a fierce debate among the computer scientists about whether NP-complete problems could be solved in polynomial time on a deterministic Turing machine. John Hopcroft brought everyone at the conference to a consensus that the question of whether NP-complete problems are solvable in polynomial time should be put off to be solved at some later date, since nobody had any formal proofs for their claims one way or the other. This is known as the question of whether P=NP. Nobody has yet been able to determine conclusively whether NP-complete problems are in fact solvable in polynomial time, making this one of the great unsolved problems of mathematics. The Clay Mathematics Institute is offering a US\$1 million reward to anyone who has a formal proof that P=NP or that P≠NP. In the celebrated Cook-Levin theorem (independently proved by Leonid Levin), Cook proved that the Boolean satisfiability problem is NP-complete (a simpler, but still highly technical proof of this is available). In 1972, Richard Karp proved that several other problems were also NP-complete (see Karp's 21 NP-complete problems); thus there is a class of NP-complete problems (besides the Boolean satisfiability problem). Since Cook's original results, thousands of other problems have been shown to be NP-complete by reductions from other problems previously shown to be NP-complete; many of these problems are collected in Garey and Johnson's 1979 book Computers and Intractability: A Guide to the Theory of NP-Completeness.[1] For more details refer to Introduction to the Design and Analysis of Algorithms by Anany Levintin. ## NP-complete problems Main article: List of NP-complete problems Some NP-complete problems, indicating the reductions typically used to prove their NP-completeness An interesting example is the graph isomorphism problem, the graph theory problem of determining whether a graph isomorphism exists between two graphs. Two graphs are isomorphic if one can be transformed into the other simply by renaming vertices. Consider these two problems: • Graph Isomorphism: Is graph G1 isomorphic to graph G2? • Subgraph Isomorphism: Is graph G1 isomorphic to a subgraph of graph G2? The Subgraph Isomorphism problem is NP-complete. The graph isomorphism problem is suspected to be neither in P nor NP-complete, though it is in NP. This is an example of a problem that is thought to be hard, but isn't thought to be NP-complete. The easiest way to prove that some new problem is NP-complete is first to prove that it is in NP, and then to reduce some known NP-complete problem to it. Therefore, it is useful to know a variety of NP-complete problems. The list below contains some well-known problems that are NP-complete when expressed as decision problems. To the right is a diagram of some of the problems and the reductions typically used to prove their NP-completeness. In this diagram, an arrow from one problem to another indicates the direction of the reduction. Note that this diagram is misleading as a description of the mathematical relationship between these problems, as there exists a polynomial-time reduction between any two NP-complete problems; but it indicates where demonstrating this polynomial-time reduction has been easiest. There is often only a small difference between a problem in P and an NP-complete problem. For example, the 3-satisfiability problem, a restriction of the boolean satisfiability problem, remains NP-complete, whereas the slightly more restricted 2-satisfiability problem is in P (specifically, NL-complete), and the slightly more general max. 2-sat. problem is again NP-complete. Determining whether a graph can be colored with 2 colors is in P, but with 3 colors is NP-complete, even when restricted to planar graphs. Determining if a graph is a cycle or is bipartite is very easy (in L), but finding a maximum bipartite or a maximum cycle subgraph is NP-complete. A solution of the knapsack problem within any fixed percentage of the optimal solution can be computed in polynomial time, but finding the optimal solution is NP-complete. ## Solving NP-complete problems At present, all known algorithms for NP-complete problems require time that is superpolynomial in the input size, and it is unknown whether there are any faster algorithms. The following techniques can be applied to solve computational problems in general, and they often give rise to substantially faster algorithms: • Approximation: Instead of searching for an optimal solution, search for an "almost" optimal one. • Randomization: Use randomness to get a faster average running time, and allow the algorithm to fail with some small probability. Note: The Monte Carlo method is not an example of an efficient algorithm, although evolutionary approaches like Genetic algorithms may be. • Restriction: By restricting the structure of the input (e.g., to planar graphs), faster algorithms are usually possible. • Parameterization: Often there are fast algorithms if certain parameters of the input are fixed. • Heuristic: An algorithm that works "reasonably well" in many cases, but for which there is no proof that it is both always fast and always produces a good result. Metaheuristic approaches are often used. One example of a heuristic algorithm is a suboptimal $\scriptstyle O(n\log n)$ greedy coloring algorithm used for graph coloring during the register allocation phase of some compilers, a technique called graph-coloring global register allocation. Each vertex is a variable, edges are drawn between variables which are being used at the same time, and colors indicate the register assigned to each variable. Because most RISC machines have a fairly large number of general-purpose registers, even a heuristic approach is effective for this application. ## Completeness under different types of reduction In the definition of NP-complete given above, the term reduction was used in the technical meaning of a polynomial-time many-one reduction. Another type of reduction is polynomial-time Turing reduction. A problem $\scriptstyle X$ is polynomial-time Turing-reducible to a problem $\scriptstyle Y$ if, given a subroutine that solves $\scriptstyle Y$ in polynomial time, one could write a program that calls this subroutine and solves $\scriptstyle X$ in polynomial time. This contrasts with many-one reducibility, which has the restriction that the program can only call the subroutine once, and the return value of the subroutine must be the return value of the program. If one defines the analogue to NP-complete with Turing reductions instead of many-one reductions, the resulting set of problems won't be smaller than NP-complete; it is an open question whether it will be any larger. Another type of reduction that is also often used to define NP-completeness is the logarithmic-space many-one reduction which is a many-one reduction that can be computed with only a logarithmic amount of space. Since every computation that can be done in logarithmic space can also be done in polynomial time it follows that if there is a logarithmic-space many-one reduction then there is also a polynomial-time many-one reduction. This type of reduction is more refined than the more usual polynomial-time many-one reductions and it allows us to distinguish more classes such as P-complete. Whether under these types of reductions the definition of NP-complete changes is still an open problem. All currently known NP-complete problems are NP-complete under log space reductions. Indeed, all currently known NP-complete problems remain NP-complete even under much weaker reductions.[2] It is known, however, that AC0 reductions define a strictly smaller class than polynomial-time reductions.[3] ## Naming According to Donald Knuth, the name "NP-complete" was popularized by Alfred Aho, John Hopcroft and Jeffrey Ullman in their celebrated textbook "The Design and Analysis of Computer Algorithms". He reports that they introduced the change in the galley proofs for the book (from "polynomially-complete"), in accordance with the results of a poll he had conducted of the Theoretical Computer Science community.[4] Other suggestions made in the poll[5] included "Herculean", "formidable", Steiglitz's "hard-boiled" in honor of Cook, and Shen Lin's acronym "PET", which stood for "probably exponential time", but depending on which way the P versus NP problem went, could stand for "provably exponential time" or "previously exponential time".[6] ## Common misconceptions The following misconceptions are frequent.[7] • "NP-complete problems are the most difficult known problems." Since NP-complete problems are in NP, their running time is at most exponential. However, some problems provably require more time, for example Presburger arithmetic. • "NP-complete problems are difficult because there are so many different solutions." On the one hand, there are many problems that have a solution space just as large, but can be solved in polynomial time (for example minimum spanning tree). On the other hand, there are NP-problems with at most one solution that are NP-hard under randomized polynomial-time reduction (see Valiant–Vazirani theorem). • "Solving NP-complete problems requires exponential time." First, this would imply P ≠ NP, which is still an unsolved question. Further, some NP-complete problems actually have algorithms running in superpolynomial, but subexponential time. For example, the Independent set and Dominating set problems are NP-complete when restricted to planar graphs, but can be solved in subexponential time on planar graphs using the planar separator theorem.[8] • "All instances of an NP-complete problem are difficult." Often some instances, or even almost all instances, may be easy to solve within polynomial time. • "If P=NP, all cryptographic ciphers can be broken." A polynomial-time problem can be very difficult to solve in practice if the polynomial's degree or constants are large enough. For example, ciphers with a fixed key length, such as Advanced Encryption Standard, can all be broken in constant time (and are thus already in P), though that constant may exceed the age of the universe. ## Notes 1. Garey, Michael R.; Johnson, D. S. (1979). In Victor Klee. . A Series of Books in the Mathematical Sciences. San Francisco, Calif.: W. H. Freeman and Co. pp. x+338. ISBN 0-7167-1045-5. MR 519066. 2. Agrawal, M.; Allender, E.; Rudich, Steven (1998). "Reductions in Circuit Complexity: An Isomorphism Theorem and a Gap Theorem". Journal of Computer and System Sciences (Academic Press) 57 (2): 127–143. doi:10.1006/jcss.1998.1583. ISSN 1090-2724 3. Agrawal, M.; Allender, E.; Impagliazzo, R.; Pitassi, T.; Rudich, Steven (2001). "Reducing the complexity of reductions". Computational Complexity (Birkhäuser Basel) 10 (2): 117–138. doi:10.1007/s00037-001-8191-1. ISSN 1016-3328 4. Don Knuth, Tracy Larrabee, and Paul M. Roberts, § 25, MAA Notes No. 14, MAA, 1989 (also Stanford Technical Report, 1987). 5. Knuth, D. F. (1974). "A terminological proposal". SIGACT News 6 (1): 12–18. doi:10.1145/1811129.1811130. Retrieved 2010-08-28. 6. Ball, Philip. "DNA computer helps travelling salesman". doi:10.1038/news000113-10. ## References • Garey, M.R.; Johnson, D.S. (1979). . New York: W.H. Freeman. ISBN 0-7167-1045-5. This book is a classic, developing the theory, then cataloguing many NP-Complete problems. • Cook, S.A. (1971). "The complexity of theorem proving procedures". Proceedings, Third Annual ACM Symposium on the Theory of Computing, ACM, New York. pp. 151–158. doi:10.1145/800157.805047. • Dunne, P.E. "An annotated list of selected NP-complete problems". COMP202, Dept. of Computer Science, University of Liverpool. Retrieved 2008-06-21. • Crescenzi, P.; Kann, V.; Halldórsson, M.; Karpinski, M.; Woeginger, G. "A compendium of NP optimization problems". KTH NADA, Stockholm. Retrieved 2008-06-21. • Dahlke, K. "NP-complete problems". Math Reference Project. Retrieved 2008-06-21. • Karlsson, R. "Lecture 8: NP-complete problems" (PDF). Dept. of Computer Science, Lund University, Sweden. Retrieved 2008-06-21. [] • Sun, H.M. "The theory of NP-completeness" (PPT). Information Security Laboratory, Dept. of Computer Science, National Tsing Hua University, Hsinchu City, Taiwan. Retrieved 2008-06-21. • Jiang, J.R. "The theory of NP-completeness" (PPT). Dept. of Computer Science and Information Engineering, National Central University, Jhongli City, Taiwan. Retrieved 2008-06-21. • Cormen, T.H.; Leiserson, C.E., Rivest, R.L.; Stein, C. (2001). (2nd ed.). MIT Press and McGraw-Hill. Chapter 34: NP–Completeness, pp. 966–1021. ISBN 0-262-03293-7. • Sipser, M. (1997). Introduction to the Theory of Computation. PWS Publishing. Sections 7.4–7.5 (NP–completeness, Additional NP–complete Problems), pp. 248–271. ISBN 0-534-94728-X. • Papadimitriou, C. (1994). Computational Complexity (1st ed.). Addison Wesley. Chapter 9 (NP–complete problems), pp. 181–218. ISBN 0-201-53082-1. • Computational Complexity of Games and Puzzles • Tetris is Hard, Even to Approximate • Minesweeper is NP-complete! • Bern, Marshall (1990). "Faster exact algorithms for Steiner trees in planar networks". Networks 20 (1): 109–120. doi:10.1002/net.3230200110 . • Deĭneko, Vladimir G.; Klinz, Bettina; Woeginger, Gerhard J. (2006). "Exact algorithms for the Hamiltonian cycle problem in planar graphs". Operations Research Letters 34 (3): 269–274. doi:10.1016/j.orl.2005.04.013 . • Dorn, Frederic; Penninkx, Eelko; Bodlaender, Hans L.; Fomin, Fedor V. (2005). "Efficient Exact Algorithms on Planar Graphs: Exploiting Sphere Cut Branch Decompositions". Proc. 13th European Symposium on Algorithms (ESA '05). Lecture Notes in Computer Science 3669. Springer-Verlag. pp. 95–106. doi:10.1007/11561071_11. ISBN 978-3-540-29118-3 . • Lipton, Richard J.; Tarjan, Robert E. (1980). "Applications of a planar separator theorem". 9 (3): 615–627. doi:10.1137/0209046 . ## Further reading • Scott Aaronson, NP-complete Problems and Physical Reality, ACM SIGACT News, Vol. 36, No. 1. (March 2005), pp. 30–52. • Lance Fortnow, The status of the P versus NP problem, Commun. ACM, Vol. 52, No. 9. (2009), pp. 78–86.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 11, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.8976884484291077, "perplexity_flag": "middle"}
http://mathhelpforum.com/calculus/211862-related-rates-angle-elevation-help.html	# Thread: 1. ## Related Rates & Angle of Elevation Help?? I'm not sure how to do these two questions. Any help with both or one is appreciated!! 1) A balloon rises into the air started at point P. AN observer, 100m from P looks at the balloon and the angle theta between her line of sight and the ground increases at a rate of 1/20 rads per second. Find the velocity of the balloon where theta = pi/4 The answer is 10m/s I'm just not sure how to get it. 2) One end of a ladder of length 5m slides down a verticle wall. When the upper end of the ladder is 3m from the ground, it has a downward velocity of 0.5m/s. Find the rate at which the angle of elevation is changing at that time. Again, the answer is 0.125 rad/s 2. ## Re: Related Rates & Angle of Elevation Help?? 1.) At time $t$, we may state: $\tan(\theta)=\frac{h}{100}$ Implicitly differentiating with respect to $t$, and solve for $\frac{dh}{dt}$ We are given the conditions: $\theta=\frac{\pi}{4}$ $\frac{d\theta}{dt}=\frac{1}{20}\,\frac{\text{rad}} {\text{s}}$ So, use these to find the value of $\frac{dh}{dt}$, which represents the vertical speed of the balloon at the given point. 2.) Let x be the distance of the base of the ladder from the wall and y be the height of the top of the ladder on the wall. We may then relate the angle of elevation to these variables with (1) $\tan(\theta)=\frac{y}{x}$ We also know by Pythagoras that: (2) $x^2+y^2=5^2$ Differentiate (1) with respect to $t$ and solve for $\frac{d\theta}{dt}$, and replace the resulting trig. function by its ratio definition. Differentiate (2) with respect to $t$ to find $\frac{dx}{dt}$. Now, you should have everything you need to answer the question.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 13, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9302710890769958, "perplexity_flag": "head"}
http://math.stackexchange.com/questions/237470/cardinality-of-infinite-sequences-of-integers	# Cardinality of Infinite Sequences of Integers I am having trouble answering a question about the cardinality of the infinite sequences of integers. I claimed that the cardinality would be $\aleph_0^{\aleph_0}$, which would be $2^{\aleph_0} = \mathfrak c$. According to my instructor, I have the right answer, but am lacking idea/intuition. My ideas start as that: An infinite sequence of integers will be denumerable and because of that, the cardinality of a particular infinite sequence will be $\aleph_0$. - So, what are you asking? – Cameron Buie Nov 14 '12 at 21:08 1 I think that almost all the questions I saw by you were already answered on the site at least once (and probably more). Have you tried using the search feature? – Asaf Karagila Nov 14 '12 at 21:09 1 Exactly what did you say to justify the conclusion that there are $\aleph_0^{\aleph_0}$ infinite sequences of integers? We can’t even guess what your instructor thought was wrong if we don’t know what you said. Was it what you wrote in the last paragraph of the question? – Brian M. Scott Nov 14 '12 at 21:12 1 @JulianPark: I had none, either, when I started here. There are many tutorials on the web. You can also right-click any $\LaTeX$, Show Math As -> TeX commands and see how it was done. – Ross Millikan Nov 14 '12 at 21:27 1 – Asaf Karagila Nov 14 '12 at 21:38 show 3 more comments	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 5, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9588996171951294, "perplexity_flag": "middle"}
http://quant.stackexchange.com/questions/3331/correlation-test-for-linear-dependence/4114	# Correlation: Test for linear dependence Setting the scene: Assume a multivariate GBM with correlation matrix $\Sigma$. Further, one want to estimate the correlation between two of the assets. Assume one has a suitable estimator of the correlation, the exact estimator might not be directly relevant, so assume e.g. standard, 60 days moving average. Identifying the issue: If one look at the time series of the correlations, it is obvious that a correlation is not constant over time, as postulated by gbm. Is there any good measure of "correlation volatility", i.e. a measure that says when a correlation seems stable? Alternatively, a measure that will quickly identify that the correlation is unrelieable? Attempt: Ive tried to look at the maximum (absolute) changes of the close the last 60 days, and then created a band around todays estimation equal to $(p_t + max, p_t-max)$, where p is the correlation estimate and max is the max absolute movement. Then I have said that if this spread is higher than some given value (say 0.15), then the correlation is "unstable". I have also tried different variation as looking at the maximum return, highest vs. lowest value etc. I am a bit vary to over-model this, so I hesistate to start giving correlation coefficient a probability distribution etc... At the same time, I find my approach a bit unsatisfactory, and wondering if there is any "well designed" tests to see whether two variables satisfy such a linear dependence that correlation is... - I have started to be a bit vary of just looking at the correlation coefficients, as they are simply a number and could be stable although the returns do not follow a lognormal (or any other assumed distribution) walk with correlation p. I guess one would have to look at the distribution as a whole, considering whether the observed data would be sampled from e.g. a bivariat normal distribution. – AdAbsurdum Sep 12 '12 at 13:35 ## 2 Answers Are you trying only to identify unstable correlations, or are you trying to incorporate time-varying correlation into your model? If the latter, you may want to check out Engle's Dynamic Conditional Correlation, which is an extent of GARCH modeling. In simplest terms, DCC models time-varying correlation parametrically via the GARCH residuals. Another, slightly more exotic option is to consider a time-varying or conditional copula, such as those introduced by Andrew Patton. See for example his paper Modelling Asymmetric Exchange Rate Dependence. Patton has also introduced goodness-of-fit measures for copulas. Note that by inspecting the parameters of such models, you could evaluate the hypothesis that correlations are stable. - Thanks. I think the second paper maybe can solve some of my issues (restated in a comment above). I will have a look at it. – AdAbsurdum Sep 12 '12 at 13:37 If you have time seria you could postulate a linear equation for them: x(t)=a*t+b giving a=x(t+1)-x(t) There are lots of methods to deal with the hypothesis that a is zero or not, including low sample statistics to boost your confidence (levels) in the final result. In addition, by b-estimation you could see the "volatility" of your correlations. -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9417949914932251, "perplexity_flag": "middle"}
http://unapologetic.wordpress.com/2009/12/14/some-sets-of-measure-zero/?like=1&source=post_flair&_wpnonce=7b62a7b6bf	# The Unapologetic Mathematician ## Some Sets of Measure Zero Here’s a useful little tool: Let $F=\{F_1,F_2,\dots\}$ be a countable collection of sets of measure zero in $\mathrm{R}^n$. That is, $\overline{m}(F_k)=0$ for all $k$. We define $S$ to be the union $\displaystyle S=\bigcup\limits_{k=1}^\infty F_k$ Then it turns out that $\overline{m}(S)=0$ as well. To see this, pick some $\epsilon>0$. For each set $F_k$ we can pick a Lebesgue covering $C_k$ of $F_k$ so that $\mathrm{vol}(C_k)<\frac{\epsilon}{2^k}$. We can throw all the intervals in each of the $C_k$ together into one big collection $C$, which will be a Lebesgue covering of all of $S$. Indeed, the union of a countable collection of countable sets is still countable. We calculate the volume of this covering: $\displaystyle\mathrm{vol}(C)\leq\sum\limits_{k=1}^\infty\mathrm{vol}(C_k)<\sum\limits_{k=1}^\infty\frac{\epsilon}{2^k}=\epsilon$ where the final summation converges because it’s a geometric series with initial term $\frac{\epsilon}{2}$ and ratio $\frac{1}{2}$. This implies that $\overline{m}(S)=0$. As an example, a set consisting of a single point has measure zero because we can put it into an arbitrarily small open box. The result above then says that any countable collection of points in $\mathbb{R}^n$ also has measure zero. For instance, the collection of rational numbers in $\mathbb{R}^1$ is countable (as Kate mentioned in passing recently), and so it has measure zero. The result is useful because otherwise it might be difficult to imagine how to come up with a Lebesgue covering of all the rationals with arbitrarily small volume. ### Like this: Posted by John Armstrong \| Analysis, Measure Theory ## 4 Comments » 1. [...] , we also have , and therefore have as [...] Pingback by \| December 15, 2009 \| Reply 2. [...] the other hand, the double integral does not exist. Yes, is countable, and so it has measure zero. However, it’s also dense, which means is discontinuous everywhere in the unit [...] Pingback by \| December 18, 2009 \| Reply 3. [...] have measure zero, and we assume that the discontinuities in each are of measure zero, their (countable) union will also have measure zero. And then so must the set of discontinuities of in have measure zero [...] Pingback by \| December 30, 2009 \| Reply 4. [...] of all, we observe that any hyperplane has measure zero, and so any finite collection of them will too. Thus the collection of all the hyperplanes perpendicular to vectors cannot fill up all of . We [...] Pingback by \| February 2, 2010 \| Reply « Previous \| Next » ## About this weblog This is mainly an expository blath, with occasional high-level excursions, humorous observations, rants, and musings. The main-line exposition should be accessible to the “Generally Interested Lay Audience”, as long as you trace the links back towards the basics. Check the sidebar for specific topics (under “Categories”). I’m in the process of tweaking some aspects of the site to make it easier to refer back to older topics, so try to make the best of it for now.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 21, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9395760297775269, "perplexity_flag": "head"}
http://mathoverflow.net/questions/47487?sort=newest	## Probability of the maximum (Levy Stable) random variable in a list being greater than the sum of the rest? ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) Given a list of identical and independently distributed Levy Stable random variables, $(X_0, X_1, \dots, X_{n-1})$, what is the is the probability that the maximum exceeds the sum of the rest? i.e.: $$M = \text{Max}(X_0, X_1, \dots, X_{n-1})$$ $$\text{Pr}( M > \sum_{j=0}^{n-1} X_j - M )$$ Where, in Nolan's notation, $X_j \in S(\alpha, \beta=1, \gamma, \delta=0 ; 0)$, where $\alpha$ is the critical exponent, $\beta$ is the skew, $\gamma$ is the scale parameter and $\delta$ is the shift. For simplicity, I have taken the skew parameter, $\beta$, to be 1 (maximally skewed to the right) and $\delta=0$ so everything has its mode centered in an interval near 0. From numerical simulations, it appears that for the region of $0 < \alpha < 1$, the probability converges to a constant, irregardless of $n$ or $\gamma$. For $1 < \alpha < 2$ it appears to go as $O(1/n^{\alpha - 1})$ (maybe?) irregardless of $n$ or $\gamma$. For $\alpha=1$ it's not clear (to me) but appears to be a decreasing function dependent on $n$ and $\gamma$. I have tried making a heuristic argument to the in the form of: $$\text{Pr}( M > \sum_{j=0}^{n-1} X_j - M) \le n \text{Pr}( X_0 - \sum_{j=1}^{n-1} X_j > 0 )$$ Then using formula's provided by Nolan (pg. 27) for the parameters of the implicit r.v. $U = X_0 - \sum_{j=1}^{n-1} X_j$ combined with the tail approximation: $$\text{Pr}( X > x ) \sim \gamma^{\alpha} c_{\alpha} ( 1 + \beta ) x^{-\alpha}$$ $$c_{\alpha} = \sin( \pi \alpha / 2) \Gamma(\alpha) / \pi$$ but this leaves me nervous and a bit unsatisfied. Just for comparison, if $X_j$ were taken to be uniform r.v.'s on the unit interval, this function would decrease exponentially quickly. I imagine similar results hold were the $X_j$'s Gaussian, though any clarification on that point would be appreciated. Getting closed form solutions for this is probably out of the question, as there isn't even a closed form solution for the pdf of Levy-Stable random variables, but getting bounds on what the probability is would be helpful. I would appreciate any help with regards to how to analyze these types of questions in general such as general methods or references to other work in this area. Note: There didn't seem to be much interest from this site so I've cross-posted to math.stackexchange.com here. - ## 1 Answer When $\alpha>1$, the LLN implies that the sum is nearly $n\E X$, and $\E X$ depends on $\delta$. If $\E X<0$ the probability tends to 1. If $\E X>0$ the probability decays as you say. When $\alpha<1$ the probability indeed depends to a constant. You can estimate $\P(X_0>X_1+\dots+X_{n-1})$, since the sum is just an independent stable variable. These events are almost disjoint. In the maximally skewed case, the variables are positive, and then the events are completely disjoint. The plimit probability can also be expressed as the probability that the maximal point in a certain non-homogenuous Poisson process is greater than the sum of the rest. -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 29, "mathjax_display_tex": 5, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9433720707893372, "perplexity_flag": "head"}
http://mathoverflow.net/questions/69642?sort=votes	## Subtract diagonal terms from the matrix to make it negative semi-definite ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) I'm reading one paper and on page 36 (48 in the pdf) it says: Let d(s, i) be the (positive) diagonal terms that need to be subtracted from the matrix to make it negative semi-definite... Could someone explain me why it's possible and how can I get the values of these terms? - 1 I think you need to tell us more about the matrix and the context. Is the matrix Hermitian and positive definite to start? – Geoff Robinson Jul 6 2011 at 15:46 The paper is master thesis on Markov Random Fields and matrix elements are potentials of the variables. here is the thesis: stat.berkeley.edu/~pradeepr/paperz/thesis.pdf – sbos Jul 6 2011 at 15:53 My quote was from page 36 (48 in the pdf). – sbos Jul 6 2011 at 15:54 This might just be my lack of expertise in this area talking, but this seems like a question much better suited for direct email with the author. – Cam McLeman Jul 6 2011 at 16:00 ## 1 Answer If $M$ is a Hermitian matrix, its eigenvalues $\lambda_j$ are all real. If you subtract $d$ from all the diagonal elements, you are changing $M$ to $M - d I$; if $d \ge \max_j \lambda_j$, all eigenvalues of $M - d I$ will be nonpositive so $M - d I$ will be negative semidefinite. - That's nice, but what would be in more general case? – sbos Jul 6 2011 at 19:02 Ok, let M be symmetrical matrix – sbos Jul 6 2011 at 19:17 More general in what way? Perhaps subtracting $d_i$ from $M_{i,i}$ for each $i$? Then if $\min_i d_i \ge \max_j \lambda_j$, the new matrix will be negative semidefinite. – Robert Israel Jul 7 2011 at 18:11	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 12, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.8776810169219971, "perplexity_flag": "middle"}
http://math.stackexchange.com/questions/52918/integral-of-a-spherically-symmetric-3-dimensional-function-over-all-space?answertab=votes	integral of a spherically symmetric 3-dimensional function over all space I'm very sorry because it may be a very basic question but I'm not able whether to solve it for sure, nor to find an answer in stackexchange or elsewhere. I have to calculate $\int \int n(\vec{r})u(\|\|\vec{r}-\vec{r}'\|\|) n(\vec{r}') d\vec{r} d\vec{r}'$ For some purpose, I have a case where $n(\vec{r})=n$ is a constant value. I thus have $n^2 \int\int u(\|\|\vec{r}-\vec{r}'\|\|) d\vec{r} d\vec{r}'$ Let emphasis that $u$ is a spherically symetric function (radial in my physicist vocabulary), so that $n^2 \int\int u(\|\|\vec{r}-\vec{r}'\|\|) d\vec{r} d\vec{r}' = n^2 \int\int u(r) d\vec{r} d\vec{r}'$ (if it was unclear before) Each integration is over all space in 3 dimensions. Here is my question: I am right to say that $n^2 \int\int u(r) d\vec{r} d\vec{r}' = 4 \pi n^2 \int u(r) r^2 dr$ ? My thought is that $u$ being independent of $\theta$ and $\psi$, the integration over these angles lead to the solid angle of the whole sphere : $4\pi$. The rest stays ... - ...and $u(r)$ is? – J. M. Jul 21 '11 at 16:02 1 Unless $u$ is essentially zero, the integral $\int u(\|\|\vec{r}-\vec{r}'\|\|)d\vec{r}d\vec{r}'$ cannot be finite. The change of variables $\vec{s}=\vec{r}'-\vec{r}$ makes the integral to $\int u(\|\|\vec{s}\|\|) d\vec{s} d\vec{r}$, and the integrant doesn't depend on $\vec{r}$. – user8268 Jul 21 '11 at 19:00 Sorry if the question was unclear: I wanted to ask if my last equation line is right (going from spherical to radial integration for a spherically symmetric function $u$). What $u$ is isn't important as this integration will be done numerically for any kind of $u$ (it's a two centers interaction potential let's say a Lennard Jones 6-12 for instance). Thank you Joriki et al! @Alice: Hi! it is indeed very linked in the sens that now that I understand this, I can understand your demonstration for my other question! Once more thank your! – max Jul 22 '11 at 17:00 1 Answer Contrary to what I wrote in a comment when I hadn't noticed yet that you went from a double integral to a single integral, this isn't right; you can't just drop one of the integrals. If I understand your notation correctly, $r=\lVert\vec r-\vec r'\rVert$ (which is slightly confusing, since $r$ is usually used to denote $\lVert\vec r\rVert$). By substituting $\vec r''=\vec r-\vec r'$, we can factor the integrals: $$\iint u(\lVert\vec r-\vec r'\rVert)\mathrm d\vec r\mathrm d\vec r'=\iint u(\lVert\vec r''\rVert)\mathrm d\vec r''\mathrm d\vec r'=\int u(\lVert\vec r''\rVert)\mathrm d\vec r''\mathrm \int \mathrm d\vec r'\;.$$ (We could just as well have kept $\vec r$ and replaced $\vec r'$; since all the integrals are over all space, it makes no difference.) We can indeed transform the left-hand integral over $\vec r''$ into a radial integral: $$\int u(\lVert\vec r''\rVert)\mathrm d\vec r''=4\pi\int u(r)r^2\mathrm d r\;.$$ But the right-hand integral over $\mathrm d\vec r'$ is infinite. So there's a fundamental problem in your setup. The function $n$ should usually be responsible for making the integral finite by decaying sufficiently quickly; the problem may be that you're taking it to be constant. - 1 – J. M. Jul 22 '11 at 17:41 @Joriki: thanks a lot for this very clear explanation. In order to answer your very last sentence "small wonder that you've divergence": I use pair potential that converges to zero at a non-infinite radius. In fact, when people in my community say LJ potential, they mean troncated LJ potential. Sorry for the misunderstanding I should have been more careful. Thanks again I am very happy with your answer. – max Jul 23 '11 at 10:00 1 @Maximilien: You're welcome. Note that "small wonder that you've divergence" is from J.M.'s comment, not from me. J.M.'s point is valid despite your explanation, since the potential converges to zero at a non-infinite radius $\lVert \vec r-\vec r'\rVert$, whereas the infinite contribution to your double integral comes from values near $\vec r-\vec r'=0$, of which the double integral contains an infinite measure since you approximated $n$ by a constant. The problem is not solved by truncating the LJ potential; you need the decay in $n$ to make the integral finite. – joriki Jul 23 '11 at 10:17 – max Jul 23 '11 at 17:28 @Maximilien: No matter how you modify the potential, as long as it's not zero everywhere, the non-zero values will get multiplied by the infinite integral over $\vec r'$. This isn't about either the short-range behaviour or the long-range behaviour of the potential; it's about the fact that if the interacting densities are constant, then for every value of $\lVert\vec r-\vec r'\rVert$ there's an infinite measure of configurations with that value contributing with constant density. – joriki Jul 23 '11 at 17:40	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 34, "mathjax_display_tex": 2, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.933088481426239, "perplexity_flag": "middle"}
http://mathoverflow.net/questions/77949/unique-structures-in-a-class-of-connected-directed-hypergraphs	## Unique structures in a class of connected directed hypergraphs ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) Edit: The problem has been slightly revised, as I discovered that one of the questions I asked has an answer in the negative. I'm working in a setting involving constraints on a system described by a k-uniform hypergraph, where the constraints are satisfiable if there is an injective function mapping each edge to one of the vertices it touches. The ways that these constraints can be satisfied is limited when that mapping is bijective, which may be due to something like feedback-loops of dependencies which have fixed points. This motivates the following problem, which I expect may have been considered before in different terms: Definitions. • A k-map on a set V is just a function $F: V \to \mathscr P(V)$, mapping each v ∈ V to some set $F(v) \subseteq V$ with cardinality k, such that $v \notin F(v)$. The hypergraph of F is simply the hypergraph H with vertex-set V, and whose edges are all sets of the form $e_v = \{v\} \cup F(v)$ for each v ∈ V. • We're interested not just in the hypergraph defined by F, but a sort of directed hypergraph: this is perhaps most easily defined using an incidence-matrix-cum-adjacency-matrix defined on $\mathbb C^V$, $$A_{u,v} = \begin{cases} 1 & \text{if $u \in F(v)$;} \\ 0 & \text{otherwise.} \end{cases}$$ • We say that F is connected if the hypergraph of F is connected. Observation. A 1-map on a set V is essentially just a function on V which has no fixed points. This corresponds to a garden-variety digraph without loops which (a) consists of multiple components, where (b) each component consists of a collection of at least two disjoint trees directed to their roots, with a directed cycle defined between the roots of the trees. The digraph for a connected 1-map contains just the one directed cycle (which happens also to be the only undirected cycle in the digraph; provided of course the directed cycle has length at least 3). This state of affairs is witnessed by the fact that $A$ has a +1-eigenspace of dimension 1, which is spanned by the indicator function for the cycle. The cycle is also the support for several other eigenvectors (each spanning another eigenspace of dimension 1), one for each of the dth roots of unity, where d is the length of the cycle. I'm interested in such unique structures — such as the unique cycle, or the nondegenerate eigenspace — but in the hypergraphs which correspond to k-maps for all k ≥ 1. For k arbitrary, it is clear that $A$ has a non-trivial k-eigenspace (as the matrix $A - kI$ is singular, which may be verified by considering the sums of its columns); however, the condition of being connected is less restrictive in the case k>1 than in the case k=1,† so that space may have dimension larger than 1. Of course, it could be that I should be looking at an algebraic object other than $A$ as defined above, or a "purely" combinatorial structure. Question. Is there any unique structure of any sort which arises in every connected hypergraph H which corresponds to a k-map? † It is not difficult to show that every {0,1}-matrix, with a null diagonal and columns each summing to the same constant, are valid adjacency matrices $A$ for some k-map. A simple 2-map whose 2-eigenspace has dimension 2 can be given by $$A = \left[\begin{array}{c\|ccc\|ccc} 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ \hline 1 & 0 & 1 & 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 1 & 0 & 0 & 0 \\ 0 & 1 & 1 & 0 & 0 & 0 & 0 \\ \hline 1 & 0 & 0 & 0 & 0 & 1 & 1 \\ 0 & 0 & 0 & 0 & 1 & 0 & 1 \\ 0 & 0 & 0 & 0 & 1 & 1 & 0 \end{array}\right]\,:$$ the blocks correspond to "strongly connected" components in the directed hypergraph. The fact that the 2-eigenspace has dimension 2 can be attributed to the fact that there are two such strongly connected components, which are closed under the action of the map F described by this matrix. (The cycle in a connected 1-map is also the unique strongly connected component closed under F.) Generalizing this example, a connected k-map may have an arbitrary number of strongly connected components of this sort, for k > 1. -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 10, "mathjax_display_tex": 2, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9595702886581421, "perplexity_flag": "head"}
http://math.stackexchange.com/questions/77806/some-question-of-nilpotent-matrix	# Some question of nilpotent matrix Let $N_{i}=\{A \in M_{n\times n}(k) \mid A^{i}=0\}$ that is, $N_i$ is the set of nilpotent matrix with order $i$. Then we get the chain $N_1 \subset N_2 \subset \cdots \subset N_n$. Is $N_i \subsetneqq N_{i+1}$ for all $i$ true? - 6 Six of your questions got answers, but you accepted none of them. Why? – Did Nov 1 '11 at 10:26 ## 2 Answers The following exercises will enhance your understanding on the question that you have asked: $\textbf{(1)}$ Prove that if there exists a positive integer $k$ such that null $A^k$ = null $A^{k+1}$, then null $A^n$ = null $A^{n+1}$ for all integers $n \geq k$. $\textbf{(2)}$ Prove that if $n =$ dim $V$, and $A$ an operator on $V$ then null $A^n =$ null $A^{n+1} =$ null $A^{n+2} = ...$ Hint: Use the fact that $V$ is of finite dimension and that no subspace of $V$ can be of a larger dimension than $V$. $\textbf{(3)}$ If $A$ is a nilpotent operator on a complex vector space $V$, prove that $A^{\text{dim} V}$ = 0. - Yes for $i<n$. Jordan normal form makes it easy to find an example of a matrix with $A^i\neq 0$, but $A^{i+1}=0$. -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 31, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.8988125324249268, "perplexity_flag": "head"}
http://nrich.maths.org/275/index?nomenu=1	If $$z_1 z_2 z_3 = 1$$ and $$z_1 + z_2 + z_3 = \frac{1}{z_1} + \frac{1}{z_2} +\frac{1}{z_3}$$ then show that at least one of these numbers must be 1.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 2, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9286105632781982, "perplexity_flag": "head"}
http://math.stackexchange.com/questions/40106/push-forward-and-pullback-in-products	# Push forward and pullback in products I am reading this Questions about Serre duality, and there is one part in the answer that I'd like to know how it works. But after many tries I didn't get anywhere. So here is the problem. Let $X$ and $B$ be algebraic varieties over an algebraically closed field, $\pi_1$ and $\pi_2$ be the projections from $X\times B$ onto $X$ and $B$, respectively. Then it was claimed that $R^q\pi_{2,} \pi_1^ \Omega_X^p \cong H^q(X, \Omega^p_X)\otimes \mathcal{O}_B$. I am guessing it works for any (quasi)coherent sheaf on $X$. Basically, I have two tools available, either Proposition III8.1 of Hartsshorne or going through the definition of the derived functors. Thank you. - ## 1 Answer Using flat base-change (Prop. III.9.3 of Hartshorne), one sees that $$R^1\pi_{2 } \pi_1^\Omega_X^p = \pi_1^* H^q(X,\Omega^p_X) = H^q(X,\Omega^p_X)\otimes \mathcal O_B.$$ - How come I didn't think about that? Thanks, Matt. I learned a lot from you on Math.SE. – Jiangwei Xue May 19 '11 at 17:23 @Jiangwei: Dear Jiangwei, You're welcome. Regards, – Matt E May 20 '11 at 3:51	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 9, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9470779299736023, "perplexity_flag": "middle"}
http://mathoverflow.net/questions/16487/using-weierstrasss-factorization-theorem/16493	## Using Weierstrass’s Factorization Theorem ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) I am trying to factorize $\sin(x)\over x$ which by Taylor series expansion and using the roots is $$a \cdot \left(1 - \frac{x}{\pi} \right) \left(1 + \frac{x}{\pi} \right) \left(1 - \frac{x}{2\pi} \right) \left(1 + \frac{x}{2\pi} \right) \left(1 - \frac{x}{3\pi} \right) \left(1 + \frac{x}{3\pi} \right) \cdots$$ Now I was told that this nasty factor $a$ conveniently becomes $1$ due to Weierstrass’s Factorization Theorem which is a transcendental generalization of the Fundamental Theorem of Algebra. My question Could you please show me how $a$ is being neutralized using this theorem? Or don't you even need this theorem to do so? - 4 Look at the limit of both sides as x tends to 0? – Kevin Buzzard Feb 26 2010 at 11:22 Ah, now I see - thank you! – vonjd Feb 26 2010 at 11:51 By the way, you might be interested to know that this is how Euler cracked the [Basel problem](en.wikipedia.org/wiki/…) – castal Feb 26 2010 at 16:52 Thank you, yes - ironically this was the place I was coming from - have a look at the discussion page below ;-) But now everything is clear and I clarified the Wikipedia article with some additional comments. – vonjd Feb 26 2010 at 16:55 But note that you should consider products of $\exp\left(\frac{x}{\pi n}\right) \left(1-\frac{x}{\pi n}\right)$ over all $n \ne 0$, which is necessary for an absolutely convergent infinite product allowing you rearrange the order of the terms. But then the $n$ and $-n$ terms, for $n>0$, combine together and the exponentials cancel! – Zen Harper Jul 23 2010 at 0:02 ## 2 Answers The value of this product for small x's is the product of $(1-x^2/(n \pi)^2)$ which, when you take logs (and due to the second power in x), behaves like the sum over n of $-x^2/(n\pi)^2$, which approaches 0 as x approaches 0. - This is already very helpful - thank you. Just for clarification: shouldn't it be "1 - sum over n of $-x^2/(n\pi)^2$"? And why are you taking logs? And even if you do the difference between $log(1-x^2)$ and $log(-x^2)$ is not neglectable (esp. not for small x because of $x^2$). Or what am I confusing here? Thank you again! – vonjd Feb 26 2010 at 10:51 And another thing: the value of this product is the product $(1-x^2/(n \pi)^2$ - this holds true not only for small x's but for all x, doesn't it? And what I still don't understand is how the $a$ becomes $1$... Thank you again! – vonjd Feb 26 2010 at 11:42 1 The point is that both $(\sin x)/x$ and the product approach 1 as $x\to0$, and this serves to determine the value of $a$. And the point of the logs is that the theory of infinite products (en.wikipedia.org/wiki/Infinite_product) is mostly equivalent to that of infinite series, the correspondence coming from taking logs. Furthermore, $\log(1+u)\sim u$ for small $u$; apply to $u=-x^2/(n\pi)^2$. – Harald Hanche-Olsen Feb 26 2010 at 12:42 @vonjd: I hope Herald clarified it – David Lehavi Feb 26 2010 at 14:16 ...although the proper setting is to consider complex x, for which logarithms are considerably more subtle than for real x. I regard taking logs in this case as more of a heuristic method than a rigorous proof (it takes a lot of effort to make it rigorous); the "proper" proof is to develop the theory of infinite products directly, similarly to the theory of infinite sums. – Zen Harper Jul 22 2010 at 23:58 show 1 more comment ### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you. The Weierstrass factorization theorem as usually stated tells you only that $a=e^{g(x)}$ for some entire function $g(x)$. Hadamard's refinement says a little more, based on the growth rate of the function. In your case, since $\left\| \frac{\sin x}{x} \right\| < \exp\left(\|x\|^{1+o(1)} \right)$ as the complex number $x$ grows, Hadamard tells you that $g(x)$ is a polynomial of degree at most $1$. Since $\frac{\sin x}{x}$ and $\prod_{n=1}^{\infty} \left(1 - \frac{x^2}{n^2 \pi^2} \right)$ are both even functions, so is $e^{g(x)}$. Thus $g(-x)-g(x)$ is a (constant) integer multiple of $2\pi i$. Hence $g(x)$ is constant, and so is $a$. Finally, as everyone else has pointed out, taking the limit as $x$ goes to $0$ shows that $a=1$. See Ahlfors, Complex analysis for more about Hadamard's refinement, which relates the "order" and "genus" of an entire function. -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 40, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9444162845611572, "perplexity_flag": "middle"}
http://mathhelpforum.com/calculus/194674-limit-integral-print.html	# Limit of an integral Printable View • December 26th 2011, 03:19 AM wudup Limit of an integral Could someone please shed some light on how I might show that $\lim_{\epsilon\to 0}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}{\epsilon\over 2\pi [(x-x')^2+(y-y')^2+\epsilon^2]^{3\over2}}f(x,y)\;dx\,dy\,\,=f(x',y')$? I am not sure what conditions there is on $f(x,y)$, though I do know that $\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}{z\over 2\pi [(x-x')^2+(y-y')^2+z^2]^{3\over2}}f(x,y)\;dx\,dy$ is well-defined (converges) for all $x,y\in R$ and $z>0$. Brainstorm: Perhaps we can show that $\lim_{\epsilon\to 0}{\epsilon\over 2\pi [(x-x')^2+(y-y')^2+\epsilon^2]^{3\over2}}$ is the Delta function $\delta(x-x')\delta(y-y')$? Or maybe changing variables? Thanks. • December 27th 2011, 11:07 AM Opalg Re: Limit of an integral Quote: Originally Posted by wudup Could someone please shed some light on how I might show that $\lim_{\epsilon\to 0}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}{\epsilon\over 2\pi [(x-x')^2+(y-y')^2+\epsilon^2]^{3\over2}}f(x,y)\;dx\,dy\,\,=f(x',y')$? I am not sure what conditions there is on $f(x,y)$, though I do know that $\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}{z\over 2\pi [(x-x')^2+(y-y')^2+z^2]^{3\over2}}f(x,y)\;dx\,dy$ is well-defined (converges) for all $x,y\in R$ and $z>0$. Brainstorm: Perhaps we can show that $\lim_{\epsilon\to 0}{\epsilon\over 2\pi [(x-x')^2+(y-y')^2+\epsilon^2]^{3\over2}}$ is the Delta function $\delta(x-x')\delta(y-y')$? Or maybe changing variables? Thanks. For z>0, let $K_z(x,y) = \frac z{2\pi (x^2+y^2+z^2)^{3/2}}$, and check that it has the key properties of a kernel function (such as the Fejér kernel). These are: (1) $K_z(x,y)\geqslant0$ (that's obviously true); (2) $\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}K_z(x,y)\,dx\,dy = 1$ (easy integral, if you express it in polar coordinates); and (3) given $\varepsilon>0$ and $r>0$ there exists $z_0>0$ such that $K_z(x,y)<\varepsilon$ whenever $z<z_0$ and $x^2+y^2\geqslant r^2.$ (That last condition essentially says that when z is small, $K_z(x,y)$ is also small except when (x,y) is close to the origin.) Once you have checked that those three conditions hold, you can calculate as follows: $\begin{aligned}\biggl\| \int_{-\infty}^{\infty}\int_{-\infty}^{\infty} &K_z(x-x',y-y')f(x,y)\,dx\,dy - f(x',y') \biggr\| \\ &= \left\| \int_{-\infty}^{\infty}\int_{-\infty}^{\infty} K_z(x-x',y-y')\bigl(f(x,y)-f(x',y')\bigr)\,dx\,dy \right\| \qquad\text{(by (2))} \\ &\leqslant \int_{-\infty}^{\infty}\int_{-\infty}^{\infty}K_z(x-x',y-y')\left\|f(x,y)-f(x',y')\right\|\,dx\,dy .\end{aligned}$ You now want to show that for small enough z, that last integral can be made arbitrarily small. To do that, split the plane into two regions $R_1 = \{(x,y) : \|x-x'\|^2 + \|y-y'\|^2 <r^2\}$ and $R_2 = \{(x,y) : \|x-x'\|^2 + \|y-y'\|^2 \geqslant r^2\}$, where r is chosen so that $\|f(x,y)-f(x',y')\|$ is small whenever $(x,y)\in R_1$ (that will require assuming that f is continuous). Then the integral over $R_1$ will be small by (2), and the integral over $R_2$ will be small by (3) provided we assume that \|f\| is integrable over $\mathbb{R}^2$ and provided that z is small enoungh. All times are GMT -8. The time now is 05:16 PM.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 32, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9455105066299438, "perplexity_flag": "head"}
http://mathhelpforum.com/algebra/160721-strategies-solving-linear-equations-step-1-2-mixed-up.html	# Thread: 1. ## Strategies for Solving Linear Equations (are step 1 and 2 mixed up?) According to my book, and some sources I've found on the internet, the common strategy for solving a linear equation is: 1) If fractions are present, multiply each side by the LCD to eliminate them. If decimals are present, multiply each side by a power of 10 to eliminate them. 2) Use the distributive property to remove parentheses. 3) Combine any like terms. 4) Use the addition property of equality to get all variables on one side and numbers on the other. 5) Use the multiplication property of equality to get a single variable on one side. 6) Check by replacing the variable in the original equation with your solution. My problem is 1) and 2). I've learned the hard way that if you multiply each side by the LCD before distributing to any fractions within parentheses, you get the wrong answer. Attached is my wrong answer with work shown. To come up with the right answer, at step 1 I needed to distribute the 1/2 and 1/3 to the parenthesis first. Is there any reason to not do step 2 in the strategy as a general rule before step 1? Why would they be in that order if it sometimes does not work? Thank you, Chris 2. Your problem is when distributing the LCD. $6(2 + 4(y+2) \neq 12 + 8(6y + 12)$ $6(2 + 4(y+2) = 12 + 8(y + 2)$ It's like 6(3x) = 18x You don't put (18)(6x) = 72x 3. So don't distribute the LCD (or factor of 10 in the case of decimals) to inside the parentheses. But would it make a difference if I just started removing all parentheses as step 1 prior to dealing with fractions/decimals with LCD/factor of 10? 4. If you removed all the parenthesis before distributing the LCD, then you multiply each term, yes. 5. Thank you!	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 2, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9040272235870361, "perplexity_flag": "middle"}
http://mathoverflow.net/revisions/84809/list	## Return to Answer 2 added 20 characters in body If I correctly understand, you are misinterpreting the meaning of the product and sum of observables. When you say "We can now define a sum and a product of observables. These are obtained by performing the two measures and then adding or multiplying their values." This cannot possibly describe the usual sum A+B and product AB of operators. For the product, it is not even hermitian unless A and B commute. Agreed, A+B is hermitian, but the spectrum of A+B does not contain the result of the sum of a measurement of A followed by a measurement of B (in either way), again unless A and B commute. For a counter-example take A=[[1 0];[0 -1]] $A=\pmatrix{1& 0\cr 0&-1}$ and B=[[0 1];[1 0]].$B=\pmatrix{0&1\cr 1&0}$. I hope I correctly understood your question. 1 If I correctly understand, you are misinterpreting the meaning of the product and sum of observables. When you say "We can now define a sum and a product of observables. These are obtained by performing the two measures and then adding or multiplying their values." This cannot possibly describe the usual sum A+B and product AB of operators. For the product, it is not even hermitian unless A and B commute. Agreed, A+B is hermitian, but the spectrum of A+B does not contain the result of the sum of a measurement of A followed by a measurement of B (in either way), again unless A and B commute. For a counter-example take A=[[1 0];[0 -1]] and B=[[0 1];[1 0]]. I hope I correctly understood your question.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.896787166595459, "perplexity_flag": "middle"}
http://mathoverflow.net/questions/92750?sort=newest	## Polynomials for addition in the Witt vectors ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) The addition of `$p$`-typical Witt vectors (`$p$` a prime number) is given by universal polynomials `$S_n=S_n(X_0,\dots,X_n;Y_0,\dots,Y_n)\in\mathbb{Z}[X_0,X_1,\dots;Y_0,Y_1,\dots]$` determined by the equalities `$\Phi_n(S_0,\dots,S_n)=\Phi_n(X_0,\dots,X_n)+\Phi_n(Y_0,\dots,Y_n)$` for all `$n\ge 0$`, where `$\Phi_n(T_0,\dots,T_n)=(T_0)^{p^n}+p(T_1)^{p^{n-1}}+\dots+p^nT_n$`. I guess that whoever sees the Witt vectors for the first time writes down explicitly `$S_0=X_0+Y_0$`, `$S_1=X_1+Y_1+\frac{1}{p}((X_0)^p+(Y_0)^p-(X_0+Y_0)^p)$`, maybe `$S_2$` if she/he is courageous, and then stops since the computation becomes extremely messy. I think that there is no reasonable explicit expression in general, but patterns seem to exist and my question is about making these patterns more precise. Before I ask, let me illustrate with `$S_2$`. It is easy to see that there exists a unique sequence of polynomials `$R_n\in\mathbb{Z}[X,Y]$`, `$n\ge 0$`, such that `$X^{p^n}+Y^{p^n}=R_0(X,Y)^{p^n}+pR_1(X,Y)^{p^{n-1}}+\dots+p^nR_n(X,Y)$`. For example `$R_0=X+Y$` and `$R_1=\frac{1}{p}(X^p+Y^p-(X+Y)^p)$`. Then: `$S_1=R_0(X_1,Y_1)+R_1(X_0,Y_0)$` `$S_2=R_0(X_2,Y_2)+R_1(X_1,Y_1)+R_1(R_0(X_1,Y_1),R_1(X_0,Y_0))+R_2(X_0,Y_0)$`. Can someone make the shape of `$S_n$` more precise, e.g. in the form `$S_n=P_0+\dots+P_n$` presumably with `$P_0=R_0(X_n,Y_n)$`, `$P_n=R_n(X_0,Y_0)$`? The intermediary `$P_i$`'s are more complicated but should be (uniquely) determined by a condition of the type "`$P_i$` is an iterated composition of `$R_i$` involving only the variables `$X_0,\dots,X_i$`". Maybe the polynomials `$P_i$` should be homogeneous w.r.t. some graduation. Any hint or relevant reference will be appreciated. Thanks! - You mean you're asking for $S_i$ in terms of $R_0$, $R_1$, ..., $R_i$? – darij grinberg Mar 31 2012 at 16:41 The point is that I don't know what "in terms of" should mean: the shape and properties you conjecture for an expression of `$S_n$` alters your ability to prove that by induction on `$n$`. For example, it is not clear to me if just assuming that the property P(`$n$`) : `$S_n$` is a polynomial in the `$R_i(X_j,Y_j)$` holds is enough to prove that P(`$n+1$`) holds. – Matthieu Romagny Mar 31 2012 at 21:13 ps : I know the question is (unfortunately) a little vague; it is part of the question to make it less vague. The complexity of the polynomials `$S_n$` is the cause of the problem, and the interest of the question. – Matthieu Romagny Mar 31 2012 at 21:14 ## 1 Answer I found a formula for `$S_n$` in terms of the previous `$S_i$`'s and the polynomials `$R_i$`, which I'm quite happy about. In fact, I need the multivariate version of the `$R_i$`, which I will construct all at the same time. Let us consider the ring of formal power series in countably many variables `$X_1,X_2,\dots$` with integer coefficients, that is `$A=\mathbb{Z}[[X_1,X_2,\dots]]$`. Note that there are several notions of power series in infinitely many variables; mine is that of Bourbaki, where the underlying module of `$A$` is just the product of copies of `$\mathbb{Z}$` indexed by (finitely supported!) multiindices. (Some people require that the homogeneous components of a power series be polynomials; this is not the case here.) Then one can see that there exists a unique sequence `$(R_n)$` of elements of `$A$` such that for all $n\ge 0$ we have `$X_1^{p^n}+X_2^{p^n}+\dots=R_0^{p^n}+pR_1^{p^{n-1}}+\dots+p^nR_n$` where on the left is the sum of all `$p^n$`-th powers of the variables. This is a straightforward application of Bourbaki, Algèbre Commutative, Chapitre IX, `$\S~1$`, no 2, prop. 2, c) (phew! reference is finished) with the endomorphism `$\sigma:A\to A$` defined by `$\sigma(X_i)=X_i^p$` for all `$i$`. Now if we have finitely many (say `$s$`) variables, then we set `$R_n(X_1,\dots,X_s)=R_n(X_1,\dots,X_s,0,0,\dots)$`. Examples: `$R_0(X_1,\dots,X_s)=X_1+\dots+X_s$` and `$R_1(X_1,\dots,X_s)=\frac{X_1^p+\dots+X_s^p-(X_1+\dots+X_s)^p}{p}$`. Assuming that the `$R_n$` are computable, I have an inductive recipe for `$S_n$` which is interesting because it shows that all the `$p$`-adic congruences implying integrality of the `$S_n$` are contained in the `$R_n$`. The recipe goes like this. For each `$i$`, the polynomial `$S_i$` is a sum of `$2i$` terms (it will be obvious below what these terms are) and assuming `$S_1,\dots,S_{n-1}$` are known then `$S_n=R_0Z_n+R_1Z_2+\dots+R_nZ_0+R_1S_{n-1}+R_2S_{n-2}+\dots+R_{n-1}S_1$` where: `$Z_j$` is short for the pair of variables `$(X_j,Y_j)$`, `$R_iZ_j$` is short for `$R_i(X_j,Y_j)$` (the bivariate `$R_i$`) and `$R_iS_j$` is the (`$2j$`-variate) polynomial `$R_i$` evaluated at the `$2j$` terms of `$S_j$`. I hope the following examples make it clear what this means, and how efficient it is: `$S_1 = R_0 Z_1 + R_1 Z_0$` `$S_2=R_0 Z_2 + R_1 Z_1 + R_2 Z_0 + R_1 (R_0 Z_1 , R_1 Z_0 )$` ```$S_3 = R_0Z_3+R_1Z_2+R_2Z_1+R_3Z_0 \\ \quad + R_1(R_0Z_2,R_1Z_1,R_2Z_0,R_1(R_0Z_1,R_1Z_0))+ R_2(R_0Z_1,R_1Z_0)$``` ```$S_4 = R_0Z_4+R_1Z_3+R_2Z_2+R_3Z_1+R_4Z_0 \\ \quad + R_1(R_0Z_3,R_1Z_2,R_2Z_1,R_3Z_0,R_1(R_0Z_2,R_1Z_1,R_2Z_0,R_1(R_0Z_1,R_1Z_0)),R_2(R_0Z_1,R_1Z_0)) \\ \quad + R_2(R_0Z_2,R_1Z_1,R_2Z_0,R_1(R_0Z_1,R_1Z_0)) \\ \quad + R_3(R_0Z_1,R_1Z_0)$``` The proof that the recipe is correct is an exercise. -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 5, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9385820031166077, "perplexity_flag": "middle"}
http://physics.stackexchange.com/questions/4372/does-entropy-apply-to-newtons-first-law-or-does-acted-upon-always-require-an/4375	# Does entropy apply to Newton's First Law or does “acted upon” always require an external factor? First law: Every body remains in a state of rest or uniform motion (constant velocity) unless it is acted upon by an external unbalanced force. This means that in the absence of a non-zero net force, the center of mass of a body either remains at rest, or moves at a constant speed in a straight line. Doesn't the law of increasing entropy affect all objects though, since they are all in the closed system of the universe at large, and therefore they are all subject to slowing down, regardless of the containing medium, given enough time? I guess what I'm curious is, can there ever be a body that will remain at uniform motion or uniform rest given that entropy must increase? - ## 2 Answers Yes, it will (in the classical picture) continue forever, even with its entropy increasing, the entropy increase just means that some of the potential energy within the body will turn into heat (=kinetic energy), however the center of mass motion is unaffected. Entropy increase says nothing about slowing down, rather the opposite. - 1 @kakemonsteret, "the entropy increase just means that some of the potential energy within the body will turn into heat (=kinetic energy)" ??? totally wrong! – user1355 Feb 1 '11 at 16:27 @sb1 WHAT is wrong. – Holowitz Feb 1 '11 at 16:28 1 Yes it does, "Entropy is a thermodynamic property that is a measure of the energy not available for useful work in a thermodynamic process," -wik – Holowitz Feb 1 '11 at 16:31 1 @sb1 No, you are confused, ideal gases dont exist – Holowitz Feb 1 '11 at 16:43 1 How exactly do you want to increase the entropy of anything, without converting potential energy into kinetic ? That is impossible – Holowitz Feb 1 '11 at 16:52 show 15 more comments If you ignore the microscopic explanation of entropy, entropy is just an internal state of a system, on par with the system's volume, or the number of particles in the system. If you have a gas with a fixed entropy (let's store it in a vacuum tube so it doesn't escape, and let's give the tube infinite insulation so it doesn't leech out any heat) out in deep space, and you throw it, it will just happily trail off in a straight line forever with a constant entropy, volume, and number of particles. You get changes in entropy and whatnot only when the system in question interacts either with another system or its environment. But it really is best thinking of these things, at least on a macroscopic scale, as internal degrees of freedom of the system. - ??? Entropy of a closed system always increases – Holowitz Feb 1 '11 at 15:58 1 @kakemonsteret: No: the entropy of a closed system must always either increase or remain constant. The Second Law reads $\Delta S \geq 0$, not $\Delta S > 0$. In particular, a system in equilibrium maintains constant entropy. – Jerry Schirmer Feb 1 '11 at 16:11 1 barring fluctuations... – Marek Feb 1 '11 at 18:34 @Marek: sure, but at that point, the second law isn't being obeyed anyway. And which is also why I say to ignore the microscopic explanation of entropy. – Jerry Schirmer Feb 1 '11 at 21:36	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9260075688362122, "perplexity_flag": "middle"}
http://math.stackexchange.com/questions/174427/how-can-i-find-the-value-of-these-types-of-integrals?answertab=active	# How can I find the value of these types of integrals? Let $R=\{(x,y,z): -2\leqslant z\leqslant xy\;\mathrm{ and }\;x^2+y^2=1\}\subset\mathbf R^3$ and consider the vector field $\mathbf F(x,y,z)=-x\mathbf i+y\mathbf j+\exp(z^2)\mathbf k.$ I want to find $$\iint_R\mathrm{curl}\;\mathbf F\;\mathrm d S,$$ but I do not know how to choose a suitable parametrisation. How do you solve these type of integrals is general? - ## 1 Answer Hint: did you try calculating the curl of $\bf F$? - 2 I found curl $\mathbf F=0$, which means that the integral is zero. – zeke Jul 23 '12 at 22:56	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 4, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9136472344398499, "perplexity_flag": "head"}
http://www.all-science-fair-projects.com/science_fair_projects_encyclopedia/Gradient	# All Science Fair Projects ## Science Fair Project Encyclopedia for Schools! Search Browse Forum Coach Links Editor Help Tell-a-Friend Encyclopedia Dictionary # Science Fair Project Encyclopedia For information on any area of science that interests you, enter a keyword (eg. scientific method, molecule, cloud, carbohydrate etc.). Or else, you can start by choosing any of the categories below. # Gradient In vector calculus, the gradient of a scalar field is a vector field which points in the direction of the greatest rate of change of the scalar field, and whose magnitude is the greatest rate of change. In the above two images, the scalar field is in black and white, black representing higher values, and its corresponding gradient is represented by blue arrows. More rigorously, the gradient of a function from the Euclidean space Rn to R is the best linear approximation to that function at any particular point in Rn. To that extent, the gradient is a particular case of the Jacobian. Contents ## Examples • Consider a room in which the temperature is given by a scalar field φ, so at each point (x,y,z) the temperature is φ(x,y,z). We will assume that the temperature does not change in time. Then, at each point in the room, the gradient at that point will show the direction in which it gets hottest. The magnitude of the gradient will tell how fast it gets hot in that direction. • Consider a hill whose height at a point (x,y) is H(x,y). The gradient of H at a point will show the direction of the steepest slope at that point. The magnitude of the gradient will tell how steep the slope actually is. The gradient at a point is perpendicular to the level set going through that point, that is, to the curve of constant height at that point. ## Formal definition The gradient is noted by: $\nabla \phi$ where $\nabla$ (nabla) is the vector differential operator del, and φ is a scalar function. It is sometimes also written grad(φ). In 3 dimensions, the expression expands to $\nabla \phi = \begin{pmatrix} {\frac{\partial \phi}{\partial x}}, {\frac{\partial \phi}{\partial y}}, {\frac{\partial \phi}{\partial z}} \end{pmatrix}$ in Cartesian coordinates. (See partial derivative and vector.) ### Example For example, the gradient of the function φ = 2x + 3y2 - sin(z) is: $\nabla \phi = \begin{pmatrix} {\frac{\partial \phi}{\partial x}}, {\frac{\partial \phi}{\partial y}}, {\frac{\partial \phi}{\partial z}} \end{pmatrix} = \begin{pmatrix} {2}, {6y}, {-cos(z)} \end{pmatrix}.$ ## The gradient on manifolds For any differentiable function f on a manifold M, the gradient of f is the vector field such that for any vector ξ, $\langle \nabla f(x), \xi \rangle := \xi f$ where ξf is the function that takes any point p to the directional derivative of f in the direction ξ evaluated at p. In other words, under some coordinate chart$\varphi$, ξf(p) will be: $\sum \xi_{x_{j}} (\partial_{j}f \mid_{p}) := \sum \xi_{x_{j}} (\frac{\partial}{\partial x_{j} }(f \circ \varphi^{-1}) \mid_{\varphi(p)}).$ ## See also 03-10-2013 05:06:04	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 7, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.8465449810028076, "perplexity_flag": "middle"}
http://claesjohnsonmathscience.wordpress.com/2012/01/24/questioning-relativity-13-time-dilation/	# Questioning Relativity 13: Time Dilation January 24, 2012 · theory of relativity Authors Wikipedia informs us • In the theory of relativity, time dilation is an actual difference of elapsed time between two events as measured by observers either moving relative to each other or differently situated from gravitational masses. An accurate clock at rest with respect to one observer may be measured to tick at a different rate when compared to a second observer’s own equally accurate clocks. This effect arises neither from technical aspects of the clocks nor from the fact that signals need time to propagate, but from the nature of space-time itself. According to Einstein’s special theory of relativity space-time event observations in two space-time coordinate systems $S:(x,t)$ and $S^\prime :(x^\prime ,t^\prime )$ moving with constant velocity $v$ with respect to each other, are to be connected by the Lorentz transformation (normalizing the speed of light to 1): • $x^\prime = \gamma (x -vt)$, $t^\prime = \gamma (t -vx)$, $\gamma =\frac{1}{\sqrt{1-v^2}}$. The Lorentz transformation imposes time dilation with the apparent effect of slowing down the rate of time with the factor $\gamma$ in a moving system from the point of view of a stationary system, as illustrated in the above picture (no acceleration despite the jet engine). This effect is mutual and perceived the same way in both systems. The question of the physical reality of time dilation including the related twin paradox has never been answered by physicists, but was defined to be a non-question after Herbert Dingle was muted in 1960s. But the question remains: Is it true that moving clocks really slow down, or is it just an illusion? Is it like two people viewing each other at distance, both having the impression that the other is smaller, while they are equally tall in reality? They answer can only be: Yes, it is an illusion. A Stubbornly Persistent Illusion in the words of Einstein. Why? Because the rate of time is measured by clocks and clocks are mechanical devices, classical or quantum mechanical, and it is unthinkable that the rate of a clock can be affected by motion with constant velocity because mechanics is invariant under translation with constant velocity. According to Leibniz Principle of Sufficient Reason, no physical event can happen without a sufficient reason. But there is no reason whatsoever for the mechanics of a clock to be affected by translation with constant velocity. And so, without reason there can be no physical time dilation, only apparent fictitious illusion. For a more precise argumentation see Many-Minds Relativity. Recall that Lorentz and Poincare who introduced the Lorentz transformation before Einstein, did not consider the transformed time $t^\prime$ to be real physical time, only a an apparent time. It was Einstein who took the “bold step” to view the transformed time $t^\prime$ as real time, against the view of Lorentz in particular. Boldness is often presented as Einstein’s strongest virtue, but boldness may border to temerity. # 4 Comments 1. ### Michael January 25, 2012 Please, I am not very expert on this topic but, as far as I know, the speed prolongs the life of particles: when they are accelerated, their life grows many times, even thousands of times. Is it a physical or mathematical effect? • ### Richard T. Fowler January 25, 2012 Michael, http://claesjohnsonmathscience.wordpress.com/article/many-minds-relativity-yvfu3xg7d7wt-5/ which I quote here: ——– The big problem with Einstein´s special theory of relativity is that it is non-physical, which can be seen from the fact that the x’ – axis is not parallel to the x – axis, in contradiction to physical situtation considered, where X is an observer with an x-axis and X’ and observer with an x’ -axis moving along the x-axis with speed v. Einstein picked up the Lorentz transformation form the physicist H. A. Lorentz, who when introducing it carefully explained that the transformed time t´ – coordinate should not be interpreted as physical time . What Einstein did was precisely to do that! And to remove the aether completely. The Lorentz transformation contain effects of length contraction and time dilation, and the question from start was if these effects are real physical effects or just mathematical conventions/definitions/agreements. This was not made clear by Einstein, and nobody else, and has caused the confusion surrounding relativity theory still today, with clocks slowing down and space ships being compressed by motion with constant velocity contrary to all logic. ——– (Author’s emphases.) And I single out Claes’ next sentence with my own boldface emphasis: “It has gone so far that the lack of logic is used as an argument that the theory is correct.” I would have inserted the word “now” as in, “is now used”. Either way, the meaning is clear. RTF • ### Rickard Norlander January 28, 2012 Michael, if you are not an expert on this topic, I advise you to first learn the mainstream theory. Only then are you ready to read about alternative theories. # The Other Blogs by CJ The Secret of Flight (click image) Dr Faustus of Modern Physics Many-Minds Relativity Many-Minds Quantum Mechanics Mathematical Physics of Blackbody Radiation Computational Thermodynamics • 147,141 hits # Follow Blog via Email Join 21 other followers # Twingly Statistics %d bloggers like this:	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 9, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9192168116569519, "perplexity_flag": "middle"}
http://math.stackexchange.com/questions/59920/curious-properties-of-33?answertab=oldest	# Curious Properties of 33 Because my explanation has so many words, I'll start with my question and then you can read the explanation if you need to: The Bernstein Hash method uses the number 33 as a multiplier. From what I've read Bernstein himself has no reasonable explanation as to why 33 has such useful properties. I'm wondering if the mathematics community has any theories on the matter. ## The explanation: I'm a software engineer and I was recently working on a blog post about a hash function I was writing. In the process of designing my hash function, I looked at a lot of implementations of other hash functions. For non-programmers, the gist is that the hashcodes produced by objects should be well distributed throughout the range of values in a 32 bit integer (-2,147,483,648 through 2,147,483,647). Let's say I have the string "ABCD." A hash function would loop through each character, get the ASCII value of it, do something to it and aggregate it into a composite hashcode. For example, in a lot of implementations, they'd take a hashcode initialized to a large prime, multiply it by another prime and the XOR it with the value of 'A' which is 65. Then, they'd take that and multiply it by the same prime and XOR that with the value of 'B'. They'd do this until the end of the string is reached. I found a clever implementation in the Java framework code that loops over each item and effectively applies this: $i = ((i \ll 5) - i) \oplus j$. I was confused at first until I worked out that $(i \ll 5) - i = 31i$. For a computer, bit shifting is faster than multiplying so this is a clever way to multiply by a prime number. So, I looked in Microsoft's .Net framework and I found that they do it a little differently. They use $(i \ll 5) + i$ instead! I couldn't figure out for the life of me why they used 33 instead of 31 because from what I understand multiplying by prime numbers is the foundation of many hashing and cryptographic functions. I found out that this technique is called the Bernstein Hash and that Bernstein himself doesn't know why 33 produces such a good distribution of values as a multiplier in hashing functions. - I wish I could distribute credit among the several answers I've gotten. They've all produced interesting speculation and they seem to be in line with the research I did when first investigating this topic. One interesting thing to note, the hash functions which use 31 or 33 as factors aren't just for strings. Does that make the entropy with regard to ASCII characters less relevant? Further, both 31 and 33 as factors involve the same bitshift operation so I don't know that the "lowest 5 bits" explanation is fundamentally at play here. – D. Patrick Sep 2 '11 at 5:49 ## 3 Answers I've always assumed that the factor 33 (which amounts to a 5-bit shift plus a addition, as you note) was chosen because 5 bits is roughly the significant content of (ASCII) text, hence the "spreading" is rather optimal for typical textual keys. I doubt something more can be said from the strictly mathematical point of view. "from what I understand multiplying by prime numbers is the foundation of many hashing and cryptographic functions." That sounds rather vague to me: for example, for linear congruential generators, the multiplicator is frequently not prime. Empirically, it's seen that 31 or 33 make very little difference. Bear in mind that hashing functions usually relax the strict requirements of cryptographic functions, in favour of simplicity and performance. Some reasonable heuristics and experimental results are given here. See also this old discussion about the 33 vs 31 thing. - Note that the experimental results you link to use addition where the question has XOR. See my answer about why that might make a difference. – joriki Aug 28 '11 at 6:50 ## Did you find this question interesting? Try our newsletter email address I'm not really familiar with this particular hash function, but I decided to try going for the source, which in this case seems to be the Usenet newsgroup comp.lang.c. In this message, Phong Vo[drew] points out that one advantage of 33 over 31 is that it is congruent to 1 modulo 4, so that $k \mapsto 33*k + c \pmod{2^n}$, for a given $c \ne 0$, is a cyclic permutation on $\mathbb Z / 2^n \mathbb Z$. In particular, this means that, if you feed the hash function constant (non-zero) input, it behaves like a full-period LCRNG. Of course, this doesn't really prove anything; it's just suggestive. Also note that this explanation doesn't directly apply to the version using XOR instead of addition to mix in the data. (Both seem to have been recommended by Bernstein at various times.) In practice, of course, I wouldn't expect much difference between the XOR and addition variants anyway; they are very similar operations, and either should mix in the input just about as well. Like leonbloy, I also suspect that part of the reason why 33 (or 31) works so well in practice may be that the typical input to such functions is often ASCII text, which tends to have most of its entropy in the lowest five bits. Thus, shifting the hash value left by five bits per round should diffuse this entropy particularly efficiently across all its bits. Also, it's worth noting that this function was never intended to be a particularly good hash; it was designed to be a fast, yet still acceptably good, hash for applications where the speed of hashing strings is a limiting factor. - I think it should be "congruent to $1$ modulo $4$"? – joriki Aug 27 '11 at 9:26 @joriki: Oops, yes, thanks! Corrected. – Ilmari Karonen Aug 27 '11 at 9:35 It seems the experimental results that leonbloy linked don't bear out the advantage of the multiplier being congruent to 1 mod 4? (See the remark near the end of my answer about that and also about why there might be a difference between XOR and addition.) – joriki Aug 28 '11 at 6:49 Ilmari and leonbloy have already said quite a bit and provided pertinent links; here are some more thoughts: First, about the primes: the accepted answer to this question at stackoverflow sounds right to me. It's not that either the multiplier or the hash table size need to be primes; they should just be coprime. If you're writing the code for both, there's no reason for them to be prime, but if you're writing the code for only one of them (which often happens in practice), you might want to choose primes, or at least numbers with large prime factors, to reduce the chances of a "prime collision" between the multiplier and the hash table size. Second, about hashcodes being "well distributed" or "optimally spread": That's only part of the story. For long strings, where injectivity is not an option, random spread is good, but for short strings, injectivity is even better than random spread. Ideally, a multiplier-based hash function should have a random spread for initial inputs but be injective with respect to the last few inputs. ASCII codes for letters of the same case differ only in the low $5$ bits. So for strings ending in lowercase letters and $32$-bit hash values, you can get injectivity with respect to the last six characters if the multiplier is greater than $32$, whereas for multipliers less than $32$ the lowest bit of the penultimate input influences at most the last $5$ bits and thus its contribution isn't linearly independent from the contributions of all the other bits. In that case you're only using half the available hash values, and the values for the last six inputs form pairs that are mapped to the same hash values. If the multiplier is greater than $32$ and the multiplication doesn't overflow, injectivity with respect to the last $6$ lowercase characters is guaranteed. This is the case up to a multiplier of $47$ (since $2^4\cdot47^5\lesssim2^{32}$), but even beyond that, the contributions from different bits are usually linearly independent (there's a good chance for $30$ random vectors in $\mathbb F_2^{32}$ to be linearly independent); the first odd multiplier for which they aren't is $95$. So it's not immediately clear why $33$ should be better for ASCII strings than any other odd multiplier greater than $32$ with remainder $1$ mod $4$ (see Ilmari's answer). However, more generally speaking, the lower the multiplier, the more final inputs get mapped injectively before they start getting scrambled by the overflow. So it could well be that the $33$ is a good trade-off between getting the lower $5$ bits of ASCII characters out of each other's way (which requires $f\ge32$) and delaying the onset of scrambling by overflow (which requires low $f$). A note on the difference between the XOR variant and the one where $j$ is added instead: My considerations above about linear independence apply only to the XOR case, in which we can treat the hash value as a vector in $\mathbb F_2^{32}$ and the hash function as an affine transform on that space. In the case of addition, we don't need the multiplier to be greater than $32$ for the hash to be injective with respect to the last six inputs, just greater than the number of lowercase letters, which is $26$, so in that case there's no particular advantage to $33$. That's borne out by the experimental results that leonbloy linked to, where $31$ (labeled "K&R") even does slightly better on words than $33$ (labeled "Bernstein"); you can see near the top of the page or by following the Bernstein link that these results refer to the case with addition. (This is consistent with the idea that lower multipliers are generally better as long as they map the lowercase letters injectively, but it seems to cast some doubt on Ilmari's point about the remainder mod $4$?) Examples where the combination of $31$ and XOR fails to be injective already occur with $3$ letters; for instance, "rox" and "rng" are both mapped to `1A607`. This is all just about the advantages of a multiplier with respect to avoiding hash collisions. Often the speed of computing the hash function is at least as important as avoiding collisions, and in that respect $33$ has the advantage over most other multipliers (though not over $31$) that it can easily be computed with a shift and an addition, as you explained. -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 40, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9543663859367371, "perplexity_flag": "middle"}
http://www.contrib.andrew.cmu.edu/~ryanod/?page_id=162	## Analysis of Boolean Functions by Ryan O'Donnell Fall 2012 course at Carnegie Mellon # Acknowledgments I would like to thank the many people who have taught me Fourier analysis, most especially Guy Kindler and Elchanan Mossel. Thanks to Luca Trevisan and Radu Grigore for an initial version of LaTeX2WP. I am working on this project while at the Computer Science Department at Carnegie Mellon University. There I am supported in part by NSF CAREER grant CCF-0747250, NSF grant CCF-1116594, a Sloan Foundation research fellowship, and an Okawa Foundation research fellowship. I also spent time working on this project while visiting the Institute for Advanced Study as a von Neumann Fellow. There I was supported in part by the NSF grants DMS-0835373 and CCF-0832797. This material is based upon work supported by the National Science Foundation under grant numbers listed above. Any opinions, findings and conclusions or recommendations expressed in this material are those of the author and do not necessarily reflect the views of the National Science Foundation (NSF). ### Recent comments • Deepak: Cool! That all makes sense. • Ryan O'Donnell: Thanks on both! • Ryan O'Donnell: 1. Fixed, thanks 2. Fixed, thanks 3. For the first two occ... • Avishay Tal: Two small comments regarding the proof of theorem 16. - In ... • Deepak: A couple of small things: Ex. 42, under equation (2), \pi... • Ryan O'Donnell: Correct, thanks! • Tim Black: I think your $j$ should be an $i$ (or vice-versa) in the sen...	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.8744304180145264, "perplexity_flag": "middle"}
http://mathoverflow.net/questions/73743?sort=votes	## countable union of closed subschemes over uncountable field ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) I am looking for a reference for the following well-known fact: Let $k$ be an uncountable field, and let $X$ be a $k$-variety. Let $Z_1, Z_2, \dots \subseteq X$ be proper closed subschemes. Then $\bigcup Z_i(k) \neq X(k)$. Thanks! - 3 You also need $k$ to be algebraically closed (otherwise one could well have $X(k) = \emptyset$). – ulrich Aug 26 2011 at 9:12 I'm glad you asked the question. You'd think this would discussed in some standard textbook, but I've never seen it. Most complex algebraic geometers use a sledge hammer (Baire category theorem) but it's certainly more elementary than that. – Donu Arapura Aug 26 2011 at 12:19 It should be possible to give a bare hands proof that given a countable collection of nonzero polynomials $f_i$ in $n$-variables, there exist a point such that $f_i(p)\not=0$ simultaneously. – Donu Arapura Aug 26 2011 at 12:26 Yes, I left out algebraically closed as a hypothesis. It's frustrating that there (doesn't seem to be) a standard place to quote such a well-known fact. – Sue Sierra Aug 26 2011 at 18:06 As luck would have it, a related question just came up in my research today: does an algebraic torus over $k$ have a dense cyclic subgroup? If $k$ is uncountable and algebraically closed, the above-mentioned fact shows the answer is yes: just choose a generator outside the (countable) union of all proper Zariski closed subgroups. But if $k$ is the (countable) algebraic closure of a finite field, the answer is no, since every element of $k^\times$ has finite order. – Michael Thaddeus Aug 28 2011 at 3:59 ## 2 Answers Suppose $\dim X>0$ and $k$ is algebraically closed and uncountable. Moreover, if a "variety" is not necessarily irreducible, the $Z_i$ are supposed to have positive codimension in $X$ (otherwise one could take the irreducible components of $X$). As in MP's answer, one can suppose $X$ is affine. By Noether's Normalization Lemma, there exists a finite surjective morphism $p: X\to \mathbb A^m_k$ with $m=\dim X$. Let $Y_i=p(Z_i)$. This is a closed subset of $\mathbb A^m_k$ of positive codimension. Moreover $\mathbb A^m_k(k)=\cup Y_i(k)$ because $k$ is algebraically closed (which implies that $Y_i(k)=p(Z_i(k))$). As $k$ is uncountable, there exists a hyperplane $H$ in $\mathbb A^m$ not contained in any $Y_i$ (note that $H\subseteq Y_i$ is equivalent to $H=Y_i$). So by induction on $m$ we are reduced to the case $m=1$, and the assertion is obvious. Without the hypothesis $k$ algebraically closed, one can show similarly that $X\ne \cup_i Z_i$. This is Exercise 2.5.10 in my book. EDIT In fact this statement is trivial because the generic points of $X$ don't belong to any of the $Z_i$'s. But the proof shows that the set of closed points of $X$ is not contained in $\cup_i Z_i$. - ### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you. I do not know a reference, but the following short argument seems to work. Assume that the dimension of $X$ is at least 1! Argue by induction on the dimension of $X$. Reduce to the case in which the subschemes are irreducible of codimension one. Shrinking $X$ if necessary, reduce also to the case in which $X$ is quasiprojective. Let $L$ be a pencil of integral divisors on $X$. Since the ground-field is uncountable, the pencil $L$ contains uncountably many elements that are integral; let $D \in L$ be an integral element of $L$ that is different from each of the subschemes you want to avoid. By the inductive hypothesis, $D$ contains a point that is not contained in any of the subschemes and you are done. You can find this stated as a hint in an Exercise V.4.15 (c) in Hartshorne. - As ulrich's correctly comments above, the only place that I did not argue (namely, the base case of the induction) is also the place where there is a missing assumption: the field should be algebraically closed! The argument sketched here requires curves over $k$ to have uncountably many points. – MP Aug 26 2011 at 9:23	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 52, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9528523087501526, "perplexity_flag": "head"}
http://math.stackexchange.com/questions/211337/example-of-two-matrices-a-b-where-rankab-minranka-rankb	# Example of two matrices A, B where rank(AB) < min(rank(A), rank(B)) Assuming their product exists, I can prove that the rank of the product of two equal rank matrices is less than or equal to the rank of either of the initial matrices. However I'm struggling to find an example of two matrices whose product has a rank that is less than and NOT equal to the minimum of either. Is there a theoretical way to simplify my search for an explicit case of this? - ## 3 Answers This isn't the most general example of what you're talking about, but Nilpotent matrices can be of help. For example, take $$A=B= \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix} \\ AB= [0]_{2}$$ Then $rank(A)=rank(B)=1,$ but their product, the zero matrix, has zero rank. More generally, if $M$ is a nilpotent matrix with $M^k =0$, then $A=M^{k-1},B=M$ will satisfy your requirement. - $$\pmatrix{1&0\\0&0}\pmatrix{0&0\\0&1}=\pmatrix{0&0\\0&0}\;.$$ - Here's a hint for the easiest possible case: Take $A = B$; can you find a $2 \times 2$ matrix such that $A^2 = 0$? More generally, this is possible for any choices for the size of $A,B$. We know that $ABx = 0$ if $x$ is in the null space of $A$. So to make the the null space of $AB$ even larger, find $A$ so that the null space of $A$ shares vectors with the range of $B$. -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 15, "mathjax_display_tex": 2, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.92718505859375, "perplexity_flag": "head"}
http://nrich.maths.org/5	### Cereal Packets How can you put five cereal packets together to make different shapes if you must put them face-to-face? ### Little Boxes How many different cuboids can you make when you use four CDs or DVDs? How about using five, then six? ### Christmas Presents We need to wrap up this cube-shaped present, remembering that we can have no overlaps. What shapes can you find to use? # Multilink Cubes ##### Stage: 2 Challenge Level: Take $36$ cubes. How many different blocks can you make? For example $6$ by $6$ by $1$, or $3$ by $6$ by $2$, or $3$ by $3$ by $4$, or $2$ by $3$ by $6$, or $2$ by $2$ by $9$, or $2$ by $1$ by $18$. Just how many different ones can you find? You could also try growing different pyramids of cubes which in turn generate different sequences of number. Look at the pictures below this table and see if you can work out how the table has been filled in. • Can you continue the table? • What patterns do you notice? • Can you explain why these patterns occur? The NRICH Project aims to enrich the mathematical experiences of all learners. To support this aim, members of the NRICH team work in a wide range of capacities, including providing professional development for teachers wishing to embed rich mathematical tasks into everyday classroom practice. More information on many of our other activities can be found here.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 19, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9430958032608032, "perplexity_flag": "middle"}

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