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http://mathoverflow.net/questions/9786?sort=votes	## Use of the word “data” not in the statistical sense [closed] ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) Occasionally I see the use of the word "data" in definitions. For instance, one definition of an exact sequence starts off by saying, "An exact sequence of abelian groups (or modules or vector spaces) is given by the data of two homomorphisms [...]" (Perrin, Algebraic Geometry). I've heard this term used in class as well once. In these instances data of course does not refer to data as used in statistics, as in data from an experiment. What is the purpose of using such a strange word in abstract mathematics? Has anyone noticed this word? - 5 This is not a real question. – Harry Gindi Dec 26 2009 at 8:56 I suppose it's not mathematical; nevertheless I do think it's important to be careful in mathematical exposition to make the text as clear and beautiful as possible. I'm not saying that the use of the word "data" is overly confusing, but I think the way definitions are written can subtly influence the understanding of the reader, and that we ought to pay attention in formulating them. – Jason Polak Dec 26 2009 at 15:42 2 I do not find this "strange" at all. – Gerald Edgar Dec 26 2009 at 18:31 1 At best, this is community wiki. – Harry Gindi Dec 26 2009 at 19:54 2 I'm closing this as "not a real question," but it could also be closed as "subjective and argumentative." Latching on to some technical definition of a common word and objecting to any other use is just silly, like objecting to somebody saying "mosquitoes are vectors for disease" or insisting that "my answer is technically more informative then yours because it uses more characters, so therefore carries more information." – Anton Geraschenko♦ Dec 26 2009 at 23:29 show 2 more comments ## 2 Answers This use seems in line -- although perhaps not identical -- with the following dictionary definition: Data: 1. Factual information, especially information organized for analysis or used to reason or make decisions. 2. Computer Science Numerical or other information represented in a form suitable for processing by computer. 3. Values derived from scientific experiments. It is often used in mathematics in the way you have identified above. Namely, when defining a mathematical structure, it gives the reader a heads up as to the fact that that the structure is "multi-sorted" and involves more than one object. In more formal language, one might say "tuple", e.g., "A topological group is a triple $(G,m,\tau)$, where G is a set, $m: G \times G \rightarrow G$ is a binary operation, and $\tau$ is a family of subsets of $G$, such that...." One could also have said "A topological group is given by the data G,m,$\tau$..." - This is sort of implicit in your answer, but perhaps it is worth pointing out that data is the plural of the latin word datum (something given). The word datum is often used for a single unit of information and so using the word data really does mean that several units of information need to be given in order to specify e.g. a short exact sequence. – Dan Petersen Dec 26 2009 at 8:36 @DP: That's useful, thanks. – Pete L. Clark Dec 26 2009 at 8:52 Thank you for your answer. Usually when I read definitions the words are either fairly primitive, or else mathematical. Like "together with", "tuple", etc. it seems to me that data is slightly less primitive in a way and thus it was a bit unsettling to read. It seems unnecessary :) – Jason Polak Dec 26 2009 at 15:36 ### 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. In mathematics, the word "data" is often used for the most general mathematical noun. One may consider a set, a function, a category, a group, or seven groups "data." But for data to be interesting, it must have some sort of structure. Thus the data must satisfy certain properties. A set-with-binary-operation consists only of data: a set S and a function f: S x S ---> S. It becomes a semi-group when we add a requirement to this data: the function must be associative. In general, people speak of "data" and "structure." These are the raw materials and craftsmanship that create the mathematical universe. - Is this in line with the shift in recent mathematics from set-theoretic thinking to a more categorical thinking where the objects we study are not necessarily formulated in terms of some set with relations? It would be interesting to trace the appearance of this word. – Jason Polak Dec 26 2009 at 15:46	{"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": 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": 0.9387982487678528, "perplexity_flag": "middle"}
http://math.stackexchange.com/questions/126013/taxicab-distance-city-block-distance-manhattan-distance?answertab=oldest	# Taxicab distance , city block distance, Manhattan distance Is defined in Wikipedia http://en.wikipedia.org/wiki/Taxicab_geometry But I saw it as another way of calculating it for vectors and similar $$d = \mathrm{num}_1 + \mathrm{num}_2 - 2 \times \mathrm{intersect}$$ • $d$= distance • $\mathrm{num}_1$ = number of values in left vector • $\mathrm{num}_2$ = number of values in right vector • $\mathrm{intersect}$ = intersection number of overlapping values Is this right ? and how the equation becomes ? - What is the "number of values" in a vector? And what is the "intersection number of overlapping values"? – Chris Eagle Mar 29 '12 at 18:12 1 It is not clear what you do. I would like it if you work out an example using your formula here. So, we'll all understand what `num1`, `num2` and `intersect` mean. – user21436 Mar 29 '12 at 18:13 We can consider vector as a SET , so num1 = total numbers in the set1 , – tnaser Mar 29 '12 at 18:16 I not dealing with points , but numbers in a sets. I want to compare the sets to each others to find show similar they are – tnaser Mar 29 '12 at 18:21 1 @tnaser: You seem to be considering the size of the symmetric difference $A\Delta B$. You might want to look up Hamming distance, which is close in spirit, and has a taxicab distance character. And intersection is quite different from symmetric difference. – André Nicolas Mar 29 '12 at 18:36 show 3 more comments ## 1 Answer I think we've clarified that you are suggesting defining the distance between two finite sets $A$ and $B$ to be the number of elements of $A$, plus the number of elements of $B$, minus twice the number of elements in the symmetric difference of $A$ and $B$. But this would make lots of distances negative. E.g., if $A=\{{1,2,3\}}$ and $B=\{{4,5\}}$ then since $A$ and $B$ are disjoint their symmetric difference is their union and your formula gives $3+2-2\times5$ which is negative 5. If I've misinterpreted what you propose to do, I suggest that instead of adding to the trail of comments you edit your question so it asks what you really want to ask. - There was a mapping in the programming which made the confusion , question answered . thanks – tnaser Mar 30 '12 at 15:25	{"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": 16, "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": 0.9428130984306335, "perplexity_flag": "head"}
http://www.physicsforums.com/showthread.php?t=148460	Physics Forums ## Canonical transformation 1. The problem statement, all variables and given/known data Consider a canonical transformation with generating function $$F_2 (q,P) = qP + \epsilon G_2 (q,P)$$, where $$\epsilon$$ is a small parameter. Write down the explicit form of the transformation. Neglecting terms of order $$\epsilon^2$$ and higher,find a relation between this transformation and Hamilton's equations of motion, by setting $$G_2=H$$ (why is this allowed?) and $$\epsilon = dt$$. 2. The attempt at a solution I think the transformation equations are $$\delta p = P - p = -\epsilon \frac{\partial G_2}{\partial q}$$ and $$\delta q =Q-q=\epsilon \frac{\partial G_2}{\partial q}$$ Quote by vanesch I guess there's a typo here: $$\delta q =Q-q=\epsilon \frac{\partial G_2}{\partial P}$$ but how can I solve the last part? Can I just say that with the use of H and dt the equations can be written as $$\dot{p}=-\frac{\partial H}{\partial q}$$ and $$\dot{q}=\frac{\partial H}{\partial P}$$ which are the Hamiltonian equations of motion? And why is this allowed? Quote by vanesch The idea is that we work in first order in $$\epsilon$$, and that you can hence replace everywhere $$P$$ by $$p$$ as the difference will introduce only second-order errrors. 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 Recognitions: Gold Member Science Advisor Staff Emeritus Ooops ! I'm terribly sorry, instead of replying, I erroneously edited your post ! Yes it should be a P, not a q. I get the relations $$Q = q + \delta q$$ and $$P = p + \delta p$$ but why is it allowed to use G = H? Recognitions: Gold Member Science Advisor Staff Emeritus ## Canonical transformation Quote by Logarythmic Yes it should be a P, not a q. I get the relations $$Q = q + \delta q$$ and $$P = p + \delta p$$ but why is it allowed to use G = H? You have the choice ! Every function G(q,P) (which is smooth enough) will generate a canonical transformation. So you may just as well use H(q,P), and then - that's the whole point - the transformation equations from (q,p) into (Q,P) give you simply the genuine time evolution where epsilon is the small time step. For an arbitrary G that isn't the case of course, you've just transformed your coordinates (q,p) in some other (Q,P). But for G = H, you've transformed the coordinates (q,p) in what they will be, a small moment later ! The reason for that is that your transformation equations you've found for an arbitrary G are what they are, and become the Hamilton equations of motion when you pick G to be equal to H. Now, strictly speaking we should write H(q,P) instead of H(q,p), but we can replace the P by p here, because they are only a small amount different. and it also preserve the fundamental poisson brackets.... its just a quick calcolus {Q,P} Thread Tools \| \| \| \| \|-----------------------------------------------\|-------------------------------\|---------\| \| Similar Threads for: Canonical transformation \| \| \| \| Thread \| Forum \| Replies \| \| \| Advanced Physics Homework \| 1 \| \| \| Advanced Physics Homework \| 8 \| \| \| Advanced Physics Homework \| 2 \| \| \| Introductory Physics Homework \| 1 \| \| \| General Physics \| 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": 17, "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": 0.905113697052002, "perplexity_flag": "middle"}
http://math.stackexchange.com/questions/tagged/powers+inequality	# Tagged Questions 2answers 103 views ### pow$(X,Y)$ $>$ pow$(Y,X)$, if $X<Y$. How can we proof following? if $X < Y$, then: $X^{Y} > Y^{X}$ , Where X, and Y are integers. Also $X,Y > 1$. Except a special case $2^{3} < 3^{2}$. I think for other ... 1answer 63 views ### Summation of powers inequality Can anyone provide a slick proof of the following? Let $0 < x \le 1$. Then $\displaystyle \sum_{k=0}^{n-1} x^k \ge \frac {1} {1 - (1 - 1/n)x}$. 1answer 36 views ### Problem understanding a proof about powers in ordered fields I am reading through a textbook on Analysis and have come across a question that I can't seem to make any headway with. A proof is outlined, but I can't make any sense out of it. The problem is as ... 2answers 433 views ### The inequality $b^n - a^n < (b - a)nb^{n-1}$ I'm trying to figure out why $b^n - a^n < (b - a)nb^{n-1}$. Using just algebra, we can calculate $(b - a)(b^{n-1} + b^{n-2}a + \ldots + ba^{n-2} + a^{n-1})$ \$ = (b^n + b^{n-1}a + \ldots + ... 1answer 499 views ### Prove variant of triangle inequality containing p-th power for 0 < p < 1 Sorry if this is a trivial question, but I am kind of stuck with proving the following inequality and have been searching for a while: \$\rho \left( \sum\limits_i^n d_i \right) \leq \sum\limits_i^n ... 1answer 237 views ### Comparing Powers of Different Bases How can I know if one power is bigger than the other when the bases are different? For example, considering $2^{10}$ and $10^{3}$ the former is the greater one, but how to prove this? Logarithms? ... 3answers 273 views ### Power function inequality Let $x$ and $p$ be real numbers with $x \ge 1$ and $p \ge 2$ . Show that $(x - 1)(x + 1)^{p - 1} \ge x^p - 1$ . I recently discovered this result. I am sure it is known, but it is new to me. It is ... 6answers 831 views ### Simple proof that $8\left(\frac{9}{10}\right)^8 > 1$ This question is motivated by a step in the proof given here. \$\begin{align*} 8^{n+1}-1&\gt 8(8^n-1)\gt 8n^8\\ &=(n+1)^8\left(8\left(\frac{n}{n+1}\right)^8\right)\\ &\geq ...	{"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": 21, "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": 0.9176872968673706, "perplexity_flag": "middle"}
http://math.stackexchange.com/questions/178214/a-group-of-order-108-has-a-proper-normal-subgroup-of-order-geq-6	# A group of order $108$ has a proper normal subgroup of order $\geq 6$. Problem: Let $G$ be a group of order $108 = 2^23^3$. Prove that $G$ has a proper normal subgroup of order $n \geq 6$. My attempt: From the Sylow theorems, if $n_3$ and $n_2$ denote the number of subgroups of order $27$ and $4$, respectively, in $G$, then $n_3 = 1$ or $4$, since $n_3 \equiv 1$ (mod $3$) and $n_3~\|~2^2$, and $n_2 = 1, 3, 9$ or $27$, because $n_2~\|~3^3$. Now, I don't know what else to do. I tried assuming $n_3 = 4$ and seeing if this leads to a contradiction, but I'm not even sure that this can't happen. I'm allowed to use only the basic results of group theory (the Sylow theorems being the most sophisticated tools). Any ideas are welcome; thanks! - Are you allowed to use group actions? – Brian Aug 3 '12 at 0:04 @Brian if you mean things like the Class Equation, then yes. – student Aug 3 '12 at 0:30 ## 1 Answer Let $\,P\,$ be a Sylow 3-sbgp. of $\,G\,$ and let the group act on the left cosets of $\,P\,$ by left (or right) shift. This action determines a homomorphism of $\,G\,$ on $\,S_4\,$ whose kernel has to be non-trivial (why? Compare orders!) and either of order 27 or of order 9 (a subgroup of $\,P\,$ ,say) , so in any case the claim's proved. - I suppose you mean $P$ is a Sylow-3 subgroup of $G$, and the action of $G$ is on the (left or right) cosets of $P$. But the kernel of the action may be $P$ or a subgroup of $P$ with order 9. – Serkan Aug 3 '12 at 0:35 Yes thanks, it was a typo and has already been corrected: it is a Sylow 3 subgroup. Whether the kernel of the induced homom. $\,\phi:G\to S_4\,$ by the action is a subgroup of $\,P\,$ of order 9 or $\,P\,$ itself we're cool since this is a proper subgroup of order greater than 6... – DonAntonio Aug 3 '12 at 2:13 @DonAntonio thanks, I understand it now. Still, the action is on the left cosets of $P$, rather than $P$ itself, right? It also follows that this homomorphism $\phi : G \rightarrow S_4$ cannot be surjective, since we would then have $G/\ker(\phi) \simeq S_4$, which can't happen (again because of the orders). – student Aug 3 '12 at 16:54 Yes, you're right. Edited and thanks. – DonAntonio Aug 3 '12 at 22:08	{"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": 39, "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": 0.939243495464325, "perplexity_flag": "head"}
http://medlibrary.org/medwiki/Signal-to-noise_ratio	# Signal-to-noise ratio Welcome to MedLibrary.org. For best results, we recommend beginning with the navigation links at the top of the page, which can guide you through our collection of over 14,000 medication labels and package inserts. For additional information on other topics which are not covered by our database of medications, just enter your topic in the search box below: Signal-to-noise ratio (often abbreviated SNR or S/N) is a measure used in science and engineering that compares the level of a desired signal to the level of background noise. It is defined as the ratio of signal power to the noise power. A ratio higher than 1:1 indicates more signal than noise. While SNR is commonly quoted for electrical signals, it can be applied to any form of signal (such as isotope levels in an ice core or biochemical signaling between cells). The signal-to-noise ratio, the bandwidth, and the channel capacity of a communication channel are connected by the Shannon–Hartley theorem. Signal-to-noise ratio is sometimes used informally to refer to the ratio of useful information to false or irrelevant data in a conversation or exchange. For example, in online discussion forums and other online communities, off-topic posts and spam are regarded as "noise" that interferes with the "signal" of appropriate discussion. ## Definition[] Signal-to-noise ratio is defined as the power ratio between a signal (meaningful information) and the background noise (unwanted signal): $\mathrm{SNR} = \frac{P_\mathrm{signal}}{P_\mathrm{noise}},$ where P is average power. Both signal and noise power must be measured at the same or equivalent points in a system, and within the same system bandwidth. If the signal and the noise are measured across the same impedance, then the SNR can be obtained by calculating the square of the amplitude ratio: $\mathrm{SNR} = \frac{P_\mathrm{signal}}{P_\mathrm{noise}} = \left ( \frac{A_\mathrm{signal}}{A_\mathrm{noise} } \right )^2,$ where A is root mean square (RMS) amplitude (for example, RMS voltage). Because many signals have a very wide dynamic range, SNRs are often expressed using the logarithmic decibel scale. In decibels, the SNR is defined as $\mathrm{SNR_{dB}} = 10 \log_{10} \left ( \frac{P_\mathrm{signal}}{P_\mathrm{noise}} \right ) = {P_\mathrm{signal,dB} - P_\mathrm{noise,dB}},$ which may equivalently be written using amplitude ratios as $\mathrm{SNR_{dB}} = 10 \log_{10} \left ( \frac{A_\mathrm{signal}}{A_\mathrm{noise}} \right )^2 = 20 \log_{10} \left ( \frac{A_\mathrm{signal}}{A_\mathrm{noise}} \right ).$ The concepts of signal-to-noise ratio and dynamic range are closely related. Dynamic range measures the ratio between the strongest un-distorted signal on a channel and the minimum discernable signal, which for most purposes is the noise level. SNR measures the ratio between an arbitrary signal level (not necessarily the most powerful signal possible) and noise. Measuring signal-to-noise ratios requires the selection of a representative or reference signal. In audio engineering, the reference signal is usually a sine wave at a standardized nominal or alignment level, such as 1 kHz at +4 dBu (1.228 VRMS). SNR is usually taken to indicate an average signal-to-noise ratio, as it is possible that (near) instantaneous signal-to-noise ratios will be considerably different. The concept can be understood as normalizing the noise level to 1 (0 dB) and measuring how far the signal 'stands out'. [note 1] ## Difference from conventional power[] In Physics power (physics) of an ac signal is defined as $\mathrm{P} = V_\mathrm{rms}I_\mathrm{rms}$ $\mathrm{P}= \frac{V_\mathrm{rms}^{2}}{R} = I_\mathrm{rms}^{2} R$ But in Signal Processing and Communication we usually assume that $R=1 \Omega$ so that usually we don't include that resistance term while measuring power or energy of a signal. This usually causes some confusions among readers but the resistance term is not significant for operations performed in signal processing. Most of cases the power of a signal would be $\mathrm{P}= V_\mathrm{rms}^{2} = \frac{A^{2}}{2}$ where 'A' is the amplitude of the ac signal. In some places people just use $P=A^2$ as the constant term doesn't affect much during the calculations. ## Alternative definition[] An alternative definition of SNR is as the reciprocal of the coefficient of variation, i.e., the ratio of mean to standard deviation of a signal or measurement:[2][3] $\mathrm{SNR} = \frac{\mu}{\sigma}$ where $\mu$ is the signal mean or expected value and $\sigma$ is the standard deviation of the noise, or an estimate thereof.[note 2] Notice that such an alternative definition is only useful for variables that are always non-negative (such as photon counts and luminance). Thus it is commonly used in image processing,[4][5][6][7] where the SNR of an image is usually calculated as the ratio of the mean pixel value to the standard deviation of the pixel values over a given neighborhood. Sometimes SNR is defined as the square of the alternative definition above. The Rose criterion (named after Albert Rose) states that an SNR of at least 5 is needed to be able to distinguish image features at 100% certainty. An SNR less than 5 means less than 100% certainty in identifying image details.[3] Yet another alternative, very specific and distinct definition of SNR is employed to characterize sensitivity of imaging systems; see signal to noise ratio (imaging). Related measures are the "contrast ratio" and the "contrast-to-noise ratio". ## SNR for various modulation systems[] ### Amplitude modulation[] Channel signal-to-noise ratio is given by $\mathrm{(SNR)_{C,AM}} = \frac{A_c^2 (1 + k_a^2 P)} {2 W N_0}$ where W is the bandwidth and ka is modulation index Output signal-to-noise ratio (of AM receiver) is given by $\mathrm{(SNR)_{O,AM}} = \frac{A_c^2 k_a^2 P} {2 W N_0}$ ### Frequency modulation[] Channel signal-to-noise ratio is given by $\mathrm{(SNR)_{C,FM}} = \frac{A_c^2} {2 W N_0}$ Output signal-to-noise ratio is given by $\mathrm{(SNR)_{O,FM}} = \frac{A_c^2 k_f^2 P} {2 N_0 W^3}$ ## Improving SNR in practice[] Recording of the noise of a thermogravimetric analysis device that is poorly isolated from a mechanical point of view; the middle of the curve shows a lower noise, due to a lesser surrounding human activity at night. All real measurements are disturbed by noise. This includes electronic noise, but can also include external events that affect the measured phenomenon — wind, vibrations, gravitational attraction of the moon, variations of temperature, variations of humidity, etc., depending on what is measured and of the sensitivity of the device. It is often possible to reduce the noise by controlling the environment. Otherwise, when the characteristics of the noise are known and are different from the signals, it is possible to filter it or to process the signal. For example, it is sometimes possible to use a lock-in amplifier to modulate and confine the signal within a very narrow bandwidth and then filter the detected signal to the narrow band where it resides, thereby eliminating most of the broadband noise. When the signal is constant or periodic and the noise is random, it is possible to enhance the SNR by averaging the measurement. In this case the noise goes down as the square root of the number of averaged samples. ## Digital signals[] When a measurement is digitised, the number of bits used to represent the measurement determines the maximum possible signal-to-noise ratio. This is because the minimum possible noise level is the error caused by the quantization of the signal, sometimes called Quantization noise. This noise level is non-linear and signal-dependent; different calculations exist for different signal models. Quantization noise is modeled as an analog error signal summed with the signal before quantization ("additive noise"). This theoretical maximum SNR assumes a perfect input signal. If the input signal is already noisy (as is usually the case), the signal's noise may be larger than the quantization noise. Real analog-to-digital converters also have other sources of noise that further decrease the SNR compared to the theoretical maximum from the idealized quantization noise, including the intentional addition of dither. Although noise levels in a digital system can be expressed using SNR, it is more common to use Eb/No, the energy per bit per noise power spectral density. The modulation error ratio (MER) is a measure of the SNR in a digitally modulated signal. ### Fixed point[] See also: Fixed point arithmetic For n-bit integers with equal distance between quantization levels (uniform quantization) the dynamic range (DR) is also determined. Assuming a uniform distribution of input signal values, the quantization noise is a uniformly distributed random signal with a peak-to-peak amplitude of one quantization level, making the amplitude ratio 2n/1. The formula is then: $\mathrm{DR_{dB}} = \mathrm{SNR_{dB}} = 20 \log_{10}(2^n) \approx 6.02 \cdot n$ This relationship is the origin of statements like "16-bit audio has a dynamic range of 96 dB". Each extra quantization bit increases the dynamic range by roughly 6 dB. Assuming a full-scale sine wave signal (that is, the quantizer is designed such that it has the same minimum and maximum values as the input signal), the quantization noise approximates a sawtooth wave with peak-to-peak amplitude of one quantization level[8] and uniform distribution. In this case, the SNR is approximately $\mathrm{SNR_{dB}} \approx 20 \log_{10} (2^n \sqrt {3/2}) \approx 6.02 \cdot n + 1.761$ ### Floating point[] Floating-point numbers provide a way to trade off signal-to-noise ratio for an increase in dynamic range. For n bit floating-point numbers, with n-m bits in the mantissa and m bits in the exponent: $\mathrm{DR_{dB}} = 6.02 \cdot 2^m$ $\mathrm{SNR_{dB}} = 6.02 \cdot (n-m)$ Note that the dynamic range is much larger than fixed-point, but at a cost of a worse signal-to-noise ratio. This makes floating-point preferable in situations where the dynamic range is large or unpredictable. Fixed-point's simpler implementations can be used with no signal quality disadvantage in systems where dynamic range is less than 6.02m. The very large dynamic range of floating-point can be a disadvantage, since it requires more forethought in designing algorithms.[9] [note 3] [note 4] ## Optical SNR[] Optical signals have a carrier frequency that is much higher than the modulation frequency (about 200 THz and more). This way the noise covers a bandwidth that is much wider than the signal itself. The resulting signal influence relies mainly on the filtering of the noise. To describe the signal quality without taking the receiver into account, the optical SNR (OSNR) is used. The OSNR is the ratio between the signal power and the noise power in a given bandwidth. Most commonly a reference bandwidth of 0.1 nm is used. This bandwidth is independent of the modulation format, the frequency and the receiver. For instance an OSNR of 20dB/0.1 nm could be given, even the signal of 40 GBit DPSK would not fit in this bandwidth. OSNR is measured with an optical spectrum analyzer. ## Notes[] 1. The connection between optical power and voltage in an imaging system is linear. This usually means that the SNR of the electrical signal is calculated by the 10 log rule. With an interferometric system, however, where interest lies in the signal from one arm only, the field of the electromagnetic wave is proportional to the voltage (assuming that the intensity in the second, the reference arm is constant). Therefore the optical power of the measurement arm is directly proportional to the electrical power and electrical signals from optical interferometry are following the 20 log rule. 2. The exact methods may vary between fields. For example, if the signal data are known to be constant, then $\sigma$ can be calculated using the standard deviation of the signal. If the signal data are not constant, then $\sigma$ can be calculated from data where the signal is zero or relatively constant. 3. Maximum possible full scale signal can be charged as peak-to-peak or as RMS. Audio uses RMS, Video P-P, which gave +9 dB more SNR for video. ## References[] 1. Michael A. Choma, Marinko V. Sarunic, Changhuei Yang, Joseph A. Izatt. Sensitivity advantage of swept source and Fourier domain optical coherence tomography. Optics Express, 11(18). Sept 2003. 2. D. J. Schroeder (1999). Astronomical optics (2nd ed.). Academic Press. p. 433. ISBN 978-0-12-629810-9 [Amazon-US \| Amazon-UK]. 3. ^ a b Bushberg, J. T., et al., (2e). Philadelphia: Lippincott Williams & Wilkins, 2006, p. 280. 4. Rafael C. González, Richard Eugene Woods (2008). Digital image processing. Prentice Hall. p. 354. ISBN 0-13-168728-X [Amazon-US \| Amazon-UK]. 5. Tania Stathaki (2008). Image fusion: algorithms and applications. Academic Press. p. 471. ISBN 0-12-372529-1 [Amazon-US \| Amazon-UK]. 6. Jitendra R. Raol (2009). Multi-Sensor Data Fusion: Theory and Practice. CRC Press. ISBN 1-4398-0003-0 [Amazon-US \| Amazon-UK]. 7. John C. Russ (2007). The image processing handbook. CRC Press. ISBN 0-8493-7254-2 [Amazon-US \| Amazon-UK]. 8. Defining and Testing Dynamic Parameters in High-Speed ADCs — Maxim Integrated Products Application note 728 9. Fixed-Point vs. Floating-Point DSP for Superior Audio — Rane Corporation technical library ## [] Content in this section is authored by an open community of volunteers and is not produced by, reviewed by, or in any way affiliated with MedLibrary.org. Licensed under the Creative Commons Attribution-ShareAlike 3.0 Unported License, using material from the Wikipedia article on "Signal-to-noise ratio", available in its original form here: http://en.wikipedia.org/w/index.php?title=Signal-to-noise_ratio • ## Finding More You are currently browsing the the MedLibrary.org general encyclopedia supplement. To return to our medication library, please select from the menu above or use our search box at the top of the page. In addition to our search facility, alphabetical listings and a date list can help you find every medication in our library. • ## Questions or Comments? If you have a question or comment about material specifically within the site’s encyclopedia supplement, we encourage you to read and follow the original source URL given near the bottom of each article. You may also get in touch directly with the original material provider. • ## About This site is provided for educational and informational purposes only and is not intended as a substitute for the advice of a medical doctor, nurse, nurse practitioner or other qualified health professional.	{"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": 22, "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": 0.8957344889640808, "perplexity_flag": "middle"}
http://www.physicsforums.com/showthread.php?t=112649	Physics Forums perfect square hi I am wondering whether there is any systematic way to determine for what values of x this function is a perfect square : f(x) = square root ( x!+1) thanks for any advice. Roger 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 Recognitions: Homework Help Science Advisor Hmm, well, thinking about it for a moment it's the same as saying: $$x! + 1 = y^4 \quad \text{for some} \quad x \, , y \, \in \mathbb{N}$$ Clearly: $$y \neq i \quad \text{for} \quad i = 1, \ldots, x$$ Therefore we must have $$x! > x^4$$, this is satisfied for all $$x > 6$$ We can reduce the problem yet further to: $$x! = y^4 - 1$$ Or: $$x! = (y - 1)(y + 1)(y^2 + 1)$$ Given that x > 6, then there must me at least a factor of 24, therefore y must be odd. So letting y = 2m + 1, giving us: $$x! = 8m(1 + m)(1 + 2m + 2m^2)$$ Also if x > 6 then 5 must be a factor of this. Take this polynomial modulo 5 and see what you get Just stuck that together, but it seems to be right EDIT: Well none of my LaTeX seems to have come out, don't know why, if a mod could help I'd be greatly appreciative Recognitions: Homework Help Science Advisor Well assuming my LaTeX never comes through properly, here is the post again: Hmm, well, thinking about it for a moment it's the same as saying: x! + 1 = y4 For some x,y natural numbers Clearly: y =\= i For: i = 1, ..., x Therefore we must have x! > x4, this is satisfied for all x > 6 We can reduce the problem yet further to: x! = y4 - 1 Or: x! = (y - 1)(y + 1)(y2 + 1) Given that x > 6, then there must me at least a factor of 24, therefore y must be odd. So letting y = 2m + 1, giving us: x! = 8m(1 + m)(1 + 2m + 2m2) Also if x > 6 then 5 must be a factor of this. Take this polynomial modulo 5 and see what you get Just stuck that together, but it seems to be right perfect square I have difficulty reading what's been written, but the first of all it's not x!=(y^4)-1 its x!=(y^2)-1 Recognitions: Homework Help Science Advisor Quote by roger I have difficulty reading what's been written, but the first of all it's not x!=(y^4)-1 its x!=(y^2)-1 Well you said: f(x) = square root ( x!+1) is a perfect square, which is the same as: square root ( x!+1) = y^2 or: x! + 1 = y^4 Please try and be more clear next time. But the same principle still applies, try and prove that this never is the case (which I am guessing is so) To be honest, I dont get what your doing. It was my mistake, it should have been square not (^4) I've re read what you wrote but , I cannot understand it. Please explain to me what your doing ? Ok, I will write it out again to avoid confusion. x!+1=y^2 obviously x=4 gives a perfect square, and so is x=5. All I'm trying to find is an easy way to find which values of x give me a perfect number, without keeping substituting the values in, that's all. In other words, you are asking: $$\text{For what values of }n \in \mathbb{N}\;\text{is }\sqrt {n! + 1} \in \mathbb{N}\;?$$ Recognitions: Homework Help Science Advisor Yeah, I've just had a hack at my method and I don't seem to be coming up with anything, it seems mine was only appropriate for powers of 4 I did a quick little computation and got: 7! + 1 = 712 However I think you will be stumped to find anymore, look near the bottom of this page: http://mathworld.wolfram.com/FactorialSums.html Thread Tools \| \| \| \| \|-------------------------------------\|----------------------------------\|---------\| \| Similar Threads for: perfect square \| \| \| \| Thread \| Forum \| Replies \| \| \| Precalculus Mathematics Homework \| 3 \| \| \| Calculus & Beyond Homework \| 7 \| \| \| Calculus & Beyond Homework \| 11 \| \| \| Calculus & Beyond Homework \| 8 \| \| \| Linear & Abstract Algebra \| 2 \|	{"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": 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": false, "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": 0.9373201727867126, "perplexity_flag": "middle"}
http://www.physicsforums.com/showthread.php?p=2765255	Physics Forums Thread Closed Page 1 of 2 1 2 > ## Does this reasoning ever reach infinity? 0<1<2<3<4<5 Does this reasoning ever reach infinity? 0<1<2<3<4<5..... What does it mean? Thanks! 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 Blog Entries: 2 Quote by Hippasos Does this reasoning ever reach infinity? 0<1<2<3<4<5..... What does it mean? Thanks! I think it means that n<n+1 as n goes towards infinity. So the chain 0<1<2<3... 1. Never ends while reaching infinity? 2. Never ends and does not ever reach infinity? ## Does this reasoning ever reach infinity? 0<1<2<3<4<5 Quote by Hippasos So the chain 0<1<2<3... 1. Never ends while reaching infinity? 2. Never ends and does not ever reach infinity? We don't really talk about reaching infinity. Infinity is a concept, not a number; you can not perform most mathematical procedures on it. (For example you cannot claim that infinity/infinity = 1) We talk about numbers as they approach infinity. I guess that's tantamount to your #2. " DaveC426913;2765460... Infinity is a concept, not a number ..." That is interesting - so mathematics is not strictly about numbers but also concepts. 1. Mathematics without infinities i.e. concepts - not possible? 2. Mathematics does/does not require concepts to exist? 3. How do we know if a concept is a mathematical one? Quote by Hippasos " DaveC426913;2765460... Infinity is a concept, not a number ..." That is interesting - so mathematics is not strictly about numbers but also concepts. 1. Mathematics without infinities i.e. concepts - not possible? 2. Mathematics does/does not require concepts to exist? 3. How do we know if a concept is a mathematical one? There are myriad mathematical frameworks out there; arithmetic using the real numbers is merely one tiny set. http://en.wikipedia.org/wiki/Number Quote by Hippasos Does this reasoning ever reach infinity? 0<1<2<3<4<5..... What does it mean? Thanks! Technically, infinity is not a number The reasoning that n < n+1 is satisfied for all numbers, I think that's what it's trying to say What about transfinite arithmetic? Georg Cantor? Quote by Hippasos Does this reasoning ever reach infinity? 0<1<2<3<4<5..... What does it mean? Thanks! It means precisely the following: "n<n+1 for every natural number n (or n=0)" In fact, you could also see it as an infinite number of statements such as "5<6", "15<16" or "0<1" So in a way, a statement is being made which applies to an infinite number of objects, but it says nothing about anything ever reaching infinity (and try to think about what you would actually be saying if it was, you will probably notice that you don't know what you mean precisely by this statement). Recognitions: Gold Member Science Advisor Staff Emeritus Well, the be precise, the opening post is vague. e.g. what reasoning is the opening poster referring to? What does he mean by reach? And what exactly is the ellipsis covering up? In a mathematical document, the intent would usually be clear from the context. I'm drawing a blank when it comes to trying to think of someplace I might see it naturally occurring, however. Quote by Hippasos Does this reasoning ever reach infinity? 0<1<2<3<4<5..... What does it mean? Thanks! Specifically: x < y < z is a mild form of 'abuse of notation' as it's called. This is because it's neither (x < y) < z nor x < (y < z), specifically, it's x < y /\ y < z. Often transitive relationships are abused in that way, technically you can't do that like you can do x + y + z, which is both (x + y) + z and x + (y + z). I guess that what you mean here, 'is there a highest number'. More 'mathematically' said: 'Does there exist a natural number n such that for any natural number m. m < n, or that statement formally, as in, completely properly and mathematically written down: $$\exists n \forall m : n \in \mathbb{N} \land m \in \mathbb{M} \land m < n$$ And this formal sentence just happens to be false. So if I interpreted your question correctly, the answer is 'no'. To show why: The natural numbers are defined as a set of 'objects' such that every object n has an object in that set called successor(n), the reverse is not true, namely, there is one object which is not a successor of another, that object is called zero conventionally. So, by axiom, each natural has a successor, and by definition of '<', each natural is lower than its successor. Therefore there doesn't exist a natural which is higher than all other naturals, a natural is never higher than its successor. 'Infinity', as said before is best avoided, it's not an object in most contexts, and typically used in another form of 'abuse of notation', typically I'd recommend and use myself terms such as 'diverges' or 'grows unbounded' in place of 'goes to infinity'. Quote by Hurkyl e.g. what reasoning is the opening poster referring to? What does he mean by reach? And what exactly is the ellipsis covering up? That's what I was trying to suggest, although you worded it better. Specifically: x < y < z is a mild form of 'abuse of notation' as it's called. So let's agree that x < y < z to mean x < y /\ y < z, x < y < z < a to mean x < y /\ y < z /\ z < a ... where I am using the ... to mean something similar to the meaning of the ... that the OP was asking about in the first place Quote by Jamma So let's agree that x < y < z to mean x < y /\ y < z, x < y < z < a to mean x < y /\ y < z /\ z < a ... Well, I'm afraid it's not that simple because of a few reasons. 1: Context free grammar, if we define it like this then mathematical notation is no longer generated by a context free grammar and violates the basic rule that x # y $z is either (x # y)$ z or x # (y \$ z). Which it is can be inferred from the associativity and the precedences of operators. 2: Formalism, where 'meaning' is defined as simple manipulations of symbols, it gets a lot more complex to define the inference if it's not context free. 3: ambiguity, the point is the 'false' is strictly smaller than 'true'. Indeed, we call the mathematical relationships of disjunction and conjunction monotonous because in disjunction the result is always equal or greater than its operants, and in conjunction it's always equal or less than its operants. So a < b < c < d would technically mean (a < b) < (c < d). Essentially implying here in a system of binary logic that the the former subpart is false, and the latter is true. This may be useless in binary logic, but in modal logic or fuzzy logic this has more implications. I'm not saying it's not possible to make it rigorous, or even to just work with it under the assumption of understanding, I'm just pointing out that technically it's as unmathematical as saying such things as 'the infinitieth digit 0.999...' Recognitions: Gold Member Science Advisor Staff Emeritus I'm not saying it's not possible to make it rigorous, or even to just work with it under the assumption of understanding, I'm just pointing out that technically it's as unmathematical as saying such things as 'the infinitieth digit 0.999...' Just to be clear -- this comment is not aimed at finite strings of chained inequalities, but instead to the infinite string, right? Quote by Hurkyl Just to be clear -- this comment is not aimed at finite strings of chained inequalities, but instead to the infinite string, right? Since I have no idea what you're talking about, probably neither. Recognitions: Science Advisor 0 < 1 /\ 1 < 2 /\ 2 < 3 ... is a reasonable interpretation of 0<1<2<3..., but what does this mean? It is supposed to represent an infinite string of symbols, but that doesn't make much sense. What you want to say can be described by using quantifiers as such: $$\forall n \in \mathbb{Z}^{+} (n-1 < n)$$. Quote by Jarle 0 < 1 /\ 1 < 2 /\ 2 < 3 ... is a reasonable interpretation of 0<1<2<3..., but what does this mean? It is supposed to represent an infinite string of symbols, but that doesn't make much sense. What you want to say can be described by using quantifiers as such: $$\forall n \in \mathbb{Z}^{+} (n-1 < n)$$. It makes as much sense as things like N := {1,2,3 ...}. And it happens to be true. The expression 0 < 1 /\ 1 < 2 /\ 2 < 3 ... is simply true, why, because true is a value, and the limit of that expression converges on that value. More formally we could write: E(0) = 0 < 1 E(n) = N(n-1) /\ n < (n+1) And then trivially, we can see that: lim (n -> inf) E(n) = true, abbreviated as simply: lim (n -> inf) E(n). True and false are objects and values, functions defined on those values are things like '/\' or '->' which are often called connectives. Thread Closed Page 1 of 2 1 2 > Thread Tools	{"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": 1, "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": 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": 0.9467066526412964, "perplexity_flag": "middle"}
http://mathoverflow.net/questions/87860/example-of-a-compact-kahler-manifold-with-non-finitely-generated-canonical-ring	## Example of a compact Kähler manifold with non-finitely generated canonical ring? ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) A celebrated recent theorem of Birkar-Cascini-Hacon-McKernan and Siu says that the canonical ring $R(X)=\oplus_{m\geq 0}H^0(X,mK_X)$ of any smooth algebraic variety $X$ over $\mathbb{C}$ is a finitely generated $\mathbb{C}$-algebra. On the other hand P.M.H. Wilson (using a construction of Zariski) gave an example of a compact complex manifold $X$ with $R(X)$ not finitely generated. However his manifold $X$ is not Kähler. Does anyone know an example of a compact Kähler manifold $X$ with $R(X)$ not finitely generated? Or is this an open problem? - 5 As far as I know this is open and I haven't heard of anyone working on this. The not completely unrelated problem of invariance of plurigenera is expected to hold for Kahler manifolds, though. (I.e. Demailly expects it to hold.) – Gunnar Magnusson Feb 8 2012 at 7:54 1 Thanks for the comment. The invariance of plurigenera in the Kahler case is also stated explicitly as a conjecture by Siu here (Conjecture 0.4) math.harvard.edu/~siu/siu_reprints/… – YangMills Feb 8 2012 at 15:31	{"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": 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": 0.18, "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": 0.9195442795753479, "perplexity_flag": "head"}
http://mathhelpforum.com/math-topics/111630-physics-3-a.html	# Thread: 1. ## physics 3 1.A spring has a length of L cm when a force of W N is applied to stretch it. When the force is increased to 6W N, the length of the spring increases to 2L cm. Find the percentage increase in the length of the spring from its original length when a force of W N is applied to stretch it. 2. Originally Posted by daphnewoon 1.A spring has a length of L cm when a force of W N is applied to stretch it. When the force is increased to 6W N, the length of the spring increases to 2L cm. Find the percentage increase in the length of the spring from its original length when a force of W N is applied to stretch it. Let the unloaded length be $l_0$, then if $k$ is the spring constant when loaded with a force $w$ we have the length is: $l(m)=l_0+k w$ Now we have: $L=l_0+kW$ and: $2L=l_0+k(6W)$ Subtract the first from the second to get: $L=k(5W)$ Therefor a loading force of ... is needed to extend the spring by $L$. CB 3. Originally Posted by CaptainBlack Let the unloaded length be $l_0$, then if $k$ is the spring constant when loaded with a force $w$ we have the length is: $l(m)=l_0+k w$ Now we have: $L=l_0+kW$ and: $2L=l_0+k(6W)$ Subtract the first from the second to get: $L=k(5W)$ Therefor a loading force of ... is needed to extend the spring by $L$. CB You have sufficient information to work out the unloaded length (and the spring constant) in terms of $L$ and $W$. You will find that the initial length $l_0$ is just a multiple of $L$ form that you can work out the percentage extension when loaded with force $W$	{"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": 21, "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": 0.9119469523429871, "perplexity_flag": "head"}
http://physics.stackexchange.com/questions/13917/reflection-at-speed-of-light/13919	Reflection At Speed of Light I have looked online to no avail. There is two competing answers and I am curious to know which one is right. Someone asked me this question. If you are traveling at the speed of light can you see your reflection in a mirror in front of you? My answer to the question is no, I would figure that in order for that to happen the light reflecting off you that would appear in the mirror must travel faster than the speed of light to actually reach the mirror (which we all know is impossible). He says the answer is yes, that it is all relative to the current frame of reference. Can anyone validate the correct answer with possible references? - – Qmechanic♦ Aug 25 '12 at 16:24 3 Answers This question cannot really be answered because you cannot travel at the speed of light. See Accelerating particles to the speed of light If you were massless, you would always travel at the speed of light. However, in that case you would not perceive the passing of time. In relativity, the time that passes for an observer depends on the proper time. The proper time for a light-like trajectory is always zero, so photons themselves do not experience the passage of time. If you travel very near to the speed of light - perhaps 99.9% light speed relative to Earth, you would still be able to view yourself normally in a mirror you carried with you. That is ensured by the principle of relativity, which states that all physical processes work the same way at any constant speed. - So if I am traveling at 99.9999% the speed of light, I will see my reflection in a mirror exactly the same way as if I were traveling at say our current speed on Earth (assuming both speeds are constant)? – John V. Aug 24 '11 at 15:25 @John V. Yes, that's right. – Mark Eichenlaub Aug 24 '11 at 16:25 Thanks. Do you know any good articles I can read more about the last comment of your answer? – John V. Aug 24 '11 at 16:37 2 – Mark Eichenlaub Aug 24 '11 at 16:47 Your friend is correct that it depends on the reference frame (see previous answers). Both of these responses are correct, the only difference being that Mark assumes you and the mirror share a reference frame and are measuring your speed relative to another frame whereas Zassou assumes that you are measuring your speed relative to the mirror (and traveling toward it). With regard to Zassou's answer, it is true that Lorentz contraction occurs, but, since you are the observer, it would appear as though mirror were "contracted like a pancake" and that it appears to experience time more slowly relative to you. From your point of view, everything would appear normal except for the absurdly fast approaching mirror. It is also worth noting that if you assume the opposite direction of travel your image would be red-shifted. - You can not travel the speed of light, you can only get close to it and the closer you get the more relativistic effects you experience, but we can continue talking about the problem you posed in the context of approaching the speed of light. As you approach the speed of light, the mirror actually gets closer from your perspective and the speed of light both approaching you from the mirror and leaving you moving toward the mirror is $c$. However, you still perceive the mirror moving toward you. So if the mirror is moving toward you at $0.99 c$ and the light from it is coming at you at $c$, then the light from the surface of the mirror only departs at a relative speed of $0.01 c$, which for the context of this discussion we'll say is not very fast. From the mirror's perspective The mirror is a part of an inertial reference frame that "observes" (which is really just a formalism for relativity, different than seeing with photons obviously) you moving toward it at $0.99 c$ and let's also note that you are length-contracted like a pancake and experience time more slowly according to the mirror. Light comes from the spaceship (I presume) at $c$ and moves toward the spaceship at $c$. The light that the spaceship emits only moves $0.01 c$ faster than the speed of the spaceship. In both cases it is agreed that the light from the spaceship emitted at a time $t$ before collision with the mirror only hits the mirror a small amount of time before the spaceship itself slams into the mirror (specifically $0.01 c t$, and yes I know this isn't an objective time measure as I have used it here). In the limit of going exactly the speed of light, then of course, no light from the spaceship is able to reach the mirror before the collision. A reflection of the spaceship in-travel is impossible in this case. So what do you see? You see reflections in the mirror from before you started your speed-of-light trip. If we talk about going close to the speed of light, like $0.99 c$ then you see a highly blue-shifted version of yourself for a short period of time (relative to the duration of the trip) right before you crash into the mirror. - What speed will your reflection appear to be going? Does length compression / time dilation apply to it, or can it move faster than c because it's not a physical thing (like the dot of a laser projected onto the moon can move faster than c) – Random832 Aug 24 '11 at 18:54 @Random That's a very good question that I could make a complete separate answer for. From the "perspective" (which isn't what is "seen", but what is interpreted by rules of relativity) of the spaceship that moves at $0.99 c$ its reflection should be approaching at $1.98 c$, but what you said about this being "false" is correct. More interestingly, the appearance due to solely the light rays (although blue-shifted) would limit to infinity velocity as the spacecraft velocity approaches c. It only sees its reflection for a short time and the reflection travels the same distance in that time. – AlanSE Aug 24 '11 at 19:14 Thanks - on further thinking about it, I managed to justify it to myself - if there's an actual ship behind the mirror, which exactly 'mirrors' (no pun intended) the first ship's actions (relative to the mirror, and simultaneously from the point of view of an observer near the mirror), I couldn't think of any consistent solution other than the reflection always lines up with the other ship. I do have some questions about blueshifting, that I will need to think through some more. – Random832 Aug 24 '11 at 20:20	{"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": 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": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9613878130912781, "perplexity_flag": "head"}
http://en.wikipedia.org/wiki/Product_type	# Product type In programming languages and type theory, a product of types is another, compounded, type in a structure. The "operands" of the product are types, and the structure of a product type is determined by the fixed order of the operands in the product. An instance of a product type retains the fixed order, but otherwise may contain all possible instances of its primitive data types. The expression of an instance of a product type will be a tuple, and is called a "tuple type" of expression. A product of types is a direct product of two or more types. If there are only two component types, it can be called a "pair type". For example, if two component types A and B are the set of all possible values that type, the product type written A × B contains elements that are pairs (a,b), where "a" and "b" are instances of A and B respectively. In many languages, product types take the form of a record type, for which the components of a tuple can be accessed by label. In languages that have algebraic data types, as in most functional programming languages, algebraic data types with one constructor are isomorphic to a product type. In the Curry-Howard correspondence, product types are associated with logical conjunction (AND) in logic. The notion directly extends to the product of an arbitrary finite number of types (a n-ary product type), and in this case, it characterizes the expressions which behave as tuples of expressions of the corresponding types. A degenerated form of product type is the unit type: it is the product of no types. In call-by-value programming languages, a product type can be interpreted as a set of pairs whose first component is a value in the first type and whose second component is a value in the second type. In short, it is a cartesian product and it corresponds to a product in the category of types. Most functional programming languages have a primitive notion of product type. For instance, the product of type1, ..., typen is written type1 `` ... `` typen in ML and `(`type1`,`...`,`typen`)` in Haskell. In both these languages, tuples are written `(`v1`,`...`,`vn`)` and the components of a tuple are extracted by pattern-matching. Additionally, many functional programming languages provide more general algebraic data types, which extend both product and sum types. The brochure for the International System of Units starts out in section 1.1 saying "The value of a quantity is generally expressed as the product of a number and a unit", and also presents the unit product of a Newton and a meter with the product notation of mathematics: Newton meter (N m or N · m). $\Gamma\vdash x : \text{Int}$ This programming language theory or type theory-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": 0.9183139801025391, "perplexity_flag": "head"}
http://math.stackexchange.com/questions/13054/how-to-show-eee79-is-not-an-integer/14885	# How to show $e^{e^{e^{79}}}$ is not an integer In this question, I needed to assume in my answer that $e^{e^{e^{79}}}$ is not an integer. Is there some standard result in number theory that applies to situations like this? Much later addendum: when I asked this, I didn't expect it to be an open question. If it is open, if someone could give an answer to that effect, I would accept it simply to close off the question. If some author has stated something similar as an open question, that would be helpful for me to know; I'm not a number theorist and my knowledge of the field is not very deep. - 9 This is likely to be open, I think. – Qiaochu Yuan Dec 5 '10 at 8:21 8 If we don't even know if $\exp(e)$ is algebraic or not... – J. M. Dec 5 '10 at 9:41 2 @Qiaochu Yuan and J.M.: I wasn't aware of that. I knew (vaguely) that there are a lot of open problems in transcendence theory, but I was hoping that just the problem of being an integer was easier. – Carl Mummert Dec 5 '10 at 12:56 7 It's not particularly feasible; the number of decimal digits is approximately $\log_{10}$ of the number, which is much larger than $10^{80}$, which is supposed to be an estimate for the number of atoms in the observable universe. – Carl Mummert Dec 6 '10 at 23:03 5 @Carl This is a really nice question. I don't think it needs to be "closed". There are some really innocuous questions running around the site just because the person who asked them never accepted an answer. But I believe that this is a question that deserves to show up from time to time. Who knows, maybe at some point someone will have something important to say about it, if not to answer it for good =) – Adrián Barquero Mar 20 '11 at 16:29 show 8 more comments ## 3 Answers if $e^{e^{e^{79}}}$ is an integer then $e^{e^{e^{e^{79}}}}$ is not an integer (otherwise $e$ would be algebraic). Perhaps your arguments make sense with this number too. - This is a good point. Unfortunately, for my example I need to know for sure whether it's an integer or not; knowing that at least one of two numbers is not an integer is not as useful. – Carl Mummert Jun 16 '11 at 0:14 The paper Chuangxun Cheng, Brian Dietel, Mathilde Herblot, Jingjing Huang, Holly Krieger, Diego Marques, Jonathan Mason, Martin Mereb, and S. Robert Wilson, Some consequences of Schanuel’s conjecture, Journal of Number Theory 129 (2009) 1464–1467, shows that $e,e^e,e^{e^e},\dots$ is an algebraically independent set, on the assumption of Schanuel's Conjecture. Maybe a close reading of that paper will suggest a way of applying the result to the 79-question. - 2 Maybe writing to the authors and telling them about this question directly could prove useful. – Adrián Barquero Jun 16 '11 at 4:38 Perhaps you could do something with the taylor series? Would be pretty nasty but youd only have to worry about the first few terms. - 11 @remus: Because it is too big perhaps. If you get something useful from Maple or Mathematica, please share. – Jonas Meyer Dec 19 '10 at 23:04 7 @Jonas, @remus: Already $e^{e^{79}}$ has approximately as many digits as one cubic kilometre of air contains molecules. So it's already virtually impossible to say how many digits the number in question has. – Hendrik Vogt Dec 20 '10 at 8:57 33 @remus: "Quite clearly not an integer"? $10^{0.30102999566398119521373889472449...}$ is 2. – Aryabhata Dec 22 '10 at 1:30 12 The probability that the number is an integer is either 100% or 0% depending on whether it is or isn't. There's no random process here; the property is fixed ahead of time and you can't compare it to rolling a dice. – Michael Burge Mar 20 '11 at 20:09 4 @Michael Burge: In fact; everything is fixed and everything is unfixed; et c'est la méme. This is why probabilistic is confusing; you cannot assert that an integer; before you see it; is a real number, which is why it could be a likelihood. – awllower Apr 2 '11 at 9:44 show 8 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": 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": 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": 0.9434800744056702, "perplexity_flag": "middle"}
http://math.stackexchange.com/questions/160998/finding-a-solution-of-a-heat-equation	# finding a solution of a heat equation Let $f\in C^0(\mathbb R^n)\cap L^\infty(\mathbb R^n)$ and $\alpha\in\mathbb R$. How can you find a solution $u$ of $$\begin{cases}\frac{\partial u}{\partial t}(x,t)-\Delta u(x,t)&=&-\alpha\cdot u(x,t) &\text{ in }]0,\infty[\times\mathbb R^n\\ \space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space u&=&f&\text{ on }\mathbb R^n\end{cases}$$ ? - ## 1 Answer Construct the solution $u$ of the usual heat equation $$\frac{\partial}{\partial t} u = \Delta u, \quad \quad u(0,x) = f(x)$$ in your favorite manner. Now consider $w(t,x) = e^{-\alpha t}u(t,x)$. Then by the product rule $$\frac{\partial}{\partial t} w(t,x) = e^{-\alpha t}\frac{\partial}{\partial t}u(t,x) - \alpha e^{\alpha t}u(t,x)$$ $$= e^{-\alpha t}\Delta u(t,x) - \alpha w(t,x) = \Delta (e^{-\alpha t}u(t,x)) - \alpha w(t,x)$$ $$= \Delta w(t,x) - \alpha w(t,x)$$ and $$w(0,x) = e^{-\alpha \cdot 0}u(0,x) = u(0,x) = f(x)$$ so that $w$ is the desired solution. For the construction of solutions to the usual heat equation, it would be easier if you imposed some decay conditions on $f$, say, $f \in L^2$. In that case, one way of doing it is by using the Fourier transform. Consider the Laplacian $\Delta: H^2 \to L^2$. Conjugating by the Fourier Transform yields $$(F\Delta F^{-1})u(x) = -\\|x\\|^2u(x)$$ and thus in Fourier space, the heat equation reads $$\frac{\partial}{\partial t}\hat{u}(t) = -\\|\xi\\|^2\hat{u}(t)$$ which yields $$\hat{u}(t) = e^{-\\|\xi\\|^2t}\hat{u}(0) = e^{-\\|\xi\\|^2t}\hat{f}$$ Transforming back yields $$u(t) = (F^{-1}e^{-\\|\xi\\|^2t}F)f$$ -	{"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": 10, "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": 0.9061759114265442, "perplexity_flag": "head"}
http://mathoverflow.net/questions/90197?sort=newest	## Transpose of unbounded operators between Banach spaces. ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) Let $X$ and $Y$ be Banach spaces, and let $L : X \rightarrow Y$ be a unbounded operator with dense domain $\operatorname{dom}(L)$. We can then talk about the transposed operator $L' : \operatorname{dom}(L') \subset Y' \rightarrow X' : y' \rightarrow y'(T\cdot)$ whose domain is given by those functionals $y'$, such that the term $y'(T\cdot)$, initially defined on $\operatorname{dom}(L)$, has bounded extension to all of $X$. If $L$ is closed and densely defined, then it is standard to show that $L'$ is closed, too. But if what the density of the domain of transpose? The proof by Reed and Simons seems in the Hilbert space case seems to use specific Hilbert space techniques. Question: Suppose $L$ is a closed densely-defined operator between Banach spaces. Is it transpose a closed densely-defined operator, too? - ## 2 Answers The transpose is closed but it may not be densely defined. For more info see Sec. 2.6 of H. Brezis: Functional Analysis, Sobolev Spaces and Partial Differential Equations, Springer Verlag, 2011 Exercise 2.22 in this book describes a closed densely defined operator whose adjoint is not dense. Here it is. Consider the Banach space $E=\ell^1$ with dual $E^=\ell^\infty$. Consider the densely defined operator $$A: D(A)\subset E\to E,$$ $$D(A)=\bigl\lbrace\; (u_n)\in\ell^1;\;\; (nu_n)\in \ell^1 \;\bigr\rbrace, \;\; A(u_n)= (nu_n).$$ Then $A$ is closed, densely defined, $A^$ is closed, but $D(A^*)$ is not dense. - ### 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. In general, the transpose need not be densely defined. For an example see S. G. Krein, Linear equations in Banach spaces. Birkhäuser, Boston, 1982. -	{"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": 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": 0.9316056370735168, "perplexity_flag": "head"}
http://mathoverflow.net/questions/83965/interesting-applications-of-the-classical-stokes-theorem/84002	## Interesting Applications of the Classical Stokes Theorem? ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) When students learn multivariable calculus they're typically barraged with a collection of examples of the type "given surface X with boundary curve Y, evaluate the line integral of a vector field Y by evaluating the surface integral of the curl of the vector field over the surface X" or vice versa. The trouble is that the vector fields, curves and surfaces are pretty much arbitrary except for being chosen so that one or both of the integrals are computationally tractable. One more interesting application of the classical Stokes theorem is that it allows one to interpret the curl of a vector field as a measure of swirling about an axis. But aside from that I'm unable to recall any other applications which are especially surprising, deep or interesting. I would like use Stokes theorem show my multivariable calculus students something that they enjoyable. Any suggestions? - 3 I suggest making this community wiki... – Igor Rivin Dec 20 2011 at 21:05 Thanks Igor, done. – Jonah Sinick Dec 20 2011 at 21:43 stokes theorem implies that the "angle form" on a sphere is not exact, [i.e. that the de rham cohomology of a sphere is non zero]. Thus corollaries include: brouwer fixed point, fundamental theorem of algebra, and absence of never zero vector fields on S^2. I gave all these applications in my first class on stokes theorem, since I myself had previously no idea what the theorem was good for. I was unable to deduce the jordan curve theorem, but perhaps that too is feasible, in the smooth case. – roy smith Mar 6 at 19:36 ## 8 Answers If you don't mind specializing Stokes theorem to Green's theorem, then one of the most practical applications is computation of the area of a region by integrating around its contour. I am old enough to have used a planimeter, a delightful physical embodiment of Green's theorem: One can also derive an (otherwise non-obvious) formula for the area of a planar polygon via Green's theorem: $A = \frac{1}{2} \sum_{i=0}^{n-1} x_i y_{i+1} - x_{i+1} y_i$. Sorry—no swirling vector fields in these examples! - This is a wonderful example! – Nilima Nigam Dec 21 2011 at 3:24 1 The shoelace formula! :) – Gjergji Zaimi Dec 21 2011 at 4:25 1 I think that the formula is used in Xfig, to compute the area enclosed in a curve. – Denis Serre Dec 21 2011 at 6:40 ### 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. In the theory of electromagnetism, the classical Stokes Theorem moves between the differential and integral forms of two of Maxwell's four equations; see https://en.wikipedia.org/wiki/Stokes%27_theorem#In_electromagnetism for discussion. Note that the integral forms may be directly interpreted using classical physical intuition, while the differential forms give us differential equations that we might solve, so it is important that we can switch between them. ETA: I think that Wikipedia's discussion is a little vague, although possibly appropriate in that context. So here is more detail, looking at Faraday's Law. In terms of physically observable quantities, the law states that the rate of change of the magnetic flux through a stationary surface is proportional to the electromotive force around the boundary of the surface. The magnetic flux is the surface integral of the magnetic field $\vec H$, and the EMF is the line integral of the electric field $\vec E$, so we have $$\oint _ { \partial S } \vec E \cdot \mathrm d \vec r = - \frac { \mathrm d } { \mathrm d t } \iint _ S \vec H \cdot \mathrm d ^ 2 \vec A$$ using standard units and sign conventions. Applying the classical Stokes Theorem on the left and using that $S$ is stationary on the right, this becomes $$\iint _ S ( \nabla \times \vec E ) \cdot \mathrm d ^ 2 \vec A = - \iint _ S \frac { \partial \vec H } { \partial t } \cdot \mathrm d ^ 2 \vec A \text ;$$ since this holds for arbitrarily small surfaces, we conclude that $$\nabla \times \vec E = - \frac { \partial \vec H } { \partial t } \text ,$$ a differential equation. (The argument in reverse is even easier, since you don't have to worry about arbitrarily small surfaces.) - The proof of Brower's fixed point theorem using Stokes' theorem is a nice application I think. - One could argue that complex function theory (the fact that analytic functions integrate to zero around contours) is an application, and a nice one. - Sure, I even consider that to be the main application. But like Joseph O'Rourke's example this uses Stokes theorem restricted to closed curves and regions and vector fields on the plane. Also, hard to fit into a multivariable calculus class. – Jonah Sinick Dec 20 2011 at 23:42 A student may also learn about the content from Stokes theorem from instances where it failed to hold as expected. For example, one has to exercise care when trying to use the theorem on domains with holes. Turn this around: the failure of Stokes to hold as expected tells you about the cohomology of the domain. I think it is possible via concrete examples to illustrate this point in a multivariate calculus class without using the more technical phraseology. A similar discussion occured at http://mathoverflow.net/questions/57025/down-to-earth-uses-of-de-rham-cohomology-to-convince-a-wide-audience-of-its-usefu For a non-standard application of the failure of the Stokes theorem, there's the odd case of the Purcell Swimmer: http://iopscience.iop.org/1367-2630/10/6/063016/fulltext/ Rendering this accessible to a multivariable calculus class may take some work, depending on your students. - 3 There's a nice book called From calculus to cohomology by Ib Madsen and Jørgen Tornehave with exactly this viewpoint. This can possibly be a used as a supplement for a strong and interested audience. – Daniel Larsson Dec 21 2011 at 8:16 Daniel, thank you for the reference! – Nilima Nigam Dec 21 2011 at 16:10 A nice application in fluid mechanics is Kelvin's circulation theorem. You could also discuss how it fails to hold, if there are obstacles in the fluid flow. In the same spirit stokes theorem is applied in the canonical formalism of classical mechanics to find the poincare-cartan integral invariants. - You can tell your students that a clever use of Stokes theorem can give you a Fields medal. Indeed, the proof that the formality map given by M. Kontsevich is a $L_\infty$-morphism, is nothing else than Stokes theorem. A detailed account of this can be found in `Deformation quantization of Poisson manifolds. Lett. Math. Phys. 66 (2003), no. 3, 157–216` or with more details in `Déformation, quantification, théorie de Lie, 123–164, Panor. Synthèses, 20, Soc. Math. France, Paris, 2005` which is in English, contrary to its title. - I find interesting that divergence theorem (that is a corollary of the general Stokes theorem on differentiable varieties) in a vectorial form (one integral for any cathesian cohordinate) give a proof of the Archimedes' principle of buoyancy on the fluid . given a body immersed on a (incompressible) fluid let $S$ its surface and for each point $p\in S$ let $\overrightarrow{n}:=(n_x(p), n_y(p), n_z(p))$ the normal versor to $S$ in $p$. then the total force on the body is the (vector) integral $\int_S\mu\cdot (L-z(p))\cdot -\overrightarrow{n}\cdot dS=$ $-(\int_S\mu\cdot (l-z(p))\cdot n_x(p) \cdot dS, \int_S\mu\cdot (l-z(p))\cdot n_x(p) \cdot dS, \int_S\mu\cdot (l-z(p))\cdot n_z(p) \cdot dS) =$ $-\mu\cdot(\int_S\ (l-z(p))\cdot \overrightarrow{i} \circ \overrightarrow{n}\circ dS, \int_S (l-z(p))\cdot \overrightarrow{j} \circ \overrightarrow{n}\circ dS, \int_S (l-z(p))\cdot \overrightarrow{k} \circ \overrightarrow{n}\circ dS)$ (where $l$ is the fluid level, and $\mu$ its density, $\overrightarrow{i}, \overrightarrow{j}, \overrightarrow{k}$ the usual cartesian versors). then from the divergence theorem (on each of the three components) and from $\nabla\circ((l-z)\overrightarrow{i})=\partial/\partial x (l-z)=0,\ \nabla\circ((l-z)\overrightarrow{j})=\partial/\partial y (l-z)=0$ $\nabla\circ((l-z)\overrightarrow{k})=\partial/\partial z (l-z)=-1$ follow that: $\int_S\mu\cdot (L-z(p))\cdot -\overrightarrow{n}\cdot dS= \mu\cdot \int_VdV\cdot \overrightarrow{k} =\mu \|V\|\cdot \overrightarrow{k}$ (where $V$ in the volume space (internal) bounded by $S$ and $\|V\|$ its measure. -	{"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": 22, "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": 0.9305667877197266, "perplexity_flag": "head"}
http://math.stackexchange.com/questions/152703/godels-incompleteness-theorem-diagonal-lemma	# Gödel's Incompleteness Theorem - Diagonal Lemma In proving Gödel's incompleteness theorem, why does he needed the Diagonal Lemma or the Fixed Point Theorem for building a formula $\phi$ that spoke about itself? Can't this formula be built this way: Diagonal Function. Let $D(x, y)$ be the diagonal function, such that $D$ returns the result obtained by replacing the formula with Gödel number $y$ for all free occurrences of $a$ in the formula with Gödel number $x$. Example. Let $\psi(a)$ be a formula that affirms that some formula with the Gödel number of $a$ is closed (has no free variables) and $k$ be it's Gödel number. Using Diagonal Function to construct a formula, by applying $D(k, k) = j$, the Gödel number $j$ will already be the Gödel number of a formula affirming that the formula itself has no free variables. I mean, $\ulcorner \psi(\ulcorner \psi(a) \urcorner) \urcorner = j$, or $\ulcorner \psi(\overline{k}) \urcorner = j$. What i am missing here? - Are you referring to a particular exposition of Gödel's work? His own original article does not name any "Diagonal Lemma" or "Fixed Point Theorem" -- its lemmas and theorems are simply numbered. – Henning Makholm Jun 2 '12 at 15:01 You are right, i'm referring to Mendelson's (Introduction to Mathematical Logic) exposition on the subject. – felipegf Jun 2 '12 at 15:05 So you're using $D(u,u)$ for what Mendelson (fourth edition, section 3.5) just calls $D(u)$? – Henning Makholm Jun 2 '12 at 15:10 I didn't wanted to define $sub$, but i wanted everyone to see that in fact $D(x,y)$ just replaces all free occurrences of a free variable $a$ (in the $sub$ predicate the Gödel number of the free variable is passed as an argument) in the Gödel number of a formula $x$ for the formula (whose Gödel number is) $y$. In fact, $D(u, u)$ is equivalent to $D(u)$. – felipegf Jun 2 '12 at 15:18 ## 1 Answer If I understand your notation correctly, the formula (whose Gödel number is) $j$ does not assert that $j$ itself has no free variables, merely that $\psi(a)$ (which is different from $j$ itself) has no free variables. Which, incidentally, is false because it has $a$ as a free variable. - You are right, thank you very much. Do you have any reference on why the diagonal lemma is named after Cantor's diagonal argument? – felipegf Jun 4 '12 at 4:05	{"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": 0.3, "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": 0.9049021005630493, "perplexity_flag": "head"}
http://mathhelpforum.com/advanced-algebra/75293-find-subspaces-show-linear-transformation-permutes-subspace.html	Thread: 1. Find the subspaces and show a linear transformation permutes subspace Let $V = K^2$ be a vector space over a field finite field K, in which \|K\| = 4. Show that V has exactly five 1-dimensional subspaces, say $U_1,...,U_5$ and that a non-singular linear transformation of V permutes these five subspaces. 2. Also, the question goes on to "associate to $a$ the map $\sigma_a\$ in $S_5$ (where $S_5$ is the symmetric group of 5 elements under composition). This seems very unclear to me - what does it actually mean? I've got to show that the map $\sigma: a \rightarrow \sigma_a$ defines a monomorphism from SL(2,4) into $S_5$ which I think would be manageable provided I understood what was meant by the "corresponding permutation" in the first place!	{"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": 8, "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": 0.9529584646224976, "perplexity_flag": "head"}
http://mathhelpforum.com/pre-calculus/896-math-quiz-question-help.html	# Thread: 1. ## Math Quiz question help ok, since my teacher isnt doing a very good job at explaining this we need to know for quizzes, I was wondering if you guys can help me. here are some of the questions om stuck on 1.Find the limit if it exists Lim log (x-4) x->4 5 2.Find the derivatives g(x)=ax+b/cx+d, where a, b, c, and d are constants with ad - bc /= 0 3. h(x)= x - x/\3 thx 2. 1. $ \lim_{x \rightarrow 4} \log_5 (x-4) = - \infty $ You try to find log (0) which is $- \infty$ 2. $ g'(x) = \dfrac{ad - bc}{(cx + d)^2} $ It's a simple fraction either you use the formula for fractions with terms above and under the fraction, or you split it up into a multiplication between the upper terms and the rest of the fraction. And you calculate them separately. Both ways work rather good. Don't let you be confused by the a,b,c and d. Look at them if they were simple numbers. All you need to do is watch out for the x. 3. $ h'(x) = 1 -3x^2 $ It's a simple polynom. Derivate term by term. That's it. 3. thx for the help alot.	{"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": 4, "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": 0.9376875758171082, "perplexity_flag": "middle"}
http://mathoverflow.net/questions/124919/angles-of-a-triangle-in-hilbert-spaces-and-in-metric-spaces	## Angles of a triangle, in Hilbert spaces and in metric spaces ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) http://mathoverflow.net/questions/124899/solid-angles-of-a-tetrahedron Does the statement "the side opposite the largest angle has the largest length (and similarly for smallest angle)" hold in a) Hilbert spaces? b) metric spaces? (where angle is defined by the construction in: A new notion of angle between three points in a metric space by Andrea Mondino http://arxiv.org/abs/1302.0520), c) any other spaces where angles and lengths can be defined? d) in rational trigonometry the statement would be about spreads and quadrances in any field with any quadratic form, but how would you order the spreads and quadrances of a triangle in an arbitrary field? In b) we have metric but no algebra, and in d) we have algebra but no metric. - 5 In a Hilbert space, each triangle lies in a 2-dimensional subspace where lengths and angles are exactly as in Euclidean $\mathbb R^2$, so Euclidean facts about triangles remain true for triangles in Hilbert space. – Andreas Blass Mar 19 at 12:53	{"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": 1, "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": 0.8622756600379944, "perplexity_flag": "middle"}
http://cstheory.stackexchange.com/questions/16582/l-0-estimation-not-in-a-stream	# $L_0$ estimation, not in a stream What is the current state of the art for $L_0$ estimation when not in a stream? I assume that the total number of (not necessarily distinct) objects is very large so you can only see a sample of them. I found http://en.wikipedia.org/wiki/Good%E2%80%93Turing_frequency_estimation but that seems rather older than I was expecting. - 2 Good-Turing estimation does more though: it estimates the frequencies of items. It sounds like you only want the number of distinct items. – Suresh Venkat♦ Feb 21 at 20:06 @SureshVenkat Yes you are right. – motl7 Feb 21 at 23:06 ## 1 Answer For sampling-based algorithms, nearly tight upper and lower bounds are known for the query complexity of estimating the number of distinct elements in a sequence. Charikar, Chowdhuri, Motwani and Narsayya showed that multiplicative approximation of the number of distinct elements to within a factor $\alpha$ can be achieved with $q= O(n/\alpha^2)$ queries into the sequence. Here's the algorithm. Sample $q$ items uniformly at random from the sequence. Let $d$ be the number of distinct items in the sample, and let $f_1$ be the number of items that appear exactly once in the sample. Then, output $d + \sqrt{n/q} \cdot f_1$. The analysis is also very clean. Assume without loss of generality that each element in the sequence is an integer in $[k]$, where $k$ is the number of distinct elements. Let $p_i$ be the probability of selecting $i$ when sampling uniformly from the sequence. Then, the expected output of the estimator is: $$\mathbb{E}[d + \sqrt{n/q}\cdot f_1] = \sum_{i=1}^k (1 - (1-p_i)^q) + \sqrt{n/q} \cdot p_i q (1-p_i)^{q-1}$$ Now, the point is that for any valid $p_i$, the value inside the summation is between $\Omega(\sqrt{q/n})$ and $O(\sqrt{n/q})$. (This uses the fact that $1/n \leq p_i < 1$.) So, the expected output of the algorithm is within $O(\sqrt{n/q})$ of $k$. Charikar et al.'s estimator can also be used to get an additive approximation of the number of distinct elements. One can estimate the number of distinct elements upto $\pm \beta n$ by making $O((1-2\beta) n)$ samples. This was the state-of-the-art for a long while until STOC '11 when Greg and Paul Valiant showed that it's possible to reduce the sample complexity and time complexity to $O(\frac{n}{\beta^2 \log n})$. Their algorithm is based on a very general result that, given a probability distribution $D$, constructs a distribution $D'$ using $O(n/\log n)$ samples such that $D$ and $D'$ are close to each other in relative earthmover distance. So, one can make sure that the support size of $D'$ and the support size of $D$ are only an additive $\beta n$ away from each other. As for lower bounds, Charikar et al. showed that for $\alpha$-multiplicative approximation, $\Omega(n/\alpha^2)$ queries are needed. The proof stems from the observation that $\Omega(n/\alpha^2)$ queries are needed to distinguish $n$ identical items from the same sequence with $\alpha^2$ unique elements inserted in random locations. For additive approximation, however, this bound is sort of trivial; for constant $\beta$, this would only give a constant lower bound on the query complexity. Raskhodnikova, Ron, Shpilka, and Smith showed a nearly linear bound even for additive approximation. Specifically, they prove that estimating to within additive error $n/23$ requires $n/2^{\sqrt{\log n} \log \log n}$ queries. The Valiants in their above-cited paper showed a lower bound of $\Omega(n/\log n)$ for estimating to any additive error less than $n/4$, implying that their upper-bound is tight upto constant factors. - Thanks very much. I suppose if you are worried about worst case input distributions and guaranteed bounds that is what you get. What about MLEs or biased estimators or if you have some useful prior? – motl7 Feb 22 at 8:51 1 – arnab Feb 22 at 21:14 Thanks very much. – motl7 Feb 23 at 15:29	{"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": 39, "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": 0.9328336715698242, "perplexity_flag": "head"}
http://mathoverflow.net/questions/93958?sort=newest	## Simple Equivariant homology [no borel-Moore] ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) Hey. I'm working with Bredon's equivariant cohomology. At some point I need to compute the $4$th equivariant cohomology group of $S^1 \x D^3$ relatively to its boundary for the antipodal action of $\mathbb{Z}_2$. I found a paper that just use "equiviriant poincaré duality" to bring the problem to compute the $0$th equivariant homology which is said to be $\mathbb{Z}_2$. This sounds trivial, but Bredon does not define what equivariant homology is, just cohomology, hence he does not talk of Poincare duality either (At least not in the 30 first pages of his paper "Equivariant Cohomology Theories" I use.). I'm trying to search for a nice definition of equivariant homology, but every paper I find use concept I don't master well or at all (Vector bundles, groupoids, Borel-Moore Homology). I also tried to design my own definition of equivariant homology, simply letting $H_0$ be the quotient of equivariant chains $H_0 = C_0^{\mathbb Z_2}/\partial(C_1^{\mathbb Z_2})$ with $C_i^{\mathbb Z_2} = \lbrace c \in C_i\| (1+\mathbb Z) \curvearrowright c = c \rbrace$. But using the relative exact sequence of chain, one finds $C_0^{\mathbb Z_2}(S^1\times D^3, S^1 \times S^2)$ is a subgroup of $C_0(S^1\times D^3, S^1 \times S^2)$, which is $0$, so $H_0^{\mathbb Z_2}(S^1\times D^3, S^1 \times S^2) \approx 0$, not $\mathbb Z_2$ ! Many thanks for any help ! - 1 mathoverflow.net/questions/42548/… – stankewicz Apr 13 2012 at 15:15 Yeah, I've already found it. But I'm rather asking "what is equivariant homology ?". – laerne Apr 13 2012 at 16:33 ## 1 Answer I'm afraid this is not an easy subject to get into. There is no problem defining Bredon homology. Maybe first in print in a 1975 memoir of Soren Illman. A more recent summary is in my Equivariant homotopy and cohomology theory''. However, just that won't help you. Tautologically, Poincar\'e duality is about duality. A manifold M embedded in Euclidean space is (after adding a disjoint basepoint) Spanier-Whitehead dual to the Thom complex of its normal bundle. Equivariantly, you must start with an embedding of M into a representation, and then to understand Poincar\'e duality you must use $RO(G)$-graded Bredon homology and cohomology, which is only available for those coefficient systems that extend to Mackey functors. Even then, equivariant orientation theory is very subtle. There is a long paper by Costenoble, Waner, and myself that explains what is going on conceptually and geometrically. Unfortunately, unlike nonequivariantly, unless one soups up cohomology to deal with equivariant fundamental groupoids, orientations are not (as far as is known) definable in purely homological terms. This is a fascinating area, still in its infancy, but not an easy one. - Huuu... This is not encouraging, especially since I don't know what a Mackey functor and most of those concepts are... – laerne Apr 13 2012 at 16:31	{"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": 14, "mathjax_display_tex": 0, "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": 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": 0.938766360282898, "perplexity_flag": "middle"}
http://mathhelpforum.com/advanced-algebra/50009-need-help-permutation-proof.html	# Thread: 1. ## need help in this permutation proof Can anyone help me in this proof ? If A and B are in S_n, then $ABA^{-1}$ is the permutation that has the same cycle structure as B and that is obtained by applying A to the symbols in B. 2. Hello, Prove it when the length of A is 2 and the B is cyclic: A=(i j), B=(m_1 m_2... m_k). Consider several cases depending on whether i, j are included in B. For example, if i=m_n and j doesn't occur in B, ABA^{-1}=(m_1 m_2... m_{n-1} j m_{n+1} ... m_k). Bye. 3. Originally Posted by deniselim17 Can anyone help me in this proof ? If A and B are in S_n, then $ABA^{-1}$ is the permutation that has the same cycle structure as B and that is obtained by applying A to the symbols in B. Let $\sigma = (a_1, ... , a_k)$ (where $k\geq 1$). Let $\tau$ be a permutation. Prove that $\tau \sigma \tau^{-1} = (\tau(a_1), ... , \tau(a_k))$. Now let $\theta$ be any permuation. Then we can write $\theta = \sigma_1\sigma_2 ... \sigma_m$ where $\sigma_i$ are disjoint cycles. Therefore, $\tau \theta \tau^{-1} = (\tau \sigma_1 \tau^{-1})(\tau \sigma_2 \tau^{-1}) ... (\tau \sigma_m \tau^{-1})$. But each $\tau \sigma_i \tau$ has the same structure as $\sigma_i$ by above. Therefore, $\theta$ and $\tau \theta \tau^{-1}$ both have the same number of disjoint cycles in their cyclic decompositition. Therefore $\theta$ and $\tau \theta \tau^{-1}$ have the same structrure. #### Search Tags View Tag Cloud Copyright © 2005-2013 Math Help Forum. All rights reserved.	{"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": 16, "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": 0.9480597376823425, "perplexity_flag": "head"}
http://mathoverflow.net/questions/60655/subspaces-of-a-subfactor/60661	## Subspaces of a Subfactor ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) Is the following true? Let $\mathcal N \subset \mathcal M$ be a subfactor. There is a bijective correspondence between the ultraweakly closed subspaces of $\mathcal M$ that are bimodules over $\mathcal N'\cap \mathcal M$, and the ultraweakly closed subspaces of $\mathcal N$. If the statement is false, is there a simple way to modify it to make it true? I am particularly interested in the case that $\mathcal M$ is of type $\mathrm I$. If $\mathcal V \subseteq \mathcal N$ is ultraweakly closed, then $\mathcal V (\mathcal N' \cap \mathcal M)$ is a bimodule over $\mathcal N'\cap \mathcal M$. If the subfactor admits a conditional expectation, then this function is injective. (Edit. Jesse points out that the conditional expectation should be ultrweakly continuous.) Edit. Steven points out that the statement as written is trivially true by a counting argument. Of course, I'm asking about the function $\mathcal V \mapsto \mathcal V \mathcal (\mathcal N'\cap \mathcal M)$, or something similarly natural. He also notes that irreducible subfactors are a counterexample to the bijectivity of $\mathcal V \mapsto \mathcal V(\mathcal N' \cap\mathcal M )$ in general. This leaves a single concrete question: Let $\mathcal N \subseteq \mathcal B (\mathcal H)$ be a factor. Is the function $\mathcal V \mapsto \mathcal V \mathcal N'$ a bijection between the ultraweakly closed subspaces of $\mathcal N$ and the ultraweakly closed subspaces of $\mathcal B(\mathcal H)$ that are $\mathcal N'$ bimodules? - 2 What is $\mathcal V (\mathcal N' \cap \mathcal M)$? Is this the algebraic span of $\mathcal V$ and $(\mathcal N' \cap \mathcal M)$ or are you taking the closure in some topology? – Jesse Peterson Apr 5 2011 at 12:31 1 This is a nice question, Andre. Thanks for asking! If N is a type I factor, we should be able to decompose B(H) as N⊗N′. Of course N′ is also type I. We can get the identity as a strong-operator limit of finite rank operators in N′, should it be infinite dimensional. The compact operators feel like the only obstruction for doing what you want in this case. Your map should be bijective in the type I case... – Jon Bannon Apr 6 2011 at 1:40 1 Thanks, and why is the the function $\mathcal V \mapsto \mathcal V (\mathcal N' \cap \mathcal M)$ 1-1 if there is a conditional expectation from $\mathcal M$ to $\mathcal N$? For instance, what if $\mathcal N = R$ is the amenable $II_1$ factor and $\mathcal M = \mathcal B(L^2 R)$? If $\mathcal V$ is any irreducible subfactor in $R$ then doesn't it follow from von Neumann's bicommutant theorem that $\mathcal V ( R' \cap \mathcal B(L^2R) ) = \mathcal B(L^2R)$? – Jesse Peterson Apr 6 2011 at 6:03 1 @ Jesse: Let $E: \mathcal M \rightarrow \mathcal N$ be a conditional expectation. For all $m \in \mathcal N' \cap \mathcal M$ and $n \in \mathcal N$, $nE(m)= E(nm) = E(mn)= E(m)n$, so $E(m) \in \mathcal N' \cap \mathcal N = \mathbb C$. It follows that $E(\mathcal V(\mathcal N' \cap \mathcal M)) = \mathcal V E(\mathcal N'\cap \mathcal M) = \mathcal V$. Thus, the function $\mathcal V \mapsto \mathcal V (\mathcal N' \cap \mathcal M)$ is injective. I may be missing something. – Andre Apr 6 2011 at 20:51 2 A good reference might be Section IV.2 of Blackadar's "Operator algebras" (ams.org/mathscinet-getitem?mr=2188261). Of particular interest to you might be Theorem IV.2.2.3 which shows that if there is an ultraweakly continuous conditional expectation from $\mathcal B(\mathcal H)$ to $\mathcal N$, then $\mathcal N$ is type I. – Jesse Peterson Apr 7 2011 at 5:45 show 15 more comments ## 1 Answer I suppose the statement is true because the set of ultraweakly-closed subspaces of $N$ has the same cardinality as the set of $N^\prime\cap M$-bimodular subsets of $M$. You probably want an explicit description of the correspondence, preferably such that the $N^\prime\cap M$-bimodule that corresponds to $V\subset N$ is exactely $V(N^\prime\cap M)$. This is certainly not true in general. For an irreducible subfactor $N\subset M$, i.e. $N^\prime\cap M=\C$, the condition that $V\subset M$ is an $N^\prime\cap M$-bimodule is empty, while there are strictly more subsets of $M$. Of course, the bicommutant theorem shows that there are no irreducible subfactors of type I factors. I do not immediately see a counterexample in the type I case. - Thank you for the answer. Irreducible subfactors came up as a counterexample to one of my previous questions, but I managed to completely forget about them! Since I'm interested primarily in the type $\mathrm I$ case, I'd like to wait until that part of the question is resolved. – Andre Apr 5 2011 at 9:12	{"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": 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": 0.9187872409820557, "perplexity_flag": "head"}
http://mathoverflow.net/questions/37239/dense-cyclic-subgroup/37241	## Dense cyclic subgroup ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) Does anyone know a continuous group (not necessarily locally compact) with dense cyclic subgroup other than a torus? - Anything procyclic. Someone below already listed the $p$-adic integers, although I'm not sure why this example wasn't one in which you're interested. Another is the absolute Galois group of a finite field of size $q$, which is topologically generated ($=$ equal to the closure of the abstract subgroup generated) by the Frobenius automorphism $x\mapsto x^q$. This is just $\hat{\mathbb{Z}}$, the product over all primes $p$ of $\mathbb{Z}_p$. – Keenan Kidwell Aug 31 2010 at 12:11 Take your favorite group, and give it the indiscrete topology. Then the identity element comprises a dense cyclic subgroup. Presumably you don't want this. – Theo Johnson-Freyd Aug 31 2010 at 15:36 ## 4 Answers You already have some examples in the other answers. Groups which have a dense cyclic subgroup are called Monothetic groups. In the article "On monothetic groups" by P.R. Halmos and H. Samelson, you can find many of their properties, such as Every compact connected separable (abelian) group is monothetic. - ### 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. First, it is clear the group has to be abelian. Now, if you assume that $G$ is locally compact, then by the classification you can decompose $G$ as $G={\mathbb R}^n \times H$ where $H$ has a compact open subgroup. Clearly, there can be no ${\mathbb R}^n$ factor, so $G$ has a compact open subgroup. Now, suppose $G$ is itself compact and topologically generated by $g$. Then any character $\chi$ in the dual of $G$ vanishing on $g$ will be identically zero. So, the map $\chi \mapsto \chi (g)$ is injective, hence the dual is a subgroup of $U(1)$. Conversely, you can also see that if $\Gamma$ is a subgroup of $U(1)$ (considered with the discrete topology) then the dual of $\Gamma$ has a dense cyclic subgroup. By taking various subgroups you can, for instance, get the $p$-adic integers, or the n-torus. - How about the Bohr compactification of the infinite cyclic group? - How about the infinite cyclic group itself with the discrete topology? Or p-adic integers? - I care about is not a question of existence obvious examples like this. In some sense and for some reasons p-adic integers for me are not much better than infinite dimensional torus (IDT). It is easy to understand, that forgetting about topology, p-adic integers realized subgroup of IDT. Need something unlike torus. Sorry for my english. – Bad English Aug 31 2010 at 11:09	{"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": 24, "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": 0.9327758550643921, "perplexity_flag": "head"}
http://mathhelpforum.com/calculus/37236-stationary-points-revisited.html	# Thread: 1. ## stationary points revisited I have this function and I am looking for the stationary points: $<br /> f(x,y) = x^4 + 2x^2y^2 + y^4 + 2x^2 - 2y^2 + 6<br />$ I have the two partial derivatives: $<br /> 4x(x^2 + y^2 + 1) = 0, (E.1)<br />$ $<br /> 4y(y^2 + x^2 - 1) = 0, (E.2)<br />$ I have already found the stationary points: $(0,0), (0,1)$ and $(0,-1)$ For $(E.2)$ can we set: $<br /> y^2 + x^2 = 1<br />$ and thus have: $<br /> ( \sqrt{1/2}, \sqrt{1/2} )<br />$ or is this useless because it doesn't satisfy $(E.1)$ ? 2. Two things: 1) I'd have to agree that it is of no value because it does not satisfy (E.1) 2) Why didn't you wonder about the other three really obvious suspects (not to mention the infinitely many others) from that last errant exercise? 3. 1. Yeah i figured it was a dumb question, just making sure. 2. Huh? 4. Not a dumb question. It shows you were on the right track. $x^{2} + y^{2} = 1$ defines a circle. How many points are there on a circle? 5. oooohh... does that mean (0,0) isn't a solution? 6. or wait no. sorry. whoops. 7. now i am confused 8. Please be unconfused. You got it right in the first place. It was only talking about the circle that became confusing. Rule out the circle and you're done. Of course, (0,1) and (0,-1) were identified before you started playing with the circle, so keep those two.	{"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": 10, "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": 0.9730071425437927, "perplexity_flag": "middle"}
http://mathhelpforum.com/calculus/51598-lebesgue-integration.html	# Thread: 1. ## Lebesgue Integration Let $(S,\Sigma,\mu)$ be a measure space. Prove that if $f$ is a Lebesgue integrable function, then for each $\varepsilon > 0$ there exists a $\delta > 0$ such that $\int_A \mid f \mid \mathrm{d} \mu$ for each $A \in \Sigma$ with $\mu(A) < \delta$. 2. Originally Posted by syme.gabriel Let $(S,\Sigma,\mu)$ be a measure space. Prove that if $f$ is a Lebesgue integrable function, then for each $\varepsilon > 0$ there exists a $\delta > 0$ such that $\int_A \mid f \mid \mathrm{d} \mu \color{red}<\varepsilon$ for each $A \in \Sigma$ with $\mu(A) < \delta$. Suppose on the contrary that for some $\varepsilon>0$ there exists, for each n=1,2,3,..., $E_n\in\Sigma$ such that $\mu(E_n)<2^{-n}$ but $\int_{E_n}\|f\|\,d\mu\geqslant\varepsilon$. Let $E = \limsup_nE_n = \bigcap_nF_n$, where $F_n = \bigcup_{k=n}^\infty E_k$. Then $\mu(E)\leqslant\sum_{i=n}^\infty\mu(E_i)<2^{-n+1}$ for all n, and hence $\mu(E)=0$. On the other hand it follows from Dominated Convergence or some such theorem that $\int_E\|f\|\,d\mu = \lim_n\int_{F_n}\|f\|\,d\mu \geqslant\limsup_n\int_{E_n}\|f\|\,d\mu\geqslant\var epsilon$. That is a contradiction since clearly \|f\| cannot have a nonzero integral over a null set. The idea of this proof comes from the fact that if we define a new measure $\nu$ on $\Sigma$ by $\nu(A) = \int_A\|f\|\,d\mu\ (A\in \Sigma)$, then $\nu\ll\mu$. There is then a standard theorem (Theorem 30B in Halmos's Measure Theory) saying that absolutely continuous measures have this "continuity" property. I just took Halmos's proof and re-wrote it for this situation.	{"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": 27, "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": 0.9435908794403076, "perplexity_flag": "head"}
http://mathhelpforum.com/geometry/106639-left-right-coordinate-space.html	# Thread: 1. ## left or right in coordinate space Hi all. Im new to the forum, and came across it on the hunt for some maths brains better than mine. I'm a hobbyist programmer and have hit a barrier. Eternally grateful if you can help. In the diagram there are two points creating a line in a coordinate space. This is for Flash, so the origins of x and y are in the top left corner, but hopefully this shoudn't matter. When point B moves, I get a third point (C). I need a reliable method for determining whether point B has moved left or right of the original line. All I know are the coordinates for the two orginal points and the new position of A. I'm happy with finding distances between points, angles etc - it's the application in this context I'm finding tricky. If anyone has an idea of where might I start with this, I'd be grateful. Many thanks. Attached Thumbnails 2. You will need to know the coordinates of two points on the line separating "left" and "right". It appears that a and b are initially on that line so save those coordinates as, say $(x_1, y_1)$ and $(x_2, y_2)$. Then the equation of that line is $y= \frac{y_2- y_1}{x_2- x_1}(x- x_1)+ y_1$ or, equivalently, $y- \frac{y_2- y_1}{x_2- x_1}(x- x_1)- y_1= 0$. Points to the left of that line will have $y- \frac{y_2- y_1}{x_2- x_1}(x- x_1)- y_1< 0$, points to the right will have $y- \frac{y_2- y_1}{x_2- x_1}(x- x_1)- y_1> 0$. That is, to determine whether a point is left or right of that line, calculate $y- \frac{y_2- y_1}{x_2- x_1}(x- x_1)- y_1$ and see if it is positive or negative. 3. Wow, thank you for such a fast and detailed response. I'm going to plug this in as soon as possible. Cheers.	{"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": 7, "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": 0.9370242953300476, "perplexity_flag": "head"}
http://mathoverflow.net/questions/55014?sort=votes	some questions on Riemann surface Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) There are several puzzling questions on Riemann surface for me: Q.1 Definition of Riemann surface can be given in at least two ways: Def.1) it is a complex one dimensional manifold; Def.2) for each $a\in \mathbb{C}$, consider collection of germs at $a$, of analytic functions, and give a topology on it. Are these really equivalent definitions? or Def.2 is more general than Def.1? Q.2 When we say a group $G$ is an automorphism group of a compact Riemann surface, how is the action? (for ex. what is description of action of of PSL(2,7) on a genus 3 Riemann surface? In the book of Thomas Breuer, I couldn't see any description of action of a group on a Riemann surface; he has given computational methods to investigate groups.) Q.3 The automorphisms of a compact Riemann surface can always be lifted to universal cover? Q.4 If a group $G$ acts on a compact Riemann surface $X_g$, of genus $g$, then $X_g/G$ is also a compact Riemann surface of some genus $h$ and $g,h$ are related by Riemann-Hurwitz formula. Can anyone suggest some good reference for this relation? (here, I would like to see this Riemann Hurwitz relation topologically; many books describe it using algebraic geometry techniques). (I went through many books on Riemann surface for these questions; but not understood many things) - 2 I can recommend for all those questions the book "Lectures on Riemann surfaces" by Otto Forster. One doesn't have to know any algebraic geometry to understand the book. – Someone Feb 10 2011 at 13:20 Regarding Q4: You can find a purely topological derivation of Riemann-Hurwitz in paragraph 21 of Prasolov & Sossinskys "Knots, links, braids and 3-manifolds". – bavajee Feb 10 2011 at 14:13 Your definition 2 does not seem to define anything at all, since you didn't specify additional conditions on the topology. Do you plan to choose some set-theoretic bijection between the points $a \in \mathbb{C}$ and the points on your Riemann surface? – S. Carnahan♦ Feb 10 2011 at 15:54 The first one is explained pretty well in Weyl's "The Idea of a Riemann Surface" (though the terminology is somewhat old-fashioned). – arsmath Feb 11 2011 at 12:08 2 Answers Q1. There are two DIFFERENT notions of Riemann surface in the literature. a) One-dimensional complex analytic manifold (coming from the book of Weyl). b) Riemann surface "spread over the plane (or over the Riemann sphere)". Your second definition, the set of germs with an appropriate topology on it, formalizes this second notion. Older books seem to understand Riemann surfaces in the sense of the second definition. Sometimes a) was called an "abstract Riemann surface" in these books. For most mathematicians with modern training the "Riemann surface of log z" and the "Riemann surface of arccos z" are meaningless expressions because these are the same as the plane, in the sense of definition a). The formal relation between a) and b) is the following. "A Riemann surface spread over the plane" is a pair (S,f), where S is an abstract Riemann surface and f is a holomorphic function from S to C. (If f is meromorphic, we have a Riemann surface spread over the sphere.) Here is another way to say this. Let S be a Riemann surface in the sense a). It has a set of charts $\phi_j: U_j\to D_j$ from the elements of an open covering U to discs D in the plane. The correspoddence maps $\phi_k\circ\phi_j^{-1}$ on $D_j\cap D_k$ must be conformal. Now let us require that these correspondence maps be IDENTITY maps of $D_j\cap D_k$. Then we obtain notion b). This is an additional structure on a Riemann surface in the sense a) which is sometimes called a flat structure. If you look carefully (say, on the example of arccos) you will see that the two definitions of a Riemann surface in the sense b) that I gave are not exactly equivalent. More about this in my survey "Geometric theory of meromorphic functions", and in the preprint of Biswas and Perez Marco, Log Riemann Surfaces. - 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. Q1: Use the manifold definition, going back to Weyl. The other definition comes out of the theory of analytic continuation. (And is somewhat puzzling historically - I'm not quite sure how the Poincaré-Volterra theorem fits in, but these days you'd probably want to read this material in terms of sheaf theory, to which it was one of the inputs.) Q2: G acts on the field of meromorphic functions, is one way to look at it. These are holomorphic mappings of the surface to itself, described by some algebraic mappings in fact. Q3: I think so, by "abstract nonsense". Q4: The quotient is to be treated carefully, since quotients of manifolds are not always manifolds. But in terms of the function field this can be seen as Galois theory, and X is a ramified (usually) covering of the quotient curve. The topological explanation of the Euler characteristic in the Riemann-Hurwitz formula is intuitively clear: just look at what happens under the k-th power map on the unit complex disc, in terms of a simple triangulation, to see how ramification affects coverings. - For Q3: If H is the upper half-plane, S the Riemann surface, and pi:H->S, then composing pi with any automorphism gives a holomorphic map from H to S. Since this map takes a simply-connected surface to S (in particular the image of pi_1 of H is trivial), it must lift to a holomorphic map to H. – Jonah Feb 11 2011 at 16:48 "lifting" the holomorphic map is easy; we apply a topological criteria; but is it easy to show that the lift is "holomorphic"? – Martin David Feb 12 2011 at 3:12 Sure. Say the automorphism of the surface you start with is called f, and the lifted automorphism is called F, and pick x in H. Pick a small enough neighborhood of F(x) so that the restriction of the (holomorphic) covering map is an analytic isomorphism. Now it is clear, since f compose pi is the same as pi compose F, that F can be written, in a suitably small neighborhood of x, as the composition of holomorphic maps. – Jonah Feb 12 2011 at 9:05	{"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": 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": 0.9323309063911438, "perplexity_flag": "head"}
http://math.stackexchange.com/questions/88503/computing-the-minimal-polynomial-of-an-invertible-matrix-whose-minimal-and-chara?answertab=active	# Computing the minimal polynomial of an invertible matrix whose minimal and characteristic polynomial coincide Let $M$ be an invertible $n \times n$ matrix whose minimal polynomial $q_M$ and characteristic polynomial are equal. Suppose as well that the $M$ has entries in an algebraically closed field. How do we show that the minimal polynomial of $M^{-1}$ is given by $q_M(0)^{-1} x^n q_M(\frac{1}{x})$? ($x$ is the indeterminate in the polynomial ring over the base field $k$? - 2 You're aware of the relationship between the roots of a polynomial $p(x)$ and the "reciprocal polynomial" $x^n p(1/x)$? Note that the extra factor is needed so that the minimal polynomial is monic (as a hint, why is $p(0)$ the same as the constant term of $p(x)$?). – J. M. Dec 5 '11 at 8:10 It would seem that the minimal polynomial being equal to the characteristic polynomial is a red herring: it isn't needed. Also your title is strange, as it provokes the reply "well, then compute the characteristic polynomial!". – Marc van Leeuwen Dec 5 '11 at 12:44 ## 1 Answer Let $f(x) = q_M(0)^{-1} x^n q_M(1/x)$. Note that $f$ is a polynomial of degree $n$ and $f(M^{-1}) = 0$. Similarly, if $g(M^{-1}) = 0$ where $g$ is a polynomial of degree $k < n$ with nonzero constant term, then $h(x) = x^k g(1/x)$ would be a polynomial of degree $k$ such that $h(M) = 0$, contradicting minimality of $q_M$. -	{"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": 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": 0.9471390247344971, "perplexity_flag": "head"}
http://mathoverflow.net/questions/107437/eigenvalues-of-a-special-block-matrix-associated-with-strongly-connected-graph/107555	Definition Let $G=(V,E,A)$ be a strongly connected directed graph, where $V={1,2,...,n}$ denotes the vertex set, $E$ is the edge set, and $A$ is the associated adjacency matrix with $0-1$ weighting, that is $a_{i,j}=1$ if $(j,i)\in E$, and $a_{i,j}=0$ otherwise. $B$ and $D$ are two diagonal matrices, where $b_{ii}=\sum_{j=1}^na_{i,j}$ and $d_{ii}=\sum_{j=1}^na_{j,i}$. In other words, the diagonal entries of $B$ are the row sum of $A$, and the diagonal entries of $D$ are the column sum of $A$. Problem Now define a new matrix $$M = \begin{bmatrix} B-A, & -A \\ A-B, & D \end{bmatrix}\in \mathbb{R}^{2n\times 2n}$$ Since the column sum of $M$ are identical zeros, zero must be one of its eigenvalue. Can I claim that the rest eigenvalues all have positive real parts? I tried many numerical examples, the rest eigenvalues all have positive real parts. Anyone can help prove or disprove the above claim? (Gershgorin Circle Theorem does not apply here because $M$ is not diagonally dominate) Some facts: Both $(B-A)$ and $(D-A)$ have exactly one zero eigenvalue and all the rest eigenvalues lie in the open right half complex plane because the directed graph is strongly connected. In particular, $(B-A)$ is called the Laplacian matrix of the graph. - 3 Answers the answer was essentially correct but i believe H should be defined as H11=H21=H22=1 and H12=0. the resulting matrix HMH^T is block upper triangular with B-A and D-A on the diagonal, hence is hurwitz stable. although the eigenvalues change, the inertia theorem guarantees that their stability type cannot change, i.e. they cannot cross the imaginary axis. - 1 But $M$ is non-symmetric. I don't think you can apply the inertia theorem. There ought to be some way to fix the argument, but I haven't figured it out yet. – Felix Goldberg Sep 19 at 20:22 Hi, Felix. Do you have any suggestions on how to handle this problem? All the answers posted here are not correct. – Zhang Changhe Sep 25 at 9:31 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 I have a counterexample. Try this: $A=\begin{bmatrix}1 & 1 & 0 & 1 \\ 0 & 1 & 1 & 0 \\ 1 & 0 & 1 & 0 \\ 0 & 1 & 0 & 1 \end{bmatrix}$. When I construct $M$ I get a pair of conjugate complex eigenvalues in its spectrum. - Hi, Felix. Thanks again. I calculate the eigenvalues of M by the following matlab code %% A = [1 1 0 1; 0 1 1 0; 1 0 1 0; 0 1 0 1]; B = diag([3,2,2,2]); D = diag([2,3,2,2]); M = [B-A, -A; A-B, D]; eig(M) %% The result is -0.0000 0.3501 + 0.8344i 0.3501 - 0.8344i 1.3820 2.0000 2.7600 3.6180 3.5398 %% Yes, the eigenvalues have a pair of conjugate complex eigenvalues. But all the eigenvalues have positive real parts except the zero eigenvalue. So this is not a counterexample. – Zhang Changhe Jan 2 at 11:36 @ZhangChanghe: Oops, my bad. I'll look again. – Felix Goldberg Jan 2 at 16:46 I think your claim is true. Taking your matrix $M$ and multiplying from the left by $$H=\begin{bmatrix} I, & 0 \\ -I, & I \end{bmatrix},$$ and from the right by its transpose, $$H^t=\begin{bmatrix} I, & -I \\ 0, & I \end{bmatrix},$$ the eigenvalues don't change, yet the matrix you get is $$HMH^t=\begin{bmatrix} B-A, &B \\ 0, & D-A \end{bmatrix}.$$ The eigen values of the above matrix are those of $B-A$ and those of $D-A$ and are all non-negative as you already explained. - 1 Something seems to be amiss here: $HMH^{T}$ has the same inertia as $M$ but the eigenvalues do change. Maybe you meant $H^{-1}$, which is also a nice block matrix? – Felix Goldberg Sep 19 at 13:20 Sorry, I wasn't completely awake when writing my comment. It has several quite embarrassing mistakes.. – puzne Sep 19 at 16:40	{"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": 30, "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": 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": 0.9124969244003296, "perplexity_flag": "head"}
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How can I ... 0answers 15 views ### Consider filtering nn images by separable mm filters.What are the computation time for filtering using FFT;2-D convolution;by two 1-D convolutions Consider filtering square nn images by square, separable mm filters. a) Give general equations for the computation time for the following approaches to this 1) filtering using FFT 2) by 2-D ... 1answer 104 views ### How to get fundamental frequency of a signal using autocorrelation? I am trying to get the fundamental frequency of a signal that only has a single pitch. I coded out the autocorrelation function using FFT and already got the autocorrelation result. Unfortunately, I ... 1answer 132 views ### What is the phase and magnitude response of white noise? I would like to create white noise in the frequency domain, and then transform it to the time domain using python. 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I would like someone to explain to me the relationship between the frequency of the wave being processed, the number of point (usually called NFFT), the Nyquist ... 1answer 43 views ### Parameter extraction from DFT results, possibly with leakage effects Let's start with a simple temporal law, that I want to analyze in the frequency domain : $x(t)=\frac{1}{2}(A+Bcos(2\pi \nu_0 t+\phi_0))$, with : $A=4$ $B=10$ $\nu_0=10^8$ $\phi_0=10^8$ I made a ... 1answer 50 views ### FFT function isn't working on imported signals greater than a second I have an fft function in matlab/octave that works great on an imported mono wav/audio signal that is 1 second long, but when the signal imported is more than 1 second the frequency values become ... 1answer 35 views ### Transform Length and event duration Does the length of a Fourier Transform affect the ability to discern a small event in the data? This is a tricky question to ask, let me clarify. I don't care about when the event happened and I ...	{"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": 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": 0.9243167638778687, "perplexity_flag": "middle"}
http://mathhelpforum.com/discrete-math/173047-prove-equal-powers-probably-fairly-primitive-print.html	# Prove equal powers (probably fairly primitive) Printable View • March 1st 2011, 06:06 AM Pranas Prove equal powers (probably fairly primitive) Hello. My discrete mathematics course has just started, so I'm not very good even at probably fairly easy things: Prove if $\displaystyle \[G = A \times B\]$ is a bijection between finite sets $\displaystyle \[A\]$ and $\displaystyle \[B\]$ then $\displaystyle \[\left\| A \right\| = \left\| B \right\|\]$. It looks pretty obvious because of the nature of this bijection, but I just can't seem to figure out notations to prove it in a mathematical way. Thanks. • March 1st 2011, 06:15 AM emakarov A bijection is a function, but if A is nonempty and \|B\| > 1, then A x B is not a function. If there is no mistake in the question, then the fact that A x B is a bijection implies that \|A\| = \|B\| = 1 or \|A\| = \|B\| = 0, so \|A\| = \|B\|. All times are GMT -8. The time now is 12:02 AM.	{"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": 4, "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": 0.9451910257339478, "perplexity_flag": "middle"}
http://stats.stackexchange.com/questions/10021/new-mean-calculation-in-multivariate-gaussian	# New mean calculation in multivariate gaussian Given a Gaussian distribution $N(\mu_1,\sigma_1^2)$, i would like to choose another mean $\mu_2$ which is $2\sigma_1$ away from $\mu_1$. In this case our new mean $\mu_2=\mu_1\pm 2\sigma_1$. How do we calculate the new mean($\mu_2$) in multivariate case? I mean to say, when your multivariate Gaussian distribution is $N(\mu_1,\Sigma_1)$ and my $\Sigma_1$ is symmetric positive definite matrix. i.e $\left[ \begin{array}{cc} \sigma_x^2 & \sigma_{xy} \\ \sigma_{yx} & \sigma_y^2 \end{array} \right]$. - I am not sure that I understand even the univariate case. By $\mu_2=\mu_1\pm 2\sigma_1$ you mean that both answers are ok? – mpiktas Apr 27 '11 at 10:24 1 what is the context of this problem? As it stands now, it is impossible to answer, since in multivariate case the means are vectors and the distance between them is a number, but the covariance matrix is not. – mpiktas Apr 27 '11 at 10:26 ## 1 Answer In the bivariate case you can substitute the two points ($\mu_2=\mu_1\pm 2\sigma_1$) with an isodensity ellipse: http://www.stat.psu.edu/online/courses/stat505/05_multnorm/06_multnorm_revist.html . Your $2\sigma_1$ criterion seems a bit arbitrary, but it includes 95.44997% of the random variable. So you may want to use the 95.44997% isodensity ellipse in the bivariate case, too. The principle axes of this rotated ellipse are the eigenvectors of the covariance matrix, see http://web.as.uky.edu/statistics/users/viele/sta601s08/multinorm.pdf . You can generalise this to more than 2 dimensions. -	{"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": 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": 0.8879522681236267, "perplexity_flag": "head"}
http://physics.stackexchange.com/questions/50153/eutectic-systems-behavior-near-100-0-composition-and-low-temperature/51167	# Eutectic systems behavior near 100%-0% composition and low temperature I can't understand why in this phase diagram , in the area where the system is approaching the 100% $\alpha$ or 100% $\beta$ composition, the state is just a pure solid. I mean, where does the other component go? If I am interpreting it right, there's no liquid-phase mixture left, and there's neither any solid rest from the other substance. - ## 2 Answers This used to confuse me too! The key is that the $\alpha$ and $\beta$ phases are not necessarily pure component A or B. For example $\alpha$ is a single phase which consists of mostly component A with small amounts of component B dissolved in it - it is a single "A-rich" phase. The B atoms are distributed throughout the A crystal but do not form into separate regions. If you looked into a microscope you would see only one kind of crystal throughout. In contrast, in the $\alpha + \beta$ region you will see two different kinds of crystal one where there is mostly A with B dissolved in it and one where there is B with a little A dissolved in it. - Thanks! That completely answers my question. However, what will happen if we were to lower even more the temperature of the almost-pure sample? According to the diagram it would turn into $\alpha$ + $\beta$, does it mean that the atoms of the low-concentration component would "get closer" until we could see some of them with the naked eye? How could this even happen at an atomic level if the high-concentration element particles have already organized into a solid body? – carllacan Jan 14 at 21:07 At 100 \% there is only one component present, right? So there is no missing component. I am guessing that your problem lies in the "approaching 100 %" part of the question. Now how can that happen. Either one is fishing one component out of the system, leaving the other alone, or one is adding huge amounts of the other component making the percentage of the first smaller and smaller. In either case only one component is left and the other component has either been removed from the solution or is present at zero relative concentration. - Suppose we have a number of mixtures of $\alpha$ and $\beta$. If we pick one with a 99% concentration of, say $\alpha$ at a temperature above $\alpha$'s fusion point. When the temperature drops to below the L+$\alpha$ - Solid $\alpha$ equilibrium line, what happens? According to the diagram we have pure solid $\alpha$. Where does the other component go? – carllacan Jan 14 at 4:07 1 It doesn't mean that you've only got $\beta$ material, what it means is that you've got a material which has the structure of a pure $\beta$ material with some impurities of $\alpha$ material in a given proportion. If you looked at a good chunk of it with the naked eye (except in some particular cases), you would never tell the difference between a pure $\beta$ material and this non-pure alloy. This case is akin to the case of the aliovalent impurities in a semiconductor. It's not exactly, rigorously the same idea, but it's a similar one and helps you getting it right. – user17581 Jan 14 at 10:26	{"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": 0.9513107538223267, "perplexity_flag": "head"}
http://mathoverflow.net/questions/19648/method-of-moments-and-laplace-transform-from-shepp-and-lloyd/19737	method of moments and Laplace transform from Shepp and Lloyd Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) Again from the Shepp and Lloyd paper "ordered cycle lengths in a random permutation", I found this puzzling equality. This one might require access to the paper itself since it's quite a mouthful: In equation (15), they claimed it is straightforward that if there is an $F_r$ such that $$\int_0^1 \exp(-y/\xi) dF_r(\xi) = \int_y^{\infty} \frac{E(x)^{r-1}}{(r-1)!} \frac{\exp(-E(x) -x)}{x} dx$$ then $F_r$ will have moments $G_{r,m}$. Here $$G_{r,m} = \int_0^{\infty} \frac{x^{m-1}}{m!} \frac{E(x)^{r-1}}{(r-1)!} \exp(-E(x)-x) dx$$ and $$E(x) = \int_x^{\infty} \frac{e^{-y}}{y} dy$$ which is related to the thread http://mathoverflow.net/questions/19526/reference-request-for-a-well-known-identity-in-a-paper-of-shepp-and-lloyd It looks to me like some sort of Laplace transform, but I can't manage to get the algebra to work, because of the inverse exponent $y/\xi$ with respect to $\xi$. I will be happy enough if one can tell me why we are looking at the transform $\int_0^1 \exp(-y/\xi) dF_r(\xi)$ instead of the usual moment generating function $\int_0^1 \exp(-y \xi) dF_r(\xi)$, or maybe it's a typo? - I fixed your LaTeX. We don't support those fancy environments around here! – Qiaochu Yuan Mar 28 2010 at 19:08 Thanks. I wasn't aware. – John Jiang Mar 28 2010 at 19:20 1 Answer This is a general fact: assume that $X$ is a positive random variable and that, for a given nonnegative function $g$, $\displaystyle E(\mathrm{e}^{-y/X})=\int_y^{\infty}g(x)\mathrm{d}x$ for every positive $y$. Then $\Gamma(s+1)E(X^s)=\displaystyle\int_0^{\infty}x^sg(x)\mathrm{d}x$ for every positive $s$. To prove this, integrate the equality $\displaystyle\int_0^{\infty}\mathrm{e}^{-y/x}y^{s-1}\mathrm{d}y=\Gamma(s)x^s$ over $x>0$ with respect to the distribution of $X$ and change the order of integration in the LHS. - C'est Chouette! Merci Beaucoup, Mon ami! – John Jiang Mar 29 2010 at 16:05 De rien, tout le plaisir est pour moi. :-) – Didier Piau Mar 29 2010 at 16:24	{"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": 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": 0.3, "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": 0.8693495988845825, "perplexity_flag": "middle"}
http://physics.stackexchange.com/questions/22664/what-are-the-physycal-meaning-of-universal-constants-such-as-the-magnetic-elect/22672	# What are the physycal meaning of universal constants such as the magnetic, electric and gravitational constants? I know that: $G=6.67300 × 10^{-11}\dfrac{Nm^2}{kg^2}$(Gravitational constant) $K_e=9×10^9\dfrac{Nm^2}{C^2}$(Electric constant or Coloumb's constant) $k_m=1×10^{-7}\dfrac{Ns^2}{C^2}$(Magnetic constant) But what are the meaning of this values? Would the universe be much different with a little change on this values? What if we hadn't known them? - 2 Closely related to Deriving or justifying fundamental constants and to a lesser degree to units and nature. The short-short version is that the numeric values of dimensional constants have no meaning at all (because they depend of the system of units you choose). However this does not apply to dimensionless constants like the Fine Structure Constant $\alpha(Q^2 = 0) \approx 1/137$. – dmckee♦ Mar 21 '12 at 19:32 – Qmechanic♦ Mar 21 '12 at 20:00 – 0x5f3759df Mar 21 '12 at 21:47 This isn't a duplicate, because it asks about the electric constants, one of which is "too round" to be accidental, and the other is the exact speed of light squared, ignoring the power of 10. – Ron Maimon Mar 22 '12 at 3:03 ## 3 Answers The three constants you give illustrate the arbitrariness of units. The magnetic constant $$K_m = 10^{-7} {N\over A^2}$$ serves to define the Ampere. The definiton of the Ampere implied by this (defined) constant is that the force between two long wires carrying one Ampere of current at a separation of 1 meter is 10^{-7} Newtons per unit length. If you change this constant, you redefine the Ampere. the other constant you mention is $$K_e = 8.996 \times 10^9 {Nm^2\over C^2}$$ This is interesting, because if you just look at the number, forgetting the power of 10, it's the square of the speed of light. The reason for this is that electric and magnetic effects are related by relativity, and the ratio of $K_e$ and $K_m$ is the speed of light squared. Because we choose units where $K_m$ is a power of 10, the $K_e$ is then the same as the square of the speed of light. Since the speed of light, like $K_m$, is defined, thereby setting the standard definition of the meter, the electric constant $K_e$ is also defined--- it is $10^{-7}$ times the exact speed of light squared. If you vary it, you can only do so in conjunction with varying the definition of the speed of light, and therefore the meter. The gravitational constant, in an ideal world, would define the unit of mass. But gravity is too weak to measure accurately enough, so we use a block of metal in a vault in Paris for now to define what a "Kilogram" is. This will probably change at some point. But "G" is also a constant that is philosophically incapable of varying, since it defines the system of units. If you want to ask a sensible question, you should ask them of quantities that don't depend on the units. In practice, you ignore $G,\hbar,c,k_b,K_m$ (your stuff, swapping out $K_e$ for c, plus Planck's constant and Boltzmann's constant). These you set to 1 in a reasonable system of units, and then all other quantities become dimensionless and meaningful, so you can ask about why they have the values that they do. - The fine-structure constant is a good example of a more sensible question, I guess.. – Manishearth♦ Mar 22 '12 at 3:04 @Manisheath: that, and the mass of the hydrogen atom in Planck units, or the force between two H-atoms at a distance of 10^{25} Planck distances, etc. – Ron Maimon Mar 22 '12 at 3:19 – Manishearth♦ Mar 22 '12 at 3:25 Explaining the meaning of universal constants has been discussed by other posts. Many of these universal constants are set by attempting to keep as many quantities and units equal to 1 as possible. Having the number 1 in calculations simplifies the math. If we did not know these universal constants, we would eventually measure them with experiments or prove them from other fundamental equations. - These three fundamental constants do matter and the universe would be much different when they would be far off the curretn values. Let us look at G as an example. There is a fundamental property called mass. Mass is a property of objets like particles. Mass manifests itself in two ways: 1) as a force that is required to accelerate a massive object, and 2) a force that acts on a massive object when it is exposed to a gravitational field. $G$ defines and scales the second manifestation: gravity. If you expose an object with mass $m_1$ to a gravitational field produced by a second object $m_2$, the force acting on $m_1$ is: $$F = G \frac {m_1 \cdot m_2} {d^2}$$ Where $d$ is the distance between the two objects (their center of mass, to be more accurate). This means that if $G$ suddenly became 10 times larger than it is today, your weight would increase for a factor of ten, making your life quite miserable. And this would even be the least problem. Note 1: Many people and unfortunately many theoretical Physicists do not make the difference between the value of a constant (which in case of G would be $6.67300\cdot10^{−11} \frac {Nm^2}{kg^2}$) and the numerical value of a constant (which in case of G would be $6.67300\cdot10^{−11}$ when using SI units). This lack of rigour results in various misunderstandings. Note 2: It is NOT true that the value of a constant depends on the units. Note 3: It is true that the numerical value of a constant depends on the units. If you increase the unit U for a factor of $a$, the numerical value $n$ must compesate this by decrasing for a factor of $a$. However, this is extremely trivial and was not part of your question. Note 4: Many people belive that it is possible to "scale" a constant to 1. In case of $G$, this would mean $G = 1$. However, this is a misconception. It is only possible to change the numerical value of a constant to 1, which is written as $\{G\} = 1$. This is trivially done by adapting the unit. In case of $G$, the unit U has to be multiplied by a factor of $6.67300\cdot10^{−11}$ for the numerical factor $n = \{G\}$ to become 1. Note 5: Here is the correct nomenclature (for any system of units): $$G = n \times U = \{G\} \times [G]$$ For the SI system we get: • numerical factor $= n = \{G\} = 6.67300 \cdot 10^{−11}$ • unit = U $= [G] = \frac {Nm^2}{kg^2}$ • hence $G= 6.67300 \cdot 10^{−11} \times \frac {Nm^2}{kg^2}$ For a system of natural units: • numerical factor $= n = \{G\} = 1$ • unit = U $= [G] = 6.67300 \cdot 10^{−11}\frac {Nm^2}{kg^2} = G$ • hence $G = 1 \times G$ This means: in natural units $G$ becomes a unit (not in all systems of natural units, but in many). However, it does not magically "disapear". Note 6: Natural units is one subject where Wikipedia got it completely wrong. - -1: I am sorry for downvoting a new user, but when the numerical factor is made equal to 1, G disappears. It is a philosophical thing, but it is clarified by considering the situation where you measure height in feet and horizontal lengths in meters, and you try to do a rotation. The number of units is a construct of macroscopic physics, it has nothing to do with nature. – Ron Maimon Mar 26 '12 at 7:28 "..., but when the numerical factor is made equal to 1, G disappears." - As I have shown in Note 5, this is not true. The numerical factor is 1, but the unit is still there. – Vera Metro Mar 26 '12 at 10:59 You have not shown it, because when the numerical factor is made 1, G disappears. You have not shown anything different. You should sit down and think about it before saying silly things. – Ron Maimon Mar 26 '12 at 16:32 I have shown that if the numerical factor $= n = \{G\} = 1$, then the unit becomes U $= [G] = 6.67300 \cdot 10^{−11}\frac {Nm2}{kg2}$ which is $G$. Thus, G does not disapear, it becomes a unit. I am well aware that this is in contrast with your believe-system and that it is extremely difficult to overcome one's believe-system. But I also think you are smart enough to achieve this. Sit back and take your time to really think about it. In the end it is very trivial. – Vera Metro Mar 26 '12 at 23:02 Think of it this way: another constant quantity, the kg prototype in Paris, also has the numerical factor 1 (per definition). Its mass is $m = 1$kg. If your statement were true, the quantity would also "disappear" and become 1. However, I hope we both agree that the mass of the prototype kg is $m = 1$kg. – Vera Metro Mar 26 '12 at 23:10 show 8 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": 42, "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": 0.9460347890853882, "perplexity_flag": "head"}
http://math.stackexchange.com/questions/296796/existence-and-uniqueness-theorem/296807	# Existence and Uniqueness Theorem I had a question about how to do one of these problems. So here's the question: Given this equation $y'=\frac{-\cos(t)y(t)}{(t+2)(t-1)}+t$, find if the initial conditions $y(0)=10, y(2)=-1, y(-10)=5$ exist. So I think the first step is just to take the partial derivative with respect to y which gives me: $$y''=\frac{-\cos(t)y'(t)}{(t+2)(t-1)}$$ So the 1'st equation doesn't exist at $t=-2,1$ and the partial derivative doesn't exist at $t=-2,1$ ....so do I conclude that all the initial values exists since none of them are $y(-2)$ or $y(1)$. Don't really know how to do this whole existence and uniqueness thing....so am I right or completely off track? - The initial conditions exist, because you have written them down. The question is whether there exist unique solutions, given the DE and the initial conditions. – Christian Blatter Feb 7 at 11:46 ## 1 Answer What you've done looks perfectly fine. Here's a general outline of what you do when you're looking to find where solutions exist for the first-order differential equation $y'+ p(t) y = g(t)$. 1. Write the differential equation in the form: $y'=f(y, t)$ 2. Find $f_y = \frac{\partial}{\partial y} f$. 3. Determine points of discontinuities of both $f_y$ and $f$. At this point, if you're just looking to see if a particular initial condition ($t_0$) has a solution, just check if $t_0$ is one of the points of discontinuity. If you're looking for where the solution exists: 1. Draw a number line denoting where the discontinuities are (if possible). 2. Find where the initial condition falls on the number line. 3. If the discontinuity to the left of $t_0$ is $a$, and the discontinuity to the right of $t_0$ is $b$, then the solution exists on the interval $(a, b)$. EDIT Based on requests from comments below, here's a statement of the existence and uniqueness theorem: Let the functions $f$ and $\frac{\partial f}{\partial y}$ be continuous in some rectangle $\alpha < t < \beta$, $\gamma < y < \delta$ containing the point $(t_0, y_0)$. Then, in some interval $t_0 - h < t < t_0 + h$ contained in $\alpha < t < \beta$, there is a unique solution $y = \phi(t)$ of the initial value problem: $$\begin{array}{cc} y' = f(t,y) & y(t_0) = y_0. \end{array}$$ Source: Elementary Differential Equations and Boundary Value Problems, Boyce and DiPrima, 10th Edition, pg 70. - what happens if $f$ is discontinuous at $y=2$ but $f_y$ is discontinuous at $y=3$? What will I say about the existence of $y(2)$ and $y(3)$? – Charlie Yabben Feb 7 at 3:00 @CharlieYabben The uniqueness and existence theorem assumes $f$ and $f_y$ are continuous on an interval containing $t_0$. Thus, in your example from the previous comment, conclude that $y(2)$ and $y(3)$ do not exist. – anorton Feb 7 at 3:10 @anorton: It is a good idea to include the statement of the existence and uniqueness theorem or add a link to it. – Mhenni Benghorbal Feb 7 at 9:05 @MhenniBenghorbal I have now included one. – anorton Feb 7 at 12:16	{"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": 37, "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": 0.9182857871055603, "perplexity_flag": "head"}
http://math.stackexchange.com/questions/165883/what-should-i-call-this-commutative-monoid-of-order-three	# What should I call this commutative monoid of order three? I'm looking for a name for the monoid given by the following table: $$\begin{array}{c\|ccc}&1&a&b\\ \hline 1&1&a&b\\ a&a&1&b\\ b&b&b&b \end{array}$$ Is there a name that would be understandable to an undergraduate student who hasn't read anything about semigroups but has had a first course in algebra and knows what a semigroup/monoid is? What name would it be good to go under in a list of order-three semigroups? - 3 I'd call it $\{-1,0,1\}$, but I don't know very much about abstract algebra... – Rahul Narain Jul 2 '12 at 21:54 @RahulNarain Right. Thank you. This is really embarrassing. – user23211 Jul 2 '12 at 21:56 Call it Diego, just for kicks. – Asaf Karagila Jul 2 '12 at 22:13 @Asaf I haven't been so embarrassed for quite some time. Mocking may lead to an attempt at burying my head in the floor. That's probably safer than using sand, but still not a good idea. :) – user23211 Jul 2 '12 at 22:18 @ymar, if you prefer we can give it a Polish name, Stanislaw, perhaps. – Asaf Karagila Jul 2 '12 at 22:26 show 1 more comment ## 2 Answers Call it : $(\mathbb{Z}/3\mathbb{Z},*)$ - Oh... Thanks! The way I looked at it I did not notice that $b$ was a zero element. – user23211 Jul 2 '12 at 21:55 it was the first think i saw :) – hassan Jul 2 '12 at 21:58 In computer science I would call it $\mathbb{Z}_2$ with errors, i.e. $\langle \{0,1,\bot\},+,0\rangle$. In math I would follow hassan's idea. - Thanks. So if $G$ is a group, then $G^0$ ($G$ with zero adjoined) is called "$G$ with errors"? – user23211 Jul 2 '12 at 22:24 1 @ymar Not zero, but an element that spreads like a virus (usually denoted by $\bot$). Any operation can result in an error $\bot$ (e.g. if you divide $1$ by $0$ in some algebra containing division) and if the "error result" is passed to any other expression, it propagates, i.e. the result of that expression has to be $\bot$ as well. Indeed, that kind of structures are denoted $G^\bot$, and called "$G$ with errors" or "$G$ with error handling". This is very similar to NaNs or monad `Maybe` from Haskell language (e.g. $G^\bot \cong \mathtt{Maybe}\ G$). – dtldarek Jul 3 '12 at 6:09	{"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": 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": 0.3, "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": 0.9516905546188354, "perplexity_flag": "middle"}
http://math.stackexchange.com/questions/180915/is-the-green-tao-primes-theorem-true-or-pseudo-true/180916	Is the Green-Tao primes theorem true or pseudo-true? [closed] The Green-Tao primes theorem "The primes contain arbitrarily long arithmetic progressions" Ann Math 2008, is certainly a tour de force and extremely technical. The authors point out that they have taken precautions to not use Axiom of Choice. Choice is mentioned several times in the paper, eg p.485: In particular we shall always remain in the ﬁnitary setting of ZN , in contrast to the standard ergodic theory framework in which one takes weak limits (invoking the axiom of choice) to pass to an inﬁnite measure-preserving system. p.516: In the original ergodic theory arguments of Furstenberg this algorithm was not guaranteed to terminate, and indeed one required the axiom of choice (in the guise of Zorn’s lemma) in order to conclude the structure theorem. However, in our setting we can terminate in a bounded number of steps... In terms of constructive vs classical mathematics however AC is only one of the forbidden axioms. Another is excluded middle. But I can't follow the paper to understand whether results that invoke excluded middle are used. Kolmogorov, in "On the tertium non datur principle" writes "a proposition is pseudo-true if its double negation is true". Is the Green-Tao theorem true or pseudo-true? - 1 It is probably "pseudo-true". Many fundamental results in real analysis – such as the intermediate value theorem – are only "pseudo-true". But even if the Green–Tao proof isn't intuitionistically valid as-is, to answer the question properly one would have to either produce an intuitionistic counterexample to Green–Tao, or reprove the theorem using intuitionistic methods. – Zhen Lin Aug 10 '12 at 4:38 Excluded middle is omnipresent in mathematics. – André Nicolas Aug 10 '12 at 5:36 closed as not a real question by Will Jagy, JDH, Zhen Lin, Asaf Karagila, J. M.Aug 14 '12 at 5:00 It's difficult to tell what is being asked here. This question is ambiguous, vague, incomplete, overly broad, or rhetorical and cannot be reasonably answered in its current form. For help clarifying this question so that it can be reopened, see the FAQ. 1 Answer Concerning just classical logic and the use of AC, I am surprised to hear that precautions were taken to avoid AC, since the theorem is expressible in first-order number theory, and such statements are absolute from the set-theoretic universe $V$ to the constructible universe $L$, where the axiom of choice holds. It follows from this that the axiom of choice has no bearing on the argument, and any use of the axiom of choice could be omitted simply by pointing out this absoluteness. So actually, no precautions were necessary. The basic situation is that the arithmetic consequences of ZFC are precisely the same as the arithmetic consequences of ZF, and so one may assume AC for such purposes without loss of generality. A similar situation arises for example with Fermat's last theorem, meaning that it wouldn't have mattered whether or not Wiles had used AC in his proof. Even if he had, we could have constructed a new proof that does not use AC by arguing that we may first assume V=L and hence AC without loss of generality, since the truth of FLT is not affected by moving to L, and then carrying out the previous proof. - 4 But actually, reading the passages now, I would say that the concern was apparantly not the axiom of choice so much as the axiom of infinity, since they were concerned to remain in a finitary setting. This is a very different concern, and of course ZF is not conservative over very weak finitary systems, even PA. So it does make sense to be concerned about one's assumption if one wants to prove the system in a weak finitary system. – JDH Aug 10 '12 at 4:35 3 @alancalvitti: "finitary" methods are those that do not make essential use of infinite sets or processes. These methods were first singled out by Hilbert in his consistency program. However, like "constructive", the term "finitary" is not formal, and different authors use it with different meanings. – Carl Mummert Aug 10 '12 at 11:11 6 Oh, since you put things that way, I beg your pardon, and I'm very sorry to have engaged with you. – JDH Aug 11 '12 at 2:05 7 @alancalvitti Such confrontational responses are not perceived well here. I can only imagine that you are ostracizing those who might wish to help, like JDH. – mixedmath♦ Aug 11 '12 at 14:58 4 – Kaveh Aug 13 '12 at 20:08 show 15 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": false, "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": 0.9500937461853027, "perplexity_flag": "middle"}
http://math.stackexchange.com/questions/139897/fourier-transform-of-sinx-sinx/139961	Fourier transform of Sin(x + Sin(x)) Does anybody happen to know what the Fourier transform of $f(x) = \sin(x + \sin(x))$ is? More generally, what is the Fourier transform of $g(x) = \sin(\phi_1 x + \mu \sin(\phi_2 x))$? ($\phi_1$, $\phi_2$ and $\mu$ are constants.) I'm pretty sure I saw these written down somewhere, but Wolfram\|Alpha insists that it cannot determine the result... - 1 Answer Since $\sin(x+\sin(x))$ is an odd function with period $2 \pi$, its Fourier series is $$\sin(x+\sin(x)) = \sum_{n=1}^\infty c_n \sin(n x)$$ where $$\begin{eqnarray} c_n &=& \frac{1}{\pi} \int_{-\pi}^\pi \sin(n x) \sin(x+\sin(x)) \mathrm{d} x = \frac{2}{\pi} \int_{0}^\pi \sin(n x) \sin(x+\sin(x)) \mathrm{d} x \\ &=& \frac{1}{\pi} \int_{0}^\pi \cos((n-1)x - \sin(x) ) \mathrm{d} x - \frac{1}{\pi} \int_{0}^\pi \cos((n+1)x + \sin(x) ) \mathrm{d} x \\ &=& J_{n-1}(1) - J_{n+1}(-1) = J_{n-1}(1) + (-1)^{n} J_{n+1}(1) \end{eqnarray}$$ where the last line was evaluated using Bessel integrals. - I thought I remembered Bessel-J being in there somewhere... Is there an easy expression for the more complicated function? – MathematicalOrchid May 3 '12 at 8:54	{"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": 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": 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": 0.9229908585548401, "perplexity_flag": "middle"}
http://mathhelpforum.com/calculus/146875-finding-parametrizattion-line-integral.html	# Thread: 1. ## Finding parametrizattion for a Line Integral I've been asked to evaluate the line integral $\int (10x^4 - 2xy^3)dx - 3x^2y^2dy$ Over the curve $x^4 - 6xy^3 - 4y^2 = 0$ Between the points (0,0) and (2,1). What I'm trying to do is finding a parametrization x(t) from the given curve and then put that into the integral so I can solve it. However I can't seem to make it work. I tried solving x for y, so I can say x=t and y is a function of t. Or the other way round would work as well. But the crossterm $6xy^3$ gets in the way. And the powers of both x and y aren't right to use the ABC formula. Could anyone help me out on what to do here? Thanks in advance Edit: Apologies for the spelling mistake in the thread title. Can't edit it, it seems. 2. Can I ask to check on the sign of one of your terms in the integral i.e. is it $<br /> \int \limits_{c} (10x^4 {\color{red}{+}}\, 2xy^3)dx + 3x^2y^2dy$ 3. Ah, I did write it down incorrectly, but it's not that either. Correct version is: $<br /> \int (10x^4 - 2xy^3)dx - 3x^2y^2dy<br />$ 4. Originally Posted by AbAeterno Ah, I did write it down incorrectly, but it's not that either. Correct version is: $<br /> \int (10x^4 - 2xy^3)dx - 3x^2y^2dy<br />$ Note that $2xy^3dx + 3 x^2 y^2 dy = d(x^2y^3)$. 5. Originally Posted by Danny Note that $2xy^3dx + 3 x^2 y^2 dy = d(x^2y^3)$. Okay. So I can write that down in the integral and get $<br /> \int 10x^4 dx - d(x^2y^3)<br />$ But then what? Still doesnt help me find a parametrisation as far I can see. Or is that not what I'm supposed to do? Edit: oh wait.. so now I don't need to find a x(t) and y(t) anymore, but an x(t) and x^2y^3(t)? 6. Originally Posted by AbAeterno Okay. So I can write that down in the integral and get $<br /> \int 10x^4 dx - d(x^2y^3)<br />$ But then what? Still doesnt help me find a parametrisation as far I can see. Or is that not what I'm supposed to do? Since the vector field is conservative, the parameterization doesn't matter. So $<br /> \int \limits_c 10x^4 dx - d(x^2y^3) = <br /> \int \limits_c d(2x^5 - x^2y^3) = \left. 2x^5 - x^2y^3 \right\|_{(0,0)}^{(2,1)}<br />$ 7. Ooh.. I totally didn't see that. 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": 11, "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": 0.9541168808937073, "perplexity_flag": "middle"}
http://mathhelpforum.com/calculus/27343-determing-range-values-x-y.html	# Thread: 1. ## Determing range of values of x & y Hi If real variables x and y are related by the equation: ln(3-y) = 3ln(2x+1)-x then how do i: a). determine the range of values of x and y for which expressions on each side of equation are defined, and b). find y explicitly as a function of x, that is, express the equation in the form y=f(x) Thanks 2. Originally Posted by moolimanj Hi If real variables x and y are related by the equation: ln(3-y) = 3ln(2x+1)-x then how do i: a). determine the range of values of x and y for which expressions on each side of equation are defined, and b). find y explicitly as a function of x, that is, express the equation in the form y=f(x) Thanks to a) The ln-function is defined for positive arguments: $3-y>0~\implies~y<3$ and $2x+1>0~\implies~x>-\frac12$ To b) $\ln(3-y) = 3\ln(2x+1)-x~\implies~3-y=\frac{(2x+1)^3}{e^x}~\implies~y=3- \frac{(2x+1)^3}{e^x}~,~x>-\frac12$ 3. a) The conditions are $3-y>0$ $2x+1>0$ Now, can you solve them? b) The equation can be written as $\displaystyle\ln(3-y)=\ln(2x+1)^3-\ln e^x\Leftrightarrow\ln(3-y)=\ln\frac{(2x+1)^3}{e^x}\Leftrightarrow$ $\displaystyle\Leftrightarrow 3-y=\frac{(2x+1)^3}{e^x}\Leftrightarrow y=3-\frac{(2x+1)^3}{e^x}$	{"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": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 7, "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": 0.7817173004150391, "perplexity_flag": "middle"}
http://math.stackexchange.com/questions/207379/proving-the-limit-of-an-improper-integral-of-a-sequence-of-functions?answertab=votes	# Proving the limit of an improper integral of a sequence of functions. I was trying to prove that the following limit $$\lim_{n\to\infty}\int^{\infty}_{1}\frac{\sin{x}}{x^{n+1}}\mathrm{d}x$$ is equal to $0$. I believe that the easiest option in similar cases - and the only one I know... - is proving that that $f_{n}$ converges uniformly to $f$ on the interval of the integral. However, this is not the case here, as for $x=1$ we have $\lim_{n\to\infty}f_{n}=1$ and for $x>1$ $\lim_{n\to\infty}f_{n}=0$. I would be very thankful for thoughts on how this should be proven. - 1 Are you familiar with the monotone convergence theorem for integrals? – Christopher A. Wong Oct 4 '12 at 19:59 The problem that I see with applying it here is that our function $f$, to which $f_{n}$ converges, is a branched function... and I guess even that wouldn't be problem, if it wasn't the fact that $f(1)=1$. – Johnny Westerling Oct 4 '12 at 20:27 @JohnnyWesterling You might add where the problem comes from. I assume your definition of integral is that of Riemann? – AD. Oct 5 '12 at 5:47 @AD. Oh, its just a set of problems I have, not even in English, and I don't think it has some online source either, but it got into my hands and I thought of solving it. The question doesn't ask for more than what I wrote, and does not set forth any assumptions. – Johnny Westerling Oct 5 '12 at 14:19 ## 3 Answers Hint: What is $\int_1^\infty \dfrac{dx}{x^{n+1}}$ ? - That gives us $1/n$, which obviously tends to 0. Thus, I believe we can write that $\int^{\infty}_{1}\frac{\sin{x}}{x^{n+1}}dx\leq\int^{\infty}_{1}\frac{dx}{x^{n+1‌}}$. Now, do we need to bound our integral by something from below or is that sufficient? – Johnny Westerling Oct 4 '12 at 20:11 @JohnnyWesterling Approximate the integral of $\|\sin x/ x^{n+1}\|$. – David Mitra Oct 4 '12 at 20:16 Sorry, but I am not exactly sure what do you mean by "approximate" in this case... What I do see is that for any $x\geq{1}$ and $n\geq{1}$, we can write $0\leq\|\sin{x}/x^{n+1}\|\leq1/x^{n+1}$... – Johnny Westerling Oct 4 '12 at 20:23 @JohnnyWesterling Sorry, that wasn't what I meant to say; but I think you have the idea... – David Mitra Oct 4 '12 at 20:30 Hint: use $\triangle$ – AD. Oct 4 '12 at 20:33 show 3 more comments Your idea almost works, to proceed with it you might first consider the sequence $f_n$ on $[1+\varepsilon,\infty)$ for fixed $\varepsilon>0$. Do you see the next step? - Well, our sequence is indeed uniformly convergent on $[1+\varepsilon,\infty)$. Now we would have to check our integral on $[1,1+\varepsilon]$. Well, in this case I believe we should consider $\varepsilon\to{1}^{+}$... is that the correct direction...? – Johnny Westerling Oct 4 '12 at 20:16 sure, what can you say about $f_n$ there ? – AD. Oct 4 '12 at 20:30 Well, we know that its integral on this interval would be equal to $F_{n}(\varepsilon)-F_{n}(1)$ where $F_{n}$ is the antiderivative of $f_{n}$, which must be continuous on $[1,1+\varepsilon]$; thus, we can conclude that $\lim_{\varepsilon\to{1}^{+}}(F(\varepsilon)-F(1))=0$. Is that correct? – Johnny Westerling Oct 4 '12 at 20:37 There is a problem on when to switch limits..(btw i would stick to the less confusingr $1+\varepsilon$). – AD. Oct 4 '12 at 20:49 I would try to estimate the integral, it is like the area of a thin rectangl. – AD. Oct 4 '12 at 20:53 show 9 more comments STEP 1: NOTE THAT, SINCE $x>1$ $$\left\|\frac{sin x}{x^{n+1}}\right\|\leq \frac{1}{x^{n+1}}$$ STEP 2: TAKING INTEGRAL $$\int_{1}^{\infty}\left\|\frac{\sin x}{x^{n+1}}\right\| dx\leq \int_{1}^{\infty}\frac{dx}{x^{n+1}}=\lim_{t\to\infty}\left(\frac{x^{-n}}{-n}\right)\Big\|_{1}^t=\frac{1}{n}.$$ STEP 3: TAKING LIMIT WHEN $n\to\infty$ $$0\leq\lim_{n\to\infty}\left\|\int_{1}^{\infty}\frac{\sin x}{x^{n+1}}\mathrm{d}x\right\|\leq \lim_{n\to\infty}\int_{1}^{\infty}\left\|\frac{\sin x}{x^{n+1}}\right\| dx\leq\lim_{n\to \infty}\frac{1}{n}=0$$ Therefore $$\lim_{n\to\infty}\left\|\int_{1}^{\infty}\frac{\sin x}{x^{n+1}}\mathrm{d}x\right\|=0,$$ and so $$\lim_{n\to\infty}\int_{1}^{\infty}\frac{\sin x}{x^{n+1}}\mathrm{d}x=0$$ - $\sin{x}$ can indeed take on negative values, while for $x\geq{1}$ the denominator can't. Why would say then in step three that this integral is always greater or equal to $0$? – Johnny Westerling Nov 3 '12 at 6:15	{"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": 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": 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": 0.9477154016494751, "perplexity_flag": "head"}
http://mathhelpforum.com/calculus/159792-tough-max-min-problem-print.html	# Tough Max/Min Problem Printable View • October 15th 2010, 10:09 PM Em Yeu Anh Tough Max/Min Problem Let $F(x,y) = 2(1-x)^{\frac{1}{2}}+(x+y)^2+x-2y$ be a function that is bounded by $x=-3, y(t)=t^2, x(t)=2t+t^2$ I am to find the absolute max/min values. Completely lost! This is significantly more difficult that any examples we went through in class. • October 16th 2010, 04:39 AM HallsofIvy You appeared to have copied the problem wrong. $x= t^2+ 2t= t^2+ 2t+ 1- 1= (t+1)^2- 1$. No matter what t is, x is never less than -1 and, in particular, is never equal to -3. Since the "boudaries" given do not intersect, they cannot bound a region. • October 16th 2010, 09:49 AM Em Yeu Anh Quote: Originally Posted by HallsofIvy You appeared to have copied the problem wrong. $x= t^2+ 2t= t^2+ 2t+ 1- 1= (t+1)^2- 1$. No matter what t is, x is never less than -1 and, in particular, is never equal to -3. Since the "boudaries" given do not intersect, they cannot bound a region. Sorry, should be $x=t^2-2t.$ • October 17th 2010, 06:12 AM HallsofIvy Mathematics requires great precision. If you do not take the time and care to read the problem correctly you are are never going to get anywhere. " $x= t^2- 2t$" has exactly the same difficulty as " $x= t^2+ 2t$". $x= t^2- 2t= t^2- 2t+ 1- 1= (t- 1)^2- 1$ so that x is never less than -1. The curve given by $x= t^2- 2t$, $y= t^2$ does not intersect x= -3 so the curves you give do NOT bound a region. If it were $x= 2t- t^2= 1- (1- 2t+ t^2)= 1- (t- 1)^2$, $x= -3$ when $-3= 1- (t-1)^2$ or $(t-1)^2= 4$, $t- 1= \pm 2$, t= -1 and t= 3. At those points, y= 1 and 9. Now, find the gradient of F(x,y) and set it equal to 0 to find any points in the interior of the region at which there may be max or min. Then set x=-3 in F(x,y) to get a function in the single variable y giving the value of F on the line x= -3. You can set the derivative of that equal to 0 to find any critical points there (remember that y must be between 1 and 9 to be on that boundary). Set $x= 2t- t^2$, $y= t^2$ in F to get a function of the single variable t on the parabola. Set its derivative equal to 0 to find any critical points there (remember that t must be between -1 and 3). Evaluate the function F(x,y) at each of the critical points you have found, as well as at the vertices (-3, 1) and (-3, 9) to determine the largest and smallest values of F in that region. All times are GMT -8. The time now is 11:13 AM.	{"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": 17, "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": 0.9486417174339294, "perplexity_flag": "middle"}
http://physics.stackexchange.com/questions/12353/dalemberts-principle-and-the-term-containing-the-reversed-effective-force/12355	D'Alembert's Principle and the term containing the reversed effective force For our Classical Mechanics class, I'm reading Chapter 1 of Goldstein, et al. Now I come across Eq. (1.50). To put it in context: $$\begin{align} \sum_i{\dot{\mathbf{p}_i} \cdot \delta\mathbf{r}_i}&=\sum_i{m_i\ddot{\mathbf{r}}_i \cdot \delta{\mathbf{r}_i}}\\ &=\sum_{i,j}{m_i\ddot{\mathbf{r}}_i} \cdot \frac{\partial\mathbf{r}_i}{\partial q_j} \delta q_j \end{align}$$ Consider now the relation Eq. (1.50): $$\begin{align} \sum_{i,j}{m_i\ddot{\mathbf{r}}_i} \cdot \frac{\partial\mathbf{r}_i}{\partial q_j}&= \sum_i{\left[ \frac{d}{dt} \left( m_i\dot{\mathbf{r}}_i \cdot \frac{\partial\mathbf{r}_i}{\partial q_j} \right) - m_i\dot{\mathbf{r}}_i \frac{d}{dt} \left( \frac{\partial \mathbf{r}_i}{\partial q_j} \right) \right]} \end{align}$$ I'm at a loss for how he resolved it that way. He goes on to explain that we can interchange the differentiation with respect to $t$ and $q_j$. My question is: Why is there a subtraction in Eq. (1.50)? - 1 Answer Why is there a subtraction in Eq. (1.50)? Goldstein is using the Leibniz rule for differentiation of a product $$\frac{d (fg)}{dt}~=~\frac{d f}{dt}g + f\frac{d g}{dt}$$ with $$f=m_i\dot{\mathbf{r}}_i$$ and $$g=\frac{\partial \mathbf{r}_i}{\partial q_j}.$$ The minus is caused by moving a term to the other side of the equation. - I see. So the left-hand term in my question is actually $g\frac{df}{dt}$, first term on the right-hand is $\frac{d\left(fg\right)}{dt}$, right? – Kit Jul 17 '11 at 11:33 @Kit: Yes, you are right. – Qmechanic♦ Jul 17 '11 at 11:41	{"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": 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": 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": 0.9295858144760132, "perplexity_flag": "head"}
http://mathoverflow.net/revisions/41172/list	## Return to Answer 3 added 423 characters in body; added 1 characters in body; edited body; edited body This is all best explaied by working with groupoids in schemes. By this, I mean, groupoid objects in the category of schemes over $S$. A groupoid object $\mathcal G$ consists of two schemes $\mathcal G_0$ (the objects) and $\mathcal G_1$ (the morphisms) and a whole bunch of maps between them: • a map $s:\mathcal G_1\to \mathcal G_0$ that sends an arrow to its source. • a map $t:\mathcal G_1\to \mathcal G_0$ that sends an arrow to its target. • a map $e:\mathcal G_0\to \mathcal G_1$ that sends an object to its identity arrow. • a map $i:\mathcal G_1\to \mathcal G_1$ that sends an arrow to its inverse. • a map $m:\mathcal G_1\times_{\mathcal G_0}\mathcal G_1\to \mathcal G_1$ that composes arrows. subject to even more axioms. From now on, I will simply say "groupoid" instead of groupoid abject object in schemes. Let $G$ be a group over $S$. Then, there is a groupoid called $EG$, whose objects are $G$, and whose arrows are $G\times_S G$. The groupoid $EG$ is equivelent to $S$, viewed as a groupoid with only identity morphisms. The group $G$ acts freely on $EG$ (this is all happening in the category of schemes over $S$), and the quotient $EG/G$ is $BG$. Here, $BG$ is the groupoid with objects $S$ and morphisms $G$. If $X$ is a scheme over $S$, then $[X/G]$ is the groupoid whose objects are $X$, and whose morphisms are $X\times _S G$. This groupoid can also be described as the quotient of the (free) diagonal action of $G$ on the groupoid $X\times EG$. You're asking: "why is the construction of the classifying stack is not related to the notion of the homotopy type, whereas the construction of the classifying space relies on the fact that $EG$ has the homotopy type of a point?". There are two ways of answering your question: - One is to make the algebraic-geometric story look a little bit more like what people do in topology. That's what I did above. In particular, the fact that the groupoid $EG$ is equivalent to $S$ is the analog of the fact that $EG$ is contractible in topology. - The other is to make the topological story a little bit like what people do in algebraic geometry. Namely, instead of defining $BG$ as $EG/G$, define it as the space that represents the functor $X\mapsto$ {iso-classes of $G$-bundles over $X$} where I'm now working in the category of topological spaces and homotopy classes of maps. I should maybe add: There is a stack (over the category of $S$-schemes) associated to any groupoid object in $S$-schemes. The converse is not true for general stacks. But it is true, essentially by definition, for Artin stacks. If you restrict yourself to Deligne-Mumford stacks, then you can also assume that the groupoid is pareticularly nice, namely, that the maps $s$, $t$, $e$, $i$ and $m$ are étale. 2 added 796 characters in body; added 2 characters in body This is all best explaied by working with groupoids in schemes. By this, I mean, groupoid objects in the category of schemes over $S$. A groupoid object $\mathcal G$ consists of two schemes $\mathcal G_0$ (the objects) and $\mathcal G_1$ (the morphisms) and a whole bunch of maps between them: • a map $s:\mathcal G_1\to \mathcal G_0$ that sends an arrow to its source. • a map $t:\mathcal G_1\to \mathcal G_0$ that sends an arrow to its target. • a map $e:\mathcal G_0\to \mathcal G_1$ that sends an object to its identity arrow. • a map $i:\mathcal G_1\to \mathcal G_1$ that sends an arrow to its inverse. • a map $m:\mathcal G_1\times_{\mathcal G_0}\mathcal G_1\to \mathcal G_1$ that composes arrows. subject to even more axioms. From now on, I will simply say "groupoid" instead of groupoid abject in schemes. Let $G$ be a group over $S$. Then, there is a groupoid called $EG$, whose objects are $G$, and whose arrows are $G\times_S G$. The groupoid $EG$ is equivelent to $S$, viewed as a groupoid with only identity morphisms. The group $G$ acts freely on $EG$ (this is all happening in the category of schemes over $S$), and the quotient $EG/G$ is $BG$. Here, $BG$ is the groupoid with objects $S$ and morphisms $G$. If $X$ is a scheme over $S$, then $[X/G]$ is the groupoid whose objects are $X$, and whose morphisms are $X\times _S G$. This groupoid can also be described as the quotient of the (free) diagonal action of $G$ on the groupoid $X\times EG$. [Sorry, once again You're asking: "why is the construction of the classifying stack is not related to the notion of the homotopy type, I pressed whereas the construction of the classifying space relies on the "POST" button too early..fact that $EG$ has the homotopy type of a point?". There are two ways of answering your question: - One is to make the algebraic-geometric story look a little bit more like what people do in topology. That's what I did above. In particular, the fact that the groupoid $EG$ is equivalent to $S$ is the analog of the fact that $EG$ is contractible in topology. - The other is to make the topological story a little bit like what people do in algebraic geometry. Namely, instead of defining $BG$ as $EG/G$, define it as the space that represents the functor $X\mapsto$ {iso-classes of $G$-bundles over $X$} where I'm still writing my answer]now working in the category of topological spaces and homotopy classes of maps. 1 This is all best explaied by working with groupoids in schemes. By this, I mean, groupoid objects in the category of schemes over $S$. A groupoid object $\mathcal G$ consists of two schemes $\mathcal G_0$ (the objects) and $\mathcal G_1$ (the morphisms) and a whole bunch of maps between them: • a map $s:\mathcal G_1\to \mathcal G_0$ that sends an arrow to its source. • a map $t:\mathcal G_1\to \mathcal G_0$ that sends an arrow to its target. • a map $e:\mathcal G_0\to \mathcal G_1$ that sends an object to its identity arrow. • a map $i:\mathcal G_1\to \mathcal G_1$ that sends an arrow to its inverse. • a map $m:\mathcal G_1\times_{\mathcal G_0}\mathcal G_1\to \mathcal G_1$ that composes arrows. subject to even more axioms. From now on, I will simply say "groupoid" instead of groupoid abject in schemes. Let $G$ be a group over $S$. Then, there is a groupoid called $EG$, whose objects are $G$, and whose arrows are $G\times_S G$. The groupoid $EG$ is equivelent to $S$, viewed as a groupoid with only identity morphisms. The group $G$ acts freely on $EG$ (this is all happening in the category of schemes over $S$), and the quotient $EG/G$ is $BG$. Here, $BG$ is the groupoid with objects $S$ and morphisms $G$. If $X$ is a scheme over $S$, then $[X/G]$ is the groupoid whose objects are $X$, and whose morphisms are $X\times _S G$. This groupoid can also be described as the quotient of the (free) diagonal action of $G$ on the groupoid $X\times EG$. [Sorry, once again, I pressed on the "POST" button too early... I'm still writing my answer]	{"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": 118, "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": 0.9471263885498047, "perplexity_flag": "head"}
http://math.stackexchange.com/questions/227263/inverse-functions-in-real-analysis	# inverse functions in real analysis Can anyone help me with this? Spose $I$ is a closed interval from $0$ to $1$ and let $f$ go from $I$ to $R$ be defined by $f(x) = x$ for $x$ rational, and $f(x) = 1-x$ for $x$ irrational. Show that $f$ is injective on $I$ and that $f(f(x)) = x$ for all $x \in I$. Show that $f$ is cont. only at the point $x = 1/2$ - 1 It's strange how "Show that $f$ is injective on $I$" is before "$f(f(x))=x$", considering the former follows from the latter. – wj32 Nov 2 '12 at 2:56 ## 1 Answer First note that $x$ and $f(x)$ are either both simultaneously rational or simultaneously irrational. To prove that it is injective, let $f(x_1) = f(x_2)$. Then $f(x_1)$ and $f(x_2)$ are both rational or both irrational. Hence, either $x_1 = x_2$ or $1-x_1 = 1-x_2$. Both imply that $x_1 = x_2$. Hence, $f(x)$ is injective. If $x$ is rational, then $f(f(x)) = f(x) = x$. If $x$ is irrational, then $f(f(x)) = f(\underbrace{1-x}_{\text{irrational}}) = 1-(1-x) = x$. To show that $f$ is not continuous for all $x \neq \dfrac12$, approach $x$ by a sequence of rationals to get a limit of $x$ and a sequence of irrationals to get a limit of $1-x$. For instance, if $x \neq \dfrac12$ is rational, consider the sequence $x_n = x-\dfrac{x}{n\sqrt{2}}$, to get $$\lim_{n \to \infty} f(x_n) = 1-x \neq x = f(x)$$ For instance, if $x \neq \dfrac12$ is irrational, consider the sequence $x_n = \dfrac{\lfloor 10^n x\rfloor}{10^n}$, to get $$\lim_{n \to \infty} f(x_n) = x \neq 1-x = f(x)$$ To show that $f$ is continuous at $x = \dfrac12$. Note that $\vert f(1/2+h) - f(1/2) \vert = h$, irrespective of $h$ being rational or irrational. -	{"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": 45, "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": 0.3, "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": 0.9603086709976196, "perplexity_flag": "head"}
http://mathhelpforum.com/calculus/142763-l-hopital-s-questions.html	# Thread: 1. ## L'Hopital's Questions Hi I need some help on the following questions: 1) Use L'Hopital's rule to obtain: $\lim_{x \to 0} \frac{e^{x}-x-1}{3x^2}$ $f'(x) = \frac{3x^{2}(e^{x}-1)-(e^{x}-x-1)6x^2}{9x^4}$ $f'(x) = \frac{3x^{2}e^{x} - 6xe^{x}+3x^2+6x}{9x^4}$ $f'(x) = \frac{3x(xe^{x} - 2e^{x} +3x+2}{9x^4}$ $<br /> f'(x) = \frac{(xe^{x} - 2e^{x} +3x+2}{x^3}$ so if x=0, i get 0 however book's answer says $\frac{1}{6}$ 2) Use L'Hopital's rule to obtain: $\lim_{x \to \infty} \frac{x+2}{x^2+2x+1}$ $f'(x) = \frac{1}{2x+2} = \frac{1}{2\infty+2}$ The only problem i am having is what does $\infty$ equal to? P.S 2. Originally Posted by Paymemoney Hi I need some help on the following questions: 1) Use L'Hopital's rule to obtain: $\lim_{x \to 0} \frac{e^{x}-x-1}{3x^2}$ $f'(x) = \frac{3x^{2}(e^{x}-1)-(e^{x}-x-1)6x^2}{9x^4}$ $f'(x) = \frac{3x^{2}e^{x} - 6xe^{x}+3x^2+6x}{9x^4}$ $f'(x) = \frac{3x(xe^{x} - 2e^{x} +3x+2}{9x^4}$ $<br /> f'(x) = \frac{(xe^{x} - 2e^{x} +3x+2}{x^3}$ so if x=0, i get 0 however book's answer says $\frac{1}{6}$ This is NOT L'Hopital's rule! You have taken the limit of the derivative of f which is NOT, in general, the limit of f. (And even for that you would NOT get "0"- you would get $\frac{1}{0}$ which does not exist. Your f' does not have a limit.) L'Hopital's rule says that if, at x= a, $\frac{f(x)}{g(x)}$ is the indeterminant form " $\frac{0}{0}$" then $\lim_{x\to a}\frac{f(x)}{g(x)}= \frac{\lim_{x\to a} f'(x)}{\lim_{x\to a} g'(x)}$ provided the limits on the right exist. That is, you differentiate the numerator and denominator separately NOT using the "quotient law". Here, the numerator is $e^x- x- 1$ and the denominator is $3x^2$ which are 0 when x= 0, giving the form $\frac{0}{0}$ so we can apply L'Hopital's rule. The derivative of $e^x- x- 1$ is $e^x- 1$ and the derivative of $3x^2$ is $6x$. Those are both 0 at x= 0 so apply L'Hopital's rule again. The derivative of $e^x- 1$ is $e^x$ and the derivative of $6x$ is $6$ which are 1 and 6 at x= 0. The limit is $\frac{1}{6}$. 2) Use L'Hopital's rule to obtain: $\lim_{x \to \infty} \frac{x+2}{x^2+2x+1}$ $f'(x) = \frac{1}{2x+2} = \frac{1}{2\infty+2}$ Okay, here you did differentiate numerator and denominator separately- although it should not be labeled "f' ". The only problem i am having is what does $\infty$ equal to? P.S It doesn't equal anything- $\infty$ is not a number, just shorthand for "gets larger without bound." Here you have a fixed number, 1, over a denominator that gets larger and larger without bound- think of $\frac{1}{100}$, $\frac{1}{10000}$, $\frac{1}{10000000}$, etc. What is happening to those numbers as the denominator gets larger and larger? 3. so for question 2 the book's answer says it 0 how did the get that?? 4. Did you look at the numbers I suggested? $\frac{1}{100}$, $\frac{1}{100000}$, $\frac{1}{100000000000}$ are equal to what in decimal form? If the numerator of a fraction is a constant, and the denominator gets larger and larger, what happens to the fraction? 5. closer and closer to 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": 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": 0.3, "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": 0.9532144069671631, "perplexity_flag": "head"}
http://mathhelpforum.com/pre-calculus/86514-inverse-function.html	# Thread: 1. ## Inverse Function Is the inverse of f(x) = 9-2x^2 f-1(x) = -sqrt( (9/2) - (x/2) ) Thanks, D. 2. Originally Posted by dboyd435 Is the inverse of f(x) = 9-2x^2 f-1(x) = -sqrt( (9/2) - (x/2) ) Thanks, D. The working in your post is correct except I do not see why you can't have a plus or minus infront of that square root unless there is something about the domain you're not telling me $y = 9-2x^2$ $2x^2 = 9-y$ $x^2 = \frac{9-y}{2}$ $x = \pm \sqrt{\frac{9-y}{2}}$ $f^{-1}(x) = \pm \sqrt{\frac{9-x}{2}}$ Bear in mind that's not a function since it's not one to one - the domain would have to be restricted to make it a function. t 3. Thanks, ya, I meant to put +/- but i guess it didn't type out for some reason.	{"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": 0.9480859041213989, "perplexity_flag": "middle"}
http://mathoverflow.net/questions/81265/getting-b-t-from-its-local-times-lx-t	## Getting $B_t$ from its local times $L^x_t$ ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) Hi Given a Brownian Motion $B_t$ is it possible to reconstruct it from the knowledge of the local times $L^x_t$ ? Using occupation time formula this would mean solving for some $f$ the following equation : $$B_t=\int_{-\infty}^{+\infty}f(x)L^x_t.dx=\int_0^t f(B_s)ds$$ This seems achievable but I couldn't find out the solution or prove that there is none. By the way, if someone can achieve this reconstruction of $B_t$ from $L^x_t$ using some other device than the occupation formula I would be equally interested. Best regards - The closest $f$ I could find was $f(x)=x$ for which we get $\int_0^t B_s ds =\int_{-\infty}^{+\infty}x.L^x_tdx$. – The Bridge Nov 18 2011 at 17:03 1 Other than Yuri's answer, what's wrong with just writing $B_t=\frac{d}{dt}\int_{-\infty}^\infty xL^x_t\,dx$? – George Lowther Nov 19 2011 at 5:00 @George Lowther : What you say is true but I think it is hard to work with such an expression. Best regards – The Bridge Nov 19 2011 at 7:55 ## 2 Answers Knowing local times you can derive if the path $\gamma={(t,B_t): t\in[0,T]}$ passes through any rectangle of the following form: $[k/2^n,(k+1)/2^n]\times[j/2^n,(j+1)/2^n]$. For fixed $n$, denote by $G_n$ the union of all these visited rectangles. Since $B_t$ is uniformly continuous on $[0,T]$, we have $\gamma=\bigcap_n G_n$. - @yuri : Thank's for this very nice idea. But what if all I have is the realisation of $L^x_t$ (for any $x$ at fixed time) but no information on the Brownian path itself, do you think it is still possible to work out $B_t$'s value ? Best regards – The Bridge Nov 21 2011 at 8:41 @The Bridge: Ah, $t$ is fixed! Then no. Think about a trajectory that passes through a point 3 times before $t$ with 2 excursions between them. If you exchange the order of those two excursions you get a different trajectory that has same local times at terminal point $t$. This trajectory is "as probable as the original one". – Yuri Bakhtin Nov 21 2011 at 15:57 @Yuri : Sorry about the "fixed time" condition as this was not included as a condition in my original question (note that I accepted your answer). Regarding your comment, I am not completely convinced by the argument, because first I have all $L^x$ so if two paths coincide at a particular $x$ then at another level they should differ (at least intuitively), second the fact that many paths can lead to $B_t$ is not really a problem as long as the information included in all $L^x$ can lead to the right $B_t$. Best regards – The Bridge Nov 21 2011 at 17:25 @The Bridge: To be honest, I do not know how to explain my argument more precicely in a short message. It seems obvious that the occupation times do not change if you cut your trajectory in pieces, move them rigidly around (only along time axis) and reglue them again. – Yuri Bakhtin Nov 21 2011 at 21:24 @yuri : Hi I got your point about the "swap and glue" excursions path transform that is "invariant" for local times. The thing that I still don't clearly "see" is why does this process makes $B_t$ different after the swap and glue path transformation. Best regards – The Bridge Nov 22 2011 at 7:22 show 3 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. This paper seems to answer (something very close to) your question : Warren, J. and Yor, M. (1998), The Brownian burglar: conditioning Brownian motion by its local time process. Seminaire de Probabilites XXXII, pages 328-342. (pdf link) Abstract : Imagine a Brownian crook who spent a month in a large metropolis. The number of nights he spent in hotels A,B,C...etc. is known; but not the order, nor his itinerary. So the only information the police has is total hotel bills..... - @pgassiat : Thank's for the reference but I wasn't able to link the mathematical results of the article with my question, even though, the "down to earth" explanation of Burglar's process seemed promising in this regard. But often Yor's articles stay way too far from my level of understandings. Best Regards – The Bridge Jan 23 2012 at 8:07	{"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": 30, "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": 0.3, "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": 0.9392126202583313, "perplexity_flag": "middle"}
http://mathoverflow.net/questions/19014/finding-questions-between-functional-analysis-and-set-theory/23880	## Finding questions between functional analysis and set theory ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) Are there some good questions on functional analysis whose solution depends on tools in set theory? My major is mathematical logic, I think tools in set theory, especially infinity combinatorics and forcing, should be used to solve some questions in functional analysis. For functional analysis, I just have read the main part of Conway's textbook. In this book, I have not found such questions. - 2 "should" or "could" be used? – Yemon Choi Mar 22 2010 at 17:08 1 Tom, would you object less to the question if it asked about applying "advanced tools of set theory", or something along those lines? – Mark Meckes Mar 22 2010 at 23:19 1 I kind of disagree with the premise. Almost all of functional analysis, as customarily phrased, depends on set theory. A Banach space is defined as a set equipped with certain structure. An operator is a function with certain properties. When you take the direct sum of vector spaces, you use the set-theoretic theorem/axiom that cartesian products of sets exist. When you define "invariant subspace", you use the notion of the image of a subset under a function. I know that's not what you're asking, but I think it's a disservice to set theory to speak as if it consists only of "exotic stuff". – Tom Leinster Mar 22 2010 at 23:20 1 Mark, yes, I would object less, or not at all. But I think there's a bit more than that going on in the background, to do with social attitudes towards set theory and the extent to which it's viewed as esoteric. A MO comment isn't the place for an explanation, though. (By the way, I slightly edited my comment, i.e. deleted and replaced it, at the same time as you were writing your comment, which is why they appear out of order.) – Tom Leinster Mar 22 2010 at 23:24 It's slightly strange (for me) to read you say, Tom, that "a Banach space is defined as a set equipped with certain structure", since I've had a vague but long-standing feeling that this POV has hindered functional analysts from fully profiting from categori(c)al ideas. But this is connected to my feeling that the forgetful functor from Ban to Set is not the most profitable one to lean on... – Yemon Choi Mar 23 2010 at 16:04 ## 6 Answers One of the most elementary examples is the use of the infinite version of Ramsey's theorem to prove Rosenthals $\ell_1$ theorem. See Chapter 10 in Albiac & Kalton, Topics in Banach Space Theory. You'll find ultraproducts in Chapter 11. More model theory than set theory, but still logic. Deeper set theory (large cardinals, for example) have also been used. Check out, e.g., Todorcevic, S* on MathSci Net for recent things. - This book should be what I was searching for. Thank you. – Ant emyy Lee Mar 22 2010 at 14:44 3 Also, Ilijas Farah has several related results. – François G. Dorais♦ Mar 22 2010 at 17:12 1 Was just writing a long answer, but you beat me to a mention of Farah – Yemon Choi Mar 22 2010 at 17:16 ### 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 interesting example is A discontinuous homomorphism from C(X) without CH, by W. Hugh Woodin, which begins with the following introduction. Suppose that X is an infinite compact Hausdorff space and let C(X) be the algebra of continuous real-valued functions on X. Then C(X) is a commutative Banach algebra relative to the supnorm: \|\| f \|\| = sup{ f(p) \| p ∈ X }. A well-known question of I. Kaplansky posed around 1947 asks whether every algebra homomorphism of C{X) into a Banach algebra B is necessarily continuous. There is a discontinuous homomorphism of C(X) if and only if there is a discontinuous homomorphism of C(X, C), the C-algebra of continuous complexvalued functions on X. We prefer to deal with the real case; some of the references adopt the complex view. The question is now known to be independent of the axioms of set theory, ZFC. H. G. Dales [1] and J. Esterle [3] independently constructed discontinuous homomorphisms of C{X) for any infinite space X assuming the Continuum Hypothesis, CH. About the same time R. Solovay [7] proved that it is relatively consistent with ZFC that every homomorphism of C(X) for any space X is necessarily continuous. Solovay's result was improved [8] fairly soon thereafter, to obtain the relative consistency with ZFC + Martin's Axiom (ZFC + MA) that every homomorphism of C(X) for any space X is continuous. We refer the reader to [2] for an exposition of the latter result concerning MA, historical points and related results. After these results several questions remained. This paper is concerned with the question of whether the existence of a discontinuous homomorphism of C{X) is possible given the failure of the Continuum Hypothesis. A standard method to obtain the consistency of a proposition with the negation of the Continuum Hypothesis (¬CH) when a proof of the proposition assuming CH is known is to attempt to use MA + ¬CH in place of CH and prove the proposition. However by the result indicated above, this approach will not succeed. The main theorem of this paper, formulated using the terminology of forcing, is the following. THEOREM. Assume CH. Let P be the Cohen partial order for adding ω2 Cohen reals. Then in VP there exists a discontinuous homomorphism of C(X)for every infinite compact Hausdorff space X. - en. This is a good example. Also, do you know any questions as this style which is still open now? – Ant emyy Lee Mar 22 2010 at 14:32 2 Woodin mentions the principal open question in this area on the second page of that article, which is whether ZFC+MA+not-CH is consistent with a discontinuous homomorphism. But I would expect that any such question is hard, given that Solovay and Woodin have both worked on this topic. – Joel David Hamkins Mar 22 2010 at 14:37 There are many questions in this field worth working on. To get a good idea, you may want to consult Dales's monograph on Automatic continuity and his survey papers; for closely related material, se ethe book by Dales and Woodin on Super-real fields. – Andres Caicedo Jan 20 2011 at 5:51 The survey article Set theory and C-algebras'' by Nik Weaver might have some things along the lines you are looking for ( Bull. Symbolic Logic Volume 13, Issue 1 (2007), 1-20; see also math/0604198 on the arXiv). For a particular example of Weaver's recent work in this area, see Akemann & Weaver's paper Consistency of a counterexample to Naimark's problem We construct a C*-algebra that has only one irreducible representation up to unitary equivalence but is not isomorphic to the algebra of compact operators on any Hilbert space. This answers an old question of Naimark. Our construction uses a combinatorial statement called the diamond principle, which is known to be consistent with but not provable from the standard axioms of set theory (assuming those axioms are consistent). We prove that the statement there exists a counterexample to Naimark's problem which is generated by $\aleph_1$ elements'' is undecidable in standard set theory. There has also been some work by I. Farah on applying set-theoretical technqiues to operator-algebraic problems. - Thanks for adding this Yemon. I think Farah's work is fascinating and many of the functional analysts that I know have told me similar things. – François G. Dorais♦ Mar 22 2010 at 20:25 Ah, I was going to mention Weaver, but you beat me to it! – Matthew Daws May 7 2010 at 20:57 Although the construction of Tsirelson's space doesn't use set theory per se, in this short essay Tsirelson recounts (among other things) how his construction was inspired by forcing. - Thanks for the link, Mark. In a follow up paper to Tsirelson's, Figiel and I wrote "The 'forcing method' employed by Tsirelson..." because it reminded us of forcing as used in set theory, but we did not know that Tsirelson really had been exposed to forcing. – Bill Johnson Mar 22 2010 at 18:02 Chris Phillips and Nik Weaver wrote a paper called The Calkin Algebra has Outer Automorphisms, where they showed that the Continuum Hypothesis implies that the Calkin algebra $\mathcal{B(H)/K(H)}$ has outer automorphisms. See also this paper of Farah, McKenney, and Schimmerling, and references therein; they show that it is relatively consistent with ZFC that the Calkin algebra has only inner automorphisms, and hence the question of existence of outer automorphisms is independent of ZFC. - 1 That paper by Farah et al. dealt with generalized Calkin algebras. The consistency of all automorphisms being inner was shown in an earlier paper by Farah alone, All automorphisms of the Calkin algebra are inner (arXiv:0705.3085). – Nik Weaver Apr 19 at 1:46 @Nik, thanks for the correction. – MTS Apr 19 at 16:53 Sure, thanks for mentioning my paper. Incidentally, Ilijas gave another proof of our consistency result in his paper, and I think his proof is better. Our overall strategy seems more natural but making it work was really hard. His proof goes in an unexpected direction but it ends up being easier. – Nik Weaver Apr 19 at 18:10 Though this is a little more advanced, there is actually some very exciting research right now at the intersection of descriptive set theory, ergodic theory, and von Neumann algebras. It is quite striking that the three areas have powerful tools for looking at similar problems, and yet tend to be applicable in different cases. For a nice introduction to some of these ideas from a more set-theoretical point of view I would say check out "Topics in Orbit Equivalence" by Kechris and Miller.	{"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": 2, "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": 0.9450232982635498, "perplexity_flag": "middle"}
http://physics.stackexchange.com/questions/19254/violation-of-conservation-of-energy-and-potential-energy-between-objects	# Violation of conservation of energy and potential energy between objects I would like to clarify my question. I have numbered them to be independent questions 1. For any conservative fields, $\vec{F} = -\nabla U$. Which means the restoring force is opposite to the increasing values of the potential energy. If it so happens that $F = \nabla U$, does that mean the restoring force (not even called that anymore I guess) would be in the same direction as the increasing values of energy? Is this a violation to the first law in thermodynamics. 2. For attracting bodies, the closer the objects come together, the potential energy drops as $r \to 0^-$ and $U \to -\infty$, but for repelling bodies we get $r \to 0^-$ as $U \to \infty$ which means the potential energy actually increase. Am I right? - 1 Welcome to Physics.SE! Could you please state your question a little more clearly? – dmckee♦ Jan 8 '12 at 0:09 Hmm you are right. I can't put it all in this commentbox. But one question is, for repelling bodies, will their potential energy increase or decrease as they come together? How will this affect the electric potential? – lem Jan 8 '12 at 0:13 You should edit that into the text of the question. Given how much text you have you might want to highlight it a bit – dmckee♦ Jan 8 '12 at 0:45 I totally don't understand what you are asking. – C.R. Jan 8 '12 at 0:54 You seem to misinterpret what the equations say and how their solutions are used. The solutions at a given time and space are a snapshot. They do not contain a change in energy in any form. Taking the difference as you are doing and finding a change in energy between the two solutions, tells one that if energy conservation is not to be violated something else is taking it up or giving it in. In your example the kinetic energy changes, and becomes potential energy(repulsive), or the potential energy turns into kinetic energy (attractive), always conserving energy. – anna v Jan 8 '12 at 6:13 ## 1 Answer You are confused because the potential energy for repelling bodies is positive, and for attractive bodies is negative, but in both cases $F = -\nabla U$. The force always points towards decreasing potential energy. There is no force which is in the direction of increasing potential energy, because of the reason you state--- such a force would push things to have more kinetic energy at the same time as they get more potential energy, violating conservation of energy. Flipping the sign of U means that you flip the sign of the force, but the force is still in the direction of maximally decreasing potential. As for question 2, you are right. -	{"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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9563557505607605, "perplexity_flag": "head"}
http://physics.stackexchange.com/questions/38237/is-there-a-time-delay-during-tunnelling/45815	# Is there a time delay during tunnelling? A particle hitting a square potential barrier can tunnel through it to get to the other side and carry on. Is there a time delay in this process? - ## 2 Answers There is a roughly constant time delay during tunnelling that is roughly proportional to $\hbar$ over the barrier height. Consider for simplicity a monoenergetic particle beam of energy $E=\hbar^2k^2/2m$, trying to tunnel through a barrier of length $l$ and height $V_0$. This means that the wavefunction coming up to the barrier will go as $e^{ikx}$, and some portion of it will tunnel through and make a tunnelled wavefunction right of the barrier going as $A_\textrm{tunnel}e^{ikx}$. This is the standard treatment as given in UG texts on QM (I'm working off of Cohen-Tannoudji) and it can be expanded to accommodate a time-dependent, moving wavepacket by superposing different energies in the usual way. The tunnelling time is given by the phase difference between the incoming and the tunnelled components - that is, in the phase of $A_\textrm{tunnel}$. This phase counts the number of wave peaks and troughs that the outgoing wavefunction has been delayed by, which essentially gives the time delay you want. For a time-dependent wavepacket, the different phases between different-energy components make up the position where the wavepacket is, so it should be no surprise. In such a situation, dividing the accumulated phase by the central energy will give the time. (In a related note, it's important to realize that unless the pulse is very narrow in energy and thus quite broad in time then the barrier will be highly dispersive and can quite radically alter the pulse's shape. This is because the higher-energy components will tunnel more easily than the longer-wavelength ones.) The details of the square-barrier tunnelling are given in Cohen-Tannoudji, for example (vol I, 1977 English edition, p. 72), from which one can get the tunnelling amplitude as $$A_\textrm{tunnel}=\frac{e^{-\kappa l}}{\cosh(\kappa l)- i\frac{k^2-\kappa^2}{k\kappa}\sinh(\kappa l)}$$ where $\kappa=\sqrt{2m(V_0-E)}/\hbar$. The first important feature is that this is indeed exponentially decreasing with $l$, going as $A_\textrm{tunnel}\propto e^{-2\kappa l}$ for largish $l\gtrsim 1/\kappa$. The second important feature is that it does have a nontrivial phase. For simplicity consider the case of a pretty large barrier, for which $$A_\textrm{tunnel}\approx\frac{e^{-2\kappa l}}{1- i\frac{k^2-\kappa^2}{k\kappa}} =\frac{ik\kappa e^{-2\kappa l}}{k^2-\kappa^2+ik\kappa} =\frac{ik\kappa e^{-2\kappa l}}{(k+i\kappa)^2} = e^{-2\kappa l}\frac{k\kappa}{\|k+i\kappa\|^2}e^{i\frac{\pi}{2}-2i\arctan(\kappa/k)} .$$ For $E=V_0/2$, you get $k=\kappa$ and the phase vanishes. For $E<V_0$ the phase approaches $e^{i\pi/2}$ as $E\rightarrow0$, and similarly it approaches $e^{i\pi/2}$ as $E\rightarrow V_0$ from the left. That is, the wavefunction gets delayed/advanced by at most 1/4 of a wavelength. For a real pulse the situation is a bit more complicated because you should distinguish between a shift in the global phase, which is irrelevant, and a $k$-dependent phase that will shift the pulse. Thus it's the derivative $\partial \varphi/$ that matters (where $A_\textrm{tunnel}=\|A_\textrm{tunnel}\|e^{i\varphi}$). Since $\varphi$ is more or less a straight line from $E=0$, $\varphi=\pi/2$ to $E=V_0$, $\varphi=-\pi/2$, the upshot is that the final, tunnelled pulse is displaced to the left in space by about $$\Delta x=\frac{\partial \varphi}{\partial k}(k_0)=-\frac{\pi\hbar}{V_0}\frac{\hbar k_0}{m}$$ and this corresponds to a delay in time by $h/2V_0$. (Note that the weird scaling at "low" $V_0$ is offset by the requirement that $\kappa l\leq \sqrt{V_0}l$ be large in natural units, so the limit doesn't really make sense.) - if the delay is independent from the width of the tunnel, can't this in principle lead to FTL signal propagation? – lurscher Dec 3 '12 at 20:01 I don't think it can, since it's a further delay on the slower-than-light, nonrelativistic Schrödinger propagation of a free particle. – Emilio Pisanty Dec 4 '12 at 2:27 On the other hand, though, to be quite honest I think the maths is right but would be glad for someone else to go over it with a fine comb and to correct where necessary. – Emilio Pisanty Dec 4 '12 at 2:28 For a wave packet, it should be the barrier width divided by the packet group velocity, I think. - Well, what if the group velocity is imaginary? – Emilio Pisanty Dec 3 '12 at 17:28	{"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": 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": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9481409192085266, "perplexity_flag": "head"}
http://mathhelpforum.com/number-theory/178840-inverse-2-mod-108-quadratic-equation.html	# Thread: 1. ## Inverse of 2 mod 108 and the quadratic equation This is actually related to a problem in this thread, but is so much more basic I didn't want to interrupt any discussion that the original thread might generate. The problem is to solve $x^2 + 3x - 22 \equiv 0~\text{mod(108)}$ Since the details of solving the problem are irrelevent here I will simply state a pair of solutions to the quadratic: $x \equiv -\frac{3}{2} \pm \frac{1}{2} \cdot 23 \equiv \frac{1}{2} ( -3 \pm 23 ) ~\text{(mod 108)}$ So the solutions mod 108 are 10 and 95, which both work in the original equation. My question about all this. 1/2 has no multiplicative inverse mod 108. So even though,say, (1/2)(-3 + 23) = (1/2)(20) = (1/2)(210) = (1/2)210 is apparently equal to 10, how can we say (1/2)2 = 1? Thanks! -Dan 2. $\frac{1}{2}\times 2 \equiv 1\ (mod\ 108)$ The multiplicative inverse of 1/2 (mod 108) is 2. 3. Originally Posted by alexmahone $\frac{1}{2}\times 2 \equiv 1\ (mod\ 108)$ The multiplicative inverse of 1/2 (mod 108) is 2. I really don't want to dispute this because it obviously works, but both WolframAlpha and I agree that the equation 2x = 1 (mod 108) has no solution! -Dan Edit: I suppose you could say that this question is more Philosophically based than it is practical. 4. Originally Posted by topsquark I really don't want to dispute this because it obviously works, but both WolframAlpha and I agree that the equation 2x = 1 (mod 108) has no solution! -Dan It really depends on whether you're restricting yourself to integral solutions. For example, 2x=1 (mod 108) does have rational solutions given by x=1/2 (mod 108). 5. Originally Posted by alexmahone It really depends on whether you're restricting yourself to integral solutions. For example, 2x=1 (mod 108) does have rational solutions given by x=1/2 (mod 108). Well that would work. My apologies...I have never heard of allowing rational solutions in a modular system. $\mathbb{Z} _{108}$ contains only integers so I thought all solutions had to be in $\mathbb{Z} _{108}$? -Dan 6. Originally Posted by topsquark This is actually related to a problem in this thread, but is so much more basic I didn't want to interrupt any discussion that the original thread might generate. The problem is to solve $x^2 + 3x - 22 \equiv 0~\text{mod(108)}$ Since the details of solving the problem are irrelevent here I will simply state a pair of solutions to the quadratic: $x \equiv -\frac{3}{2} \pm \frac{1}{2} \cdot 23 \equiv \frac{1}{2} ( -3 \pm 23 ) ~\text{(mod 108)}$ So the solutions mod 108 are 10 and 95, which both work in the original equation. My question about all this. 1/2 has no multiplicative inverse mod 108. So even though,say, (1/2)(-3 + 23) = (1/2)(20) = (1/2)(210) = (1/2)210 is apparently equal to 10, how can we say (1/2)2 = 1? Thanks! -Dan The 'division' in the field of integers mod 108 is performed as 'ordinary division' and that means that n must be an even number. The equation You have written is solvable only if is a 'perfect square' mod 108 and is even. In Your case is... ... so that it is 'all right'... Kind regards $\chi$ $\sigma$	{"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": 10, "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": 0.9565876126289368, "perplexity_flag": "middle"}
http://math.stackexchange.com/questions/48230/investigations-about-the-trace-form	# Investigations about the trace form Let us (again) consider the bilinear form $\beta(A,B)=\operatorname{Tr}(AB)$ for $A,B \in \mathbb{F}^{n,n}$ (quadratic matrices over a field $\mathbb{F}$). I am interested in finding the biggest subspace $U \subset \mathbb{F}^{n,n}$ such that for all $A \in U: \beta(A,A)=\operatorname{Tr}(A^2)=0$. - What do you mean by "closed under addition and multiplication"? – Joel Cohen Jun 28 '11 at 14:50 That you can add linear combinations of matrices in $U$ and for the resulting matrix $B$ the fact $\beta(B,B)=0$ is still valid. – Listing Jun 28 '11 at 14:52 So if I understand correctly, you are looking for a generating set? – Joel Cohen Jun 28 '11 at 14:56 Yes I am looking for a basis of $U$. But if its not possible to give one, just the dimension of the maximal $U$ would already be great, too. – Listing Jun 28 '11 at 14:59 1 $Tr(A+B)$ will be $0$, because trace is additive, but $Tr((A+B)^2)$ will not be. Try basically any example. I'm hesitating to write out an answer, because this sort of thing is making it hard for me to understand what you're looking for. – David Speyer Jun 28 '11 at 17:17 show 7 more comments ## 1 Answer By coincidence, I learned very recently that the trace form on $M_n(k)$ is nondegenerate iff the characteristic of $k$ does not divide $n$. This was a mistake: it is in fact easy to see that the bilinear form $x,y \in M_n(k) \mapsto \operatorname{Trace}(xy)$ is nondegenerate for all $n \in \mathbb{Z}^+$ and all fields $k$. Indeed, if $x \neq 0$, then it has some $(i,j)$ entry nonzero, and then if you multiply on the right by the matrix $E_{ji}$ which has $(j,i)$ entry equal to $1$ and all other entries $0$, you get a matrix with nonzero trace. I was thinking of the definition of a strongly separable algebra, which is a separable $k$-algebra with nondegenerate trace form. But the trace form on an arbitrary finite dimensional algebra is the "unreduced trace", i.e, the trace of $x \bullet$ acting $k$-linearly on $A$. When $A$ is a matrix algebra (or more generally a central simple algebra), this unreduced trace is precisely $\sqrt{[A:k]}$ times the reduced trace, so when $A = M_n(k)$ it is $n$ times the usual matrix trace. Of course, when the characteristic of $k$ is divisible by $n$, this makes the unreduced trace identically zero, so $M_n(k)$ is not "strongly separable" (but the definition looks a little strange to me now, since it seems like we are focusing our attention on the wrong trace form). I will assume throughout that the characteristic of $k$ is not $2$ so that the standard algebraic theory of quadratic forms can be applied. You are asking for the maximal dimension of a totally isotropic subspace. If your quadratic form is nondegenerate, every totally isotropic subspace $U$ has an "isotropic supplement" $U'$ such that $U \cap U' = 0$, $\dim U = \dim U'$ and the quadratic form restricted to $U + U'$ is an orthogonal direct sum of $\dim U$ copies of the hyperbolic plane $\mathbb{H} = \langle 1, - 1 \rangle$. (See e.g. $\S 6$ of these notes on quadratic forms.) Therefore the dimension of a maximal totally isotropic subspace is equal to the number, say $r$, such that the Witt Decomposition of $q$ is $q \cong \bigoplus_{i=1}^r \mathbb{H} \oplus q'$, where $q'$ is anisotropic, i.e., $q'(x) = 0 \implies x = 0$. So we want to know the Witt Decomposition of the trace form. When $k = \mathbb{R}$, joriki's answer to your previous question shows that $q \cong \left(\frac{n^2+n}{2} \right) \langle 1 \rangle \oplus \left( \frac{n^2-n}{2} \right) \langle -1 \rangle \cong \left( \frac{n^2-n}{2} \right) \mathbb{H} + \left(n \right) \langle 1 \rangle$, so the number $r$ is $\frac{n^2-n}{2}$. This is different from the formula you have given -- in fact, eventually smaller -- so if I have not made a mistake then you have: you should check first of all that the subspace you have in mind is really totally isotropic. The next order of business is to compute the Witt Decomposition for the trace form over a more general field. Looking at the "matrix units" $E_{ij}$ as joriki did when $k = \mathbb{R}$ seems like a good start, but I haven't done this computation myself. - 1 @Listing: no problem. Let me emphasize that such a maximal subspace is not unique: there will be infinitely many such subspaces, all conjugate under the orthogonal group of the quadratic form. – Pete L. Clark Jun 28 '11 at 17:46 2 For an example, I think the subspace of upper triangular matrices with $0$ on the diagonal fits the bill. – Joel Cohen Jun 28 '11 at 21:58 Yes I deleted the comments. Now everything is clear, thank you. – Listing Jun 30 '11 at 10:21	{"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": 58, "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": 0.9269070029258728, "perplexity_flag": "head"}
http://mathoverflow.net/questions/46794?sort=oldest	## Generalization of the Structure theorem for artinian rings? ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) Let $A$ be a commutative ring with identity. If $A$ is a ring with only a finite set of prime ideals $p_1...p_n$ and moreover $\prod_{i=1}^n p_i^{k_i}=0$ for some k_i. Is $A$ then isomorphic to $\prod_{i=1}^nA_{(p_i)}$? - 1 Exactly on how many examples did you try this on? :) – Mariano Suárez-Alvarez Nov 21 2010 at 4:03 1 ti4: Did you ask this question math.stackexchange.com/questions/10980/… ? – Hailong Dao Nov 21 2010 at 4:03 1 Did you mean to assume that $A$ was Artinian? – Karl Schwede Nov 21 2010 at 4:28 Mariano: Perhaps I should've added that I thought it was wrong but could'nt find a convincing counterexample :) – Pandamic Nov 21 2010 at 11:31 Hailong: Yes I did, it is almost the same question, only this is a little weaker perhaps. – Pandamic Nov 21 2010 at 11:33 show 1 more comment ## 1 Answer No. Let $A$ be a DVR. It has two prime ideals: the maximal ideal $p_1=\mathfrak m\subset A$ and $p_2=(0)\subset A$. So, $p_1p_2=0$, but $A$ is not a product (of two local rings). - Perhaps I have missed something huge but in this example it seems to me that product will be the ring localized at m, i.e. itself multiplied by the trivial ring, is this not isomorphic to the original ring? – Pandamic Nov 21 2010 at 11:30 1 You probably need to think carefully about what it means to localize at the prime $(0)$. – Todd Trimble Nov 21 2010 at 11:42 Oh, I was being stupid. Thank you – Pandamic Nov 21 2010 at 12:06	{"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": 13, "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": 0.3, "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": 0.9374815821647644, "perplexity_flag": "middle"}
http://math.stackexchange.com/questions/194409/interpolation-error-for-the-exponential-function	# Interpolation error for the exponential function I'm studiyng for my exam of scientific computing, specifically to the subject of interpolation techniques, I'm stuck with this problem: How many equally spaced nodes must be taken to interpolate the function $f(x) = \exp(x)$ on the interval $[-1,1]$, so that the interpolation error is $0.5 * 10^{-4}$? I have no idea how to start or what should I do in the exercise. Any help? - ## 1 Answer You have to define the interpolation technique used. If you use linear interpolation between the nodes (which I suspect is what you are doing) you are using the first order Taylor's series. The error term is $\frac {f''(x)(x-a)^2}{2!}$. $x-a$ is at most half your node spacing, which you can choose to get the error low enough. What is the maximum value of $f''(x)$ on $[-1,1]$? - i think is 2.7182, right? – Melkhiah66 Sep 12 '12 at 5:31 @Melkhiah66: that's right, so plug that in and you can find $x-a$ from $\frac {e(x-a)^2}{2!} \lt 0.5\cdot 10^{-4}$ – Ross Millikan Sep 12 '12 at 13:02	{"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": 0.18, "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": 0.914711594581604, "perplexity_flag": "head"}
http://mathoverflow.net/questions/62302/examples-of-amenable-groups-other-than-z-n/62333	## Examples of Amenable Groups other than Z_n ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) I'm reading about amenable groups. What are explicit examples of nonabelian discrete amenable groups other than finite groups? Perhaps a group presentation or matrix representation would be useful. - 7 Shouldn't this question be community-wiki, since the goal is to have a list rather than to find a definitive answer? – Yemon Choi Apr 20 2011 at 1:02 ## 10 Answers To "specify" Alain's answer: 0) The group $\langle a,b \mid bab^{-1}=a^2\rangle$ (solvable of class 2 Baumslag-Solitar group) 1) The group of upper triangular $n\times n$ matrices with integer coefficients and 1 on the diagonal (nilpotent), $n\ge 1$. 2) The group of all permutations of $\mathbb{Z}$ with finite support (locally finite). 3) The subexp. growth groups, unfortunately, would require more space to define. But you can find them in Wiki. - ### 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. Solvable (which includes nilpotent and polycyclic); locally finite; subexponential growth. - 1 can you be more explicit? – John Mangual Apr 19 2011 at 18:09 The lamplighter group, defined as the wreath product $\mathbb{Z}/2\mathbb{Z} \wr \mathbb{Z}$, is amenable yet has exponential growth. It can be thought of as a bi-infinite sequence of street lamps, each of which can be turned on and off, and a lamplighter who moves along the sequence. The three generators of the group are to move the lamplighter right or left, and to switch the state of the lamp he is positioned in front of. With this picture in mind, it is easy to show the group has exponential growth. - Groups generated by bounded automaton are amenable: http://dx.doi.org/10.1215/00127094-2010-046 - Mustafa, the link is broken – Kate Juschenko Apr 19 2011 at 19:45 Thanks Kate i fixed it. – Mustafa Gokhan Benli Apr 19 2011 at 22:15 - All compact hausdorff groups are amenable because they have a Haar-measure which can be chosen with $\mu(G)=1$. So $O_n(\mathbb{R})$, $U_n(\mathbb{C})$, closed subgroups thereof as well as quotients by closed normal subgroups are amenable. - 2 Yes, but the original question was about discrete groups. – Alain Valette Apr 19 2011 at 21:26 Oh, right. I missed that... – Johannes Hahn Apr 20 2011 at 12:50 $\mathrm{Symm}(\mathbb{Z}) \leftthreetimes \mathbb{Z}$ - see page 4321 in http://www.cse.sc.edu/~maxal/a-g-g.pdf - Elementary amenable groups: the smallest class of groups which includes finite groups and commutative groups, and is closed under formation of homomorphic images, subgroups, group extensions (by an other element of the class) and directed unions. (See http://en.wikipedia.org/wiki/Elementary_amenable_group) - Solvable and locally finite are elementary, I guess those of subexponential growth are, too. - E.g. a solvable group is one that can be constructed from abelian groups using extensions. But now I see that locally finite groups are not elementary (for example, the free group on two generators is finitely generated, hence locally finite, but it is definitely not amenable, hence it cannot be elementary), sorry for that. - Free groups are not locally finite. I think you meant that inverse limit of finite groups need not be amenable. – Misha Apr 4 at 20:59 free groups are not inverse limits of finite groups neither (at least not in the category of groups). They are residually finite, anyway (that is, in the category of marked groups, the same as inverse limits of finite groups). – Yves Cornulier Apr 4 at 22:01 Yves, I meant that they embed in inverse limits of finite groups. – Misha Apr 4 at 23:03	{"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": 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": 0.18, "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": 0.9231602549552917, "perplexity_flag": "middle"}
http://mathoverflow.net/questions/26304/is-there-a-definition-of-analogue-weyl-group-for-lie-super-algebra/26332	## Is there a definition of analogue Weyl group for Lie super algebra? ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) I heard from some people working in Lie super algebra that there was no proper definition for Weyl group of Lie super algebra. I do not know Lie super algebra at all. But When I searched on Google, I found that it seems there still exists some definitions of Weyl group. I wonder whether there is a well-accepted definition for it. The reason I want to ask this question is that I need Weyl group for Lie super algebra to play with the geometry related to super Lie algebra. Another question is that I heard from some experts in Lie super algebra that there was no well-accepted super geometry related to Lie super algebra. However, It seems that one of the students of Manin, who is Dimitry Leites ever developed supergeometry. - The word "geometry" is terribly loose. On AG side, there is theory of flag varieties and Schubert varieties in the super-setting. Two names are Penkov and Voronov, but there may have been more people involved. Bernstein and Leites developed differential and integral calculus on supermanifolds. – Victor Protsak May 28 2010 at 22:01 Do you really mean to use the tags "lie" "super" and "algebra" or is this meant to be "lie super algebra" in which case it should perhaps be hyphenated? – José Figueroa-O'Farrill May 28 2010 at 22:05 2 Leites is an interesting mathematician, but sometimes quirky. Others already mentioned may be more reliable guides. – Jim Humphreys May 28 2010 at 22:24 1 @José: I've changed it for Shizhuo to "lie-superalgebras", since the corresponding tag is "lie-algebras". – Harry Gindi May 29 2010 at 6:26 ## 3 Answers The answer to the question in the title is affirmative. In the Dictionary of Lie superalgebras, there is an entry on the Weyl group of a classical Lie superalgebra. It is generated by reflections associated to the simple even roots, hence it is the standard Weyl group of the even subalgebra. In addition, they also mention that one can extend the Weyl group by the addition of so-called generalised Weyl transformations associated to the odd roots. They also give a couple of references. As for the geometry associated to Lie superalgebras, there is a notion of Lie supergroup (this link is the not-particularly-good wikipedia article), which stands in relation to Lie superalgebras just as their non-super counterparts. Lie supergroups are particular examples of supermanifolds, on which there is a substantial literature. - ### 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'm not an expert in this area, but I'm told that the key phrase in the superalgebra world is "Weyl groupoid" rather than Weyl group. I did not look at the construction long enough to understand it. Serganova has a paper describing foundations in a super analogue of the Kac-Moody setting, and you can find a description of the Weyl groupoid there. Edit: The situation seems to be the following: For Kac-Moody algebras, there is a unique conjugacy class of Cartan subalgebra (under automorphisms), and the Weyl group acts transitively on systems of simple roots. These properties fail to hold in the superalgebra setting. One instead can form a groupoid whose objects are finite size square matrices $A$ with integer entries (or the Lie superalgebras $g(A)$ obtained by a generators-and-relations construction), and whose morphisms from $A$ to $A'$ are superalgebra isomorphisms $g(A) \to g(A')$ that take a Cartan of $g(A)$ to a Cartan of $g(A')$. The Weyl groupoid of $g(A)$ is then the connected component of $A$ in the larger groupoid. Regarding geometry, I think Penkov has done some work with flag supermanifolds and Borel-Weil-Bott. I don't think there is much debate about the foundations of supermanifold theory, but I guess the geometric representation theory doesn't extend from the even case by rote translation of proofs. - From the point of view of super geometry, Manin and I introduced the notion of a super Weyl group in relation to the geometry of homogeneous superspaces. We constructed Schubert supercells which were labeled by elements of a super Weyl group. I am not sure whether this is a well-accepted definition; it was rather a construction done in an ad hoc way for each classical simple Lie supergroup. The results were announced in Manin, Y. I.; Voronov, A. A. Schubert supercells. Functional Anal. Appl. 18, 329-330 (1985). Voronov, A. A. Relative disposition of the Schubert supervarieties and resolution of their singularities. Functional Anal. Appl. 21, 62-64 (1987). and published in detail in Manin, Y. I.; Voronov, A. A. Supercellular partitions of flag superspaces. Current problems in mathematics. Newest results, USSR Acad. Sci., Moscow. 32, 27-70 (1988). (in Russian). English translation: J. Soviet Math. 51(1), 2083-2108 (1990). -	{"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": 0.18, "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": 0.9360240697860718, "perplexity_flag": "middle"}
http://stats.stackexchange.com/questions/4446/sum-of-squares-two-ways-how-are-they-connected?answertab=votes	# Sum of squares two ways, how are they connected? The standard sum of squares as I know it is: Σ (X-m)2 where m is the mean. I ran into a different one which can be written two ways: Σ(X2) - (ΣX)2/n = Σ(X2) - m ΣX I believe the latter is called the "correction term for the mean" (e.g. here). My algebra seems to be inadequate to show these are equivalent, so I was looking for a derivation. Thanks. - ## 2 Answers Expanding the square we get: $\sum_i(X_i-m)^2 = \sum_i(X_i^2 + m^2 - 2 X_i m)$ Thus, $\sum_i(X_i-m)^2 = \sum_i{X_i^2} + \sum_i{m^2} - 2 \sum_i{X_i m}$ Since $m$ is a constant, we have: $\sum_i(X_i-m)^2 = \sum_i{X_i^2} + n m^2 - 2 m \sum_i{X_i}$ But, $\sum_i{X_i} = n m$. Thus, $\sum_i(X_i-m)^2 = \sum_i{X_i^2} + n m^2 - 2 n m^2$ Which on simplifying gets us: $\sum_i(X_i-m)^2 = \sum_i{X_i^2} - n m^2$ Thus, we get can rewrite the rhs of the above in two ways: $\sum_i{X_i^2} - m (n m) = \sum_i{X_i^2} - m \sum_i{X_i}$ (as $n m = \sum_i{X_i}$) and $\sum_i{X_i^2} - n (m)^2 = \sum_i{X_i^2} - \frac{(\sum_i{X_i})^2}{n}$ (as $m = \frac{\sum_i{X_i}}{n}$) - Thanks. Very pretty. – telliott99 Nov 11 '10 at 15:29 Although the formula are equal, the practical difference is ease-of-calculation if you're doing it by hand. If all I had was a piece of paper and a pencil, I'd prefer the second formula--- $\sum X^2$ and $\sum X$ together take less time and are less error prone to calculate than $\sum (X - m)^2$. - 1 But note that the first formula is less prone to round-off error. – onestop Nov 12 '10 at 13:25	{"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": 14, "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": 0.9554784893989563, "perplexity_flag": "middle"}
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My general premise is that I want to investigate the transformations between two distinct sets of vertices on n-dimensional manifolds and then find applications to theoretical physics by: ...	{"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": 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": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9259381890296936, "perplexity_flag": "middle"}
http://physics.stackexchange.com/questions/18395/simple-description-of-exchange-interaction/18528	# Simple description of exchange interaction? What is a simple bare-bones description of exchange interaction between two electrons? For instance, it seems to me that the only necessary ingredients are the Coulomb interaction and the requirement that the total wavefunction be antisymmetric. - 1 Your intuition is correct. A mathematical description of how those two ingredients conspire to create exchange interactions can be found in Ashcroft & Mermin (chapter 32) [this is a pretty standard calculation and I'm sure it appears in a lot of other places too] – wsc Dec 17 '11 at 20:34 1 Its in Griffiths intro quantum textbook as well. Somewhere. – BebopButUnsteady Dec 17 '11 at 20:59 It has nothing to do with the Coulomb force, there would be an exchange interaction between two uncharged but indistinguishable bosons as well. – joseph f. johnson Dec 18 '11 at 5:41 ## 1 Answer Exchange interaction is an addition to other interactions between identical particles caused by permutation symmetry. This addition is a result of specific form of multi-particle wave function. It gives no contribution to Hamiltonian unlike "usual" interactions but appears as an additional term in equations for single-particle wave functions (e.g. Hartree-Fock equation). Interaction usually associated with energy and forces. We could find the exchange correction as a force added to Coulomb forces, but we should understand first what is force in quantum system. Let's consider two fermions with single-particle coordinate wave functions $\psi_a(x)$ and $\psi_b(x)$ and spin wave fucntions $\phi_a(s)$ and $\phi_b(s)$. The possible two-particle wave fucntions are singlet with symmetric coordinate part $$\Psi_S(x_1, x_2) = \frac{1}{\sqrt{2}}\left[ \psi_a(x_1)\psi_b(x_2) + \psi_a(x_2)\psi_b(x_1) \right]$$ and triplet with antisymmetric coordinate part $$\Psi_A(x_1, x_2) = \frac{1}{\sqrt{2}}\left[ \psi_a(x_1)\psi_b(x_2) - \psi_a(x_2)\psi_b(x_1) \right]$$ Let the two-particle Hamiltonian do not depend on spins: $$\hat{H} = \frac{\hat{\mathbf{p}}_1 + \hat{\mathbf{p}}_2}{2m} + V(x_1, x_2)$$ then the average energy of the interaction will be: $$U_S = \left<\Psi_S\right\|V\left\|\Psi_S\right> = U + U_\text{ex}$$ $$= \left<\psi_a(x_1)\psi_b(x_2)\right\|V\left\|\psi_a(x_1)\psi_b(x_2)\right> + \left<\psi_a(x_1)\psi_b(x_2)\right\|V\left\|\psi_a(x_2)\psi_b(x_1)\right>$$ $$U_A = \left<\Psi_A\right\|V\left\|\Psi_A\right> = U - U_\text{ex}$$ $$= \left<\psi_a(x_1)\psi_b(x_2)\right\|V\left\|\psi_a(x_1)\psi_b(x_2)\right> - \left<\psi_a(x_1)\psi_b(x_2)\right\|V\left\|\psi_a(x_2)\psi_b(x_1)\right>$$ The term $U_\text{ex}$ is not zero only if the particles are close enough to each other and their wave functions overlap (see picture below). In classical limit when distance $L$ is big the overlapping is zero and $U_S=U_A=U$ Let's assume that $\psi_a$ and $\psi_b$ are non-negative everywhere anv $V$ acts as Coulomb interaction (i.e. positive and decreases when the distance increases). Then $U$ and $U_\text{ex}$ are positive and energy of symmetric coordinate state (opposite spines) is higher than energy of antisymmetric coordinate state (similar spines). If the average positions of the particles are fixed the exchange interaction will put the spins same direction. The force of interaction between the particles can be defined as the generalized force corresponding to the parameter L: $$F = -\frac{\partial U}{\partial L}$$ Within our assumptions concerning $\psi_a$, $\psi_b$ and $V$ the derivative of both $U$ and $U_\text{ex}$ are negative. Hence the "usual" force is positive (repulsion) and the exchange force is positive for symmetric coordinate state and negative for antisymmetric coordinate state (attraction). So the exchange interaction for the case of two particles can be considered as additional force depending on spin configuration. For multiple particles this is more complicated. -	{"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": 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": 0.9074850082397461, "perplexity_flag": "middle"}
http://www.physicsforums.com/showthread.php?s=10ca930c2e49cc6d79006fce8f40692e&p=4322219	Physics Forums ## Colour Octet States Hey, I'm having an issue seeing how these octet states are reproduced via SU(3) transformations, in my notes it is written: "Now, the remaining 8 states in (25) mix into each other under SU(3) transformations. For example just interchange two labels such as R<->B and you'll see these mix" I'm not exactly sure what he means by interchanging the labels, equation (25) reads: $$R\bar{R},G\bar{G},B\bar{B},R\bar{G},R\bar{B},B\bar{G},B\bar{R},G\bar{R} ,G\bar{B}$$ and one of these states is a singlet (i.e. colour neutral state) which is given by $$\frac{1}{\sqrt3}(R\bar{R}+B\bar{B}+G\bar{G})$$ anyway he interchanges the labels R and B and finds you get the octet states $$\frac{1}{\sqrt2}(R\bar{R}-B\bar{B}),\frac{1}{\sqrt6}(R\bar{R}+G\bar{G}-2B\bar{B}),R\bar{G},R\bar{B},B\bar{R},B\bar{G},G\bar{R},G\bar{B}$$ But I can't exactly see how we get these states by an interchange (but then again I'm not exactly sure what is meant by an interchange of labels) - it's using the SU(3) transformations obviously but I'm not sure how I'd set up the above 'mathematically' or 'matrix-ically'! Anyway probably obvious but I'm slow, Thanks in advance! SK PhysOrg.com physics news on PhysOrg.com >> Promising doped zirconia>> New X-ray method shows how frog embryos could help thwart disease>> Bringing life into focus I was thinking maybe an interchange of R<-->B means this: $$\begin{pmatrix} R\\ G\\ B \end{pmatrix}\rightarrow \begin{pmatrix} B\\ G\\ R \end{pmatrix}$$ but even if this is true, I'm getting confused on how to act our generator matrices on the wavefunction in order to reproduce the states listed in the last post! Thanks! SK Blog Entries: 1 Recognitions: Science Advisor Yes, the interchanges are particular elements of SU(3) and will look for example like R ↔ G, under which a state RB → GB, etc. It's obvious that all 9 states will transform into each other this way, but the point the author is making is the split between the singlet state (which transforms into itself) and the remaining octet. The octet also transforms only into itself. Looking at the states you've listed, and calling the first one a, the second one b, under R ↔ G what happens is a → -a, and b → b - 3a/√3. ## Colour Octet States I'm still having trouble trying to see how exactly these 8 states transform into the second list of states under this interchange. I think I would of understood it better if the author had explicitly shown this transformation though I'm aware this is something which is probably obvious to most but I can't figure it out... Though I have just had an idea so I'll give it a try! Thanks as well for clearing up the octet transformation stuff! Recognitions: Gold Member Quote by Sekonda $$\frac{1}{\sqrt2}(R\bar{R}-B\bar{B}),\frac{1}{\sqrt6}(R\bar{R}+G\bar{G}-2B\bar{B}),R\bar{G},R\bar{B},B\bar{R},B\bar{G},G\bar{R},G\bar{B}$$ But I can't exactly see how we get these states you get the eight 33 matrices of the representation of SU (3) by exponing the eight generators of the su(3) algebra's adjoint representation. Those generators are traceless hermitian matrices. six are "Pauli like" matrices and two are diagonal: $$\begin{pmatrix} 1&0&0 \\ 0&0&0 \\ 0&0& -1 \end{pmatrix} \\ and \begin{pmatrix} 1&0&0 \\ 0&1&0\\ 0&0& -2 \end{pmatrix}$$ The first two terms come from them. (And the 33 identity gives the singlet). Recognitions: Gold Member http://www.physicsforums.com/showthread.php?t=251070It is more usual to write the diagonal generators as $$\begin{pmatrix} 1&0&0 \\ 0&-1&0 \\ 0&0& 0 \end{pmatrix} \\ and \begin{pmatrix} 1&0&0 \\ 0&1&0\\ 0&0& -2 \end{pmatrix}$$ So the first two terms would be proportional to $$R\bar R - G\bar G$$ and $$R\bar R + G\bar G -2B \bar B$$ read this for colour hypercharge and colour isospin Thanks naima for your input - didn't see this till just now. The 33 matrices of SU(3) represent the transformations or rotations of colours in colour space that are allowed, the octet states represent the 8 gluons which are permanently confined and the singlet state represents a free particle (this state linked to the transformation matrix of the identity 33). However I'm still confused on exactly what is meant when the author says 'Interchanging two labels such as R and B gives you'll see these states mix' I'm not really sure what is meant physically by interchanging two labels and how this results in the states given - however I can see the link between the matrices you have stated above and the two states which are sums of colour-anticolour states. I keep trying to show it mathematically by acting the Gell-Mann matrices on the RGB wavefunctions and see that these states arise for each SU(3) matrix but I'm not having any luck! (this is when I've interchanged the labels so that the state would be BGR with anti (RGB)) I'll keep on looking at it! Recognitions: Gold Member Quote by Sekonda Hey, I'm having an issue seeing how these octet states are reproduced via SU(3) transformations, in my notes it is written: "Now, the remaining 8 states in (25) mix into each other under SU(3) transformations. For example just interchange two labels such as R<->B and you'll see these mix" I'm not exactly sure what he means by interchanging the labels, equation (25) reads: $$R\bar{R},G\bar{G},B\bar{B},R\bar{G},R\bar{B},B\bar{G},B\bar{R},G\bar{R} ,G\bar{B}$$ and one of these states is a singlet (i.e. colour neutral state) which is given by $$\frac{1}{\sqrt3}(R\bar{R}+B\bar{B}+G\bar{G})$$ anyway he interchanges the labels R and B and finds you get the octet states $$\frac{1}{\sqrt2}(R\bar{R}-B\bar{B}),\frac{1}{\sqrt6}(R\bar{R}+G\bar{G}-2B\bar{B}),R\bar{G},R\bar{B},B\bar{R},B\bar{G},G\bar{R},G\bar{B}$$ But I can't exactly see how we get these states by an interchange (but then again I'm not exactly sure what is meant by an interchange of labels) - it's using the SU(3) transformations obviously but I'm not sure how I'd set up the above 'mathematically' or 'matrix-ically'! Anyway probably obvious but I'm slow, Thanks in advance! SK I think you misunderstand what the author wrote: He starts from (25) a list of 9 states which generate a vector space. one state (the singlet) is a sum of the first three. he says that this state is invariant by colour permutation and then writes "Now the eight remaining states ..." in fact the singlet is not belonging to the basis (25). he chooses another one (25b): $$(R\bar{R}+G\bar{G}+B\bar{B}),(R\bar{R}-G\bar{G}),(R\bar{R}+G\bar{G}-2B\bar{B}),R\bar{G},R\bar{B},B\bar{R},B\bar{G}, G \bar {R},G\bar{B}$$ When he deletes the first one he gets the 8 remainig ones (he does not say he uses colour transformation to get them) he only looks at them and says: "the remaining 8 states in (25b) mix into each other under SU(3) transformations" then he asks you to see what (25b) becomes when you exchange R <--> G and do not gives the answer. (Bill_K gave you the answer) Recognitions: Gold Member I am reading Georgi's book "Lie algebras in particle physics" and i found a curious identity (Chapter 16): for all x,y,z $$(xR\bar{R}+yG\bar{G}+zB\bar{B}) = \\ (x + y + z)(R\bar{R}+G\bar{G}+B\bar{B}) / 3 + \\ (x - y )(R\bar{R}- G\bar{G}) / 2 + \\ (x + y -2 z)(R\bar{R}+G\bar{G}-2B\bar{B}) / 6$$ x,y,z and the colors are symmetric and we see it uses 3 = 12 + 12 + 12 2 = 12 + (-1)2 6 = 12 + 12 + (-2)2 I know that this is just mathematics but is there a reason why it appears here like that? Recognitions: Science Advisor Thanks Naima, I thought I was misinterpreting the author. Thanks for being patient and explaining this to me! Very much appreciated! SK Thread Tools \| \| \| \| \|------------------------------------------\|-------------------------------------\|---------\| \| Similar Threads for: Colour Octet States \| \| \| \| Thread \| Forum \| Replies \| \| \| Astrophysics \| 3 \| \| \| Chemistry \| 8 \| \| \| Biology, Chemistry & Other Homework \| 8 \| \| \| Biology, Chemistry & Other Homework \| 9 \| \| \| Biology, Chemistry & Other Homework \| 2 \|	{"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": 14, "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": 0.947555661201477, "perplexity_flag": "middle"}
http://mathhelpforum.com/differential-geometry/83047-intermediate-value-theorem-application.html	Thread: 1. Intermediate Value theorem application Let p(x) be an odd-degree polynomial function. Prove that p(x)=0 has at least one real solution 2. Not sure if there is a formal proof of this, but one can see that as $x \to \infty$ then $p(x) \to \infty$ and as $x \to -\infty$ then $p(x) \to -\infty$. (assuming leading coefficient is positive, if negative its just the other way around) By definition any polynomial is continuous on $\mathbb{R}$ and thus by the IVT we can say there exists at least one $x \in \mathbb{R}$ such that $p(x)=0$	{"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": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 7, "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": 0.9589492082595825, "perplexity_flag": "head"}
http://mathoverflow.net/questions/62577/	## Rate of decay of variance for a tensor product Markov process (100 pt bounty for good answer by 1800 EST Fri) ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) Let $Q$ be the generator of a well-behaved (not necessarily reversible) Markov process $X$ on `$[n] = \{1,\dots,n\}$` and let $Q^\otimes = \sum_{m=1}^N I^{\otimes(m-1)} \otimes Q \otimes I^{\otimes(N-m)}$ be the tensor sum generating $N$ instances of $X$ (see this MO answer). Let the spectrum of $Q$ be `$\sigma(Q) = \{\lambda_j\}$`. It can be shown that `$\sigma(Q^\otimes) = \{\sum_{m=1}^N \lambda_{j_m}\}$`. In particular, the spectral gaps of $Q$ and $Q^\otimes$ coincide. On the other hand, the exponential decay rate of the variance of $e^{tQ^\otimes}f^\otimes$ is $N$ times the decay rate for the variance of $e^{tQ}f$ (see, e.g. here). As I understand it in the reversible case the spectral gap and (minimal) decay rate of the variance (given by the infimum of the quotient of the Dirichlet form and the variance itself) are supposed to be the same. But in tensor product land this weird factor of $N$ seems to keep popping up. I thought fleetingly that I understood this apparent contradiction, but now realize that I still don't. So: if the spectral gap of $Q^\otimes$ equals that of $Q$ and the decay rate for the variance of $e^{tQ^\otimes}f^\otimes$ is $N$ times that for $e^{tQ}f$, how can these various facts(?) be reconciled? I'm probably just being obtuse, but this issue is really annoying me, and so I'll retroactively award a bounty of 100 points for the best definitive answer that I can understand and that's provided within 24 hours of this question's posting time. - ## 2 Answers Working with a single copy of the Markov chain for now, you can simplify the expression for the Dirichlet form as $\mathcal D_Q(f)=-\sum_{j,k}p_jQ_{jk}f_jf_k$ (you obtain this from the previous expression just by noting that the row sums of the $Q$ are 0). Let $e^{(0)},\ldots,e^{(d-1)}$ be the eigenvectors of your $Q$. Assume reversibility of $Q$. Then $Q$ is self-adjoint with respect to the inner product $\langle f,g\rangle=\sum p_if_ig_i$. This guarantees that the eigenvectors are orthogonal with respect to this inner product. Let $e^{(0)}$ be the eigenvector of 1's (with eigenvalue 0). The other eigenvectors can be chosen to have length 1 with respect to the inner product. They have negative real eigenvalues $\lambda^{(1)}>\ldots>\lambda^{(d-1)}$. Notice that the Dirichlet form is very simple for $f$'s that are expressed as a sum of eigenvectors: $D_Q(\sum a_i e^{(i)})=\sum a_i^2D_Q(e^{(i)})$ (using the orthogonality properties of $Q$). Further $D_Q(e^{(i)})=-\lambda^{(i)}$ so that $D_Q(\sum a_ie^{(i)})=\sum a_i^2\|\lambda_i\|$. Also Var($e^{(i)}$) is 1 for $i\ne 0$ and 0 for $i=0$. The minimum ratio of the Dirichlet form to the variance occurs at the non-trivial eigenvector with the smallest absolute eigenvalue. Define $Q^\otimes=\sum_{i}c_i I^{\otimes(i-1)}\otimes Q\otimes I^{\otimes{n-i}}$ (this is the generalization of what's in your posting with multiple $c$'s). The eigenvectors of $Q^\otimes$ are simply the tensor products of the eigenvectors of $Q$ so that $e^{(i_1)}\otimes \cdots\otimes e^{(i_n)}$ has eigenvalue $c_1\lambda^{(i_1)}+\ldots+c_n\lambda^{(i_n)}$. The variance of each of these eigenvectors is 1 except for the eigenvector $e^{(0)}\otimes\cdots\otimes e^{(0)}$ which has variance 0. Once again the minimum ratio of the Dirichlet form to the variance occurs at the non-trivial eigenvector with the smallest absolute value. Assuming that $c_1\le c_2\le c_3\le\ldots$, this is the eigenvector $e^{(1)}\otimes e^{(0)}\otimes\cdots\otimes e^{(0)}$. In this case the ratio is given by $c_1\|\lambda_1\|$. Hopefully this resolves the discrepancy? - OK, so this shows me that there is a mistake at the very end of my previous self-answer (the non-calculational part), where I wrongly concluded that the minimum ratio was at $e^{(1)\otimes N}$ instead of at $e^{(1)} \otimes e^{(0)\otimes(N-1)}$. I was (unsurprisingly) getting the same factor of $N$ when I computed the variance of $e^{(1)\otimes N}$ and didn't see that I was making the same silly mistake twice. – Steve Huntsman Apr 22 2011 at 21:01 ...Restricting to zero mean functions probably influenced this repeated error, because while the $e^{(i)}$ have mean zero for $i > 0$ and so do $e^{(1)\otimes N}$ and $e^{(1)} \otimes e^{(0)\otimes(N-1)}$, $e^{(0)}$ most emphatically does not. Anyway, many thanks for your patient help! I'll issue the bounty and select this answer when it becomes available. – Steve Huntsman Apr 22 2011 at 21:02 ### 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. Isn't it that the spectral gap arises from eigenvectors like $e_1\otimes e_0\otimes e_0\otimes\cdots \otimes e_0$ while the variance arises from $e_1\otimes e_1\otimes \cdots \otimes e_1$ (so in the first case you get just one factor of $e^{-\lambda t}$ whereas in the second case you get $n$ such factors)? - OK, but suppose $Q$ is reversible for convenience (so that $Q^\otimes$ is too). Isn't "spectral gap = decay rate of variance" supposed to hold not only for $Q$ but for $Q^\otimes$? – Steve Huntsman Apr 22 2011 at 4:11 What happens if you compute the variance using $e_1\otimes e_0\otimes\cdots\otimes e_0$? Then you get 0 in all the terms except transitions in the first coordinate. I'm guessing what's wrong is the answer to your previous bounty Q, but I will try and think a bit more about this... – Anthony Quas Apr 22 2011 at 6:42	{"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": 0.18, "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": 0.9363994598388672, "perplexity_flag": "head"}
http://lembolies.com/energy	# Energy Research Materials This page contains a list of latest news about Energy which has been collected from credible news resources: CNN, Forbes, Bloomberg, TIME Magazine, FoxNews, BBC, New York Times, CBS, Sky News, Reuters, World Magazine, etc. Apart from News, you can also use the tabs to browse Energy images, videos, wiki information, tweets, documents and weblinks. Energy News Energy Images couldn't connect to hostcouldn't connect to host Rihanna - Take A Bow Music video by Rihanna performing Take A Bow. YouTube view counts pre-VEVO: 66288884. (C) 2008 The Island Def Jam Music Group. THE LEGEND OF ZELDA RAP [MUSIC VIDEO] WATCH BLOOPERS & MORE: http://bit.ly/ZELDAxtras DOWNLOAD THE SONG: http://smo.sh/13NrBp8 DOWNLOAD UNCENSORED SONG: http://smo.sh/WMYpsf GET LEGEND OF SMOSH T... Key & Peele: Substitute Teacher A substitute teacher from the inner city refuses to be messed with while taking attendance. FIRETRUCK! 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P!nk - Try (The Truth About Love - Live From Los Angeles) Music video by P!nk performing Try (The Truth About Love - Live From Los Angeles). (C) 2012 RCA Records, a division of Sony Music Entertainment. David Guetta - Just One Last Time ft. Taped Rai "Just One Last Time" feat. Taped Rai. Available to download on iTunes including remixes of : Tiësto, HARD ROCK SOFA & Deniz Koyu http://smarturl.it/DGJustOne... YOLO (feat. Adam Levine & Kendrick Lamar) YOLO is available on iTunes now! http://smarturl.it/lonelyIslandYolo New album coming soon... Check out the awesome band the music in YOLO is sampled from Th... Most Annoying People On The Internet Don't be these people. Mapoti See Bloopers and Behind-The-Scenes Here!: http://youtu.be/dfpo7uXwJnM Huge thank you and shout out to Dtrix: http://www.youtube... Skrillex & Damian "Jr. Gong" Marley - Make It Bun Dem [OFFICIAL VIDEO] Buy the track here: http://atlr.ec/TZ8yBf Directed by Tony T. Datis. "Energetic" redirects here. For other uses, see Energetic (disambiguation). This article . Please discuss this issue on the talk page and conform with Wikipedia's Manual of Style by replacing the section with a link and a summary of the repeated material, or by spinning off the repeated text into an article in its own right. (November 2012) Energy transformation; In a typical lightning strike, 500 megajoules of electric potential energy is converted into an equivalent amount of light energy, sound energy and thermal energy. In physics, energy is an indirectly observed quantity which comes in many forms, such as kinetic energy, potential energy, radiant energy, and many others; which are listed in this summary article. This is a major topic in science and technology and this article gives an overview of its major aspects, and provides links to the many specific articles about energy in its different forms and contexts. The question "what is energy?" is difficult to answer in a simple, intuitive way, although energy can be rigorously defined in theoretical physics. In the words of Richard Feynman, "It is important to realize that in physics today, we have no knowledge what energy is. We do not have a picture that energy comes in little blobs of a definite amount."[1] However, it is clear that energy is always an indispensable prerequisite for performing mechanical work, and the concept has great importance in natural science. The natural basic units in which energy is measured are those used for mechanical work; they always are equivalent to a unit of force multiplied by a unit of length. Other equivalent units for energy are mass units multiplied by velocity units squared. ## Forms of energy Main article: Forms of energy Energy exists in many forms: Thermal energy is partly potential energy and partly kinetic energy. In the context of physical sciences, several forms of energy have been defined. These include[need quotation to verify]: These forms of energy may be divided into two main groups; kinetic energy and potential energy. Other familiar types of energy are a varying mix of both potential and kinetic energy[need quotation to verify]. Energy may be transformed between different forms at various efficiencies. Items that transform between these forms are called transducers. The above list of the known possible forms of energy is not necessarily complete. Whenever physical scientists discover that a certain phenomenon appears to violate the law of energy conservation, new forms may be added, as is the case with dark energy, a hypothetical form of energy that permeates all of space and tends to increase the rate of expansion of the universe.[need quotation to verify] Classical mechanics distinguishes between potential energy, which is a function of the position of an object within a field, and kinetic energy, which is a function of its movement. Both position and movement are relative to a frame of reference, which must be specified: this is often (and originally) an arbitrary fixed point on the surface of the Earth, the terrestrial frame of reference[need quotation to verify]. It has been attempted to categorize all forms of energy as either kinetic or potential, but as Richard Feynman points out: These notions of potential and kinetic energy depend on a notion of length scale. For example, one can speak of macroscopic potential and kinetic energy, which do not include thermal potential and kinetic energy. Also what is called chemical potential energy is a macroscopic notion, and closer examination shows that it is really the sum of the potential and kinetic energy on the atomic and subatomic scale. Similar remarks apply to nuclear "potential" energy and most other forms of energy. This dependence on length scale is non-problematic if the various length scales are decoupled, as is often the case ... but confusion can arise when different length scales are coupled, for instance when friction converts macroscopic work into microscopic thermal energy. ## History of understanding Thomas Young – the first to use the term "energy" in the modern sense. The word energy derives from the Greek ἐνέργεια , which possibly appears for the first time in the work of Aristotle in the 4th century BCE. (Ancient Greek: ἐνέργεια "activity, operation"[2]) The concept of energy emerged out of the idea of vis viva (living force), which Gottfried Leibniz defined as the product of the mass of an object and its velocity squared; he believed that total vis viva was conserved. To account for slowing due to friction, Leibniz theorized that thermal energy consisted of the random motion of the constituent parts of matter, a view shared by Isaac Newton, although it would be more than a century until this was generally accepted. In 1807, Thomas Young was possibly the first to use the term "energy" instead of vis viva, in its modern sense.[3] Gustave-Gaspard Coriolis described "kinetic energy" in 1829 in its modern sense, and in 1853, William Rankine coined the term "potential energy". The law of conservation of energy, was first postulated in the early 19th century, and applies to any isolated system. According to Noether's theorem, the conservation of energy is a consequence of the fact that the laws of physics do not change over time.[4] Since 1918 it has been known that the law of conservation of energy is the direct mathematical consequence of the translational symmetry of the quantity conjugate to energy, namely time. It was argued for some years whether energy was a substance (the caloric) or merely a physical quantity, such as momentum. In 1845 James Prescott Joule discovered the link between mechanical work and the generation of heat. This led to the theory of conservation of energy, and development of the first law of thermodynamics. Finally, William Thomson (Lord Kelvin) amalgamated these many discoveries into the laws of thermodynamics, which aided the rapid development of explanations of chemical processes by Rudolf Clausius, Josiah Willard Gibbs, and Walther Nernst. It also led to a mathematical formulation of the concept of entropy by Clausius and to the introduction of laws of radiant energy by Jožef Stefan. During a 1961 lecture[5] for undergraduate students at the California Institute of Technology, Richard Feynman, a celebrated physics teacher and Nobel Laureate, said this about the concept of energy: There is a fact, or if you wish, a law, governing all natural phenomena that are known to date. There is no known exception to this law—it is exact so far as we know. The law is called the conservation of energy. It states that there is a certain quantity, which we call energy, that does not change in manifold changes which nature undergoes. That is a most abstract idea, because it is a mathematical principle; it says that there is a numerical quantity which does not change when something happens. It is not a description of a mechanism, or anything concrete; it is just a strange fact that we can calculate some number and when we finish watching nature go through her tricks and calculate the number again, it is the same. ## Units of measure Main article: Units of energy Energy, like mass, is a scalar physical quantity. The joule is the International System of Units (SI) unit of measurement for energy. It is a derived unit of energy, work, or amount of heat. It is equal to the energy expended (or work done) in applying a force of one newton through a distance of one metre. However energy is also expressed in many other units such as ergs, calories, British Thermal Units, kilowatt-hours and kilocalories for instance. There is always a conversion factor for these to the SI unit; for instance; one kWh is equivalent to 3.6 million joules.[6] ## Energy in various contexts ### Classical mechanics Classical mechanics Branches Formulations Fundamental concepts Core topics Scientists Main articles: Mechanics, Mechanical work, Thermodynamics, and Quantum mechanics Work, a form of energy, is force times distance. $W = \int_C \mathbf{F} \cdot \mathrm{d} \mathbf{s}$ This says that the work ($W$) is equal to the line integral of the force F along a path C; for details see the mechanical work article. Work and thus energy is frame dependent. For example, consider a ball being hit by a bat. In the center-of-mass reference frame, the bat does no work on the ball. But, in the reference frame of the person swinging the bat, considerable work is done on the ball. ### Chemistry In the context of chemistry, energy is an attribute of a substance as a consequence of its atomic, molecular or aggregate structure. Since a chemical transformation is accompanied by a change in one or more of these kinds of structure, it is invariably accompanied by an increase or decrease of energy of the substances involved. Some energy is transferred between the surroundings and the reactants of the reaction in the form of heat or light; thus the products of a reaction may have more or less energy than the reactants. A reaction is said to be exergonic if the final state is lower on the energy scale than the initial state; in the case of endergonic reactions the situation is the reverse. Chemical reactions are invariably not possible unless the reactants surmount an energy barrier known as the activation energy. The speed of a chemical reaction (at given temperature T) is related to the activation energy E, by the Boltzmann's population factor e−E/kT – that is the probability of molecule to have energy greater than or equal to E at the given temperature T. This exponential dependence of a reaction rate on temperature is known as the Arrhenius equation.The activation energy necessary for a chemical reaction can be in the form of thermal energy. ### Biology In biology, energy is an attribute of all biological systems from the biosphere to the smallest living organism. Within an organism it is responsible for growth and development of a biological cell or an organelle of a biological organism. Energy is thus often said to be stored by cells in the structures of molecules of substances such as carbohydrates (including sugars), lipids, and proteins, which release energy when reacted with oxygen in respiration. In human terms, the human equivalent (H-e) (Human energy conversion) indicates, for a given amount of energy expenditure, the relative quantity of energy needed for human metabolism, assuming an average human energy expenditure of 12,500kJ per day and a basal metabolic rate of 80 watts. For example, if our bodies run (on average) at 80 watts, then a light bulb running at 100 watts is running at 1.25 human equivalents (100 ÷ 80) i.e. 1.25 H-e. For a difficult task of only a few seconds' duration, a person can put out thousands of watts, many times the 746 watts in one official horsepower. For tasks lasting a few minutes, a fit human can generate perhaps 1,000 watts. For an activity that must be sustained for an hour, output drops to around 300; for an activity kept up all day, 150 watts is about the maximum.[7] The human equivalent assists understanding of energy flows in physical and biological systems by expressing energy units in human terms: it provides a “feel” for the use of a given amount of energy[8] ### Earth sciences In geology, continental drift, mountain ranges, volcanoes, and earthquakes are phenomena that can be explained in terms of energy transformations in the Earth's interior.,[9] while meteorological phenomena like wind, rain, hail, snow, lightning, tornadoes and hurricanes, are all a result of energy transformations brought about by solar energy on the atmosphere of the planet Earth. ### Cosmology In cosmology and astronomy the phenomena of stars, nova, supernova, quasars and gamma ray bursts are the universe's highest-output energy transformations of matter. All stellar phenomena (including solar activity) are driven by various kinds of energy transformations. Energy in such transformations is either from gravitational collapse of matter (usually molecular hydrogen) into various classes of astronomical objects (stars, black holes, etc.), or from nuclear fusion (of lighter elements, primarily hydrogen). ### Energy and life Main articles: Bioenergetics and Food energy Basic overview of energy and human life. Any living organism relies on an external source of energy—radiation from the Sun in the case of green plants; chemical energy in some form in the case of animals—to be able to grow and reproduce. The daily 1500–2000 Calories (6–8 MJ) recommended for a human adult are taken as a combination of oxygen and food molecules, the latter mostly carbohydrates and fats, of which glucose (C6H12O6) and stearin (C57H110O6) are convenient examples. The food molecules are oxidised to carbon dioxide and water in the mitochondria C6H12O6 + 6O2 → 6CO2 + 6H2O C57H110O6 + 81.5O2 → 57CO2 + 55H2O and some of the energy is used to convert ADP into ATP ADP + HPO42− → ATP + H2O The rest of the chemical energy in the carbohydrate or fat is converted into heat: the ATP is used as a sort of "energy currency", and some of the chemical energy it contains when split and reacted with water, is used for other metabolism (at each stage of a metabolic pathway, some chemical energy is converted into heat). Only a tiny fraction of the original chemical energy is used for work:[10] gain in kinetic energy of a sprinter during a 100 m race: 4 kJ gain in gravitational potential energy of a 150 kg weight lifted through 2 metres: 3kJ Daily food intake of a normal adult: 6–8 MJ It would appear that living organisms are remarkably inefficient (in the physical sense) in their use of the energy they receive (chemical energy or radiation), and it is true that most real machines manage higher efficiencies. In growing organisms the energy that is converted to heat serves a vital purpose, as it allows the organism tissue to be highly ordered with regard to the molecules it is built from. The second law of thermodynamics states that energy (and matter) tends to become more evenly spread out across the universe: to concentrate energy (or matter) in one specific place, it is necessary to spread out a greater amount of energy (as heat) across the remainder of the universe ("the surroundings").[11] Simpler organisms can achieve higher energy efficiencies than more complex ones, but the complex organisms can occupy ecological niches that are not available to their simpler brethren. The conversion of a portion of the chemical energy to heat at each step in a metabolic pathway is the physical reason behind the pyramid of biomass observed in ecology: to take just the first step in the food chain, of the estimated 124.7 Pg/a of carbon that is fixed by photosynthesis, 64.3 Pg/a (52%) are used for the metabolism of green plants,[12] i.e. reconverted into carbon dioxide and heat. ## Energy transformation Main article: Energy transformation The concept of energy and its transformations is vital in explaining and predicting most natural phenomena. One form of energy can often be readily transformed into another; for instance, a battery, from chemical energy to electric energy; a dam: gravitational potential energy to kinetic energy of moving water (and the blades of a turbine) and ultimately to electric energy through an electric generator. There are strict limits to how efficiently energy can be converted into other forms of energy via work, and heat as described by Carnot's theorem and the second law of thermodynamics. These limits are especially evident when an engine is used to perform work. Some energy transformations can be quite efficient. The direction of transformations in energy (what kind of energy is transformed to what other kind) is often described by entropy (equal energy spread among all available degrees of freedom) considerations, as in practice all energy transformations are permitted on a small scale, but certain larger transformations are not permitted because it is statistically unlikely that energy or matter will randomly move into more concentrated forms or smaller spaces. Energy transformations in the universe over time are characterized by various kinds of potential energy that has been available since the Big Bang, later being "released" (transformed to more active types of energy such as kinetic or radiant energy), when a triggering mechanism is available. Familiar examples of such processes include nuclear decay, in which energy is released that was originally "stored" in heavy isotopes (such as uranium and thorium), by nucleosynthesis, a process ultimately using the gravitational potential energy released from the gravitational collapse of supernovae, to store energy in the creation of these heavy elements before they were incorporated into the solar system and the Earth. This energy is triggered and released in nuclear fission bombs or in civil nuclear power generation. Similarly, in the case of a chemical explosion, chemical potential energy is transformed to kinetic energy and thermal energy in a very short time. Yet another example is that of a pendulum. At its highest points the kinetic energy is zero and the gravitational potential energy is at maximum. At its lowest point the kinetic energy is at maximum and is equal to the decrease of potential energy. If one (unrealistically) assumes that there is no friction or other losses, the conversion of energy between these processes would be perfect, and the pendulum would continue swinging forever. ### Conservation of energy and mass in transformation Energy gives rise to weight when it is trapped in a system with zero momentum, where it can be weighed. It is also equivalent to mass, and this mass is always associated with it. Mass is also equivalent to a certain amount of energy, and likewise always appears associated with it, as described in mass-energy equivalence. The formula E = mc², derived by Albert Einstein (1905) quantifies the relationship between rest-mass and rest-energy within the concept of special relativity. In different theoretical frameworks, similar formulas were derived by J. J. Thomson (1881), Henri Poincaré (1900), Friedrich Hasenöhrl (1904) and others (see Mass-energy equivalence#History for further information). Matter may be destroyed and converted to energy (and vice versa), but mass cannot ever be destroyed; rather, mass remains a constant for both the matter and the energy, during any process when they are converted into each other. However, since $c^2$ is extremely large relative to ordinary human scales, the conversion of ordinary amount of matter (for example, 1 kg) to other forms of energy (such as heat, light, and other radiation) can liberate tremendous amounts of energy (~$9\times 10^{16}$ joules = 21 megatons of TNT), as can be seen in nuclear reactors and nuclear weapons. Conversely, the mass equivalent of a unit of energy is minuscule, which is why a loss of energy (loss of mass) from most systems is difficult to measure by weight, unless the energy loss is very large. Examples of energy transformation into matter (i.e., kinetic energy into particles with rest mass) are found in high-energy nuclear physics. ### Reversible and non-reversible transformations Transformation of energy into useful work is a core topic of thermodynamics. In nature, transformations of energy can be fundamentally classed into two kinds: those that are thermodynamically reversible, and those that are thermodynamically irreversible. A reversible process in thermodynamics is one in which no energy is dissipated (spread) into empty energy states available in a volume, from which it cannot be recovered into more concentrated forms (fewer quantum states), without degradation of even more energy. A reversible process is one in which this sort of dissipation does not happen. For example, conversion of energy from one type of potential field to another, is reversible, as in the pendulum system described above. In processes where heat is generated, quantum states of lower energy, present as possible excitations in fields between atoms, act as a reservoir for part of the energy, from which it cannot be recovered, in order to be converted with 100% efficiency into other forms of energy. In this case, the energy must partly stay as heat, and cannot be completely recovered as usable energy, except at the price of an increase in some other kind of heat-like increase in disorder in quantum states, in the universe (such as an expansion of matter, or a randomization in a crystal). ### Transformation with the age of the universe As the universe evolves in time, more and more of its energy becomes trapped in irreversible states (i.e., as heat or other kinds of increases in disorder). This has been referred to as the inevitable thermodynamic heat death of the universe. In this heat death the energy of the universe does not change, but the fraction of energy which is available to do work through a heat engine, or be transformed to other usable forms of energy (through the use of generators attached to heat engines), grows less and less. In a slower process, radioactive decay of these atoms in the core of the Earth releases heat. This thermal energy drives plate tectonics and may lift mountains, via orogenesis. This slow lifting represents a kind of gravitational potential energy storage of the thermal energy, which may be later released to active kinetic energy in landslides, after a triggering event. Earthquakes also release stored elastic potential energy in rocks, a store that has been produced ultimately from the same radioactive heat sources. Thus, according to present understanding, familiar events such as landslides and earthquakes release energy that has been stored as potential energy in the Earth's gravitational field or elastic strain (mechanical potential energy) in rocks. Prior to this, they represent release of energy that has been stored in heavy atoms since the collapse of long-destroyed supernova stars created these atoms. In another similar chain of transformations beginning at the dawn of the universe, nuclear fusion of hydrogen in the Sun also releases another store of potential energy which was created at the time of the Big Bang. At that time, according to theory, space expanded and the universe cooled too rapidly for hydrogen to completely fuse into heavier elements. This meant that hydrogen represents a store of potential energy that can be released by fusion. Such a fusion process is triggered by heat and pressure generated from gravitational collapse of hydrogen clouds when they produce stars, and some of the fusion energy is then transformed into sunlight. Such sunlight from our Sun may again be stored as gravitational potential energy after it strikes the Earth, as (for example) water evaporates from oceans and is deposited upon mountains (where, after being released at a hydroelectric dam, it can be used to drive turbines or generators to produce electricity). Sunlight also drives many weather phenomena, save those generated by volcanic events. An example of a solar-mediated weather event is a hurricane, which occurs when large unstable areas of warm ocean, heated over months, give up some of their thermal energy suddenly to power a few days of violent air movement. Sunlight is also captured by plants as chemical potential energy in photosynthesis, when carbon dioxide and water (two low-energy compounds) are converted into the high-energy compounds carbohydrates, lipids, and proteins. Plants also release oxygen during photosynthesis, which is utilized by living organisms as an electron acceptor, to release the energy of carbohydrates, lipids, and proteins. Release of the energy stored during photosynthesis as heat or light may be triggered suddenly by a spark, in a forest fire, or it may be made available more slowly for animal or human metabolism, when these molecules are ingested, and catabolism is triggered by enzyme action. Through all of these transformation chains, potential energy stored at the time of the Big Bang is later released by intermediate events, sometimes being stored in a number of ways over time between releases, as more active energy. In all these events, one kind of energy is converted to other types of energy, including heat. ## Conservation of energy Main article: Conservation of energy Energy is subject to the law of conservation of energy. According to this law, energy can neither be created (produced) nor destroyed by itself. It can only be transformed. Most kinds of energy (with gravitational energy being a notable exception)[13] are subject to strict local conservation laws as well. In this case, energy can only be exchanged between adjacent regions of space, and all observers agree as to the volumetric density of energy in any given space. There is also a global law of conservation of energy, stating that the total energy of the universe cannot change; this is a corollary of the local law, but not vice versa.[5][14] Conservation of energy is the mathematical consequence of translational symmetry of time (that is, the indistinguishability of time intervals taken at different time)[15] - see Noether's theorem. According to Conservation of energy the total inflow of energy into a system must equal the total outflow of energy from the system, plus the change in the energy contained within the system. This law is a fundamental principle of physics. It follows from the translational symmetry of time, a property of most phenomena below the cosmic scale that makes them independent of their locations on the time coordinate. Put differently, yesterday, today, and tomorrow are physically indistinguishable. This is because energy is the quantity which is canonical conjugate to time. This mathematical entanglement of energy and time also results in the uncertainty principle - it is impossible to define the exact amount of energy during any definite time interval. The uncertainty principle should not be confused with energy conservation - rather it provides mathematical limits to which energy can in principle be defined and measured. Each of the basic forces of nature is associated with a different type of potential energy, and all types of potential energy (like all other types of energy) appears as system mass, whenever present. For example, a compressed spring will be slightly more massive than before it was compressed. Likewise, whenever energy is transferred between systems by any mechanism, an associated mass is transferred with it. In quantum mechanics energy is expressed using the Hamiltonian operator. On any time scales, the uncertainty in the energy is by $\Delta E \Delta t \ge \frac { \hbar } {2 }$ which is similar in form to the Heisenberg uncertainty principle (but not really mathematically equivalent thereto, since H and t are not dynamically conjugate variables, neither in classical nor in quantum mechanics). In particle physics, this inequality permits a qualitative understanding of virtual particles which carry momentum, exchange by which and with real particles, is responsible for the creation of all known fundamental forces (more accurately known as fundamental interactions). Virtual photons (which are simply lowest quantum mechanical energy state of photons) are also responsible for electrostatic interaction between electric charges (which results in Coulomb law), for spontaneous radiative decay of exited atomic and nuclear states, for the Casimir force, for van der Waals bond forces and some other observable phenomena. ## Applications of the concept of energy Energy is subject to a strict global conservation law; that is, whenever one measures (or calculates) the total energy of a system of particles whose interactions do not depend explicitly on time, it is found that the total energy of the system always remains constant.[16] • The total energy of a system can be subdivided and classified in various ways. For example, it is sometimes convenient to distinguish potential energy (which is a function of coordinates only) from kinetic energy (which is a function of coordinate time derivatives only). It may also be convenient to distinguish gravitational energy, electric energy, thermal energy, and other forms. These classifications overlap; for instance, thermal energy usually consists partly of kinetic and partly of potential energy. • The transfer of energy can take various forms; familiar examples include work, heat flow, and advection, as discussed below. • The word "energy" is also used outside of physics in many ways, which can lead to ambiguity and inconsistency. The vernacular terminology is not consistent with technical terminology. For example, while energy is always conserved (in the sense that the total energy does not change despite energy transformations), energy can be converted into a form, e.g., thermal energy, that cannot be utilized to perform work. When one talks about "conserving energy by driving less," one talks about conserving fossil fuels and preventing useful energy from being lost as heat. This usage of "conserve" differs from that of the law of conservation of energy.[14] In classical physics energy is considered a scalar quantity, the canonical conjugate to time. In special relativity energy is also a scalar (although not a Lorentz scalar but a time component of the energy-momentum 4-vector).[17] In other words, energy is invariant with respect to rotations of space, but not invariant with respect to rotations of space-time (= boosts). ## Energy transfer A system can transfer energy to another system by simply transferring matter to it (since matter is equivalent to energy, in accordance with its mass). However, when energy is transferred by means other than matter-transfer, the transfer produces changes in the second system, as a result of work done on it. This work manifests itself as the effect of force(s) applied through distances within the target system. For example, a system can emit energy to another by transferring (radiating) electromagnetic energy, but this creates forces upon the particles that absorb the radiation. Similarly, a system may transfer energy to another by physically impacting it, but in that case the energy of motion in an object, called kinetic energy, results in forces acting over distances (new energy) to appear in another object that is struck. Transfer of thermal energy by heat occurs by both of these mechanisms: heat can be transferred by electromagnetic radiation, or by physical contact in which direct particle-particle impacts transfer kinetic energy. Because energy is strictly conserved and is also locally conserved (wherever it can be defined), it is important to remember that by the definition of energy the transfer of energy between the "system" and adjacent regions is work. A familiar example is mechanical work. In simple cases this is written as the following equation: $\Delta{}E = W$ () if there are no other energy-transfer processes involved. Here $E$ is the amount of energy transferred, and $W$ represents the work done on the system.[dubious ] More generally, the energy transfer can be split into two categories: $\Delta{}E = W + Q$ () where $Q$ represents the heat flow into the system. There are other ways in which an open system can gain or lose energy. In chemical systems, energy can be added to a system by means of adding substances with different chemical potentials, which potentials are then extracted (both of these process are illustrated by fueling an auto, a system which gains in energy thereby, without addition of either work or heat). Winding a clock would be adding energy to a mechanical system. These terms may be added to the above equation, or they can generally be subsumed into a quantity called "energy addition term $E$" which refers to any type of energy carried over the surface of a control volume or system volume. Examples may be seen above, and many others can be imagined (for example, the kinetic energy of a stream of particles entering a system, or energy from a laser beam adds to system energy, without either being either work-done or heat-added, in the classic senses). $\Delta{}E = W + Q + E$ () Where E in this general equation represents other additional advected energy terms not covered by work done on a system, or heat added to it. Energy is also transferred from potential energy ($E_p$) to kinetic energy ($E_k$) and then back to potential energy constantly. This is referred to as conservation of energy. In this closed system, energy cannot be created or destroyed; therefore, the initial energy and the final energy will be equal to each other. This can be demonstrated by the following: $E_{pi} + E_{ki} = E_{pF} + E_{kF}$ () The equation can then be simplified further since $E_p = mgh$ (mass times acceleration due to gravity times the height) and $E_k = \frac{1}{2} mv^2$ (half mass times velocity squared). Then the total amount of energy can be found by adding $E_p + E_k = E_{total}$. ## Energy and the laws of motion In classical mechanics, energy is a conceptually and mathematically useful property, as it is a conserved quantity. Several formulations of mechanics have been developed using energy as a core concept, as below; ### The Hamiltonian The total energy of a system is sometimes called the Hamiltonian, after William Rowan Hamilton. The classical equations of motion can be written in terms of the Hamiltonian, even for highly complex or abstract systems. These classical equations have remarkably direct analogs in nonrelativistic quantum mechanics.[18] ### The Lagrangian Another energy-related concept is called the Lagrangian, after Joseph Louis Lagrange. This is even more fundamental than the Hamiltonian, and can be used to derive the equations of motion. It was invented in the context of classical mechanics, but is generally useful in modern physics. The Lagrangian is defined as the kinetic energy minus the potential energy. Usually, the Lagrange formalism is mathematically more convenient than the Hamiltonian for non-conservative systems (such as systems with friction). ### Noether's Theorem Noether's (first) theorem (1918) states that any differentiable symmetry of the action of a physical system has a corresponding conservation law. Noether's theorem has become a fundamental tool of modern theoretical physics and the calculus of variations. A generalization of the seminal formulations on constants of motion in Lagrangian and Hamiltonian mechanics (1788 and 1833, respectively), it does not apply to systems that cannot be modeled with a Lagrangian; for example, dissipative systems with continuous symmetries need not have a corresponding conservation law. ## Energy and thermodynamics ### Internal energy Internal energy is the sum of all microscopic forms of energy of a system. It is the energy needed to create the system. It is related to the potential energy, e.g., molecular structure, crystal structure, and other geometric aspects, as well as the motion of the particles, in form of kinetic energy. Thermodynamics is chiefly concerned with changes in internal energy and not its absolute value, which is impossible to determine with thermodynamics alone.[19] ### The first law of thermodynamics The first law of thermodynamics asserts that energy (but not necessarily thermodynamic free energy) is always conserved[20] and that heat flow is a form of energy transfer. For homogeneous systems, with a well-defined temperature and pressure, a commonly used corollary of the first law is that, for a system subject only to pressure forces and heat transfer (e.g., a cylinder-full of gas), the differential change in the internal energy of the system (with a gain in energy signified by a positive quantity) is given as $\mathrm{d}E = T\mathrm{d}S - P\mathrm{d}V\,$, where the first term on the right is the heat transferred into the system, expressed in terms of temperature T and entropy S (in which entropy increases and the change dS is positive when the system is heated), and the last term on the right hand side is identified as work done on the system, where pressure is P and volume V (the negative sign results since compression of the system requires work to be done on it and so the volume change, dV, is negative when work is done on the system). This equation is highly specific, ignoring all chemical, electrical, nuclear, and gravitational forces, effects such as advection of any form of energy other than heat and pV-work. The general formulation of the first law (i.e., conservation of energy) is valid even in situations in which the system is not homogeneous. For these cases the change in internal energy of a closed system is expressed in a general form by $\mathrm{d}E=\delta Q+\delta W$ where $\delta Q$ is the heat supplied to the system and $\delta W$ is the work applied to the system. ### Equipartition of energy The energy of a mechanical harmonic oscillator (a mass on a spring) is alternatively kinetic and potential. At two points in the oscillation cycle it is entirely kinetic, and alternatively at two other points it is entirely potential. Over the whole cycle, or over many cycles, net energy is thus equally split between kinetic and potential. This is called equipartition principle; total energy of a system with many degrees of freedom is equally split among all available degrees of freedom. This principle is vitally important to understanding the behavior of a quantity closely related to energy, called entropy. Entropy is a measure of evenness of a distribution of energy between parts of a system. When an isolated system is given more degrees of freedom (i.e., given new available energy states that are the same as existing states), then total energy spreads over all available degrees equally without distinction between "new" and "old" degrees. This mathematical result is called the second law of thermodynamics. ### Oscillators, phonons, and photons This section . Please improve it by verifying the claims made and adding inline citations. Statements consisting only of original research may be removed. (August 2009) In an ensemble (connected collection) of unsynchronized oscillators, the average energy is spread equally between kinetic and potential types. In a solid, thermal energy (often referred to loosely as heat content) can be accurately described by an ensemble of thermal phonons that act as mechanical oscillators. In this model, thermal energy is equally kinetic and potential. In an ideal gas, the interaction potential between particles is essentially the delta function which stores no energy: thus, all of the thermal energy is kinetic. Because an electric oscillator (LC circuit) is analogous to a mechanical oscillator, its energy must be, on average, equally kinetic and potential. It is entirely arbitrary whether the magnetic energy is considered kinetic and whether the electric energy is considered potential, or vice versa. That is, either the inductor is analogous to the mass while the capacitor is analogous to the spring, or vice versa. 1. By extension of the previous line of thought, in free space the electromagnetic field can be considered an ensemble of oscillators, meaning that radiation energy can be considered equally potential and kinetic. This model is useful, for example, when the electromagnetic Lagrangian is of primary interest and is interpreted in terms of potential and kinetic energy. 2. On the other hand, in the key equation $m^2 c^4 = E^2 - p^2 c^2$, the contribution $mc^2$ is called the rest energy, and all other contributions to the energy are called kinetic energy. For a particle that has mass, this implies that the kinetic energy is $0.5 p^2/m$ at speeds much smaller than c, as can be proved by writing $E = mc^2$ √$(1 + p^2 m^{-2}c^{-2})$ and expanding the square root to lowest order. By this line of reasoning, the energy of a photon is entirely kinetic, because the photon is massless and has no rest energy. This expression is useful, for example, when the energy-versus-momentum relationship is of primary interest. The two analyses are entirely consistent. The electric and magnetic degrees of freedom in item 1 are transverse to the direction of motion, while the speed in item 2 is along the direction of motion. For non-relativistic particles these two notions of potential versus kinetic energy are numerically equal, so the ambiguity is harmless, but not so for relativistic particles. ### Quantum mechanics Main article: Energy operator In quantum mechanics energy is defined in terms of the energy operator as a time derivative of the wave function. The Schrödinger equation equates the energy operator to the full energy of a particle or a system. In results can be considered as a definition of measurement of energy in quantum mechanics. The Schrödinger equation describes the space- and time-dependence of slow changing (non-relativistic) wave function of quantum systems. The solution of this equation for bound system is discrete (a set of permitted states, each characterized by an energy level) which results in the concept of quanta. In the solution of the Schrödinger equation for any oscillator (vibrator) and for electromagnetic waves in a vacuum, the resulting energy states are related to the frequency by the Planck equation $E = h\nu$ (where $h$ is the Planck's constant and $\nu$ the frequency). In the case of electromagnetic wave these energy states are called quanta of light or photons. ### Relativity When calculating kinetic energy (work to accelerate a mass from zero speed to some finite speed) relativistically - using Lorentz transformations instead of Newtonian mechanics, Einstein discovered an unexpected by-product of these calculations to be an energy term which does not vanish at zero speed. He called it rest mass energy - energy which every mass must possess even when being at rest. The amount of energy is directly proportional to the mass of body: $E = m c^2$, where m is the mass, c is the speed of light in vacuum, E is the rest mass energy. For example, consider electron-positron annihilation, in which the rest mass of individual particles is destroyed, but the inertia equivalent of the system of the two particles (its invariant mass) remains (since all energy is associated with mass), and this inertia and invariant mass is carried off by photons which individually are massless, but as a system retain their mass. This is a reversible process - the inverse process is called pair creation - in which the rest mass of particles is created from energy of two (or more) annihilating photons. In this system the matter (electrons and positrons) is destroyed and changed to non-matter energy (the photons). However, the total system mass and energy do not change during this interaction. In general relativity, the stress-energy tensor serves as the source term for the gravitational field, in rough analogy to the way mass serves as the source term in the non-relativistic Newtonian approximation.[17] It is not uncommon to hear that energy is "equivalent" to mass. It would be more accurate to state that every energy has an inertia and gravity equivalent, and because mass is a form of energy, then mass too has inertia and gravity associated with it. ## Measurement This section requires expansion. (January 2013) A schematic diagram of a Calorimeter - An instrument used by physicists to measure energy. In this example is it is X-Rays. Because energy is defined as the ability to do work on objects, there is no absolute measure of energy. Only the transition of a system from one state into another can be defined and thus energy is measured in relative terms. The choice of a baseline or zero point is often arbitrary and can be made in whatever way is most convenient for a problem. For example in the case of measuring the energy deposited by X-rays as shown in the accompanying diagram, conventionally the technique most often employed is calorimetry. This is a thermodynamic technique that relies on the measurement of temperature using a thermometer or of intensity of radiation using a bolometer. ### Energy density Main article: Energy density Energy density is a term used for the amount of useful energy stored in a given system or region of space per unit volume. For fuels, the energy per unit volume is sometimes a useful parameter. In a few applications, comparing, for example, the effectiveness of hydrogen fuel to gasoline it turns out that hydrogen has a higher specific energy than does gasoline, but, even in liquid form, a much lower energy density. ## Notes and references 1. Richard Feynman, in The Feynman Lectures on Physics (1964) Volume I, 4-1 2. Harper, Douglas. "Energy". Online Etymology Dictionary. Retrieved May 1, 2007. 3. Smith, Crosbie (1998). The Science of Energy – a Cultural History of Energy Physics in Victorian Britain. The University of Chicago Press. ISBN 0-226-76420-6. 4. Lofts, G; O'Keeffe D; et al. (2004). "11 — Mechanical Interactions". Jacaranda Physics 1 (2 ed.). Milton, Queensland, Australia: John Willey & Sons Australia Ltd. p. 286. ISBN 0-7016-3777-3. 5. ^ a b Feynman, Richard (1964). The Feynman Lectures on Physics; Volume 1. U.S.A: Addison Wesley. ISBN 0-201-02115-3. 6. Ristinen, Robert A., and Kraushaar, Jack J. Energy and the Environment. New York: John Wiley & Sons, Inc., 2006. 7. "Retrieved on May-29-09". Uic.edu. Retrieved 2010-12-12. 8. "Earth's Energy Budget". Okfirst.ocs.ou.edu. Retrieved 2010-12-12. 9. These examples are solely for illustration, as it is not the energy available for work which limits the performance of the athlete but the power output of the sprinter and the force of the weightlifter. A worker stacking shelves in a supermarket does more work (in the physical sense) than either of the athletes, but does it more slowly. 10. Crystals are another example of highly ordered systems that exist in nature: in this case too, the order is associated with the transfer of a large amount of heat (known as the lattice energy) to the surroundings. 11. Ito, Akihito; Oikawa, Takehisa (2004). "Global Mapping of Terrestrial Primary Productivity and Light-Use Efficiency with a Process-Based Model." in Shiyomi, M. et al. (Eds.) Global Environmental Change in the Ocean and on Land. pp. 343–58. 12. "E. Noether's Discovery of the Deep Connection Between Symmetries and Conservation Laws". Physics.ucla.edu. 1918-07-16. Retrieved 2010-12-12. 13. ^ a b 14. "Time Invariance". Ptolemy.eecs.berkeley.edu. Retrieved 2010-12-12. 15. Berkeley Physics Course Volume 1. Charles Kittel, Walter D Knight and Malvin A Ruderman 16. ^ a b Misner, Thorne, Wheeler (1973). Gravitation. San Francisco: W. H. Freeman. ISBN 0-7167-0344-0. 17. I. Klotz, R. Rosenberg, Chemical Thermodynamics - Basic Concepts and Methods, 7th ed., Wiley (2008), p.39 18. Kittel and Kroemer (1980). Thermal Physics. New York: W. H. Freeman. ISBN 0-7167-1088-9. ## Further reading • Alekseev, G. N. (1986). Energy and Entropy. Moscow: Mir Publishers. • • Ross, John S. (23 April 2002). "Work, Power, Kinetic Energy". Project PHYSNET. Michigan State University. • Smil, Vaclav (2008). Energy in nature and society: general energetics of complex systems. Cambridge, USA: MIT Press. ISBN 0-262-19565-8. • Walding, Richard, Rapkins, Greg, Rossiter, Glenn (1999-11-01). New Century Senior Physics. Melbourne, Australia: Oxford University Press. ISBN 0-19-551084-4. News Documents Don't believe everything they write, until confirmed from LEMBOLIES site. ### What is LEMBOLIES? It's a social web research tool that helps anyone exploring anything. ### Updates: Stay up-to-date. Socialize with us! We strive to bring you the latest from the entire web.	{"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": 31, "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": 0.9246591925621033, "perplexity_flag": "middle"}
http://physics.stackexchange.com/questions/46180/can-a-deformable-object-swim-in-curved-space-time	# Can a deformable object “swim” in curved space-time? [duplicate] Possible Duplicate: Swimming in Spacetime - apparent conserved quantity violation It is well known that a deformable object can perform a finite rotation in space by performing deformations - without violating the law of conservation of angular momentum since the moment of inertia can be changed by the deformations of the object, see e.g. this Phys.SE question. It is also well known that in flat space-time, it is not possible for a deformable object to displace it's center of gravity by performing deformations, see e.g. this Phys.SE question. However in curved space-time can a deformable object swim through space by performing deformations? - – Qmechanic♦ Dec 7 '12 at 15:52 ## marked as duplicate by Qmechanic♦, Manishearth♦, Emilio Pisanty, Sklivvz♦Dec 10 '12 at 3:48 This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question. ## 1 Answer Yes, you can swim through space, but only if space is curved - in the vicinity of a gravitating body (which creates curvature of space-time) it is possible for an isolated body to move by only executing internal motions of parts of the body. The reason this is possible is that the center of gravity of an object is not well defined in a curved space-time. Therefore a set of deformations of the object can result in a net displacement of the not well defined center of mass A deformable object with no angular momentum can perform a rotation without violating the law of conservation of angular momentum. A simple way of imagining this is to think of 4 weights (red) on 2 tracks (black) that are connected by a motor (green) that can rotate the two rods around their centers of mass as seen as this picture: Starting with this object stationary in space and with no angular momentum, move two weights closer to the pivot point. Then rotate the rods with the green motor - the rod with the weight far apart will rotate through a smaller angle than the weights that are closer together since the moment of inertia of the two rods are different. Then reverse the positions of the weights on the two rods and reverse the angle of rotation of the green motor. When you finish this maneuver there will be an overall change of the angle of the object in space. However at the end of the maneuver the center of mass of the object will be in the same position and the total angular momentum will still be zero - thus the angular momentum is still conserved. You can actually do this yourself using your arms and legs and a swiveling desk chair. In the same way that angular momentum was conserved in the rotation case, linear momentum will be conserved for the displacement in curved space case - if the object was stationary before the maneuver it will be stationary after the maneuver and thus there is no violation of linear momentum conservation. Even though there is a displacement of the center of mass caused by the maneuver there is no continuing motion of the center of mass. However this kind of displacement maneuver is only possible in curved space-time - in flat space-time this kind of displacement is impossible since the center of mass is well defined whereas the center of masss is not well defined in curved space-time. In "ordinary" curved space-time, such as the curvature caused by the Earth, the effects are very small, but they are non-zero. For example for a meter sized object performing meter sized movement sequences in the vicinity of the Earth, the distance the center of mass would move is only $10^{-34}$ meters for each sequence of moves (this distance is only about 6 Planck lengths!). A blog that gives a very readable and perhaps understandable description of the effect is here: http://www.science20.com/hammock_physicist/swimming_through_empty_space The blog explains: ...[the] key to this swimming in empty space is the fact that the concept of center-of-mass is ill-defined in non-Euclidean space. Non-Euclidean space swimming is geometric in nature and entirely determined by the sequence of shapes assumed. In several ways this swimming is similar to the mechanism by which a cat falling upside-down rotates itself during free fall. Physicists refer to geometric phases to describe these effects... And gives a 2 dimensional example of swimming: Imagine a two-dimensional three-legged creature moving frictionlessly over the surface of a sphere. Let's say this creature is positioned at the equator with one leg pointing east and two legs pointing along the lines of longitude towards the north and south poles. The swim stroke consists of four moves. First the tripodal creature extends its two longitudinal legs, and subsequently it extends the eastern leg. To complete the stroke, it retracts the longitudinal legs and finaly retracts the eastern leg. As a result of each such 'swimstroke' the creature moves a wee bit westward. Why is this? Key is that when the eastern leg extends, the longitudinal legs are extended away from the equator, whilst when the eastern leg retracts, the they are closer to the equator. If the creature keeps its longitudinal legs all the time oriented along the sphere's lines of longitude, the backreaction to the eastern leg extending translates into a smaller movement at the tip of the longitudinal legs, and a larger movement at the base of the longitudinal legs located at the equator. The reverse is true for the backreaction to the eastern leg retracting. As a result, the westward movement of the base of the longitudinal legs along the equator is larger than its eastward movement. A video that shows how 3 connected weights can swim on a two dimensional sphere is available here: http://www.iop.org/EJ/mmedia/1367-2630/8/5/068/movie1.avi . Here is a frame from that video: This two dimensional example also shows the difficulty with defining the center of mass in curved space. For example, in our 3 dimensional space where the 2 dimensional sphere is embedded, we can see that the center of mass of these three masses will be inside the sphere, yet we are assuming the 3 mass object is only in a 2 dimensional curved space. Since the 2-D center of mass must be on the surface of the sphere, the best we can do is to project the 3-D center of mass onto the surface of the sphere. In particular, when the two masses are closer to the two poles, the center of mass is deeper inside the sphere. So when the 3rd equator mass moves while the center of mass is deeper inside the sphere, the projected center of mass on the surface of the sphere will move a larger distance than when the other two weights are closer to the equator and the center of mass is closer to the surface. That is how the net displacement of the 2-D center of mass is achieved. A link to the very technical original research article is: http://dspace.mit.edu/bitstream/handle/1721.1/6706/AIM-2002-017.pdf?sequence=2 - Did you ask a question and then answered it yourself? – MBN Dec 8 '12 at 15:04 1 Yes I did and that is perfectly acceptable. When you are typing the question it gives you an option where you can "[_] Answer your own question - share your knowledge Q&A style" – FrankH Dec 8 '12 at 15:38	{"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": 1, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9300970435142517, "perplexity_flag": "head"}
http://mathoverflow.net/revisions/118451/list	Return to Answer 1 [made Community Wiki] As a set theorist, I feel some obligation to offer an answer here. First, the difficulties students may have in proving set theoretic containments like the one you mention above or in constructing $\epsilon$-$\delta$ proofs is not a matter of them struggling with set theory but rather of them struggling with something new: constructing a proof. In the case of $\epsilon$-$\delta$ proofs, a large part of this difficulty is in understanding quantifiers and how they work (for instance "for all ... exists ..." is not the same as "exists ... for all ...". This is because math is not a trivial subject to learn and some difficulty is required as the students minds stretch and grow. Surely there is no way around this. That really has nothing to do with set theory so far. In spite of a common misconception, set theorists do no actually care how it is that ordered pairs are defined. Or how exactly one codes the notion of a function. Set theory is not in competition with category theory, in spite of what category theory thinks. A very good analogy is provided by computer science: set theory is machine language (or maybe better: a low level language like C) and category theory is object oriented programming. While object oriented programming may provide a useful way of thinking about how to write a program, there still needs to be a machine language for the computer to run on. Moreover, there are occasionally things for which it is just better (or even necessary) to code in a low level language. Set theory provides an exact standard by which to discuss questions like "is there a subset of the real line which is uncountable but not of cardinality $\|\mathbb{R}\|$" (Hilbert's First Problem) or, maybe better, "is there an almost free, non free group?" (Whitehead's problem) or "If $h$ is a homomorphism a commutative Banach algebra into $C[0,1]$, is $h$ continuous?" (the negation being Kaplanski's conjecture). With the exception of the first question, these were asked, to my knowledge at least, without any thought that there was a foundational issue involved. Surely these are questions which could reasonably be asked regardless of how one sets up their foundations. To my knowledge category theory has never resolved these questions; set theory has in as satisfactory a manner possible (or at least until we adopt a more complete set of axioms). Now, one can argue at length about whether such questions are asked in poor taste or whether we should allow them to be asked at all. Readers interested in the question of "why care about set theory" should take a look at this (which might have been titled "why care about the uncountable").	{"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": 8, "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": 0.9668211340904236, "perplexity_flag": "middle"}
http://physics.stackexchange.com/questions/30435/when-does-the-norm-of-quasi-eigenvectors-matter-in-calculations-for-which-phy	# When does the “norm of quasi-eigenvectors” matter in calculations? For which physical results are these even used? Which physical system in nonrelativistic quantum mechanics is actually described by a model, where the norm of the "position eigenstate" (i.e. the delta distribution as limit of vectors in the Hilbert state) matters? >What are some actual end-of-calculation quantum mechanical results, which are supposed to mirror some experimental values, where the constant $c$ in the explicit representation of the distribution $c\ \delta(x-y)$, which representas a sharp localization used in calculations, matters? If I write $c$ and I don't set it to 1, how does it enter the result in the end? If you want to say the norm emerges as a limit of normed states, then justify the norm for the calculation to the point of extracting the specific physical values the theory is supposed to compute. Also, I guess the "if you would position-localize" statement is a little handwaving, or does this have any practical parallel. (Given that it's not really possible, is it? Especially once relativity comes into play.) I ask because on the one hand the momentum and position eigenstates/operator are the one you learn first, but in comparison to spins, there bring some mathematical difficulties. And in a quantum mechanics course, most example calculations will be about bounded problems. I guess there might be non-relativistic scattering processes where a delta is actually used to compute some scattering angle. But nonrelativistic scattering is eighter taught in mathematical physic courses, which focus on the structure of the theory (and I don't know practical physical results coming from there), or they are just the results obtained as classical limits of QFT calculations. - ## 1 Answer Most expectations $\langle A\rangle:=\psi^* A\psi/\psi^*\psi$ are undefined in unnormalizable states. Thus the basis for doing statistics is gone. In particular, in an (unnormalizable) position eigenstate, you cannot speak sensibly about any observable except functions of position and operators commuting with the position operator (such as spin). But once this is enough, one would describe the system in a position-independent way, and the psoition observable is completely eliminated. An example is the position of the center of mass, which is eliminated by restricting to the rest frame, before bound state calculations can be done. -	{"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": 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": 0.9347691535949707, "perplexity_flag": "middle"}
http://mathoverflow.net/revisions/43153/list	## Return to Answer 2 deleted 10 characters in body Dear Tim: I am not sure perhaps I understand misunderstood your question. , but I think any reasonably technical field has a few examples. I heard algebraic geometry is such a field, so I looked up a couple of sources and found these tworight away: 1) Let $f: R\to S$ be a local homomorphism of local rings. Suppose $R$ is regular, $S$ is Cohen-Macaulay, $f$ is finite and $\text{dim} R = \text{dim} S$. Then $f$ is flat. This is a basic result that is used quite frequently. I don't think you can drop any of the hypotheses, they are all basic definitions and independent of each other. 2) A morphism of schemes $f: X \to S$ is quasi-finite if it is locally of finite type, quasi compact and has finite fibres. Again, I don't think you can drop any of the hypotheses. I am reasonably sure EGA has a few more results like this (-: 1 [made Community Wiki] Dear Tim: I am not sure I understand your question. I think any reasonably technical field has a few examples. I heard algebraic geometry is such a field, so I looked up a couple of sources and found these two right away: 1) Let $f: R\to S$ be a local homomorphism of local rings. Suppose $R$ is regular, $S$ is Cohen-Macaulay, $f$ is finite and $\text{dim} R = \text{dim} S$. Then $f$ is flat. This is a basic result that is used quite frequently. I don't think you can drop any of the hypotheses, they are all basic definitions and independent of each other. 2) A morphism of schemes $f: X \to S$ is quasi-finite if it is locally of finite type, quasi compact and has finite fibres. Again, I don't think you can drop any of the hypotheses. I am reasonably sure EGA has a few more results 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": 14, "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": 0.970832347869873, "perplexity_flag": "head"}
http://mathoverflow.net/questions/83977/second-conjugate-operators-to-operators-on-c-0	## Second conjugate operators to operators on $c_0$ ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) I posted my question at MS but unfortunately it is still without a response, so let me ask it here. We can think about a bounded operator $T\colon c_0\to c_0$ as a double-infinite matrix $[T_{mn}]_{m,n\geq 1}$ which acts on a sequence $a=[a_1, a_2, a_3, \ldots ]\in c_0$ in the same way as usual (finite) matrices act on vectors ($n$-tuples of scalars), i.e. $$Ta= [T_{mn}][a_n] = \left[ \sum_{n=1}^\infty T_{mn}a_n\right]$$ Suppose $a=[a_1, a_2, a_3, \ldots ]\in \ell^\infty = (c_0)^{}$. Does the following formula still hold: $$T^{}a= [T_{mn}][a_n] = \left[ \sum_{n=1}^\infty T_{mn}a_n\right]$$ EDIT: Second question deleted, since I wanted to ask, in fact, about something else. - The math.stackexchnage question is: math.stackexchange.com/q/93030/1438 I posted a more "simple minded" answer over there... – Matthew Daws Dec 21 2011 at 14:58 ## 1 Answer Yes. The reason is that the unit vector basis for $c_0$ is a shrinking basis, which means that the biorthogonal functionals to the basis are a Schauder basis for $c_0^* = \ell_1$. This implies that the unit vector basis for $c_0$ is a weak$^$ Schauder basis for $c_0^{} = \ell_\infty$, which means that the expansion of a vector in $\ell_\infty$ in terms of the unit vector basis must converge in the weak$^$ topology. This is enough, since the adjoint of an operator on $\ell_1$ is weak$^$ continuous on $\ell_\infty$. I'll take this opportunity to say something more about a space with shrinking bases. R. C. James, who introduced the concept, realized that if $(e_n)$ is a shrinking basis for a space $X$, then `$X^{}$` can be represented as the space of sequences $(a_n)$ s.t. the partial sums of the series $\sum a_n e_n$ are uniformly bounded. Once he realized this, he very quickly constructed the space we now call J, which is isomorphic to its second dual but is non reflexive; in fact, $\dim J^{*}/J =1$. You can read about the basics of basis theory in many books; Albiac-Kalton is a good choice. -	{"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": 21, "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": 0.9522382020950317, "perplexity_flag": "head"}
http://physics.stackexchange.com/questions/32502/can-quantum-mechanics-really-be-the-same-as-underlying-deterministic-theory	# Can quantum mechanics really be the same as underlying deterministic theory? I am perplexed by recent papers by 't Hooft giving an explicit construction for an underlying deterministic theory based on integers that is indistinguishable from quantum mechanics at experimentally accessible scales. Does it mean that it is deterministic complexity masquerading as quantum randomness? http://arxiv.org/abs/1204.4926 - – Qmechanic♦ Jul 21 '12 at 15:47 – Curious George Jul 24 '12 at 1:04 – Curious George Aug 14 '12 at 17:28 ## 4 Answers I think at least some readers should have noted by now that many of these arguments, particularly the more pathetic ones, are questions of wording rather than physics. Once you made your model simple enough, you can map anything onto anything. Now this was my starting point: if a system is sufficiently trivial, you can do anything you like. Now how can we subsequently generalize some such very simple results into something more interesting? This has been the ground rule of my approach. I am not interested at all in "no-go" theorems, I am interested in the question "what can one do instead?" I admit that I cannot solve the problems of the universe, I haven't found the Theory of Everything. Instead of pathetically announcing what you shouldn't do, I try to construct models, step by step. I now think I have produced some models that are worth being discussed. They may perhaps not yet be big and complicated enough to describe our universe, but it may put our questions concerning the distinctions between quantum mechanics and classical theories in some new perspective. Clearly, if a system is too simple, this distinction disappears. But how far can one go? Remember that cellular automata can become tremendously complex, and quantum mechanical models also. How far can we go relating the two? This is how you should look at my papers. I happen to think that the question is very important, and one can go a lot further in relating quantum models to classical ones than some people want us to believe. And is a calculation wrong if someone doesn't like the wording? - I agree that the models are interesting, and the program is important, but the main problem is that you have not shown that quantum mechanics in a beable basis is equivalent to classical statistics on a CA, and this might not be true (I think it isn't true without further restrictions). I will ask a direct question, as your interesting answers and comments have allowed one to sharpen one's thinking about this. – Ron Maimon Aug 14 '12 at 13:21 – Curious George Aug 14 '12 at 17:29 Current (experimental and theoretical) wisdom on deterministic approaches to quantum nondeterminism just say that any deterministc theory underlying quantum mechanics must be nonlocal. Research then goes on discussing the precise nature of this nonlocalness or ruling out certain versions. On the other hand, there are those who construct nonlocal deterministic theories that somehow reduce to QM. A lot of work goes into Bohmian mechanics, which however has difficulties to recover realistic quantum field theory. The paper by t'Hooft pursues a different approach, based on discreteness. However, his results are currently very limited, just reproducing the harmonic oscillator. - Your first phrase may be too categorical. For instance, it is my understanding that there has been no loophole-free experimental evidence of nonlocality. On the other hand, nonlocality is an extraordinary claim, so it needs an extraordinary proof. – akhmeteli Jul 22 '12 at 1:53 1 t'Hooft doesn't "reproduce" anything--- he has a linear space which is a quantum state space, he just truncates the evolution to be deterministic on a special basis. The actual dynamics is still quantum, it isn't really an automaton--- you can't imagine that the states are classical underneath. – Ron Maimon Jul 22 '12 at 9:03 "However, his results are currently very limited, just reproducing the harmonic oscillator" He does have two follow up papers showing equivalence of a bosonic filed theory in 1+1 dimensions and also a "superstring" like solution for undelying deteministic lattice (arxiv.org/abs/1207.3612 and arxiv.org/abs/1205.4107) However, the arguements there seem to me to be extension of the arguemnt for the first paper . I could not find any loophole in the mapping.. hence the question.. what precisely does the scheme violate? – user10733 Jul 22 '12 at 16:18 I do agree that there arent many results especially anthing that would distinguish the model from others Thank you a lot for your comment . -- Ash – user10733 Jul 22 '12 at 16:18 1 @akhmeteli: Particle nonlocality is an ordinary experimental fact (already visible in interference experiments); it is only the precise form it takes in theory that is in doubt. So I don't agree that it needs extraordinary verification. The philosophical problems go away once one gives up the particle picture; then there is nothing counterintuitive left that would need extraordinary attention. – Arnold Neumaier Jul 22 '12 at 17:48 show 11 more comments It is certainly possible that QM is based on a deterministic physical mechanism. The no-go theorems like Bell's theorem or the "Free will theorem" of Conway and Kochen are not effective against deterministic hidden variable theories because they require non-determinism as one of their assumptions. There are still many phisicist claiming that determinism has been disproven but they are commiting the logical fallacy . However, it is too early to say if 't Hooft is on the right track. - 4 -1: This is incorrect. The Bell theorem is effective against deterministic hidden variable theories which are local. The argument from superdeterminism is ridiculous, it is not science, it is conspiracy theory. – Ron Maimon Jul 22 '12 at 9:02 t'Hooft's papers are not valid. They make a mistake, which is that they assume that just because the discreate time evolution operator in a quantum system is a permutation in some basis, that the quantum theory is then a classical theory. t'Hooft considers discrete time quantum systems where the time-evolution in some basis is a discrete permutation. So that if you have a 3 state system you permute 1 to 2 to 3. He then analyzes the space of all superpositions of these three states, and discovers he can recover quantum mechanics. He then declares "quantum mechanics is equivalent to a classical determnistic system". This is just plain wrong. I suppose t'Hooft is thinking that if you start out in some basis state, you stay in a basis state forever, just permuting the basis state, and therefore this must be a classical deterministic system. But the point is that the state space includes all sorts of quantum superpositions of the basis states, and these other states, the non-basis states, are superpositions not by classical probability, but by quantum amplitudes. If you have quantum amplitudes, even if the basis states evolve by permutation, the theory can obviously reproduce quantum mechanics, because it is quantum mechanics. In fact, here is a theorem: Given any finite dimensional quantum mechanical Hamiltonian H, there exists a permutation system which includes this Hamiltonian in an approximation, acting on a subspace of the states. The proof: diagonalize H to an N by N diagonal matrix with N eigenvalues, and approximate the N energies by rational numbers with enormous prime denominators, $p_i/q_i$ $1<i<N$, and take a unit time-step. Multiply all the q_i's together and call the product Q. Then the exponential of t times the Hamitlonian is periodic with period Q time steps. Consider now a state-space whose basis is labelled by an N-tuple integers from 1 to Q. Let the permutation Hamiltonian take the basis element (a_1,....a_n) to $a_i\rightarrow a_i + s_i$ where $s_i$ is the product of all the q's except q_i, and the $Z_{q_i}$ multiplicative inverse of $p_i$. This permutation Hamiltonian has to property that it's eigenvalues include a subset with $p_i/q_i$. Project to this subspace, and call this your quantum system. This process, or anything resembling it, cannot be called a "deterministic system" in any way. There are still states which are superpositions. If you have a true classical system, the state is described by a probability distribution on the unknown starting state, not by probability amplitudes for superpositions of the unknown current state. The moment you describe states by superpositions, you are not getting quantum mechanics out, you are putting it in. This is the reason t'Hooft is able to derive mathematical results that are quantum mechanical, he is using quantum mechanics, but with a restriction that it reduces to a permutation on one basis. This doesn't explain why we see superpositions of electronic spins in nature, it doesn't produce these superpositions from ignorance of classical values, it puts in the superpositions by hand. I like t'Hooft's motivations and admire his independent thinking, but this is not valid stuff. It doesn't do what he claims it does. To call the statement that these are classical models misleading is charitable. - 't Hooft thinks there is an ontologically preferred basis and superpositions of those basis states are not real. The Bohmian treatment of measurement interactions shows that you can have a preferred basis and still describe other observables correctly - though the Bohmians also still have superpositions in their ontology, as states of the pilot wave (unless they take the "nomological" path, and treat the specific pilot wave of an individual system as a dynamical law rather than a physical state)... – Mitchell Porter Jul 22 '12 at 9:59 @MitchellPorter: I know what he says, but this is ridiculous--- you have to tell me how come a laboratory electron is described by a superposition. It doesn't help to say that there is a basis in which the Hamiltonian is a permutation. If we don't know which basis element out universe is in, we describe that with a probability distribution, not with amplitudes. Then there is no reason that the electron in the lab is described with amplitudes. It is just wrong. – Ron Maimon Jul 22 '12 at 10:01 In Bohmian mechanics with a specific pilot wave, you can substitute the specific pilot wave into the equations of motion for the classical objects of the theory, and you end up with a pseudo-classical theory in which a classical equation of motion is augmented by a nonlocal potential. It must be possible to explain superposition in terms of this nonlocal potential, because the theory is still identical to Bohmian mechanics with a specific pilot wave, but no-one has ever taken this route and exhibited what such explanations look like... – Mitchell Porter Jul 22 '12 at 10:19 In the paper after this one, 't Hooft constructs his alleged mapping from a CA to a QFT. The CA is really simple but the mapping is a little nontrivial; at least, I haven't grasped the essence of it yet. It will be hard to say anything concrete about how or even whether 't Hooft can account for observed superpositions, until someone understands this further stage of his recent work. – Mitchell Porter Jul 22 '12 at 10:21 @MitchellPorter: It took me a long time to understand it, because it is clearly wrong, and I tried to make a true map from CA to QM. What he is doing is what I described--- he takes a QM system and transforms it to a case where it turns into a permutation on a basis, and when he can do this (which is always) he declares he has gotten QM out of a classical automaton. The declaration is false, the method is producing an "t'Hooft quantum automaton" not a classical automaton since it includes superposition states a-priori. – Ron Maimon Jul 22 '12 at 19:18	{"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": 0.3, "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": 0.9470763802528381, "perplexity_flag": "middle"}
http://mathhelpforum.com/geometry/2566-test-4-14-geometry-picture-included-please-help-asap-please-print.html	# Test on 4/14. Geometry. Picture included Please help ASAP please Printable View • April 13th 2006, 04:49 PM AirForceOne http://img.photobucket.com/albums/v2...e/441fd7bf.gif The ones I know are know are correct but no well understood: a) 1:1 because both triangles have the same base? So 16:16 = 1:1 right? b) same method as ^ d) 9:16 because using the similar triangles area ratio formula. The method of similar trianges i used was AA because of the parallel lines right? The ones I don't understand at all but I know the answers. I'm all confused: c) The answer is 1:1. I think you can assume that the height of both triangles are the same? How can it be 1:1 if there are no other measurements of the triangles? e) The answer is 3:4. I figured maybe you do simplfy 12/16 to get 3:4 but I have no clue why because the the triangles that the question is asking for doesn't have 12 or 16? • April 13th 2006, 09:21 PM earboth 1 Attachment(s) Quote: Originally Posted by AirForceOne ... The ones I don't understand at all but I know the answers. I'm all confused: c) The answer is 1:1. I think you can assume that the height of both triangles are the same? How can it be 1:1 if there are no other measurements of the triangles? e) The answer is 3:4. I figured maybe you do simplfy 12/16 to get 3:4 but I have no clue why because the the triangles that the question is asking for doesn't have 12 or 16? Hello, let H be the height of the trapezoid and h the height of triangle(ZYP) (I#ve attached a diagram to demonstrate, what I'll calculate). Then you get the proportion: $\frac{h}{16}=\frac{H-h}{12}$. Solve for h and you'llget h = 4/7*H. to e.: You get the area of triangle(XPY) by: $A_{\Delta XPY}=A_{\Delta WXY}-A_{\Delta WXP}$ $\frac{1}{2} \cdot 12 \cdot H-\frac{1}{2} \cdot 12 \cdot \frac{3}{7} \cdot H=\frac{1}{2} \cdot 12 \cdot H \cdot \frac{4}{7}$ That means: $\frac{A_{\Delta XPY}}{A_{\Delta WXP}}=\frac{\frac{1}{2} \cdot 12 \cdot H \cdot \frac{4}{7}}{\frac{1}{2} \cdot 12 \cdot \frac{3}{7} \cdot H}=\frac{4}{3}$ to d.: As I've shown above you can calculate the areas of the triangles in question by calculating the differences of two triangles. Unfortunately I'm a little bit in a hurry to complete the problem, but I'm certain that you now know how to handle the problem. Greetings and Happy Easter to you. EB • April 13th 2006, 10:19 PM earboth Quote: Originally Posted by AirForceOne ... The ones I don't understand at all but I know the answers. I'm all confused: c) The answer is 1:1. I think you can assume that the height of both triangles are the same? How can it be 1:1 if there are no other measurements of the triangles? ... Hello, to c.: as you've demonstrated: $A_{\Delta ZYW}=A_{\Delta ZYX}$ (same base, same height). Thus $A_{\Delta ZYW}-A_{\Delta ZYP}=A_{\Delta ZYX}-A_{\Delta ZYP}$ $A_{\Delta WZP}=A_{\Delta XYP}$. Thus the ratio is 1:1. Greetings EB All times are GMT -8. The time now is 05:02 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": 7, "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": 0.9431257247924805, "perplexity_flag": "middle"}
http://mathoverflow.net/questions/71877/on-a-claim-of-ramanujan-in-his-lost-notebook/71922	## On a claim of Ramanujan in his “Lost Notebook”. ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) As I was flipping through the scanned version of Ramanujan's "Lost Notebook" in our library, I came across a result which caught my attention. And as any excited teenager would do, I immediately photographed that page and put it up here. The result in the photo is as follows. If $S(N)$ be the number of ways in which $N$ can be expressed as the sum of 2 squares, then the max. order of $S(N)$ $$= \sqrt{\text{max order of }\ d(N^2 +aN+b)} \cdot e^{O(\log N)^{1/2 + \epsilon}}$$ Has anyone come across any such result before? Is this result true? - and $d$ is ... – Ricky Demer Aug 2 2011 at 17:57 1 $d$ must be the number of divisors of a positive integer, $d(n) = \sigma_0(n).$ Reading from Hardy and Wright, fifth edition paperback, section 16.7. Then the "average order" of $d(n)$ is section 18.2, $r(n)$ defined 16.9, then average order and max order in section 18.7. Oh, H+W $r(n)$ is $S(N)$ above. Estimates on $r(n)$ are called the Gauss Circle Problem, upper bounds on $d(n)$ are mathoverflow.net/questions/43103/… – Will Jagy Aug 2 2011 at 19:22 ## 1 Answer From Hardy and Wright, fifth edition paperback, I am going to call your $S(N)$ their $r(n).$ Chapter 18 is called The Order of Magnitude of Arithmetical Functions. In section 18.1, Theorem 317 $$\limsup \frac{d(n) \log \log n}{\log n} \; = \; \log 2$$ with the comment Thus the true `maximum order' of $d(n)$ is about $$2^{\log n / \log \log n}$$ Then in section 18.7, There is also a theorem corresponding to Theorem 317; the maximum order of $r(n)$ is $$2^{ \frac{\log n}{2 \log \log n} }$$ So, if one takes the square root of the first, the result is the second. It is a bit unclear what is meant by the numbers $a,b.$ Anyway, there has been considerable work on both items, as separate estimation problems, see http://mathoverflow.net/questions/43103/what-is-the-lower-bound-for-highly-composite-numbers/43105#43105 and, for example, http://mathoverflow.net/questions/19079/what-is-the-status-of-the-gauss-circle-problem where the interior of the circle is included. - Thank you for the answer sir. – Koundinya Vajjha Aug 3 2011 at 9:46	{"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": 17, "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": 0.908343493938446, "perplexity_flag": "middle"}
http://mathoverflow.net/questions/74119?sort=votes	## What are categorical models of W-types in intensional type theory? ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) I'm familiar with container functors and older work by Dybjer on categorical models for W-types in the extensional theory, but I was looking for some similar semantics in the intensional case. - ## 1 Answer Unless I've not understood your question correctly (sometimes people mean different things by the distinction between intensional and extensional), then I think the answer is: the semantics for W-types in intensional type theory are exactly the same as the semantics for W-types in extensional type theory. A model of type theory (comprehension category, category with attributes, et cetera) is essentially just a Grothendieck fibration $p:E\to B$ which comes equipped with certain structure. Being a model of W-types just means that $p$ comes equipped with certain extra structure. As I understand it, the difference between intensional and extensional type theory has to do with which axioms are satisfied by the identity types. (In terms of $p$ this is just whether it is equipped with one or another kind of structure for interpreting identity types.) Whether or not you are able to interpret W-types is again simply a question of whether $p$ has a further kind of additional structure and is a prior independent of which kind of identity types $p$ is able to interpret. In practical terms, what this means is that the question of whether your model supports the interpretation of W-types has everything to do with your fibration $p$ and formulated in this way the question is completely independent of whether the model is intensional or extensional. E.g., if your fibration $p$ comes from something like a class of display maps and you want to interpret W-types as initial algebras for polynomial endofunctors, then you will need to verify that these initial algebras land in the class of display maps. - Huh - perhaps I'm coming at this from the wrong direction in the literature. So the semantics in the extensional case, as I learned it, is given by certain equivalence classes of functors on the category of types. For the extensional theory interpreted in LCCCs these are all of the form $\Sigma a:A. Ba \rightarrow X$ for some choice of types A & B. It sounds like there's a more general interpretation of W-types from the fibred perspective that I simply didn't know. Could you name a good place to start digging further into the kind of models you're talking about? Thanks! – Caylee Aug 31 2011 at 14:45 I expect there to be more interaction between the intensional aspects of a model and the W-types. For example, what can be said about the identity types of a W-type? Does it ever exceed the h-level of its constituent parts (the branching types)? – Andrej Bauer Aug 31 2011 at 15:11 @Andrej: Of course there are questions about how W-types interact with identity types in the intensional setting, but as far as the semantics is concerned this does not matter. The only way I can see that this could affect the semantics is if you have in mind a different notion of W-type from the usual one (e.g., requiring the existence of propositional equalities instead of definitional equalities in the conversion rule, or something along these lines). – Michael A Warren Aug 31 2011 at 18:03 1 @Caylee: What I have in mind are the semantics of type theory which are a bit more general than the LCCC semantics that one sometimes first encounters. For this I would look at the books by Thomas Streicher ("Semantics of Type Theory") and Bart Jacobs ("Categorical Logic and Type Theory") on semantics of type theory. There are also some papers by Andy Pitts ("Categorical logic") and Martin Hofmann ("Syntax and semantics of dependent types") on these matters which I highly recommend. – Michael A Warren Aug 31 2011 at 18:13	{"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": 8, "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": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9334810376167297, "perplexity_flag": "head"}
http://mathoverflow.net/questions/92061?sort=newest	## Square of a continuous map ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) Recently a student asked me the following (elementary looking) question : If $T$ is an invertible linear transformation of some finite-dimensional space $E$ into itself which factorizes as $T = f \circ f$ where $f : E \mapsto E$ is continuous, must $T$ have positive determinant ? Of course this is trivially true if $f$ is itself linear. It is also an easy exercise to show that this also holds when $f$ behaves locally like a linear transformation, that is, when it is $C^1$ : $T$ then factorizes as $T = df_{f(0)} \circ df_0$, and since $x \mapsto \det df_x$ keeps a constant sign, we're done. When $f$ is only continuous, this certainly still holds but I suspect this requires rather deep properties of continuous maps (unless I missed something obvious ...) with which I'm not very familiar. Hence two questions : 1) Is there an "elementary" proof of this ? (in which case I apologize for this question) 2) Does this property sound obvious to experts ? That is, is there some two-lines proof of this with a sufficient background ? If yes, what would be good references (books for example) to acquire this background ? - Isn't f being $D^1$ enough instead of $C^1$? You get the derivative of T from the chain rule. – Zsbán Ambrus Mar 24 2012 at 17:21 2 Is there are problem to extend your f on sphere? because here living obvious topological resons, but same time you can approximate f by smooth automorphism of S^n and saying what you already say. – Bad English Mar 24 2012 at 21:33 ## 1 Answer The first relevant fact about $f$ is that it is a proper map. In such a situation the topological (Brouwer) degree of $f$ is well-defined, and by the product rule $\operatorname{deg}(T)= \operatorname{deg}(f\circ f)= \operatorname{deg}(f) \operatorname{deg}(f)$. For an invertible linear transformation, the topological degree is the sign of the determinant, which proves your claim. - 4 @Pietro Majer: Actually, how do you see that $f$ is proper directly? Instead, one can observe that $f$ is 1-1 and onto and, hence, a homeomorphism by invariance of domain theorem. (Only then I see why $f$ is proper.) Now, of course, the degree argument applies. On the other hand, how do you explain this to an unprepared undergraduate without introducing homology? I can imagine telling him/her about orientation and orientation-reserving/reversing maps and indicating why reversing orientation twice amounts to preserving orientation, etc. – Misha Mar 24 2012 at 15:54 2 Yes, $f$ is proper because it's a homeo, and it is a homeo just because $f\circ f$ is so, as explained in David Cohen's answer (now deleted). Topological degree can be introduced by several tools, and it is sometimes taught at undergraduate level. There could be a more elementary proof though. – Pietro Majer Mar 24 2012 at 21:52	{"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": 22, "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": 0.944459855556488, "perplexity_flag": "head"}
http://mathoverflow.net/questions/107804/a-continuous-linear-functional-on-l-inftyr-that-vanishes-on-cr	## A continuous linear functional on $L^\infty(R)$ that vanishes on $C(R)$. ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) Intuitively, I want to construct the functional F in this way: $$F(f)=\lim_{x\rightarrow 0+}f(x)-\lim_{x\rightarrow 0-}f(x)$$ for $f\in L^\infty$. I know this is not well defined so I'd like to find a way to use this idea. Maybe find an extension using Hahn-Banach, etc. - How about the following: try averaging f over [0,1/N], averaging it over [-1/N,0], then taking the difference. That gives you a functional $F_N$. Now take a Banach limit or similar as $N\to\infty$ – Yemon Choi Sep 21 at 21:51 Yes, define $F(f)$ for all piecewise continuous functions for which that limit exists, and extend using Hahn-Banach. That works. – George Lowther Sep 21 at 22:08 This looks like homework; voting to close. – unknown (google) Sep 21 at 23:15 2 It does not look like homework to me. It looks like a student who has been told that such functionals exist by the Hahn-Banach theorem and wonders about a specific example. Which, for a student learning functional analysis, is a perfectly reasonable question. That this is not the right forum for it is another matter. – Michael Renardy Sep 21 at 23:32 2 "This looks like homework; voting to close. – unknown (google)" In all honesty, if you want to vote to close, start with choosing a unique username! – fedja Sep 22 at 3:32 show 1 more comment ## 1 Answer I am not sure what is your question. But you can construct a linear functional in this way. The limit does not exist for some $f$ in $L^\infty$. But you can use the Banach limit. Consider the bounded linear functional on $C(R)$ defined as the limit as $x\to 0$. By Hahn-Banach this has an extension to a bounded linear functioal on $L^\infty(-\infty,0)$. and also has an extension to $L^\infty(0,\infty)$. Now define your functional as you propose, using these Banach limits. But all this is a kind of trivial nonsense. Hahn Banach says you from the beginning that you have (very many) linear functionals on $L^\infty$ which are zero on $C$. This is a pure existence theorem. But your "construction" really adds nothing to it, because the existence of a Banach limit is again a pure existence theorem. -	{"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": 0.9274523854255676, "perplexity_flag": "middle"}
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Suppose one finds, or builds, a constellation of several black holes arranged in a circle. ...	{"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": 0.9228429198265076, "perplexity_flag": "middle"}
http://math.stackexchange.com/questions/91141/terminology-for-properties-of-functions	# Terminology for properties of functions. So I was wondering, is there a name for a function whose output is always less than or equal to its input ($f(x)≤x$)? I know there is a name for functions that satisfy $x_1<x_2\rightarrow f(x_1)<f(x_2)$ (monotonic) so I figured there would be a name for $f(x)≤x$. Does anyone know what it is? A good example of this would be the greatest integer function $\lfloor x\rfloor$. Thank you. - A function satisfying "$x_1<x_2\rightarrow f(x_1)<f(x_2)$" is called strictly increasing, not monotonic. – Chris Eagle Dec 13 '11 at 15:58 1 I've seen functions satisfying "$x \le f(x)$" called inflationary, progressive, or extensive. Thus one could use deflationary or regressive for your property, but I haven't actually seen such usage. – Chris Eagle Dec 13 '11 at 16:04 – Hautdesert Dec 13 '11 at 21:29	{"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": 0.9363619089126587, "perplexity_flag": "middle"}
http://mathoverflow.net/questions/12147/how-can-i-tell-if-y-is-a-function-of-x-in-a-random-sample	## How can I tell if y is a function of x in a random sample? ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) I have some data and believe that a given metric is a function of another metric. I have the values of both metrics and many different sets of these values. Can I tell if one is a function of the other through some simple exercise like a regression? I'm not sure if the function is linear. I'm not a math expert so apologies if this is a trival question. Edit: Here's my (Anton's) interpretation of the question. If I misunderstood, I hope gitkin corrects it. Given a bunch of data points `$\{(x_i,y_i)\}$` in the plane, I can find the line best fitting the data. Then I can compute the coefficient of determination $R^2$ to see how good the fit is. More generally, given a model $y=f(x)$ (where $f$ may not be linear), I can do various things to determine how well the model fits the data. Is there some way to determine if there exists a model $y=f(x)$ fitting the data well? In other words, is there a way to measure your confidence that the $x$ values completely determine the $y$ values (in some reasonable way) in the system you've sampled? Intuitively, you should somehow vary over all possible functions $f$, measure how much the model $y=f(x)$ fails to explain the data, add some penalty depending on the complexity of $f$ relative to the size of the sample,† and return the lowest value you get. Is there a precise, theoretically justified way to do this? † e.g. the penalty should be very high if $f$ is a polynomial of degree comparable to the number of data points. - 1 poor use of terminology on my part. It's just a value. "property was quantitatively measured" is exactly what I meant. – gitkin Jan 17 2010 at 22:38 1 I think a version of this question, rephrased in the language of statistics, could be very good. As it is currently written, I suspect there will be votes to close, although I hope it is not, in that Douglas's answer below sufficiently interprets the question mathematically (although not deeply so). – Theo Johnson-Freyd Jan 17 2010 at 23:58 1 I disagree. This is not a real question. It is asking for a way to fit a curve to data points on a 2d-graph. Of course, you can construct a function that hits almost every point, but having a perfect or near-perfect fit curve doesn't help you prove anything. The whole point of modeling data like this is to see a clear trend and compute how effective your estimate is. If the function is too complicated, all that you get is a really strange looking interpolated graph. – Harry Gindi Jan 18 2010 at 4:15 1 I'm going to close this as "not a real question" because it's too ambiguous. We've resolved the confusion about "metric," but now I don't see what you mean by "function" or "the values of both metrics." If you have all the values, then x is a function of y if no value of y has two different values of x, but I suspect this isn't what you're looking for. Since you used the stats tag, you probably have samples. Are you asking if there's a dependence between the two? If you edit the question to ask a precise question, I'm happy to consider reopening it. – Anton Geraschenko♦ Jan 18 2010 at 7:41 1 Although you can interpolate when there is no relation, nonmathematicians including physicists are likely to look at that mess and say, "That's not a function!" Perhaps they want the function to be $K$-Lipschitz for some $K$ which isn't too large, or $\alpha$-Holder continuous for some $\alpha$ which isn't too small. – Douglas Zare Jan 18 2010 at 12:07 show 4 more comments ## 4 Answers If you are only interested in correlation between the two feature values, then there are a lot of ways to compute it (simple correlation, rank correlation, linear or nonlinear regression, etc.). If you are interested in causality, a few places to look at are: Granger causality and NIPS workshops on causality: 2008, 2009 - 1 I'm not sure why no one is upvoting this. To my limited knowledge, these look like very useful links. – David Speyer Jan 19 2010 at 13:22 I wish I understood that notion of causality well enough to say whether it fit the question. – Douglas Zare Feb 2 2010 at 6:27 ### 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. Metric is a technical term in mathematics, but I'll ignore the usual technical meaning. In practice, I would plot the points $(metric_1,metric_2)$. Decide whether you would call the graph a function, whether you can predict the value of one from the other. A linear regression will only detect linear functions perfectly since the linear correlation will be +1 or -1. You can detect any increasing or decreasing function with a Spearman rank correlation coefficient, or just sort by one metric and see if that sorts the other. This will not detect a relationship which is not monotone like $metric_1 = \sin(metric_2)$. If you have a good guess that something like this is the case, you might try testing the rank correlation of $metric_1$ and $\sin(metric_2)$. - I believe there is another technique which may be useful, which is to consider the topology of set of points within r of a data point as you increase r. It's too bad the question is closed. – Douglas Zare Jan 30 2010 at 10:47 I do have a little to add at a much lower level. The first step is to plot lots of points and see if you still believe one quantity is determined by the other. Next, rewrite the pairs as $(x_i, y_i)$ where you believe the $x$ value may determine the $y$ value (you might need to switch the order of every pair). One necessary condition is that there be no repeated $x_i.$ Finally and hardest, you really need to GUESS a functional relationship. As long as your function is determined by a (small) finite number of quantities the method of least squares can be applied. If you think you have a sine wave, you define the general curve by constants $A,B,C$ in the function $f(x) = A \sin (B x + C).$ Least squares says you minimize $\sum_{i} ( y_i - f(x_i) )^2,$ which is a process involving your data pairs and something called partial derivatives. You should get individual help with this process, it is commonly taught just for lines (regression). If the best curve matches the data points very well perhaps you have it. Finally, the reason you are absolutely required to guess a function eventually is that, under the assumption that there is a dependence (no repeated $x_i$) there are infinitely many mathematical functions $g(x)$ that satisfy all $y_i = g(x_i)$ exactly, for example $g(x)$ can be a polynomial of high degree. What you really want is a function that will be deemed reasonable in your line of work. - Strictly a function is a mapping that assigns a unique value in a set B to every point in a set A. So the only way (in this strict sense) in which your second "metric" will not be a function of the first metric is if there are two datapoints for which the second metric gives different values but your first metric gives the same value. However I suspect you have in mind the looser non-mathematical sense in which you can write the second "metric" (the quotation marks are because metric has a particular technical meaning in mathematics) as a closed-form function of the first metric plus some reasonably well-behaved error term. In this case a suitable linear regression may help you with your problem but only once you make some assumptions about the form of the functional relationship. - 4 I also suspect this question will ultimately be closed as insufficiently mathematical/suitable for MathOverflow. – Tom Smith Jan 17 2010 at 22:35	{"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": 29, "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": 0.9404465556144714, "perplexity_flag": "head"}
http://math.stackexchange.com/questions/187794/why-did-my-friend-lose-all-his-money/187808	# Why did my friend lose all his money? Not sure if this is a question for math.se or stats.se, but here we go: Our MUD (Multi-User-Dungeon, a sort of textbased world of warcraft) has a casino where players can play a simple roulette. My friend has devised this algorithm, which he himself calls genius: • Bet 1 gold • If you win, bet 1 gold again • If you lose, bet double what you bet before. Continue doubling until you win. He claimed you will always win exactly 1 gold using this system, since even if you lose say 8 times, you lost 1+2+4+8+16+32+64+128 gold, but then won 256 gold, which still makes you win 1 gold. He programmed this algorithm in his favorite MUD client, let it run for the night. When he woke up the morning, he was broke. Why did he lose? What is the fault in his reasoning? - 38 The last clause should read "continue doubling until you win or have no money". – Chris Eagle Aug 28 '12 at 8:31 19 It's called a martingale betting strategy and it works well if you have infinite wealth which was probably not the case... – johnny Aug 28 '12 at 8:33 11 To make the claim "you will always win exactly 1 gold" approximately true (almost certainly, and assuming no limit on funds and height of bets), your second clause should be "if you win, stop" – Marc van Leeuwen Aug 28 '12 at 9:49 6 Assume you keep doing this, say, 30 times in a row. Where do you take 2^30 money from? Heck, I'm not partiucularly poor, but I don't usually have that much in my pocket... – Damon Aug 28 '12 at 13:30 6 It's a shame that we're at a place where a MUD has to be described as a "text-only World of Warcraft" – Gareth Aug 28 '12 at 19:25 show 9 more comments ## 3 Answers Suppose, for simplicity, that the probability of winning one round of this game is $\frac{1}{2}$, and the probability of losing is also $\frac{1}{2}$. (Roulette in real life is not such a game, unfortunately.) Let $X_0$ be the initial wealth of the player, and write $X_t$ for the wealth of the player at time $t$. Assuming that the outcome of each round of the game is independent and identically distributed, $(X_0, X_1, X_2, \ldots)$ forms what is known as a martingale in probability theory. Indeed, using the bet-doubling strategy outlined, at any time $t$, the expected wealth of the player at time $t + 1$ is $$\mathbb{E} \left[ X_{t+1} \middle\| X_0, X_1, \ldots, X_t \right] = X_t$$ because the player wins or loses an equal amount with probability $\frac{1}{2}$ in each case, and $$\mathbb{E} \left[ \left\| X_t \right\| \right] < \infty$$ because there are only finitely many different outcomes at each stage. Now, let $T$ be the first time the player either wins or goes bankrupt. This is a random variable depending on the complete history of the game, but we can say a few things about it. For instance, $$X_T = \begin{cases} 0 & \text{ if the player goes bankrupt before winning once} \\ X_0 + 1 & \text{ if the player wins at least once} \end{cases}$$ so by linearity of expectation, $$\mathbb{E} \left[ X_T \right] = (X_0 + 1) \mathbb{P} \left[ \text{the player wins at least once} \right]$$ and therefore we may compute the probability of winning as follows: $$\mathbb{P} \left[ \text{the player wins at least once} \right] = \frac{\mathbb{E} \left[ X_T \right]}{X_0 + 1}$$ But how do we compute $\mathbb{E} \left[ X_T \right]$? For this, we need to know that $T$ is almost surely finite. This is clear by case analysis: if the player wins at least once, then $T$ is finite; but the player cannot have an infinite losing streak before going bankrupt either. Thus we may apply the optional stopping theorem to conclude: $$\mathbb{E} \left[ X_T \right] = X_0$$ $$\mathbb{P} \left[ \text{the player wins at least once} \right] = \frac{X_0}{X_0 + 1}$$ In other words, the probability of this betting strategy turning a profit is positively correlated with the amount $X_0$ of starting capital – no surprises there! Now let's do this repeatedly. The remarkable thing is that we get another martingale! Indeed, if $Y_n$ is the player's wealth after playing $n$ series of this game, then $$\mathbb{E} \left[ Y_{n+1} \middle\| Y_0, Y_1, \ldots, Y_n \right] = 0 \cdot \frac{1}{Y_n + 1} + (Y_n + 1) \cdot \frac{Y_n}{Y_n + 1} = Y_n$$ by linearity of expectation, and obviously $$\mathbb{E} \left[ \left\| Y_n \right\| \right] \le Y_0 + n < \infty$$ because $Y_n$ is either $0$ or $Y_{n-1} + 1$. Let $T_k$ be the first time the player either earns a profit of $k$ or goes bankrupt. So, $$Y_{T_k} = \begin{cases} 0 && \text{ if the player goes bankrupt} \\ Y_0 + k && \text{ if the player earns a profit of } k \end{cases}$$ and again we can apply the same analysis to determine that $$\mathbb{P} \left[ \text{the player earns a profit of $k$ before going bankrupt } \right] = \frac{Y_0}{Y_0 + k}$$ which is not too surprising – if the player is greedy and wants to earn a larger profit, then the player has to play more series of games, thereby increasing his chances of going bankrupt. But what we really want to compute is the probability of going bankrupt at all. I claim this happens with probability $1$. Indeed, if the player loses even once, then he is already bankrupt, so the only way the player could avoid going bankrupt is if he has an infinite winning streak; the probability of this happening is $$\frac{Y_0}{Y_0 + 1} \cdot \frac{Y_0 + 1}{Y_0 + 2} \cdot \frac{Y_0 + 2}{Y_0 + 3} \cdot \cdots = \lim_{n \to \infty} \frac{Y_0}{Y_0 + n} = 0$$ as claimed. So this strategy almost surely leads to ruin. - 9 – Henning Makholm Aug 28 '12 at 11:40 @HenningMakholm Oh wow, this actually is a thing? Thanks! – Konerak Aug 28 '12 at 12:18 7 Yes This is called the gambler's ruin problem. So even for a fair game if you play long enough you will lose all your money. If the game favors the house you will lose faster. See Feller's probability theory books for a more detailed account of the problem. – Michael Chernick Aug 28 '12 at 12:25 This betting strategy is very smart if you have access to infinite wealth or can go into infinite debt. In reality however, you will eventually lose all or most of your money. Say your friend had $k$ gold at the beginning. I assume that this simple roulette has a probability of both win and loss equal to $0.5$. First, let's see how many times you need to lose in a row in order to lose all your wealth. \begin{align} 1 + 2 + 2^2 + 2^3 + ... + 2^n &\geq k \\ 2^{n+1} - 1 &\geq k \\ n &\geq \log_{2}(k+1) - 1 \end{align} So even if you start with $10000$ gold, after $13$ lost bets you are broke and in debt. Continuing this example, the probability of this happening in a one shot-game is a mere $0.02$%. However, if you keep the algorithm running all night for $8$ hours betting every $5$ seconds, your chances of having a losing streak of $13$ in a row go up to $29.61$%. Assuming that you cannot go into debt and $12$ losses is the most you can handle, then with the same data the chance of losing most of your money goes up to $50.5$%. - 1 I think I get this! So, the probability of losing 13 times is low (1/2^13), but since the other (1-1/2^13) times where he won, he only won 1 gold, and the one time he lost, he lost all his gold, the expected value is low? – Konerak Aug 28 '12 at 9:15 Not exactly. If you try this strategy right now only once until first win or all money lost, the probability that you lose is $\frac{1}{2^{13}}$. However if you start an algorithm, all that matters is the probability of losing $13$ times in a row while your algorithm is running and the probability of this happening is much higher. – johnny Aug 28 '12 at 9:23 Somehow you have to factor in the effect that if you do manage to win $2^{13}$ times without going bankrupt, then your tolerance for runs of bad luck goes up to fourteen. I don't think that it changes the verdict, but it would be nice to see how to take that into account. – Jyrki Lahtonen Aug 28 '12 at 9:24 That is a fair point! I'm glad @ZhenLin worked out the formal details for that. – johnny Aug 28 '12 at 9:40 I suppose he was betting on a red/black or even/odd system. Now, there is a "0" which is green/not even or odd, so if you hit that zero, you lose. There are 36 numbers you can hit, plus that 0 . Even if you chose red or black, even or odd, there are bigger chances to lose than win. - Roulette tables have a zero and a double-zero, to give the casino its only mathematical advantage. Are you sure this answer true for the MUD as given? Keep in mind, one wasn't given. – Joshua Shane Liberman Aug 28 '12 at 15:56 ## protected by J. M.Aug 29 '12 at 10:28 This question is protected to prevent "thanks!", "me too!", or spam answers by new users. To answer it, you must have earned at least 10 reputation on this site.	{"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": 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": 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": 0.9568861126899719, "perplexity_flag": "middle"}
http://math.stackexchange.com/questions/34101/conditional-expectation-ea-mid-b-b-and-eb-mid-a-a-implies-a-b	# Conditional expectation: $E[A \mid B] = B$ and $E[B \mid A] = A$ implies $A = B$ Given two $L^1$ random variables $A$, $B$ satisfying $E[A \mid B] = B$, $E[B \mid A] = A$. The claim is that $A = B$. So is this easy? I don't really see it.. If $A$ and $B$ are in $L^2$ it's clear, for then $E[AB] = E[A^2] = E[B^2]$, thus $E[(A-B)^2] = 0$. Can one now use an approximation argument? - ## 1 Answer There is no approximation argument (that I know) that allows to deduce the general case from the $L^2$ case. Rather one should go back to the very definition of conditional expectations and note that $A\leqslant B$ almost surely if and only if the events $$C_x=[A>x\geqslant B]$$ have probability zero for every real number $x$, or for sufficiently many values of $x$. Here is a proof along these lines. First recall that $\mathrm E(A\|B)=B$ means that $\mathrm E(Au(B))=\mathrm E(Bu(B))$ for (at least) every bounded measurable function $u$. Similarly, $\mathrm E(B\|A)=A$ means that $\mathrm E(Bv(A))=\mathrm E(Av(A))$ for (at least) every bounded measurable function $v$. In particular, for every real number $x$, $$\mathrm E(A;B\leqslant x)=\mathrm E(B;B\leqslant x),\qquad \mathrm E(B;A>x)=\mathrm E(A;A>x).$$ Next, introduce the events $$D_x=[A> x,B> x],\qquad F_x=[A\leqslant x,B\leqslant x].$$ Then $[B\leqslant x]=C_x\cup F_x$ and $[A>x]=C_x\cup D_x$ and both these unions are disjoint hence $$\mathrm E(A-B;C_x)+\mathrm E(A-B;F_x)=0=\mathrm E(A-B;C_x)+\mathrm E(A-B;D_x).$$ Summing up these two equalities and using the fact that $\mathrm E(A-B;C_x)\geqslant0$ because $A-B>0$ on $C_x$, one gets $$\mathrm E(A-B;D_x\cup F_x)\leqslant0.$$ The hypothesis we started from is symmetric with respect to $(A,B)$ hence the conclusion above holds if one replaces $(A,B)$ by $(B,A)$. Then $A-B$ becomes $B-A$ and $D_x$ and $F_x$ do not change. This proves that $\mathrm E(B-A;D_x\cup F_x)\leqslant0$, hence $\mathrm E(A-B;D_x\cup F_x)=0$, which implies $$\mathrm E(A-B;C_x)=0.$$ But $A>B$ almost surely on $C_x$, hence the random variable $(A-B)\mathbf{1}_{C_x}$ is almost surely nonnegative. This proves that the event $[A>B]\cap C_x=C_x$ has probability zero. Now, the event $[A>B]$ is the countable union of the events $C_x$ on the rational numbers $x$, hence $[A>B]$ has probability zero. By symmetry, the event $[A<B]$ has probability zero as well, hence $A=B$ almost surely. - Thanks a lot for the detailed answer! – john May 1 '11 at 18:50	{"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": 49, "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": 0.9483915567398071, "perplexity_flag": "head"}
http://mathematica.stackexchange.com/questions/tagged/distributions+random	# Tagged Questions 1answer 98 views ### Safe values of $\mu$ and $\sigma$ when randomly sampling from a Log-Normal Distribution? I believe I'm obtaining overflow errors when randomly sampling from a log-normal distribution with the command: RandomVariate[LogNormalDistribution[μ, σ], 1] ... 1answer 143 views ### Calculate variance of random walk? How can I symbolically calculate the variance of the following random walk in Mathematica? Given several discrete random variables such that $p(Z_i=1-2k)=p$, where $k$ is a small real number, and ... 2answers 138 views ### Colors associated to parts of a 3D distribution of points Suppose we have a long list of random points (cartesian coordinates) in 3D space. coords := ...; Until now, I was associating colors in a radial way, from the ... 1answer 166 views ### Radial Random Walk I'm trying to generate a spherical distribution of radial random walk points in 3D space. The following code works, but the random walk lines aren't radial. Why ? Where is my mistake ? ... 3answers 146 views ### MixtureDistribution in Mathematica 7 Is it possible to use a mixture of several distributions in Mathematica 7.0? Consider for example the following code : ... 1answer 243 views ### Distribution of random points in 3D space to simulate an Orion-like Nebula [closed] I'm creating 3D nebulae models for Celestia, a free OpenGL astronomy software. The following question is in part related to another one I already asked on this forum (see this question, for which I ... 1answer 151 views ### How to reject or repulse some points in a 3D cloud of particles? I'm having an issue with a 3D distribution of points. The distribution is really nicely looking in Mathematica, but it is used in an OpenGL astronomy program (Celestia) to generate a nebula, with ... 1answer 288 views ### RandomVariate returns values outside the support of a PDF Let $X$ be a random variable with pdf: dist = ProbabilityDistribution[1/(Abs[x]*Log[Abs[x]]^2), {x, -E^-2, E^-2}] Here are some pseudo-random drawings from it: ... 2answers 167 views ### Generating a range of numbers according to some rules I'm pretty new to Mathematica, and I'm mainly a programmer so I don't have a lot of knowledge about maths. I want to generate a set of UNIQUE incremental numbers (series) according to the following ... 3answers 614 views ### How to generate a RandomVariate of a custom distribution? I'm trying to generate a pseudorandom variate out of a custom distribution. Suppose I want define a custom distribution, and for the sake of simplicity I define a Poisson distribution (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": 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": 0.8745969533920288, "perplexity_flag": "middle"}
http://physics.stackexchange.com/questions/41174/does-every-wavenumber-of-ir-result-in-a-different-kind-of-vibration?answertab=oldest	# Does every wavenumber of IR result in a different kind of vibration? Does every wavenumber of IR result in a different kind of vibration? If that is true, what if a molecule absorb 2 different wavenumbers (which cause different rocking and symmetrical stretching for example ) in the same time? - ## 1 Answer In most cases, one can expect different vibrational modes to have different energies, but that is not always the case. Two or more modes may be degenerate, i.e. have the same energy, although they do not represent the same vibrational motion. This often follows from the symmetry of the molecule and can be deduced from group theory. For example, in ammonia, there are two cases of two-fold degeneracies (see on this page). As to the second part of your question, molecules may indeed absorb radiation that does excite two (or even more) different vibrations at the same time. The spectral signal you get from such an absorption is called a combination band. Similarly, you can also excite the same vibrational mode with two quanta, wich is called an overtone. The energy of the radiation that is absorbed is in both cases roughly equal to the sum of each individual excitation ("roughly", because anharmonicity becomes important here). - How will it vibrate when absorbing 2 radiations exciting 2 vibrational mode? – Abdelrahman Esmat Oct 19 '12 at 5:51 You get a "mixture" of the vibrational motions. A simplified analogy: Imagine a ball that has two springs attached to either side: It will swing from left to right with a certain frequency $\omega_1$. You then attach two weaker springs to the top and bottom; now, the ball can also move up and down with another frequency $\omega_2$. If you "excite" both vibrations at the same time, your motions superpose. For springs of the same strengths, i.e. $\omega_1 = \omega_2$, this motion is elliptic, otherwise it will be more wave-like. (But keep in mind this is only a very simple model! Read on below.) – Antimon Oct 19 '12 at 10:01 The thing is that you don't even have to excite a molecule to get such a "mixing" of vibrational motions. Molecules always vibrate, because quantum-mechanical oscillators have a zero-point energy. So even if your molecule is in it's vibrational ground state (all vibrational quantum numbers are 0), the nuclei are still in motion. Also note: All this refers to the harmonic oscillator assumption, where different vibrations don't influence each other. In reality, when you excite one vibration, it may "give" energy into another one through anharmonic coupling. – Antimon Oct 19 '12 at 10:06	{"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": 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": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9274201393127441, "perplexity_flag": "middle"}
http://math.stackexchange.com/questions/35561/visualizing-operators-on-mathbbcn?answertab=votes	# Visualizing Operators on $\mathbb{C}^n$ I am trying to get some better intuition about operators on complex inner product spaces. When we identify $\mathbb{C}^n$ with $\mathbb{R}^{2n}$, is there a nice geometric interpretation for the resulting operators on $\mathbb{R}^{2n}$? Ideally, this characterization would give a geometric construction for the adjoint generalizing conjugation as reflection in the real line. Also, "seeing" the polar decomposition would be nice. - ## 1 Answer Unitary operators on $\mathbb{C}^n$ are realized as operators which are both orthogonal and symplectic on the correspoding $\mathbb{R}^{2n}$ (with a symplectic structure compatible to the original complex structure) . This is due to the 2-out-of-3 property of the unitary group. Consequently, the polar decomposition expressed on $\mathbb{R}^{2n}$ has the structure: $A = U P$ where $U$ is an orthogonal and symplectic matrix on $\mathbb{R}^{2n}$ and $P$ is a positive $2n \times 2n$ matrix with multiplicity 2 eigenvalues. This can be seen easily through the realization of the polar decomposition through the singular value decomposition as given for example in the following Wikipedia page . -	{"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": 0.9134198427200317, "perplexity_flag": "head"}
http://stats.stackexchange.com/questions/40343/calculating-the-probability-of-the-nth-sample-in-the-birthday-problem-w-o-knowin	# Calculating the probability of the Nth sample in the birthday problem w/o knowing the previous samples If we draw two random variates from a discrete uniform distribution $[1, D]$, the probability that the samples are distinct is $(D-1)/D$. Explanations of the birthday problem state that if we sample a third time, the probability that this third sample is distinct from the first two is $(D-1)(D-2)/D^2$. Doesn't this assume that the first two samples were distinct? If they are not distinct, then the probability that the third sample is distinct is $(D-1)/D$. The solution to the birthday problem requires us to know the probability that the $n^\textrm{th}$ random variate is distinct from the previous samples. How can we calculate the probability that the $n^\textrm{th}$ sample is distinct from the previous samples without knowing the values of these previous samples? - Welcome to the site, @Dan. Is this homework? If it is, you need to add the 'homework' tag. – gung Oct 13 '12 at 21:11 2 @gung This is not homework, it is for my own edification. I saw the problem in David Mackay's textbook on information theory and was not satisfied with his explanation (or other similar explanations on the net). – Dan Oct 13 '12 at 21:13 ## 2 Answers In the usual context of the birthday problem, if the first two values are not distinct the problem is solved and the third and subsequent values are irrelevant. What is confusing here is that you have written Explanations of the birthday problem state that if we sample a third time, the probability that this third sample is distinct from the first two What is really meant here (and is a big part of the unintuitive nature of the birthday problem), is "the probability that this third sample is distinct from the first two, given that they are distinct from each other". - The probability all $3$ are distinct is $\frac{(D-1)(D-2)}{D^2}$. The probability the third is not equal to the first two is $\frac{(D-1)^2}{D^2}$. Imagine choosing $C$ first. Then there are $D-1$ choices for $A$ so that $A\ne C$, and $D-1$ choices for $B$ so that $B\ne C$. If you are given that the first two are distinct, the conditional probability that the third is different from both is $\frac{D-2}{D}$. You are correct that if you are given the first two are equal, the conditional probability that the third is different is $\frac{D-1}{D}$. If $D=10$, these are $72\%, 81\%, 80\%,$ and $90\%,$ respectively. The $81\%$ is naturally a weighted average between $80\%$ and $90\%$, weighted by the probabilities that the first two are distinct or not. The $72\%$ can also be viewed as a weighted average between $81\%$ and $0$, weighted by the conditional probability that the first two are distinct from each other given that they are distinct from the third. - Because this post has been upvoted, SE construes its deletion as removing potentially useful information; I have therefore undeleted it. – whuber♦ Mar 30 at 15:05	{"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": 27, "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": 0.9690632820129395, "perplexity_flag": "head"}
http://stats.stackexchange.com/questions/30171/boosted-ar-for-time-series-forecasting	# Boosted AR for time series forecasting? I have time series data recorded at multiple locations, stored in a matrix $Y$. I have fit a Vector Autoregressive Model to it which forecasts the data pretty well on a test set. However, if I plot the residuals there is definitely some time series structure left; ideally the residuals would look like noise. I have been learning about boosted trees (package 'gbm' in R) that fits a response vector $Y$ to some input features $X$, pulls the residuals $E = Y - f(X)$, and then fits $E$ to the same input features $X$ again (repeat until the process starts to overfit). It's my belief that a tree won't be able to model time series structure well, so I am wondering if the same process can be done using an AR model as the weak learner. After fitting my VAR model, I take the residuals and fit an AR process to it, and forecast it using the 'forecast' package in R. I then take my VAR forecasts and add them to my AR forecasts to get a final prediction. The error at some locations decrease, but for most locations the error increases. Below are the plots of the residuals having only fit the VAR model - there is definitely some structure that can still be explained using some model. Any suggestions? Update: The time series represent energy consumption at various locations. During the night we expect energy consumption to be very low and flat, during the day there may be high volatility so the variance is definitely not constant. My first step was to remove the daily seasonal component using STL (Seasonal Trend Decomposition Using Loess). After that, I fit my VAR model, and subsequently another AR model to the residuals. The time series after removing the Seasonal Component look like this: The error terms from the VAR look like - Can you show us the original time series? It looks like there might be an increase in variability and perhaps some other nonstationary effect. But it could be some short term blip due to an intervention also. It would help to put this into context. What do these series represent? Is there something that you would expect would happen? – Michael Chernick Jun 10 '12 at 12:16 ## 1 Answer You probably need to include some nonlinear regressors into your AR formulation. Thus instead of regressing the vector $x_{t+1}$ against some $x_{t-i}$ ($i=0,1,...$) only (which gives standard AR), regress against $x_{t-i}$ and $x_{t-i}^2$ ($i=0,1,...$) or even $x_{t-i}x_{t-j}$ ($i\le j$). To get the most useful result you probably need to use subset selection to identify the most useful nonlinear terms. (All operations on vectors pointwise, but you may also consider products of different components.) - 3 Without further elaboration, this is better placed as a comment. – cardinal Jun 10 '12 at 16:12 @cardinal: I added details – Arnold Neumaier Jun 11 '12 at 8:39	{"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": 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": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9444652795791626, "perplexity_flag": "middle"}
http://mathhelpforum.com/calculus/197465-dot-vs-cross.html	# Thread: 1. ## dot vs. cross What is the purpose of the dot product as opposed to the purpose of the cross product? 2. ## Re: dot vs. cross Originally Posted by delgeezee What is the purpose of the dot product as opposed to the purpose of the cross product? the result of a dot product of two vectors is a scalar ... work, a scalar quantity of energy, is the dot product of force and displacement vectors $W = \vec{F} \cdot \vec{\Delta x}$ the result of a cross product of two vectors is another vector ... torque, a vector quantity, is the cross product of radius and force vectors $\vec{\tau} = \vec{r} \times \vec{F}$ try a google search ... you'll find many more examples and applications. 3. ## Re: dot vs. cross Originally Posted by delgeezee What is the purpose of the dot product as opposed to the purpose of the cross product? There is a very simple answer and distinction between the two operations. The dot product operation maps pairs of vectors to a scalar. The cross product operation maps pairs of vectors to a vector.	{"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": 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": 0.9345458149909973, "perplexity_flag": "middle"}
http://math.stackexchange.com/questions/142821/matrix-for-rotation-around-a-vector	# Matrix for rotation around a vector I'm trying to figure out the general form for the matrix (let's say in $\mathbb R^3$ for simplicity) of a rotation of $\theta$ around an arbitrary vector $v$ passing through the origin (look towards the origin and rotate counterclockwise). This is inspired by a similar problem which asked me to find the matrix for a rotation of $120^\circ$ around the vector $v=\begin{bmatrix}1&1&1\end{bmatrix}^\top$. However, in this case I was able to cheat a little since the transformation corresponds to a rotation of the vertices. So even though I found a solution, I'm not satisfied with my methodology. Is there a general form for rotation around an arbitrary vector in $\mathbb R^3$? A reference would be perfectly acceptable. Thanks. - 4 Look up the Rodrigues rotation formula. – J. M. May 8 '12 at 21:14 That's exactly what I'm looking for. Thanks. – chris May 8 '12 at 21:35 ## 1 Answer To settle this question: one can use the Rodrigues rotation formula to construct the rotation matrix that rotates by an angle $\varphi$ about the unit vector $\mathbf{\hat u}=\langle u_x,u_y,u_z\rangle$ (and if your vector is not a unit vector, normalization does the trick). Letting $$\mathbf W=\begin{pmatrix}0&-u_z&u_y\\u_z&0&-u_x\\-u_y&u_x&0\end{pmatrix}$$ the Rodrigues rotation matrix is constructed as $$\mathbf I+\left(\sin\,\varphi\right)\mathbf W+\left(2\sin^2\frac{\varphi}{2}\right)\mathbf W^2$$ where $\mathbf I$ is an identity matrix. Conventionally, the scalar multiplying the $\mathbf W^2$ term above is written as $1-\cos\,\varphi$, but this version is more prone to subtractive cancellation when $\varphi$ is near $2k\pi$ ($k$ is an integer), so the expression with the sine is more numerically sound. -	{"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": 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": 0.9413239359855652, "perplexity_flag": "head"}
http://math.stackexchange.com/questions/152462/inverse-of-transformation-matrix/152479	# Inverse of transformation matrix I am preparing for a computer 3D graphics test and have a sample question which I am unable to solve. The question is as follows: For the following 3D transfromation matrix M, find its inverse. Note that M is a composite matrix built from fundamental geometric affine transformations only. Show the initial transformation sequence of M, invert it, and write down the final inverted matrix of M. $M =\begin{pmatrix}0&0&1&5\\0&3&0&3\\-1&0&0&2\\0&0&0&1\end{pmatrix}$ I only know basic linear algebra and I don't think it is the purpose to just invert the matrix but to use the information in the question to solve this. Can anyone help? Thanks - What are the fundamental affine transformations you are referring to? permutation matrices and scaling matrices, maybe? – rschwieb Jun 1 '12 at 13:36 translation-, scaling- and rotation-matrices – Steven Jun 1 '12 at 14:22 – hardmath Jun 1 '12 at 14:45 ## 3 Answers $$\begin{pmatrix}0&0&1&5\\0&3&0&3\\-1&0&0&2\\0&0&0&1\end{pmatrix}$$ $$=\begin{pmatrix}0&0&1&0\\0&1&0&0\\1&0&0&0\\0&0&0&1\end{pmatrix}\begin{pmatrix}-1&0&0&2\\0&3&0&3\\0&0&1&5\\0&0&0&1\end{pmatrix}$$ $$=\begin{pmatrix}0&0&1&0\\0&1&0&0\\1&0&0&0\\0&0&0&1\end{pmatrix}\begin{pmatrix}-1&0&0&0\\0&3&0&0\\0&0&1&0\\0&0&0&1\end{pmatrix}\begin{pmatrix}1&0&0&-2\\0&1&0&1\\0&0&1&5\\0&0&0&1\end{pmatrix}$$ $$=\begin{pmatrix}0&0&1&0\\0&1&0&0\\1&0&0&0\\0&0&0&1\end{pmatrix}\begin{pmatrix}-1&0&0&0\\0&3&0&0\\0&0&1&0\\0&0&0&1\end{pmatrix}\begin{pmatrix}1&0&0&-2\\0&1&0&0\\0&0&1&0\\0&0&0&1\end{pmatrix}\begin{pmatrix}1&0&0&0\\0&1&0&1\\0&0&1&5\\0&0&0&1\end{pmatrix}$$ $$=\begin{pmatrix}0&0&1&0\\0&1&0&0\\1&0&0&0\\0&0&0&1\end{pmatrix}\begin{pmatrix}-1&0&0&0\\0&3&0&0\\0&0&1&0\\0&0&0&1\end{pmatrix}\begin{pmatrix}1&0&0&-2\\0&1&0&0\\0&0&1&0\\0&0&0&1\end{pmatrix}\begin{pmatrix}1&0&0&0\\0&1&0&1\\0&0&1&0\\0&0&0&1\end{pmatrix}\begin{pmatrix}1&0&0&0\\0&1&0&0\\0&0&1&5\\0&0&0&1\end{pmatrix}$$ The inverse of each factor individually is easy, so you can just compute those, then multiply in the reverse order to find the inverse of the matrix. (Remember for matrices $(AB)^{-1}=B^{-1}A^{-1}$, that is what I mean by reverse the factors.) Is this what you call fundamental affine transformations? The first factor is permutes the rows, the second rescales them, then the last three are operations of the same type which produce a linear combination of of two rows. - @Steven Also, be sure to double check my work above. It's hard to get everything right when there is so much TeX, and my comp was displaying it extremely slowly so I might have some corrections to make. – rschwieb Jun 1 '12 at 13:56 Thanks for your super fast answer. Can you explain to me how you you computed the transformation sequence of M? I don't quite get that. And are you sure that there isn't an easier way? Don't get me wrong, I don't want to insult you or anything like that but the task is from a former test and the time one has for this question is 4 minutes. That's why I think there might be a faster way. – Steven Jun 1 '12 at 14:08 @Steven At each step I just mentally factored it by asking myself "what row operation do I need?". With a little practice it goes faster. You would also have to have good practice with inverting those things. There could be a faster way, but I haven't been sitting in your class so I'm not very sure what your teacher has in mind. – rschwieb Jun 1 '12 at 14:27 @Steven No, that's OK I totally understand about wanting a quicker method :) I'll do my best! – rschwieb Jun 1 '12 at 14:36 Note that the 4x4 matrix is said to be " a composite matrix built from fundamental geometric affine transformations". So you need to separate the 3x3 matrix multiplication from the affine translation part. – hardmath Jun 1 '12 at 14:48 show 3 more comments Here $4\times4$ matrix $M$ represents an affine transformation in 3D. It does so by conveniently combining a $3\times3$ matrix $P$ and a translation $v$ in a way that allows the affine transformation $Pu + v$ to be computed by a single matrix multiplication: $$M \begin{pmatrix} u \\ 1 \end{pmatrix} = \begin{pmatrix} Pu + v \\ 1 \end{pmatrix}$$ where $M = \begin{pmatrix} P & v \\ 0 & 1 \end{pmatrix}$. It follows that "undoing" the affine transformation can be accomplished by multiplying by $M^{-1}$: $$M^{-1} = \begin{pmatrix} P^{-1} & -P^{-1}v \\ 0 & 1 \end{pmatrix}$$ Given that $M = \begin{pmatrix} 0 & 0 & 1 & 5 \\ 0 & 3 & 0 & 3 \\ -1 & 0 & 0 & 2 \\ 0 & 0 & 0 & 1 \end{pmatrix}$, one computes by any of a variety of ways: $$M^{-1} = \begin{pmatrix} 0 & 0 & -1 & 2 \\ 0 & ^1/_3 & 0 & -1 \\ 1 & 0 & 0 & -5 \\ 0 & 0 & 0 & 1 \end{pmatrix}$$ - I'll take a stab at a second method too, to see if that's what your teacher had in mind. The idea is to use this post and/or this post, because: $$M=\begin{pmatrix}0&0&1&0\\0&1&0&0\\1&0&0&0\\0&0&0&1\end{pmatrix}\begin{pmatrix}-1&0&0&2\\0&3&0&3\\0&0&1&5\\0&0&0&1\end{pmatrix}$$ $$=\begin{pmatrix}0&0&1&0\\0&1&0&0\\1&0&0&0\\0&0&0&1\end{pmatrix}\begin{pmatrix}-1&0&0&0\\0&3&0&0\\0&0&1&0\\0&0&0&1\end{pmatrix}\begin{pmatrix}1&0&0&-2\\0&1&0&1\\0&0&1&5\\0&0&0&1\end{pmatrix}$$ $$=\begin{pmatrix}0&0&1&0\\0&1&0&0\\1&0&0&0\\0&0&0&1\end{pmatrix}\begin{pmatrix}-1&0&0&0\\0&3&0&0\\0&0&1&0\\0&0&0&1\end{pmatrix}\left[\begin{pmatrix}1&0&0&0\\0&1&0&0\\0&0&1&0\\0&0&0&1\end{pmatrix}+\begin{pmatrix}0&0&0&-2\\0&0&0&1\\0&0&0&5\\0&0&0&0\end{pmatrix}\right]$$ You can see that the brackets have a unit plus a nilpotent $B$ ($B^2=0$). I think the link above might be able to help you invert that right half quickly. Yes, of course here $(I+B)^{-1}=I-B$. So we know: $$M^{-1}=\begin{pmatrix}1&0&0&2\\0&1&0&-1\\0&0&1&-5\\0&0&0&1\end{pmatrix}\begin{pmatrix}-1&0&0&0\\0&1/3&0&0\\0&0&1&0\\0&0&0&1\end{pmatrix}\begin{pmatrix}0&0&1&0\\0&1&0&0\\1&0&0&0\\0&0&0&1\end{pmatrix}$$ which isn't so hard to compute. Multiply the left pair first, then with the final matrix on the right to get this in only a minute: \begin{pmatrix}0&0&-1&2\\0&1/3&0&-1\\1&0&0&-5\\0&0&0&1\end{pmatrix} - The $\frac{-1}{3}$ is a sign error, as you will see by checking the product with $M$. Otherwise my answer agrees with yours. – hardmath Jun 3 '12 at 14:38 @hardmath Yup, looks like that crept in... will correct it now. That day the TeX was rendering realllly slowly and every time I changed something it took 15-20 seconds to come up. So I lost patience... – rschwieb Jun 3 '12 at 14:44	{"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": 16, "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": 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": 0.9407587647438049, "perplexity_flag": "middle"}
http://mathhelpforum.com/statistics/191052-finding-probability-function-r-remainder-when-divided-distribution.html	# Thread: 1. ## Finding a probability function R (the remainder when divided by a distribution) Hi everyone, I am really unsure of how to do this question. Let X have a geometric distribution with f(x) = p(1-p)^x for x = 0,1,2,.... Find the probability function of R, the remainder when X is divided by 4. I don't know where to start, but I'm suspecting that you would construct a geometric series with r, and find an expression for f(r)... Not really sure. 2. ## Re: Finding a probability function R (the remainder when divided by a distribution) If $0\le r < 4$, then $P(R = r) = P(X = r) + P(X = 4 + r) + P(X = 8 + r) + \dots =$ $p(1 - p)^r + p(1 - p)^{4+r} + p(1 - p)^{8+r} + \dots$, which is a geometric series. 3. ## Re: Finding a probability function R (the remainder when divided by a distribution) So taking that geometric series, I can find the sum of the geometric series,which will express P (R = r)? That is, we have P (R = r) = p(1-p)^r/ 1-((1-p)^4+r)) ? THanks! 4. ## Re: Finding a probability function R (the remainder when divided by a distribution) Originally Posted by KelvinScale That is, we have P (R = r) = p(1-p)^r/ 1-((1-p)^4+r)) ? There is no +r in the exponent: $P(R = r) =\frac{p(1-p)^r}{1-(1-p)^4}$.	{"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": 0, "codecogs_latex": 4, "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": 0.9219643473625183, "perplexity_flag": "middle"}
http://math.stackexchange.com/questions/40930/what-is-the-name-for-the-archetypical-example-of-a-test-function-varphix-e	# What is the name for the archetypical example of a test function, $\varphi(x)=e^{1/(x^2-1)}$? $$\varphi(x)=e^{1/(x^2-1)}$$ This function (on the interval $\quad]\!-1,1[ \,\,\,$, outside of it simply $\equiv0$) is used as the typical example of a test function / bump function, I have so far seen it it every book that covers $\mathcal{C}_0^\infty$ functions. But it's usually not called any specific name, though it does seem to have one, at least I heard it being called by some name recently, but forgot it. I'd greatly like to know a name for this function, both for my computer functions library and for ease when writing proofs where a test function is required, and you can quickly reassure its existence with a simple "like the ...-function". Friedrichs'sche Glättungsfunktion is in fact the name I was looking for! - 6 In German it is sometimes called Friedrichs'sche Glättungsfunktion (roughly: Friedrichs's mollifying function) to honour its use in Friedrichs's work on differential equations. You can find a discussion and references on the Wikipedia-page on mollifiers. I don't know how "official" that name is, however. – t.b. May 23 '11 at 21:34 Usually you don't need an explicit formula for your bump function right? You just need to know that it exists. This functions allows you to prove that such $C^\infty$ bump functions exist. – Jonas Teuwen May 23 '11 at 23:41 ## 1 Answer - Ahm... it would be a bit unfair to accept this answer now, considering Theo already gave the desired answer in the comments and bumb function refers to general functions of this type, not specifically $e^{1/(x^2-1)}$. – leftaroundabout Sep 27 '11 at 21:10 Your problem, not mine. In fact, I could not care less... – Did Sep 27 '11 at 21:14 This is interesting in the sense that I hadn't considered that html links would contribute to message length. Hmm. – mixedmath♦ Sep 29 '11 at 22:00 @DidierPiau : Yes it is his problem, It has happened to me too! To choose between 2 correct answers is never fair. Why multiple ansers are not allowed? after all both answers are correct. – Arjang Nov 26 '11 at 13:44 2 The other is not an answer. – timur Dec 3 '11 at 19:47 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": 6, "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": 0.9444892406463623, "perplexity_flag": "middle"}
http://en.wikipedia.org/wiki/Gas_constant	# Gas constant Values of R [1] Units (V P T −1 n−1) 8.3144621(75)[2] J K−1 mol−1 5.189×1019 eV K−1 mol−1 0.08205746(14) L atm K−1 mol−1 1.9858775(34) cal K−1 mol−1 1.9858775(34)×10−3 kcal K−1 mol−1 8.3144621(75)×107 erg K−1 mol−1 8.3144621(75) L kPa K−1 mol−1 8.3144621(75)×103 cm3 kPa K−1 mol−1 8.3144621(75) m3 Pa K−1 mol−1 8.3144621(75) cm3 MPa K−1 mol−1 8.3144621(75)×10−5 m3 bar K−1 mol−1 8.205746×10−5 m3 atm K−1 mol−1 8.205746×10−2 m3 atm K−1 kg-mol−1 82.05746 cm3 atm K−1 mol−1 84.78402×10−6 m3 kgf/cm2 K−1 mol−1 8.3144621(75)×10−2 L bar K−1 mol−1 62.36367(11)×10−3 m3 mmHg K−1 mol−1 62.36367(11) L mmHg K−1 mol−1 62.36367(11) L Torr K−1 mol−1 6.132440(10) ft lbf K−1 g-mol−1 1,545.34896(3) ft lbf R−1 lb-mol−1 10.73159(2) ft3 psi R−1 lb-mol−1 0.7302413(12) ft3 atm R−1 lb-mol−1 1.31443 ft3 atm K−1 lb-mol−1 998.9701(17) ft3 mmHg K−1 lb-mol−1 1.986 Btu lb-mol−1 R−1 The gas constant (also known as the molar, universal, Mendeleev's constant[3][4] or ideal gas constant, denoted by the symbol R or R) is a physical constant which is featured in many fundamental equations in the physical sciences, such as the ideal gas law and the Nernst equation. It is equivalent to the Boltzmann constant, but expressed in units of energy (i.e. the pressure-volume product) per temperature increment per mole (rather than energy per temperature increment per particle). The constant is also a combination of the constants from Boyle's law, Charles's law, Avogadro's law, and Gay-Lussac's law. Physically, the gas constant is the constant of proportionality that happens to relate the energy scale in physics to the temperature scale, when a mole of particles at the stated temperature is being considered. Thus, the value of the gas constant ultimately derives from historical decisions and accidents in the setting of the energy and temperature scales, plus similar historical setting of the value of the molar scale used for the counting of particles. The last factor is not a consideration in the value of the Boltzmann constant, which does a similar job of equating linear energy and temperature scales. The gas constant value is $R=8.314\,4621(75)~\frac{\mathrm{J}}{\mathrm{mol~K}}$[2] The two digits in parentheses are the uncertainty (standard deviation) in the last two digits of the value. The relative uncertainty is 9.1×10−7. Some have suggested that it might be appropriate to name the symbol R the Regnault constant in honor of the French chemist Henri Victor Regnault, whose accurate experimental data was used to calculate the early value of the constant; however, the exact reason for the original representation of the constant by the letter R is elusive.[5] [6] The gas constant occurs in the ideal gas law, as follows: $PV = nRT = m R_{\rm specific} T \,\!$ where P is the absolute pressure (SI unit pascals), V is the volume of gas (SI unit cubic metres), n is the chemical amount of gas (SI unit moles), m is the mass (SI unit kilograms) contained in V, and T is the thermodynamic temperature (SI unit kelvins). The gas constant is expressed in the same physical units as molar entropy and molar heat capacity. ## Dimensions of R From the general equation PV = nRT we get R = PV/nT or (pressure × volume) / (amount × temperature). As pressure is defined as force per unit area, we can also write the gas equation as R = [(force/area) × volume] / (amount × temperature). Again, area is nothing but (length)2 and volume is equal to (length)3. Therefore, R = [force / (length)2] (length)3 / (amount × temperature). Since force × length = work, R = (work) / (amount × temperature). The physical significance of R is work per degree per mole. It may be expressed in any set of units representing work or energy (such as joules), other units representing degrees of temperature (such as degrees celsius or Fahrenheit), and any system of units designating a mole or a similar pure number that allows an equation of macroscopic mass and fundamental particle numbers in a system, such as an ideal gas (see Avogadro's number). ## Relationship with the Boltzmann constant The Boltzmann constant kB (often abbreviated k) may be used in place of the gas constant by working in pure particle count, N, rather than amount of substance, n, since $\qquad R = N_{\rm A} k_{\rm B},\,$ where NA is the Avogadro constant. For example, the ideal gas law in terms of Boltzmann's constant is $PV = N k_{\rm B} T.\,\!$ where N is the number of particles (molecules in this case). ## Measurement As of 2006, the most precise measurement of R is obtained by measuring the speed of sound ca(p, T) in argon at the temperature T of the triple point of water (used to define the kelvin) at different pressures p, and extrapolating to the zero-pressure limit ca(0, T). The value of R is then obtained from the relation $c_\mathrm{a}^2(0, T) = \frac{\gamma_0 R T}{A_\mathrm{r}(\mathrm{Ar}) M_\mathrm{u}},$ where: • γ0 is the heat capacity ratio (5/3 for monatomic gases such as argon); • T is the temperature, TTPW = 273.16 K by definition of the kelvin; • Ar(Ar) is the relative atomic mass of argon and Mu = 10−3 kg mol−1.[1] ## Specific gas constant Rspecific for dry air Units 287.058 J kg−1 K−1 53.3533 ft lbf lb−1 °R−1 1716.49 ft lbf slug−1 °R−1 Based on a mean molar mass for dry air of 28.9645 g/mol. The specific gas constant of a gas or a mixture of gases (Rspecific) is given by the molar gas constant, divided by the molar mass (M) of the gas/mixture. $R_{\rm specific} = \frac{R}{M}$ Just as the ideal gas constant can be related to the Boltzmann constant, so can the specific gas constant by dividing the Boltzmann constant by the molecular mass of the gas. $R_{\rm specific} = \frac{k_{\rm B}}{m}$ Another important relationship comes from thermodynamics. This relates the specific gas constant to the specific heats for a calorically perfect gas and a thermally perfect gas. $R_{\rm specific} = c_{\rm p} - c_{\rm v}\$ where cp is the specific heat for a constant pressure and cv is the specific heat for a constant volume.[7] It is common, especially in engineering applications, to represent the specific gas constant by the symbol R. In such cases, the universal gas constant is usually given a different symbol such as R to distinguish it. In any case, the context and/or units of the gas constant should make it clear as to whether the universal or specific gas constant is being referred to.[8] ## U.S. Standard Atmosphere The U.S. Standard Atmosphere, 1976 (USSA1976) defines the gas constant R* as:[9][10] $R^* = 8.314\,32\times 10^3 \frac{\mathrm{N\,m}}{\mathrm{kmol\,K}}.$ The USSA1976 does recognize, however, that this value is not consistent with the cited values for the Avogadro constant and the Boltzmann constant.[10] This disparity is not a significant departure from accuracy, and USSA1976 uses this value of R* for all the calculations of the standard atmosphere. When using the ISO value of R, the calculated pressure increases by only 0.62 pascal at 11 kilometers (the equivalent of a difference of only 0.174 meter or 6.8 inches) and an increase of 0.292 Pa at 20 km (the equivalent of a difference of only 0.338 m or 13.2 in).[citation needed] ## Individual gas constants Individual gas constants in base SI units of J/kg.K can also be derived for any gas species, by making use of their molar mass.[11] An average value could also be derived for gas mixtures. Use of individual gas constants may make it more difficult to follow the workings of a calculation, as the relevant values tend to be less well known, and less intuitive, than the fixed value of the universal gas constant, and the well-known values of gas molecular masses. ## References 1. ^ a b Mohr, Peter J.; Taylor, Barry N.; Newell, David B. (2008). "CODATA Recommended Values of the Fundamental Physical Constants: 2006". 80 (2): 633–730. Bibcode:2008RvMP...80..633M. doi:10.1103/RevModPhys.80.633. Direct link to value. 2. ^ a b 3. Д. И. Менделеев. О сжимаемости газов (Из лаборатории С.-Петербургского Университета). стр. 309-352. Журнал русского химического общества и физического общества. Том 6. 1874 г. [1] 4. Jensen, William B. (July 2003). "The Universal Gas Constant R". J. Chem. Educ. 80 (7): 731. Bibcode:2003JChEd..80..731J. doi:10.1021/ed080p731. 5. 6. Anderson, Hypersonic and High-Temperature Gas Dynamics, AIAA Education Series, 2nd Ed, 2006 7. Moran and Shapiro, Fundamentals of Engineering Thermodynamics, Wiley, 4th Ed, 2000 8. "Standard Atmospheres". Retrieved 2007-01-07. 9. ^ a 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": 9, "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": 0.7909436225891113, "perplexity_flag": "middle"}
http://mathoverflow.net/revisions/19470/list	Return to Answer 2 added 1 characters in body I feel it's more informative to have the thought process that leads to the example rather than just be told some magic formula that works, however simple that formula might be. Here is a way of explaining it. We'll decide that our function is going to be discontinuous at (0,0). In order to ensure that the partial derivatives there are defined we'll try the simplest thing and make the function zero on the two axes. How can we make sure that it is discontinuous at (0,0)? Well, a simple way might be to make the function equal to 1 on the line x=y. (If you object that that step was unmotivated, then read on -- it will become clear that I could have chosen pretty well any function and the argument still works.) How are we going to make sure that the function has partial derivatives everywhere? We could do it by trying to give it a nice formula everywhere. We care particularly about how the function behaves when you keep x or y constant, so let's see what happens if we try to choose the nicest possible dependence on y for each fixed x. If we do that then we'll be tempted to make the function linear. Since f(x,0)=0 and f(x,x)=1, this would tell us to choose f(x,y)=y/x. Unfortunately, that doesn't work: we also need the function to tend to zero for fixed y as x tends to zero. But at least this has given us the idea that we would like to write f as a quotient. What properties would we need of g and h if we tried f(x,y)=g(x,y)/h(x,y)? We would want g(x,0)=g(0,y)=0, g(x,x)=h(x,x), and h(x,y) is never 0 (except that we don't mind what happens at (0,0). We also want g and h to be nice so that the partial derivatives will obviously exist. The simplest function that vanishes only if (x,y)=(0,0) is $x^2+y^2$. The simplest function that vanishes when x=0 or y=0 but not when $x=y\ne 0$ is $xy$. Multiply that by 2 to get 1 down the line x=y and there we are. Now suppose we had wanted the value to be, say, $e^{1/x}$ at (x,x), for so that the function is wildly unbounded near (0,0). Then we could just multiply the previous function by $\exp((2/(x^2+y^2))^{1/2})$. The main point I want to make is that we could just as easily have chosen many other functions. For example, $\sin(x)\sin(y)/(x^4+y^4)$ vanishes on the axes and clearly does not tend to zero down the line x=y (in fact it tends to infinity). 1 I feel it's more informative to have the thought process that leads to the example rather than just be told some magic formula that works, however simple that formula might be. Here is a way of explaining it. We'll decide that our function is going to be discontinuous at (0,0). In order to ensure that the partial derivatives there are defined we'll try the simplest thing and make the function zero on the two axes. How can we make sure that it is discontinuous at (0,0)? Well, a simple way might be to make the function equal to 1 on the line x=y. (If you object that that step was unmotivated, then read on -- it will become clear that I could have chosen pretty well any function and the argument still works.) How are we going to make sure that the function has partial derivatives everywhere? We could do it by trying to give it a nice formula everywhere. We care particularly about how the function behaves when you keep x or y constant, so let's see what happens if we try to choose the nicest possible dependence on y for each fixed x. If we do that then we'll be tempted to make the function linear. Since f(x,0)=0 and f(x,x)=1, this would tell us to choose f(x,y)=y/x. Unfortunately, that doesn't work: we also need the function to tend to zero for fixed y as x tends to zero. But at least this has given us the idea that we would like to write f as a quotient. What properties would we need of g and h if we tried f(x,y)=g(x,y)/h(x,y)? We would want g(x,0)=g(0,y)=0, g(x,x)=h(x,x), and h(x,y) is never 0 (except that we don't mind what happens at (0,0). We also want g and h to be nice so that the partial derivatives will obviously exist. The simplest function that vanishes only if (x,y)=(0,0) is $x^2+y^2$. The simplest function that vanishes when x=0 or y=0 but not when $x=y\ne 0$ is $xy$. Multiply that by 2 to get 1 down the line x=y and there we are. Now suppose we had wanted the value to be, say, $e^{1/x}$ at (x,x), for so the function is wildly unbounded near (0,0). Then we could just multiply the previous function by $\exp((2/(x^2+y^2))^{1/2})$. The main point I want to make is that we could just as easily have chosen many other functions. For example, $\sin(x)\sin(y)/(x^4+y^4)$ vanishes on the axes and clearly does not tend to zero down the line x=y (in fact it tends to infinity).	{"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": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "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": 0.9648241400718689, "perplexity_flag": "head"}
http://mathhelpforum.com/calculus/2382-derivative-invert-tan.html	# Thread: 1. ## Derivative of invert tan? I read from the rules that the derivative of (tan^-1)x is (1/x^2+1), but I cannot figure out why and I can't find any explantations on the web. Anyone have any ideas? Thank you. 2. Originally Posted by tttcomrader I read from the rules that the derivative of (tan^-1)x is (1/x^2+1), but I cannot figure out why and I can't find any explantations on the web. Anyone have any ideas? Thank you. Correction, you cannot find an explanation on the web EXCEPT for this site. Anyway, since it is inverse we have, $-1<x<1$ $\tan(\arctan x)=x$ Now, take the derivative of both sides, (use chain rule), $\frac{d\arctan x}{dx}\cdot \sec^2(\arctan x)=1$. Thus, $\frac{d\arctan x}{dx}=\cos^2{\arctan x}$ But, $\cos^2(\arctan x)=\frac{1}{1+x^2}$ Thus, $\frac{d\arctan x}{dx}=\frac{1}{1+x^2}$ ----------------------- Note, I assumed that the derivative of the arctangent exists. Which is true, because if a bijective function is differenciable then so is its inverse. But I did not bother with the proof here. It happens to be true. Q.E.D. 3. One thing. The $x \in (-1, 1)$ assumption is unnecessary. 4. Originally Posted by albi One thing. The $x \in (-1, 1)$ assumption is unnecessary. Of course it is, the function, $f(x)=x$ is not differenciable at $x=\pm 1$ On the interval $x\in[-1,1]$ because that is the interval for which the arc-tangent was defined. And as you know endpoints are NEVER differenciable because by definition of derivative it means that the quotient limit exists. But the problem is that this has only one sided differenciability which implies the function IS NOT differenciable. 5. Originally Posted by ThePerfectHacker Of course it is, the function, $f(x)=x$ is not differenciable at $x=\pm 1$ On the interval $x\in[-1,1]$ because that is the interval for which the arc-tangent was defined. And as you know endpoints are NEVER differenciable because by definition of derivative it means that the quotient limit exists. But the problem is that this has only one sided differenciability which implies the function IS NOT differenciable. No. You have missed my point. The arc-tangent is the inverse of tangent restricted to the [-pi/2; pi/2] interval. It is: $<br /> f: \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \rightarrow (-\infty, +\infty)<br />$ $<br /> x \mapsto f(x) = \tan x<br />$ and arc-tan is defined: $<br /> \arctan x = f^{-1}(x)<br />$ Cause that during the inverse the domain and counterdomain are swapped, the arc-tangent domain is (-infinity, +infinity) 6. Originally Posted by albi No. You have missed my point. The arc-tangent is the inverse of tangent restricted to the [-pi/2; pi/2] interval. It is: $<br /> f: \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \rightarrow (-\infty, +\infty)<br />$ $<br /> x \mapsto f(x) = \tan x<br />$ and arc-tan is defined: $<br /> \arctan x = f^{-1}(x)<br />$ Cause that during the inverse the domain and counterdomain are swapped, the arc-tangent domain is (-infinity, +infinity) Your right I do not know what I was thinking. But at least see how I made that mistake, I was thinking about inverse sine or cosine. 7. Thanks for the solution, now I understand how the derivative of arctanx equals to cos^2(arctanx). But would you please explain why cost^2(arctan(x)) equals to 1/(x^2+1)? Thank you, sir. $<br /> \cos^2 x + \sin^2 x = 1<br />$ Let $x \neq \frac{\pi}{2} + k \pi, \quad k \in \mathbb{Z}$ Dividing the equation by $\cos^2 x$ we get: $<br /> 1 + \tan^2 x = \frac{1}{\cos^2 x}<br />$ Thus: $<br /> \cos^2 x = \frac{1}{1 + \tan^2 x} <br />$ Finally: $<br /> \cos^2 (\arctan x) = \frac{1}{1 + \tan^2 (\arctan x)} = \frac{1}{1 + x^2}<br />$	{"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": 26, "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": 0.9469389915466309, "perplexity_flag": "middle"}
http://mathoverflow.net/questions/22583/why-torsion-is-important-in-cohomology/22640	## Why torsion is important in (co)homology ? ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) I've once been told that "torsion in homology and cohomology is regarded by topologists as a very deep and important phenomenon". I presume an analogous statement could be said in the context of algebraic geometry. In this community wiki question I would like to gather examples, in geometrical fields such as algebraic topology and algebraic geometry, of phenomena that manifest themselves by the presence of torsion in (co)homology groups and whose trace is consistently lost if we simply disregard the torsion part of those groups. As guidelines for the answers: Which kind of information is lost disregarding torsion in (co)homology? (provide examples) What does the torsion part of (co)homology tell us about the geometric object involved? (provide examples) Here "(co)homology" should be understood in any relevant sense, from singular cohomology of cw complexes to étale cohomology of algebraic varieties and so on and so forth. It may well be true that the algebro geometric examples have nothing to do, conceptually, with the topological ones: I'm not interested in a unifying pattern per se, but if such a unifying pattern does appear in some answers, well, it's just good. - 7 I generally regard torsion in (co)homology as a sign that one should be computing K-theory instead, which has less of it. – Allen Knutson Apr 26 2010 at 14:22 1 I agree in principle but this is not always true, the real projective spaces form one example. – Torsten Ekedahl Apr 26 2010 at 14:48 I would say that forgetting all about torsion really is tensoring with rationals, and then any generalized (co)homology theory comes from the usual one with Q coefficients by shifts and sums/products. Stable homotopy tensored with Q is usual homology with rational coefficients, for instance. – BS Jun 4 2010 at 10:05 3 Sometimes singularities of varieties can only be measured if we allow for torsion. For example, the complex locus of y^2= xz is rationally smooth at the origin, but local homology with integer coefficients reveals nonzero homology where there wouldn't be if it was smooth at the origin. I'd post this as an answer but I really don't know what I'm talking about- we just had this as an exercise in my algebraic topology class last week. – Dylan Wilson Nov 11 2010 at 23:24 ## 11 Answers In their paper "Some Elementary Examples of Unirational Varieties Which are Not Rational", Artin and Mumford show that the torsion in $H^3(V, Z)$ of a non singular projective 3-fold $V$ is a birational invariant. This is great because it gives a cohomological obstruction to rationality (there is no torsion in the cohomology of projective space). They they are able to show that certain conic bundles over rational surfaces are not rational by exhibiting such torsion (their conic bundles are unirational, hence the title). The paper is very nice. - ### 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 first place that one sees that torsion is deep is in the homotopy groups of spheres, which, mod torsion, are described completely by a theorem of Serre. However the torsion part of the homotopy groups of spheres is very complicated. If we work rationally, that is, if we forget about torsion, then lots of cohomology theories tend to be the same. (There's a general theorem of this sort, but I've forgotten the precise statement.) For example, singular cohomology and K-theory are isomorphic, rationally, via Chern character. - 5 I have to disagree with you about that one...first place that I saw torsion (and also, note, the question is about (co)homology more than homotopy, I think), was in the cohomology of the real projective plane, and I recall using it to give a proof of nonorientability...though admittedly, I haven't thought about $\mathbb{RP}^n$ in awhile. – Charles Siegel Apr 26 2010 at 15:23 2 Ok, sure. But that example not really very "deep", is it? (Also, I know that homotopy is closely related to (co)homology... but I will wait for others more expert than I to elaborate on the connection.) – Kevin Lin Apr 27 2010 at 10:03 See for example en.wikipedia.org/wiki/… and en.wikipedia.org/wiki/… – Kevin Lin Apr 28 2010 at 9:35 [[ Sorry I missed that the question was also concerned with the question in an algebraic topological context. This answer is only concerned with algebraic geometry.]] I think the first question is much easier to answer. mdeland has given the Artin-Mumford non-rationality example as one answer. Another is the Atiyah-Hirzebruch example of an even-degree torsion class of a smooth projective variety which is not algebraic, showing that an integral version of the Hodge conjecture is false. This gives examples (and there are others) where torsion can be used to show something about an algebraic variety which one couldn't show without (actually I would say that it is more a question of integral versus rational cohomology even without torsion one can exploit that certain cohomology classes are not divisible by some particular integer). I would say that gives an answer to the first question. The second is of a very different nature. In algebraic topology torsion (and more general integral cohomology again versus rational cohomology) are enormously important for understanding the homotopy type of a space. Take as an example the spheres. Rationally their homotopy theory is trivial but integrally you have highly non-trivial homotopy groups (this non-triviality does not reflect itself in the cohomology of the spheres but is closely related to spaces derived from the spheres, the pieces of the Postnikov tower). Of course algebraic varieties (over $\mathbb C$, but that is not essential) give homotopy types too but it not always clear what the homotopy type of an algebraic variety tells you about the algebro-geometric structure of the variety (unless you somehow incorporate algebraic topology under algebraic geometry...). There are some examples though: The torsion in the second cohomology group comes directly from the fundamental group and in particular give you abelian étale covers of the variety. The torsion in the third cohomology group tells you about the Brauer group of the variety and in particular corresponds (for some definition of "corresponds") to projective fibrations over the variety. The correspondence is quite indirect however. I would for instance love to know the least relative dimension of a projective fibration over an Enriques surface which realises the element of order $2$ in the third cohomology group or even better a geometric construction of any such fibration. In higher cohomological degrees the situation is even worse (unless one chooses the above incorporation option, higher algebraic stacks could be said to do that). - I just wanted to add two more examples about torsion in cohomology groups of low degree that came into my mind reading the above (great) answers: • Any torsion element in $H^2(M, \mathbb{Z})$ for a space $M$ can be realized as the first Chern class of a complex flat line bundle. • Similar to this, you may know that elements in $H^3(M, \mathbb{Z})$ correspond (up to some equivalence) to twists in twisted K-theory. Now, if that class is torsion, you get a very nice description of twisted K-theory via modules over bundle gerbes. Or, if you don't like twisted K-theory, the torsion elements in $H^3(M,\mathbb{Z})$ correspond to (stable equivalence classes) of those bundle gerbes, which allow a (finite dimensional) bundle gerbe module. - Following up on Charles' comment to Kevin's answer, torsion can be helpful in determining whether or not a manifold is orientable: `$H_{n-1} (M; Z)$` is torsion-free when M is orientable and has torsion subgroup Z/2 when M is non-orientable. For surfaces, this means orientability can be detected from H_1, which is quite nice. On the other hand, you don't really need to pay attention to torsion to see the difference between orientability and non-orientability. A closed (connected) n-manifold M is orientable iff `$H_n (M; Z) = Z$`, and non-orientable iff `$H_n (M; Z) = 0$`. The same statements hold with integral coefficients replaced by real coeffients. This is all in Hatcher's section on Poincare Duality. - An orientable closed 3-manifold $M$ with $rank(H_1(M,\mathbb{Z}/p\mathbb{Z}))\geq 3$ has infinite fundamental group, by a result of Shalen & Wagreich (one may also deduce this now from the Geometrization theorem, but their theorem gives more information, such as the $p$-completion of $\pi_1(M)$ is infinite). Of course, if $b_1(M)=0$, then this is undetected by rational cohomology. - There are some very important 'torsion motivic' statement: the calculation of Suslin's homology, Milnor and Bloch-Kato conjecture (proved by Voevodsky). Also, the proof of the latter statements uses algebraic cobordism and motivic cohomology operations, which do not work integrally. Also, I believe that Steenrod's operations should be important for algebraic topology, but I do not know anything about that. - 4 Indeed, they are terribly important. For example, most of the techniques computing (2-primary, stable) homotopy groups of spheres use Steenrod operations. – Lennart Meier Apr 27 2010 at 8:15 Inspired by Ulrich Pennig's answer, I'll mention that Chern-Weil theory tells us that the Chern classes of a flat bundle over a manifold are always trivial in rational cohomology. But quite often they are non-trivial in integral cohomology, and hence provide a method of distinguishing between flat bundles. For instance, over a non-orientable surface, there are precisely two isomorphism types of flat vector bundles in each dimension (one being the trivial bundle), distinguished by their first Chern class in `$H^2 (S; \mathbb{Z}) = \mathbb{Z}/2$`. - You mean line bundles? – Qfwfq May 5 2010 at 9:23 1 These statements are true for bundles of any dimension. The fact about bundles over surfaces can be seen in several ways; here's one: Note that the classifying map `$S\to BU(n)$` can be assumed to land in the 2-skeleton of `$BU(n)$`, which is just `$S^2$`, regardless of n (I'm thinking of the standard CW structure on the Grassmannian). So the classifying map really lands in `$CP^\infty$`, and hence bundles over S all have the form `$L\oplus \epsilon^k$`, where L is a line and `$\epsilon^k$ is trivial. – Dan Ramras May 5 2010 at 16:33 The fact that both the trivial and non-trivial bundle over S admit a flat connection takes more work. Melissa Liu and Nan-Kuo Ho have some papers about this. – Dan Ramras May 5 2010 at 16:34 Thanks, Dan . – Qfwfq May 12 2010 at 17:43 The Hantsche obstruction to embedding a 3-manifold $M$ in a homology 4-sphere is a $\mathbb Q/\mathbb Z$-valued bilinear form on the torsion subgroup of $H_1(M;\mathbb Z)$. If you were to use (co)homology with rational coefficients this would be invisible to you. If you're less fussy about using the integers in your discussion of torsion, the Alexander polynomial is a torsion invariant of the homology of a covering space of knots and links. This time the ring is the ring of single-variable Laurent polynomials with integer coefficients. - Integer pontrjagin classes are diffeomorphism invariant, while rational pontrjagin classes are homeomorphism invariant, due to Novikov. Also there are examples where two smooth manifolds are homeomorphic but with different integer pontrjagin classes. And of course the cohomology of the manifolds need to have some torsion in order to make this work. See for example, Matthias kreck and Wolfgang Lueck's book, The novikov conjecture - geometry and algebra, pp 29- 31 - Suppose $G$ is a split semisimple $\mathbf{Q}$-group and $\Gamma \subset G(\mathbf{Q})$ is a lattice. Conjectures due to Ash and his collaborators (elaborating on earlier work of Serre) predict a fairly precise correspondence between continuous representations $\rho: \mathrm{Gal}(\overline{\mathbf{Q}}/\mathbf{Q})\to\widehat{G}(\overline{\mathbf{F}}_p)$ and "Hecke eigenclasses" in $H^{\ast}(\Gamma,\overline{\mathbf{F}}_p)$. See for example this paper where the conjecture is elaborated very precisely for $\mathrm{GL}_n/\mathbf{Q}$, and this paper for a more general prediction. The really remarkable thing here is that for may groups $G$ - say, if $G(\mathbf{R})$ does not admit discrete series - there should be a serious paucity of non-torsion characteristic zero homology, and the classes predicted by Galois representations will often not be the mod-$p$ reduction of some characteristic zero class! So these genuine torsion classes should be tied rather intimately to Galois representations - that seems pretty remarkable to me! -	{"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": 27, "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": 0.9370181560516357, "perplexity_flag": "head"}
http://math.stackexchange.com/questions/184285/i2-does-not-retract-into-comb-space	$I^2$ does not retract into comb space In Hatcher's book on Algebraic Topology in Chapter 0 (where the Homotopy Extension Property is discussed) the claim is being made that there is no retraction of $I^2$ into $I \times \{0\} \cup A \times I$ where $A = \{0, 1, \frac 12 , \frac 13, \ldots \}$. However I couldn't come up with a proof. One can define a retraction if there is only a finite number of "teeth in the comb". It's not clear to me why continuity fails (presumably) at 0 if one tries to extend this retraction to the infinite case. I hope someone can help out. - 2 Answers If $f$ is a retraction, then it must map $(0,1)$ to $(0,1)$. Since it is continuous, there must a ball $A$ centered on $(0,1)$ such that $f(A)$ is contained in the ball $B$ of radius $\frac 12$ centered on $(0,1)$. However $A$ must contain a point of the form $(\frac 1n,1)$ for some $n$ (which also maps to itself under $f$), and then the points $(\frac tn,1)$ for $0\le t\le 1$ all lie in $A$ too. Then $t\to f(\frac tn,1)$ is a path in the comb from $(0,1)$ to $(\frac 1n,1)$. Since there is no such path in the comb that doesn't wander outside $B$, $f$ cannot exist. - Let $C$ be the comb space, and suppose that $r:I^2\to C$ is a retraction. Let $\pi:I^2\to I$ be the projection to the first coordinate, and define $f:I\to I:x\mapsto\pi\big(f(\langle x,1\rangle)\big)$; clearly $f$ is continuous, and $f(x)=x$ for $x\in\{0\}\cup\{1/n:n\in\Bbb Z^+\}$. For $n\in\Bbb Z^+$ let $p_n=\left\langle\frac1n,1\right\rangle$. Let $B_n$ be an open ball centred at $p_n$ of radius less than $\frac1n-\frac1{n+1}$; since $r$ is continuous, there is an open ball $B_n'$ of some radius $\epsilon_n$ centred at $p_n$ such that $r[B_n']\subseteq B_n$. It follows that $f(x)=\frac1n$ for all $x\in I$ such that $\left\|x-\frac1n\right\|<\epsilon_n$. Now consider $x\in\left(\frac1{n+1},\frac1n\right)$. If $x<\frac1{n+1}+\epsilon_{n+1}$, then $f(x)<x$, and if $x>\frac1n-\epsilon_n$, then $f(x)>x$. By continuity of $f(x)-x$ there must be a point $x_n\in\left[\frac1{n+1}+\epsilon_{n+1},\frac1n-\epsilon_n\right]$ such that $f(x_n)=x_n$. This is possible, however, iff $r(\langle x_n,1\rangle)=\langle x_n,0\rangle$. For $n\in\Bbb Z^+$ let $q_n=\langle x_n,1\rangle$. Then $$\begin{align} &\langle q_n:n\in\Bbb Z^+\rangle\to\langle 0,1\rangle\;,\\ &\langle p_n:n\in\Bbb Z^+\rangle\to\langle 0,1\rangle\;,\\ &\langle r(q_n):n\in\Bbb Z^+\rangle\to\langle 0,0\rangle\;,\text{ and}\\ &\langle r(p_n):n\in\Bbb Z^+\rangle\to\langle 0,1\rangle\;, \end{align}$$ which is impossible. -	{"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": 56, "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": 0.9608114957809448, "perplexity_flag": "head"}
http://mathhelpforum.com/advanced-algebra/140799-hermitian-positive-definite-matrix-invertibility-print.html	# Hermitian positive definite matrix and invertibility Printable View • April 22nd 2010, 04:15 PM math8 Hermitian positive definite matrix and invertibility Suppose A is a Hermitian positive definite matrix split into $A = C + C^{*} + D$ where $D$ is also Hermitian positive definite. We show that $B=C+ \omega ^{-1} D$ is invertible. Consider the iteration $x_{n+1} = x_{n} + B^{-1} (b-Ax_{n})$ , with any initial iterate $x_{0}$ . Prove that $x_{n}$ converges to $x= A^{-1}b$ whenever $0< \omega < 2$. I suppose to show invertibility, we need to show $det(B) \neq 0$. But I am not sure how to show that. Also for the convergence, do we show $lim_{n \rightarrow \infty } \left\\| x_{n}-x \right\\|= 0$ ? If yes, how? • April 22nd 2010, 06:36 PM Bruno J. You want to show that $B$ is invertible, but $B^{-1}$ appears in the expression which you plan on using to do that... Are you sure you copied the problem properly? • April 23rd 2010, 10:32 AM math8 yes, the problem is copied properly. I think we need to first show that B is invertible (i.e. B inverse exists) to be able to use the expression for $x_{n+1}$ . All times are GMT -8. The time now is 02:36 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": 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": 0.9112724661827087, "perplexity_flag": "head"}
http://math.stackexchange.com/questions/54292/find-the-isomorphism-between-two-table-representations-of-the-same-finite-field	# Find the isomorphism between two table representations of the same finite field I am given two multiplication and addition tables for a finite field (i.e. the tables are for different naming of the elements of the field) and I want to find the isomorphism between the two representations. One idea is to map a primitive element of the first field to a primitive element of the second field and define the rest of the mapping according to their powers, but I am not sure addition is preserved in this method. - 4 Addition will be preserved for some choice of primitive element, but not all (because different primitive elements may have different characteristic polynomials). How about just starting with the multiplicative identity and going from there? – Qiaochu Yuan Jul 28 '11 at 15:36 Do you also know the addition tables? – Zhen Lin Jul 28 '11 at 15:37 3 If your fields have size $q = p^n$, for $p$ a prime, is it practical for you to factorize $x^q - x$ into irreducibles in ${\rm GF}(p)[x]$. Find an irreducible factor of degree $n$, and then identify a root of it in each of your (pairs of) tables. – Geoff Robinson Jul 28 '11 at 15:46 1 Extending @Geoff's idea, since I presume your tables are computed by some form of polynomial multiplication modulo some polynomial, you must already know what the minimal polynomials of the generator on both sides are. Then it suffices to find a root of one minimal polynomial in the other field to obtain the isomorphism. If your field isn't too large, this is a little less impractical than trying to factor $x^q - x$. – Zhen Lin Jul 28 '11 at 16:11 2 @Michael: If $p$ is a prime, and $n$ is a positive integer, there is, as you know, a unique field of order $p^n$ up to isomorphism. But given two fields of order $p^n$, there are exactly $n$-isomorphisms from one to the other. This is because a field of order $p^n$ has a cyclic automorphism group of order $n$, generate by $\alpha \to \alpha^p$. – Geoff Robinson Jul 28 '11 at 21:56 show 4 more comments ## 1 Answer The following is a collection of the ideas posted in the comments together with some of my own thinking. This is too long to fit into a comment, and I think this works with reasonable complexity even assuming that you have no other information available than the addition and multiplication tables for the two fields. I'm all ears to suggestions for improvements and better solutions. I think that it is highly likely that my approach can be improved. 1. First we know the characteristic $p$ and the degree of extension $n$ over the prime field simply from the size of the tables. So at this point we know that both fields are isomorphic to $GF(p^n)$, and have cyclic Galois groups generated by the Frobenius map $F:x\mapsto x^p$. We can also easily identify the elements of the prime field, as spotting $0$ and $1$ is easy, and generating $2=1+1,3=2+1,\ldots,p-1$ is also straightforward. 2. Partition both fields into disjoint unions of orbit under the Galois group. The orbit of an element $x$ consists of the elements of the form $F^i(x)$ for $i=0,1,\ldots$. Note that for some integer $d$ we have $F^d(x)=x$, and furthermore it always happens that the smalles such $d$ is a factor of $n$. Here the exponent of $F$ denotes iteration, so $F^2(x)=F(F(x))$, $F^3(x)=F(F^2(x))$ et cetera. Also note that in the case $p=2$ the Frobenius map is just squaring, and its value can be immediately read from the multiplication table. If you want to spend less time with this step, I will disclose that it suffices to find one orbit of a maximal size $n$ from one of the fields. 3. Next I will locate a vector space basis (over the prime field) for one of the fields. With the partition of step 2 at hand (even to the extent described in the last sentence of step 2) this is easy. Let $\alpha$ be an element in an orbit of maximal size $n$. Then $\alpha$ generates the field as an extension over the prime field, and we can use $\mathcal{B}=\{1,\alpha,\alpha^2,\ldots,\alpha^{n-1}\}$. Identifying all these elements using the multiplication table is straightforward. 4. Next I describe a method for finding the coordinates of a given element with respect to the basis $\mathcal{B}$. The method depends on the fact that the bilinear form $(x,y)=Tr(xy)$ is non-degenerate. Here $$Tr(x)=x+F(x)+F^2(x)+\cdots+F^{n-1}(x)=\sum_{i=0}^{n-1}F^i(x),$$ and (this is important) its values belong to the prime field, and can thus be identified. The non-degeneracy of this bilinear form means that $0$ is the only element $y$ with the property: $(x,y)=0$ for all $x$. So given an element $\beta$ we want to find elements $x_i,i=0,\ldots,n-1$ of the prime field such that $\beta=\sum_i x_i\alpha^i$. The coordinates are the (unique) solution to the linear system of equations given by $$(\beta,\alpha^j)=\sum_i x_i(\alpha^i,\alpha^j),$$ where $j=0,1,\ldots,n-1$. Such a system can be solved by e.g. Gaussian elimination. IOW it has complexity that is polynomial in $n$ and thus gives a marked improvement to brute force search of all combinations - at least if $n$ is a bit larger. If $p^n$ is small, then brute force may be better, because you avoid the computation of $n(n+1)/2$ traces - your call. [Edit: non-degeneracy of the trace form manisfests itself here in the helpful way that the matrix $((\alpha^i,\alpha^j))_{i,j=1}^n$ is non-singular.] 5. Next we determine the minimal polynomial $m(x)$ of $\alpha$. This is easy, because all we need to do is to apply the method of step 4 to $\beta=\alpha^n$. 6. The remaining task is to locate a zero of the polynomial $m(x)$ from the other field. We know in advance that all the zeros form a Frobenius orbit of size $n$, so we only need to check such elements, one representative from each orbit untill we get lucky. Constructing the isomorphism is then easy. I really hope that there is something better than step 6. That has exponential complexity as a function of $n$ as does step 2. May be doing steps 3-5 to both fields helps? We only need one full size orbit to do steps 3-5, and a random element is more likely than not to have a full size orbit, so we may not need the full partition of step 2. This is unsatisfactory, so hopefully we can continue this as a group effort. The main contribution of my answer is the method of finding a monomial basis and the related minimal polynomial using the trace-form. -	{"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": 58, "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": 5, "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": 0.9335896372795105, "perplexity_flag": "head"}

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