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http://mathhelpforum.com/trigonometry/135244-volume-fuel-versus-height-fuel-column-cylinderical-horizontal-fuel-tank.html	# Thread: 1. ## Volume of Fuel versus Height of Fuel column in a Cylinderical Horizontal Fuel tank Hi Consider a Cylindrical Fuel tank of a Truck. The Cylinder is mounted horizontally which means the axis of the cylinder is horizontal to the ground. Let us say the truck is filled with Fuel to the Full Fuel tank level and the truck starts its long journey. The truck moves consuming Fuel as it travels. Let us assume that the truck travels till such time the Fuel in the Fuel tank is emptied. The cross section of the Fuel column is a Full circle when the Fuel is full and it is half circle when the Fuel is at half fuel tank capacity. The cross section of Fuel column is a segment of the circle at any point in time. Let us say "h" is the height of Fuel column from the horizontal Bottom. The height of fuel column is equal to Diameter of the cylinder "D" when the fuel column is Full and is Zero when fuel is emptied. Let us have "L" as length of the cylinder. I want to know for a given "h" what is the "volume of the Fuel" in the Fuel tank. I also want to use the formula in excel sheet and draw a graph between "h" and "volume of Fuel". can someone help me? 2. Let r be the radius of the circle. You need to compute the area of a segment of the circle, which can be determined by calculating (if $h \leq r$) $\theta = \arccos \left(\frac{r-h}{r}\right)$. Then the area of the segment is $\theta r^2 - (\sin \theta)(r-h)r$, where $\theta$ is in radians. If $h \geq r$, then $\theta = \arccos \left(\frac{h-r}{r}\right)$, and the area of the segment is $(\pi - \theta)r^2 + (\sin \theta)(r-h)r$. 3. Thanks icemanfan	{"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": 0.9398950338363647, "perplexity_flag": "head"}
http://math.stackexchange.com/questions/218372/how-do-i-calculate-a-multiplication-table-for-gf8	# How do i calculate a multiplication table for GF(8)? Could you please provide the steps involved in calculating a multiplication table for GF(8)? - – Jyrki Lahtonen Nov 12 '12 at 12:48 ## 1 Answer Take an irriducibile polynomial of degree $3$ over the field $\Bbb F_2$ with two elements, e.g. $P(X)=X^3+X+1$. Then you know that $$\Bbb F_8=\Bbb F_2[X]/(P(X))$$ and that its 8 elements are represented by the 8 polynomials of degree $\leq2$. Thus you can construct the multiplication of $\Bbb F_8$ simply by multiplying these polynomials and taking the result modulo $P(X)$. - Thank you so much. What i do not understand is that in the text book i have the entry for a multiplication of 2(010) and 4(100) as 3 in GF(8). Could you explain in this context? – phoenix Oct 21 '12 at 23:55 I am not sure what you (or the textbook) mean with that notation. – Andrea Mori Oct 21 '12 at 23:58 – phoenix Oct 22 '12 at 0:00 1 @phoenix: Looks like "010" represents $0\cdot\tau^2 + 1\cdot\tau + 0\cdot1$, and "100" represents $1\cdot\tau^2 + 0\cdot\tau + 0\cdot1$. – Niel de Beaudrap Oct 22 '12 at 0:46 As Andrea says, the elements of the field are (represented by) the polynomials of degree at most 2. Your source chooses a different representation; the polynomial $ax^2+bx+c$ is represented by the bit-string $abc$ (or maybe $cba$, I didn't read closely enough to tell). – Gerry Myerson Oct 22 '12 at 0:47 show 2 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": 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": 0.906246542930603, "perplexity_flag": "middle"}
http://math.stackexchange.com/questions/180234/what-is-the-sum-of-sum-limits-i-1nik-pi	# What is the sum of $\sum\limits_{i=1}^{n}i^k p^i$? (I've asked similar question, but this is much more complicated, I think) What is the sum of $\sum\limits_{i=1}^{n}i^k p^i$? Interpretation (why is it important ) : $f(k,n,p)=\sum\limits_{i=1}^{n}i^k p^i$ if n goes to plus infinity, then $f(1,n,1/2)$ is average length or series of heads while tipping symmetric coin, and $f(2,n,1/2)-f(1,n,1/2)^2$ is the variance of that length, so for another k we get k-th moment - These quantities are well-studied when $p = 1,$ and there are inductive formulae for them. – Geoff Robinson Aug 8 '12 at 10:36 ## 2 Answers Well, starting again with $f(p)=\sum_{i=1}^n p^i=\frac {p(p^n-1)}{p-1}$ Consider the theta operator $\Theta=p\frac d{dp}$ then your answer is $\Theta^k(f(p))$ (compute the derivative, multiply by $p$, repeat $k$ times). - We use the follwing denotions • $(i)_r$ - lower factorial, • $S(n,k)$ - Stirling numbers of the second kind Recall that each power of a number can be expressed in terms of its lower factorials $$i^k=\sum\limits_{r=0}^k S(k, r)(i)_r$$ So we can write $$\begin{align} f(k,n,p)=\sum\limits_{i=1}^n i^k p^i&= \sum\limits_{i=1}^n \left(\sum\limits_{r=0}^k S(k, r)(i)_r\right) p^i\\ &=\sum\limits_{i=1}^n \sum\limits_{r=0}^k S(k, r)(i)_r p^i \\ &=\sum\limits_{i=1}^n \sum\limits_{r=0}^k S(k, r) p^r \frac{d^r}{dp^r}(p^i)\\ &=\sum\limits_{r=0}^k \sum\limits_{i=1}^n S(k, r) p^r \frac{d^r}{dp^r}(p^i)\\ &=\sum\limits_{r=0}^k S(k, r) p^r\frac{d^r}{dp^r}\left(\sum\limits_{i=1}^n p^i\right)\\ &=\sum\limits_{r=0}^k S(k, r) p^r\frac{d^r}{dp^r}\left(\frac{p^{n+1}-p}{p-1}\right) \end{align}$$ For each fixed $k$ one may compute this sum. -	{"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": 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": 0.9234929084777832, "perplexity_flag": "middle"}
http://mathhelpforum.com/differential-equations/131912-newtonian-mechanics-movement-particles.html	# Thread: 1. ## Newtonian Mechanics - the movement of a particles A particle of mass m which moves along a horizontal straight line with a velocity of v encounters a resistance of av + b(v^3), where a and b are constants. If there is no other force beside the resistance acting on the particle and its initial velocity is U , show that the particle will stop after it has moved a distance of m(ab)^(-1/2)tan^(-1)(U(b/a)^1/2). Also show that the velocity is 1/2U after time (m/2a)ln(4a+b(U^2))/(a+b(U^2))) to show the distance when the particle stops : i use : F = ma => m (dv/dt) = - (av+b(v^3)) i let v = y and t = x to make it less confusion ....(1) thus m(dy/dx) = -(ay+b(y^3)) dy/dx = -(ay/m)-(b(y^3)/m) dy/dx + (a/m)y = -(b/m)(y^2) ... (2) (y^-3)(dy/dx) + (a/m)(y^-2) = (-b/m) ===> this is Bernoulli equation let v = (y^-2) thus, dv/dx = -2(y^-3) (dy/dx) => -1/2(dv/dx) + (a/m)v= -b/m searching for integrating factor, miu(x) = exp(integrating(a/m)dx) thus, miu (x) = exp(ax/m) miu(x) x (2) : d/dx (exp(ax/m).v) = exp(ax/m).(-b/m) exp(ax/m).v = (-b/m) integrating (exp(ax/m) dx) v = (-b/a) + c(exp-(ax/m)) but v = (y^-2) = (v^-2) thus, v^-2 = (-b/a) + c(exp-(ax/m)) v^2 = (1/c(exp-(ax/m))) - (a/b) v = sqrt ((1/c(exp-(ax/m))) - (a/b)) applying initial condition where v_0 = u, x=0, t=0 c = 1/u^2 + b/a thus v = dx/dt = sqrt ((1/(1/u^2 + b/a)) - a/b) when particle stop dx/dt = 0 then 0 = sqrt ((1/(1/u^2 + b/a)) - a/b) . . . x = m/a ln \|(1/ab)+1\| why i can't get the answer given which is m(ab)^(-1/2)tan^(-1)(U(b/a)^1/2)??????? where i'm going wrong?????? p/s : i don't know to use SYNTAX. sorry =p 2. Originally Posted by bobey A particle of mass m which moves along a horizontal straight line with a velocity of v encounters a resistance of av + b(v^3), where a and b are constants. If there is no other force beside the resistance acting on the particle and its initial velocity is U , show that the particle will stop after it has moved a distance of m(ab)^(-1/2)tan^(-1)(U(b/a)^1/2). Also show that the velocity is 1/2U after time (m/2a)ln(4a+b(U^2))/(a+b(U^2))) Hi Bobey, This problem is not so difficult as it seems. It takes some care in the calculation, so let's go over it with some large steps. I assume you will do the intermediate calculations yourself. The DE you obtained is correct, it is: $m\frac{dv}{dt}=-av-bv^3$ It can be solved using the method of Bernoulli but it is easier to use the method of separation of variables. This gives: $\frac{dv}{v\left(a+bv^2\right)}=-\frac{dt}{m}$ Partial fractions is the way to solve the first integral and this gives after the integration: $\frac{1}{a}ln(v)-\frac{1}{2a}ln\left(a+bv^2\right)=-\frac{t}{m}+C_1$ With C1 the integration constant which can be determined from the initial condition at t=0 is v=U, plugging in gives now: $\frac{\frac{v}{U}}{\sqrt{\frac{a+bv^2}{a+bU^2}}}=e ^{-\frac{a}{m}t}$ From this we can obtain the answer from the second question. Indeed, setting v=U/2 and derive t from this gives: $t=\frac{m}{2a}ln\left(\frac{4a+bU^2}{a+bU^2}\right )$ To answer the first question, we need to rewrite the solution of the DE as follows: $v=\frac{dx}{dt}=\frac{\sqrt{a}}{\sqrt{\left(\frac{ a+bU^2}{U^2}\right)e^{\frac{2a}{m}t}-b}}$ This can be integrated and the result is: $x+C_2=\frac{m}{\sqrt{ab}}atan\left(\sqrt{\frac{a+b U^2}{bU^2}e^{\frac{2a}{m}t}-1}\right)$ Using the boundary condition at t=0 is x=0 and plugging in we get: $x=\frac{m}{\sqrt{ab}}\left[atan\left(\sqrt{\frac{a+bU^2}{bU^2}e^{\frac{2a}{m} t}-1}\right)- atan\left(\frac{\sqrt{a}}{\sqrt{b}U}\right)\right]$ The particle will come to rest for t going to infinity as can be seen from the resulting equation from the DE. Plugging this in gives: $x_{\infty}=\frac{m}{\sqrt{ab}}\left[\frac{\pi}{2}- atan\left(\frac{\sqrt{a}}{\sqrt{b}U}\right)\right]$ which can be rewritten using pi/2=atan(T)+atan(1/T) as: $x_{\infty}=\frac{m}{\sqrt{ab}}atan\left(\frac{\sqr t{b}U}{\sqrt{a}}\right)$ The solution you were looking for. It is an exercise requiring careful calculations. Take your time for this and make sure you go over every step before going to the next one. As a final remark the DE can indeed be solved by the method of Bernoulli, it gives exactly the result as with the proposed method. I checked it :-) Hope this helps and do come back if you need more info. Coomast	{"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": 0.891438901424408, "perplexity_flag": "middle"}
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So I've read the beginning of ... 2answers 885 views ### Do the physics in the FlyBoard video make sense? If you haven't seen the video of the FlyBoard, please have a look: http://www.youtube.com/watch?v=Cd6C1vIyQ3w&feature=youtube_gdata_player Yes, it's amazing, but do the physics make sense or is ... 4answers 166 views ### Can a balloon be used as an anchor point for a pulley? For a physics/ engineering contest, I want to use a large balloon as an anchor point for a pulley. This would allow me to raise and drop masses. However, in testing, when I pull on the pulley 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": 49, "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": 0.9085862636566162, "perplexity_flag": "middle"}
http://mathoverflow.net/questions/70307?sort=newest	## Jacobsthal function related to squares ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) The ordinary Jacobsthal function $j$ is defined by setting $j(n)$ as the smallest number $m$, such that for each consecutive $m$ numbers in the integers, there is at least one of the numbers comprime to $n$. There are estimes for $j$, for example $$j(n) \ll \log ^2 (n)$$ and it is conjectured by Jacobsthal, that we could improve this to $$j(n) \ll ( \log (n) / \log ( \log (n)) )^2.$$ See for example http://www.tcnj.edu/~hagedorn/papers/JacobPaper.pdf . We define now $h(n)$ as the smallest number $m$, such that for each square $a \in \mathbb{N}$, the sequence $$a+1, a+1, \ldots, a+m$$ contains at least one number comprime to $n$. My question is now, are there any good upper estimates for $h(n)$ ? Clearly, $h(n) \leq j(n)$ holds, so we could take the above estimate, but I am searching for something stronger, perhaps like $$h(n) \leq C \log(n),$$ where $C=2$ or so. - There are still a lot of quadratiic residues for many numbers. My guess is that the upper bound will not drop significantly, and may not drop at all. The analysis becomes more challenging for me because the answer may depend on the multiplicity of prime factors in the number. If -1 is not a quadratic residue and there are about half as many qrs as nqrs less than n, then by symmetry I expect the upper bound to be the same. Even in the general case I expect there to be no provable decrease in the upper bound. Gerhard "Ask Me About Jacobsthal's Function" 2011.07.14 – Gerhard Paseman Jul 14 2011 at 15:56 Also, Erdos has for almost all n that j(n) is not far from a bound like log(n)log(log(n)) in a 1962 paper. You might review that paper to see how much carries over to your situation. Gerhard "Email Me About System Design" Paseman, 2011.07.14 – Gerhard Paseman Jul 14 2011 at 18:38 I would like to know the motivation for choosing squares. Is it an approach to study primes of the form $a^2+1$? Also, I should mention that I do not see a way to provide any tight bounds on $h(m)$ without making assumptions on $m$ like "Suppose every prime factor of $m$ has -3 as a nqr..." Gerhard "Ask Me About System Design" Paseman, 2011.07.15 – Gerhard Paseman Jul 15 2011 at 18:53 Do we have to start with a given n? I have a function that when given m will create an n that will be prime to all a in those ranges. – Fred Kline Dec 11 2011 at 8:39 ## 1 Answer I take back what I said in the comments about the bound not shrinking. I am convinced that one can get much tighter bounds, and that the tighter bounds will depend massively on the quadratic residues for each of the primes involved. To set the stage, I change notation slightly. I use $m$ for the argument to $j()$, and I use $n$ for the number of distinct prime factors of $m$. I also insist $m > 1$. One of the more accessible results is that $j(m) = j(k)$ if $m$ and $k$ have the same prime factors. Further if the prime factors are all larger than $n$, then $j(m)= n+1$, so there is a certain uniformity in the analysis and results for a large and simply stated class of numbers. (Advertisement; I am working on similar statements where $n$ is replaced by something like $\sqrt(n)$. Email me for more detail.) Many of the standard bounds for $j(m)$ can be expressed in terms of $n$. When I posted my comments above, I thought something similar would be true for this version. However, checking a few examples leads me somewhere completely different. For $m$ a prime power, $j(m)=2$. The same holds for the new variation if and only if $-1$ is a qr of the prime dividing $m$. However, things change when $n>1$. $h(6)=j(6)=4$, but this does not hold for all numbers of the form $2p$. $h(m)$ can vary from 2 to 4 depending on $m$ even if $n$ is restricted to 2. To make things interesting, I computed $h(385)$, which is bounded above by 4. There are 6 values of a mod 385 to show $j(385)=4$. To get $h(385)=4$, I had to find a square which was 4 mod 7, 9 mod 11, and 4 mod 5. The square of 47 fit the bill, but I could imagine different primes with qrs that would not nicely fit in the set {-3,-2,-1}, so $h(m)$ will not always reach 4 or higher when $n=3$; it will depend on getting qrs which are small negative numbers. For higher $n$, you may not achieve $h(m)>n$ without choosing the $n$ prime factors carefully. And this analysis is using only squarefree $m$; I do not know what happens in the general case. EDIT 2011.07.29 I let the presence of quadratic residues rattle me into thinking that $h(m)$ would depend on the multiplicity of prime factors of $m$. It does not. As is the case for $j(m)$, the value of $h(m)$ depends only on the set of prime factors of $m$, and so there is no "general case": it suffices to assume $m$ is squarefree. END EDIT 2011.07.29 I am having a few challenges showing upper bounds without having to worry about quadratic residues. This problem is a can of worms I am not ready to tackle. Certainly nothing like the uniform results involving sufficiently large prime factors will hold. One approach involves "tiling" a candidate interval with appropriate primes, and then solving $n$ many quadratic congurences simultaneously. I don't know enough about quadratic residues to see any clean looking results here. It is an interesting variation on which I welcome other viewpoints. Gerhard "Jacobsthal's Function Is Tough Already" Paseman, 2011.07.15 -	{"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": 57, "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": 0.93562912940979, "perplexity_flag": "head"}
http://mathhelpforum.com/discrete-math/8921-induction-irrational-proves.html	# Thread: 1. ## induction,irrational and proves...... please try to solve these questions. Attached Files • discrete math.doc (23.5 KB, 56 views) 2. Originally Posted by m777 please try to solve these questions. 1a) What the heck is the definition of Q? 1b) Any even integer is of the form $2n$ where n is some integer. (This is how an even number is defined.) Now square this: $(2n)^2 = 4n^2 = 2(2n^2)$ Now, since n is an integer, so is $n^2$. Again, since 2 and [tex]n^2 are integers so is $2n^2$. Thus $2(2n^2)$ is an integer it is also an even integer. -Dan 3. Hello, m777! Here's some help . . . Use mathematical induction to prove: . . $1 + \frac{1}{2} + \frac{1}{4} + \cdots + \frac{1}{2^{n-1}} \;=\;2\left(1 - \frac{1}{2^n}\right)$ for any positive integer $n$. Verify $S(1)\!:\;1\:=\:2\left(1 - \frac{1}{2}\right) \:=\:1$ . . . true! Assume $S(k)\!:\;1 + \frac{1}{2} + \frac{1}{4} + \cdots + \frac{1}{2^{k-1}} \:=\:2\left(1 - \frac{1}{2^k}\right)$ Add $\frac{1}{2^k}$ to both sides: . . $\underbrace{1 + \frac{1}{2} + \frac{1}{4} + \cdots + \frac{1}{2^{k-1}} + \frac{1}{2^k}} \:=\:2\left(1 - \frac{1}{2^k}\right) + \frac{1}{2^k}$ . . This is the left side of $S(k+1)$ The right side is: . $2 - \frac{2}{2^k} + \frac{1}{2^k} \;=\;2 - \frac{1}{2^k} \;=\; 2 - \frac{2}{2^{k+1}} \:=\:2\left(1 - \frac{1}{2^{k+1}}\right)$ . . which is the right side of $S(k+1)$ . . . We're done! 4. Hello again, m777! 3) Prove by induction that $5^n - 2^n$ is divisible by 3. Verify $S(1)\!:\;5^1 - 2^1 \:=\:3$ is divisible by 3 . . . true! Assume $S(k)\!:\;5^k - 2^k$ is divisible by 3. . . Then: . $5^k - 2^k \;=\;3a$ for some integer $a$. Add $4\!\cdot\!5^k - 2^k$ to both sides: . . $5^k + 4\!\cdot\!5^k - 2^k - 2^k \;=\;3a + 4\!\cdot\!5^k - 2^k$ We have: . $5\!\cdot\!5^k - 2\!\cdot\!2^k\;=\;3a + 3\!\cdot\!5^k + 5^k - 2^k$ $\text{Then: }5^{k+1} - 2^{k+1} \;= \;\underbrace{3(a + 5^k)}_{\text{div by 3}} + \underbrace{(5^k - 2^k)}_{\text{div by 3}}$ Therefore: . $5^{k+1} - 2^{k+1}$ is divisible by 3 . . . We're done! 5. ## reply to topsquark Hello, Topsquark. In part(a) of question no 1 i have done this so far. Q(a,b)= o if a<b. Q(a-b,b)+1 if b is less than equal to a. so Q(3,4)=0 SINCE 3<4 GIVEN Q(a,b)= Q(a-b,b)+1) if b is less than equal to a NOW Q(14,3)=Q(11,3)+1 =Q(8,3)+1 =Q(5,3)+1 =Q(2,3)+1 SO AT THAT POINT 2 IS LESS THAN 3. THEREFORE =0+4 =4 IS THE ANSWER 6. To prove $\sqrt{3}$ is irrational. Assume, $\sqrt{3}=\frac{a}{b}$ Thus, $3=\frac{a^2}{b^2}$ Thus, $3b^2=a^2$. Prime decompose $a$. Prime decompose $b$. Note, $a^2$ has an even number of prime factors. Note, $b^2$ has an even number of prime factors. But, $3b^2$ has an odd number of primes factors. Thus, we have a contradiction because both sides are note equal by the uniquess part of the fundamental theorem of arithmetic. 7. This is a very general proof. If p is a positive integer that is not a perfect square then $\sqrt p$ is irrational. Suppose that it is rational, $\sqrt p = \frac{a}{b}$ where a & b are integers. Let $T = \left\{ {n \in Z^ + :n\sqrt p \in Z^ + } \right\}\quad \Rightarrow \quad b \in T \not= \emptyset.$ By well ordering there is a first term in T, call it K. Using the floor function and because p is not a square we have the following: $\left\lfloor {\sqrt p } \right\rfloor < \sqrt p < \left\lfloor {\sqrt p } \right\rfloor + 1\quad \Rightarrow \quad 0 < \sqrt p - \left\lfloor {\sqrt p } \right\rfloor < 1.$ $0 < K\left( {\sqrt p - \left\lfloor {\sqrt p } \right\rfloor } \right) < K.$ But $K\left( {\sqrt p - \left\lfloor {\sqrt p } \right\rfloor } \right)$ is an integer less than K. But $\left[ {K\left( {\sqrt p - \left\lfloor {\sqrt p } \right\rfloor } \right)} \right]\sqrt p$ is an integer which contradicts the minimumality of K. 8. YOu guys are awesome!!!	{"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": 40, "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": 0.878528356552124, "perplexity_flag": "middle"}
http://physics.stackexchange.com/questions/21195/when-are-maximum-velocity-and-acceleration-acheived-in-simple-harmonic-motion/21196	When are Maximum Velocity and Acceleration acheived in Simple Harmonic Motion? Im trying to get my head around SMH out of curiosity because it seems simple yet I'm not getting the concept behind some ideas. For a SMH equation : $$x=a \sin(\omega t+\phi)$$ • Under what conditions will each of the two ( velocity and acceleration ) be at maximum ? we know that : $$x'=-\omega \cos(\omega t + \phi)$$ $$x''=-\omega^2 \sin(\omega t + \phi)$$ Also : • how would it be different for $x = a \cos(\omega t - \phi)$ ? - I don't think this necessarily needs to have the `homework` tag. – David Zaslavsky♦ Feb 19 '12 at 19:42 2 Answers The easiest way to determine maximum and minumums of a function is to set the derivative equal to zero. Thus, in this case, setting the equation for acceleration equal to zero and solving for the variables of interest will give you what you want. Thus, in this case we have the equation for position: $$x = a \sin(\omega t + \phi)$$ One way to see the maximum and minimum is to plot the function. I find that playing around with the constants (in this case $a, \omega,$ and $\phi$) helps you see what effect they have. You can see that $a$ is an amplitude that determines how large the swings are from the origin. The termn $\omega$ tells you something about how quickly the position moves back and forth, thus it is known as the "angular frequency." Finally, $\phi$ tells you where you are when $t=0$ (or any other time for that matter) and is called the "phase." For the position function, it is obvious that the value is maximized when the value of the sine function equals $1$ and, from what you learned in trig class, this happens when $\omega t + \phi = \pm \pi/2$ depending on the sign of $a$. Since $\omega$ and $\phi$ are constants set by the conditions and $t$ is the variable, the position is maximum when $t = (\pm \pi/2 - \phi)/\omega$, again depending on the sign of $a$. ps- sorry for the curt answer the first time, but Wikipedia has a good article on this topic and there are tons of resources out there for it. - Its not really a homework question as I made it clear in the first sentence. I want to see how it works out differently because there are many websites online that just confused me. Usually people here answer with a way that I can directly relate to and thus understand the idea better. – Fendi Feb 19 '12 at 16:34 Okay, did my answer work? or are you still confused? – AdamRedwine Feb 19 '12 at 16:40 I still don't get it. Can you provide an example or explain more thoroughly please ? – Fendi Feb 26 '12 at 18:10 If you give me more of a description of what you are thinking, it would be easier to give you a helpful answer. This might be a discussion better suited to the chat room. – AdamRedwine Feb 27 '12 at 13:28 $$x'=-\omega \cos(\omega t + \phi)$$ maximum value of cosine is 1 implies max value of $x'$ is $\|w\|$. Same for $x''$. -	{"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": 18, "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": 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": 0.9545990824699402, "perplexity_flag": "head"}
http://math.stackexchange.com/questions/293636/warehouse-location-problem-as-an-integer-progam-instead-of-a-mixed-integer-progr/293954	# Warehouse Location Problem as an integer progam instead of a mixed-integer program Given a set of costumers $M = \{1, \dots , m \}$ and a set of of factories $N = \{1, \dots , n\}$ we have • $c_{ij} \geq 0$ costs to deliver to costumer $i \in M$ from factory $j \in N$ • $F_j \geq 0$ fixed costs to run factory $j \in n$ We try to minimize the costs. Therefor we model the problem as a mixed-integer program: • $y_j \in \{0,1\}$ and $y_j = 1$ iff we decide to run factory $j$ (integer variable) • $x_{ij}$ the fraction of the consumption of costumer $i$ which is delivered by factory $j$ (non-integer variable) We add restistrictions to make it correct: • $\sum_{j \in N} x_{ij} = 1$ for each $i \in M$ to ensure that factories delivering to $i$ add up to the whole consumption • $x_{ij} \leq y_j$ for each $i \in M, j \in N$ to ensure that if a factory delivers it is open • $x_{ij} \geq0$ for each $i \in M, j \in N$ The function to minimize is $$\sum_{j \in N} F_jy_j + \sum_{i \in M}\sum_{j \in N}c_{ij}x_{ij}.$$ This is denoted by Warehouse Location Problem (WLP) modeled as a mixed-integer program. My questions is concerning the solution of such problems. I think that if there is a optimal solution to an instance of a WLP then there is a optimal solution such that • $y_i \in \{0,1\}$ by definition but also • $x_{ij} \in \{0,1\}$. Because in the optimal solution costumer $i$ will always provided by one cheapest factory $j$. As there is no limit on the capacity of the factories. There is no need to split up. Is that correct? So the problem could be modeled equivalently as a integer program as well and we would obtain solutions which are optimal? I have taken the definition from a book. Why is it modeled like this? For performance reasons? I thought mixed-integer programs are harder to solve then integer programs. - ## 1 Answer You're right that in the uncapacitated version of the warehouse location problem the value of $x_{ij}$ will be $0$ or $1$ at the optimal solution. Of course, in the capacitated version this might not be the case. It is generally somewhat easier to solve mixed-integer programs than integer programs, though. Linear programs are easy; integer programs are hard. The more you can make your problem look like a linear program, then, the easier it will be to solve. In particular, any variables that don't absolutely have to be integers you should leave as reals. That's the reason for the lack of restriction on the $x_i$'s in the warehouse location problem. To expand on this some, the major techniques for solving integer programs -- Gomory's cutting plane algorithm and branch-and-bound -- actually spend a decent amount of time solving the linear program relaxations of restrictions of the original integer program. Any variables that don't have to be integers cut down on the running time of the algorithm, since you don't have to generate cutting planes that exclude fractional solutions involving them (in Gomory's algorithm) or generate branches involving them (in branch-and-bound). -	{"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": 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": 0.9342723488807678, "perplexity_flag": "middle"}
http://mathoverflow.net/questions/25764/consolidation-aftermathematics-of-fads	## Consolidation: Aftermathematics of fads ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) From Frank Quinn's THE NATURE OF CONTEMPORARY CORE MATHEMATICS: "Mathematics has occasional fads, but for the most part it is a long-term solitary activity. In consequence the community lacks the customs evolved in physics to deal with the aftermathematics of fads. If mathematicians desert an area no one comes in afterwards to clean up. Lack of large-scale cleanup mechanisms makes mathematical areas vulnerable to quality control problems. There are a number of once-hot areas that did not get cleaned up and will be hard to unravel when the developers are not available. Funding agencies might watch for this and sponsor physics-style review and consolidation activity when it happens." Can you give examples of such once-hot areas in need of consolidation ? - 2 This "long-term solitary activity" is more collaborative than ever. This quotation seems to be an extract from a very long polemic, and unsuitable as an MO topic. – Robin Chapman May 24 2010 at 10:50 4 @RC: Eh, I've seen worse. The excerpt is from a polemic (which gives me hope that it will be interesting to read), but the question itself does not strike me as especially polemical. – Pete L. Clark May 24 2010 at 11:33 4 This probably should be community wiki since you are asking for a list of examples. – Grétar Amazeen May 24 2010 at 13:23 4 Can representation theory be regarded as an example? My impression from hearing experts talk about it is that there was a big push in maybe the 40's or 50's, driven in part by considerations from physics, to classify all unitary representations of this and that. This seems to have generated a massive, nearly impenetrable literature of which only a handful of experts have a big picture view. The pattern seems to be that algebro-geometric methods have just replaced all the analysis from back then. I would be interested if someone more informed could elaborate on or qualify this in an answer. – Paul Siegel May 25 2010 at 23:34 4 A digression re the apposite pun "aftermathematics": pace "Fermat's Last Tango", the English word "aftermath" has nothing to do with mathematics; the "math" at the end of "aftermath" is an archaic derivative of "mow" (presumably as in grow > growth, long > length, etc.). – Noam D. Elkies Aug 23 2011 at 14:47 show 1 more comment ## 6 Answers Since Quinn's article is a long opinion piece which he says is 90% complete and welcomes comments, it seems entirely appropriate to contact him for clarification on this point. He would probably be happy to tell you more. One example that springs immediately to my mind is the classification of finite simple groups. This was, by a safe margin, the largest scale collaborative activity in the history of mathematics, taking place over a decade or so. The accounts I have read describe Aschbacher, Thompson and (especially) Gorenstein as acting like army generals overseeing a war: they had the most insight into the global structure of the argument and they used it to apportion and subcontract various pieces of the proof. So far as I can think of at the moment, it is much more usual for a visionary mathematician (e.g. Langlands, Thurston, Hamilton) to lay out a program which other mathematicians are then inspired to work on as they see fit than to have this kind of explicit top-down organization. The rest of the story is well-known: in the early 80's Aschbacher, Thompson and Gorenstein were photographed on an aircraft carrier in front of a victory banner (figuratively speaking of course) and all the other group theorists shouted hurrah and cleared out. But certain key parts of the argument had never been published in any form, as a small number of mathematicians (e.g. Serre) spent the next 20 years reminding the community. It seems fair to say that the finite group theorists cleared out a little too early. I don't really know why or exactly what motivated the recent moderate resurgence of interest in the classification, including the 2004 (!) publication of a two-volume work completing the quasi-thin case (a mere 1300 additional pages were required). In the last few years it seems that there has been "the right amount" of tidying up these massive argument by those involved in the "second generation" and "third generation" classification efforts. See http://en.wikipedia.org/wiki/Classification_of_finite_simple_groups and the references therein for more details. Especially highly recommended is Aschbacher's 2004 Notices article http://www.ams.org/notices/200407/fea-aschbacher.pdf which, in addition to being gracefully written and informative, is admirably forthright. - I know Richard Lyons at Rutgers was (as of a couple of years ago) still working on cleaning things up and making it as human readable as possible. I don't know how much of that's done, though. – Charles Siegel May 24 2010 at 15:52 1 People were aware of the Aschbacher-Smith preprint on quasi-thin groups for several years before its publication, and I think it was more of a cause of the resurgence of interest than it was "included" in the resurgence. Also, the second-generation project was already underway long before this resurgence. – S. Carnahan♦ May 24 2010 at 16:47 11 Some of the "clearing out" you mention is simply young finite group theorists' inability to get jobs once math departments decided the main problem in their field had been resolved. I think this is an effect that is less internal to the field than you suggest. – S. Carnahan♦ May 24 2010 at 16:53 @Scott: you may well be right. I hope I was clear that I was offering no explanation whatsoever for the phenomenon. On the other hand, that the main problem in one's field has just been solved and there is seemingly relatively little left to do does seem like an unusual phenonemon, and thus somewhat internal to finite group theory, n'est-ce pas? Or can you think of other instances of this? – Pete L. Clark May 24 2010 at 20:12 3 I remember an (apocryphal?) story about a student coming to Richard Borcherds (Scott C's advisor) with a proposal for a thesis problem, and Richard saying "oh, all the interesting problems there are solved". The student then asked, "But what about <<what Scott C is doing>>?", and Richard demurred that that had been the last interesting problem. – Scott Morrison♦ May 27 2010 at 6:20 show 2 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. In the seventies and eighties of the preceding century, existence and classification of vector bundles on projective space $\mathbb P^n$ were all the rage, with contributions from such luminaries as Artin, Atiyah, Hartshorne and Mumford among many others. I have the feeling that not much progress has been made since. For example, as far as I know, Hartshorne's apparently naïve question "Does there exist an indecomposable algebraic vector bundle of rank 2 on $\mathbb P^n_k \; ?\:"$ is still open for all fields $k$ and all integers $n\geq 6$. Update[Next day] My colleagues André Hirschowitz and Arnaud Beauville, who are well informed about these questions, have allowed me to report that they feel quite confident that Hartshorne's question is still unsolved. - Although in this case, it is not clear to me, if poor documentation stopped the work in the area or if people are just afraid that the problem is too hard for them to solve. – Konrad Voelkel Jun 20 2010 at 9:52 See also Angelo's answer here mathoverflow.net/questions/20444/… – Martin Brandenburg Aug 23 2011 at 8:29 And this question adresses directly the statement: mathoverflow.net/questions/13990/… – Martin Brandenburg Aug 23 2011 at 8:30 @Martin Brandenburg: thanks for the links, Martin. – Georges Elencwajg Aug 23 2011 at 9:57 In email, Frank Quinn mentions that Surgery theory from the 1970s and 80s has a mostly primary literature aimed at other experts, a lack of textbooks, and now has few new people working on it. To me, this seems like a similar problem to properly documenting a computer program as you go along so that others (and your future self) can understand it, otherwise coming back to it can require the same or greater effort to go through it, as was required to create it in the first place, but the temptation is to skimp on that, and just plough ahead. - 1 Novikov has written about the "Period of Decay" in the 70s and 80s as well here arxiv.org/abs/math-ph/0004012 – jc May 25 2010 at 20:56 6 +1 for the computer program analogy. This is a problem in software engineering that took decades for people to realise how important it is (i.e. a good ratio of documentation per code). How to do this the best way is an ongoing area of research. – Konrad Voelkel Jun 20 2010 at 9:49 It's funny you mention that physics can deal with this, because, as a physicist, I see the opposite of this all the time. I was actually just having a discussion with a friend the other day about how physics is desperately in need of cleaning up, organization, and consolidation! I think that a lot of mathematicians have this (wrong) impression of physics, though, because they tend to get their physics from books with titles like "Quantum Mechanics for Mathematicians." (Let me assure you most physicists would find such books largely incomprehensible!) A big problem with the way physicists are currently educated is that there's a lot of needless redundancy, with topics presented in completely different ways, and with different methods to solve identical problems in different contexts for purely historical reasons. For example, most physicists never realize that many of the tools they use in field theory are identical to tools used in general relativity. If they learn both, most have to learn the same tools twice and never realize they're identical because they look so different! - 3 It is a way different culture. Quinn is a Hilbert fan, duplication in presentation is lèse-Bourbaki, and the argument being put forward makes sense in those terms too (no Bourbaki Seminar on an area, need it exist at all?) Well, I may be too harsh here, but as Tom Körner once explained to me, the academic politics of euthanasia for topic areas is quite different from that of natural death. Physics does have better surveys, and it also has a turnover period of c. 3 years rather than a decade. Classical analysis and "natural death" conservatism go together, in a debate active since 1945. – Charles Matthews May 26 2010 at 13:45 2 This is an interesting answer. I had the following (somewhat tangential) reaction to it: learning "different methods to solve identical problems" does not necessarily sound like a waste of time to me at all! If anything it may be less of a waste of time than applying the same method to superficially different problems. But from the next paragraph it seems that you mean that physicists learn superficially different methods several times as if from scratch without realizing their essential similarity. Yes, that sounds like a waste of people's time. – Pete L. Clark Aug 23 2011 at 15:48 One thing that comes directly to mind is the calculus of variations, in the classical sense, where the point is to get rigorous results by mathematical analysis. Now, there are probably several typical kinds of objection here. Firstly the area really is not dormant: physicists use it in the same fashion as ever; there are various kinds of "variational formalism" discussed, for example in soliton theory; and mathematicians have "gone round" this area by the use of Morse theory and moment maps, to break out of the traditional formulation into areas of geometry. But in terms of identifying a "break in tradition" (I don't entirely agree with Quinn's framing of the issue, but it is real enough when those who wrote the papers are no longer around) I would guess there is no line of textbooks that continues from the early twentieth century treatments. Few people may know what was considered important in that line of development. I'm aware of work on variational problems (e.g. the Plateau problem) that is pretty much current, but that illustrates one tendency, to make a given problem into a theory of its own. Anyway, do mathematicians in general know why Jesse Douglas got a Fields Medal in 1936? How many could read his papers? - It may be unpopular to say this, but the theory of subfactors needs a consolidating account. (I don't think of this theory as a fad, but I do think it may be in danger of being difficult to unravel without some consolidating effort.) This is a major reason why I asked the question here.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 4, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9610602855682373, "perplexity_flag": "middle"}
http://math.stackexchange.com/questions/11306/how-to-compute-the-change-in-the-angle-between-two-unit-norm-vectors-as-the-el	# Motivation Suppose that $u \in \mathbb{R}^d$ is a unit-norm vector, $\\|u\\| = 1$, $a, b, c$ are some positive constants and $\xi \in [0,1]$ is another constant (usually chosen close to 1). I am interested in solving the following problem $$\sup_{v \in \mathbb{R}^d}\ (\xi + (1-\xi)\\|v\\|_1^2)\left(\sqrt{ \frac{a \langle u, v \rangle^2 + b}{\xi + (1-\xi)\\|v\\|_1^2}} - c \right)$$ subject to $\\|v\\| = 1$. # Question While any suggestions on how to find an optimal $v$ are welcome, I am specifically interested in the following question. How to find a vector $v$ that maximizes $\langle v, u \rangle$ and satisfies $\\|v\\|_2 = 1$ and $\\|v\\|_1 = x$? A related question (that may be an easier one): given a vector $v_1$ that maximizes $\langle v_1, u \rangle$ and satisfies $\\|v_1\\|_2 = 1$ and $\\|v_1\\|_1 = x$ and another vector $v_2$ that maximizes $\langle v_2, u \rangle$ and satisfies $\\|v_2\\|_2 = 1$ and $\\|v_2\\|_1 = x + \delta$, how to find the change $\langle v_1, u \rangle - \langle v_2, u \rangle$ as a function of $\delta$? - Kuhn-Tucker would work. – Jonas Teuwen Nov 22 '10 at 11:14 3 As a side remark about your first sentence: it is generally not a good idea to start a clause with a comma separated list of variables. Especially if the clause itself is preceded by a comma. I mentally stumbled a bit when I tried to parse that. – Willie Wong♦ Nov 22 '10 at 11:52 Geometrically it is quite easy to see that for the maximizer $v$, we can get it by: WLOG assuming $u$ is in the first orthant. Assume $u$ is not proportional to $w = (1,1,\ldots,1)$ (else all solutions to $\\|v\\|_2 = 1$ and $\\|v\\|_1 = x$ in that orthant are maximizers). Then $v$ is the unique intersection of the hypersurface $\\|v\\|_2 = 1$ with the hyperplane $\\|v\\|_1 = x$ and the 2-plane spanned by $u$ and $w$. – Willie Wong♦ Nov 22 '10 at 12:05 @Willie: It's not so simple. The intersection you describe may lie outside the positive orthant. – Rahul Narain Nov 22 '10 at 16:47 ## 1 Answer Your question doesn't say what $x$ is. Also is $u$ given and fixed? Assuming it is, your first problem can be written $$\min_{v_1, v_2}\; \langle v_1 - v_2, u\rangle \;\; \text{s.t.} \;\; \lVert v_1 - v_2\rVert^2_2 = 1, \quad \sum_i v_{1,i} + \sum_i v_{2,i} = 1, \ (v_1, v_2) \geq 0,$$ where $v_{1,i}$ and $v_{2,i}$ are the individual components of $v_1$ and $v_2$. Th e idea is to split $v = v_1 - v_2$ with $v_1 \geq 0$ and $v_2 \geq 0$. This problem is smooth and can be solved using any method for smooth optimization (SQP, augmented lagrangian, interior-point method, etc.). On paper, you can write the KKT conditions and go from there. -	{"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": 41, "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": 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": 0.9244539737701416, "perplexity_flag": "head"}
http://mathoverflow.net/questions/18928/union-of-closed-subschemes-with-the-structure-sheaf-over-it/19009	## Union of closed subschemes with the structure sheaf over it ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) Elementary commutative algebra fact: for two proper ideals I and J of a commutative ring R, we have $V(IJ)=V(I\cap J)=V(I)\cup V(J)$. Closed subschemes are related to sheaves of ideals. There is operation of intersection and product between sheaves of ideals, which is similar to the affine case. I see in many places that the structure sheaf over $V(I)\cup V(J)$ are defined to be $R/I\cap J$ rather than $R/IJ$. Why should the structure sheaf be define in that way? - +1,cute question – Shizhuo Zhang Mar 22 2010 at 2:00 ## 7 Answers Because the first one is the right answer in the case of affine varieties, and the second one is not. Indeed, $R/I$, $R/J$ are nilpotent-free implies $R/I\cap J$ is nilpotent-free, but not so for $R/IJ$. - ### 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. As with all definitions, there is no "proof" that the adopted definition is the right one but only a feeling that it better corresponds to our intuition. In the case at hand, taking $R/IJ$ as structure sheaf would make union schemes nonreduced for no good reason. For example take $R=k[x,y,z], I=(y,z), J=(x,z)$. Geometrically you are describing the union $U$ of the $x$-axis and the $y$-axis in affine three space $\mathbb A^3_k$ . It should have a reduced structure, correctly described by $I\cap J$, whereas $I.J=(xy,zx, zy, z^2)$ would make the function $z$ nilpotent but not zero on $U$, which feels wrong since $U$ should be a closed subscheme of the plane $z=0$. A more brutal objection to the idea of defining the union of two subschemes by the product of their ideals is that a subscheme $U$ would practically never be equal to its union with itself, since in general$I\neq I^2$ : we would have (almost always) $$U\neq U \cup U$$ That looks bad! - Thanks for your answer. I guess that the following fact makes the $I\cap J$ better: $R/I\cap J$ is reduced if both $R/I$ and $R/J$ are reduced. We 'love' reduced scheme more than non-reduced scheme. – 7-adic Mar 21 2010 at 18:51 Dear 7-adic, it is not just a question of loving reduced schemes. See my answer below for an elaboration. – Emerton Mar 21 2010 at 22:03 (i) The union of two closed subschemes should be the smallest closed subscheme containing the two given ones; if you like, an initial object in the category of all closed subschemes containing the given two. In the case when the two given closed subschemes are $V(I)$ and $V(J)$, this is is $V(I\cap J)$. (The operation $V$ interchanges closed sets and ideals, and is order reversing.) (ii) In function-theoretic terms, a function $f$ (i.e. an element of the ring $A$) should vanish on $V(I) \cup V(J)$ (i.e. lie in the ideal cutting out $V(I)\cup V(J)$) if and only it vanishes on both $V(I)$ and $V(J)$ (i.e. lies in $I$ and $J$), which happens if and only if $f$ lies in $I\cap J$. Thus we are forced to define $V(I)\cup V(J) = V(I\cap J)$, if union is to have anything like its usual meaning. - I'm still learning AG, so I am open to feedback on this answer: If we think about it in category theoretic terms: The union of closed subschemes should be a coproduct in the category of closed subschemes of $\operatorname{Spec} R$ (with closed immersions). Thus, in the affine case, you need the product in the category of ideals of $R$ (with inclusions). The product in this category is clearly intersection of ideals, not product of ideals. - In an ordered set, viewed as a category, the product (resp. coproduct) of two objects is their "inf" (resp. "sup"). Check! You are probably misled by the terminology "product of ideals". – Laurent Moret-Bailly Nov 21 2011 at 19:50 Emerton explained well that $V(I\cap J)$ is the natural scheme structure; I'd just like to add that if $V(I)$ and $V(J)$ are divisors, it is also reasonable to use the scheme structure $V(IJ)$ on their union, i.e. their sum as divisors. So both versions have their merits. - If I and J are "coprime" (i.e., if I+J = R) then it's true that the I∩J=IJ, so the two definitions are the same. This is true for example if I and J are distinct primes. To illustrate the kind of thing that happens when there are common factors, consider this silly example: R = ℝ[x,y], I = J = (x). So the union of the two varieties of these ideals is again the y-axis, and the question you need to ask is what do you want as the ring of functions on the union--do you want R/I∩J = ℝ[x,y]/(x) or do you want R/IJ = ℝ[x,y]/(x2). the issue is that in the world of algebraic geometry, this choice matters, since an algebraic variety is not only defined by its points, but also it's ring of functions. it seems reasonable that you would want X ∪ X = X for any variety X, so you'd better choose the first choice. as a side note, the second choice is an example of a "nonreduced" variety, which i guess you wouldn't usually(?) see in a first course on algebraic geometry. it has more functions on it: one "full" dimension plus one "infinitesimal" dimension (i.e., with only linear functions in that direction). - 1 @maxmoo : It is not true that distinct primes are "coprime": take $I=(x)$ and $J=(y)$ in $k[x,y]$. These are distinct prime ideals, but they are definitely not "coprime". – Georges Elencwajg Mar 21 2010 at 16:16 hmmm yes that's a good point, better change "This is true" to "It's also true"? – Max Flander Mar 21 2010 at 16:35 1 Still not true: if $\mathfrak{p} \subset \mathfrak{q}$, the intersection and product are different. But even if you rule that out, it's no good: $R=k[x,y,z]/(xz-y^2)$, $\mathfrak{p}=(x,y)$, $\mathfrak{q}=(y,z)$. Then the product is $(xy,xz,yz)$ and the intersection is $(y)$. – Graham Leuschke Mar 21 2010 at 18:39 Relevant in comparing $IJ$ and $I \cap J$ is Atiyah-MacDonald, exer. 1.13, (iii), at the top of p. 9. It asserts that if $I, J$ are prime then $$IJ \subset rad(IJ) = I \cap J.$$ (Note: prime ideals are radical ideals, as are the intersections of radical ideals.) Two instructive examples (1): $V(y) \cup V(y)$ as mentioned earlier, where $I = J$. Who wants that the coordinate ring of the x-axis to be to be $K[x,y]/(y^2)$? (2): the variety $V$ which is the union of the $z$-axis ($V_1$) and the $xy$-plane ($V_2$). $I = (x,y)$ and $J = z$ are the corresponding prime ideals. $I \cap J = I J = (zx, zy)$ is the ideal of polynomial functions which vanishes on $V$ corresponding to the decomposition of $V$ into $V_1$ and $V_2$ and the decomposition of $IJ$ into primes. -	{"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": 75, "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": 0.9547663331031799, "perplexity_flag": "head"}
http://en.wikipedia.org/wiki/Denominator	# Fraction (mathematics) (Redirected from Denominator) A cake with one fourth (a quarter) removed. The remaining three fourths are shown. Dotted lines indicate where the cake may be cut in order to divide it into equal parts. Each fourth of the cake is denoted by the fraction ¼. A fraction (from Latin: fractus, "broken") represents a part of a whole or, more generally, any number of equal parts. When spoken in everyday English, a fraction describes how many parts of a certain size there are, for example, one-half, eight-fifths, three-quarters. A common, vulgar, or simple fraction (for example $\tfrac{1}{2}$, $\tfrac{-8}{5}$, and 3/17) consists of an integer numerator, displayed above a line (or before a slash), and a non-zero integer denominator, displayed below (or after) that line. The numerator represents a number of equal parts and the denominator indicates how many of those parts make up a whole. For example, in the fraction 3/4, the numerator, 3, tells us that the fraction represents 3 equal parts, and the denominator, 4, tells us that 4 parts make up a whole. The picture to the right illustrates $\tfrac{3}{4}$ or 3/4 of a cake. Numerators and denominators are also used in fractions that are not simple, including compound fractions, complex fractions, and mixed numerals. Fractional numbers can also be written without using explicit numerators or denominators, by using decimals, percent signs, or negative exponents (as in 0.01, 1%, and 10−2 respectively, all of which are equivalent to 1/100). An integer such as the number 7 can be thought of as having an implied denominator of one: 7 equals 7/1. Other uses for fractions are to represent ratios and to represent division.[1] Thus the fraction 3/4 is also used to represent the ratio 3:4 (the ratio of the part to the whole) and the division 3 ÷ 4 (three divided by four). In mathematics the set of all numbers which can be expressed in the form a/b, where a and b are integers and b is not zero, is called the set of rational numbers and is represented by the symbol Q, which stands for quotient. The test for a number being a rational number is that it can be written in that form (i.e., as a common fraction). However, the word fraction is also used to describe mathematical expressions that are not rational numbers, for example algebraic fractions (quotients of algebraic expressions), and expressions that contain irrational numbers, such as √2/2 (see square root of 2) and π/4 (see proof that π is irrational). ## Forms of fractions ### Common, vulgar, or simple fractions A common fraction (also known as a vulgar fraction or simple fraction) is a rational number written as a/b or $\tfrac{a}{b}$, where the integers a and b are called the numerator and the denominator, respectively.[2] The numerator represents a number of equal parts, and the denominator, which cannot be zero, indicates how many of those parts make up a unit or a whole. In the examples 2/5 and 7/3, the slanting line is called a solidus or forward slash. In the examples $\tfrac{2}{5}$ and $\tfrac{7}{3}$, the horizontal line is called a vinculum or, informally, a "fraction bar". #### Writing simple fractions In computer displays and typography, simple fractions are sometimes printed as a single character, e.g. ½ (one half). See the article on Number Forms for information on doing this in Unicode. Scientific publishing distinguishes four ways to set fractions, together with guidelines on use:[3] • special fractions: fractions that are presented as a single character with a slanted bar, with roughly the same height and width as other characters in the text. Generally used for simple fractions, such as: ½, ⅓, ⅔, ¼, and ¾. Since the numerals are smaller, legibility can be an issue, especially for small-sized fonts. These are not used in modern mathematical notation, but in other contexts; • case fractions: similar to special fractions, but with a horizontal bar, thus making them upright. An example would be $\tfrac{1}{2}$, but rendered with the same height as other characters; • shilling fractions: 1/2, so called because this notation was used for pre-decimal British currency (£sd), as in 2/6 for a half crown, meaning two shillings and six pence. While the notation "two shillings and six pence" did not represent a fraction, the forward slash is now used in fractions, especially for fractions inline with prose (rather than displayed), to avoid uneven lines. It is also used for fractions within fractions (complex fractions) or within exponents to increase legibility; • built-up fractions: $\frac{1}{2}$. This notation uses two or more lines of ordinary text, and results in a variation in spacing between lines when included within other text. While large and legible, these can be disruptive, particularly for simple fractions or within complex fractions. ### Ratios A ratio is a relationship between two or more numbers that can be sometimes expressed as a fraction. Typically, a number of items are grouped and compared in a ratio, specifying numerically the relationship between each group. Ratios are expressed as "group 1 to group 2 ... to group n". For example, if a car lot had 12 vehicles of which • 2 are white, • 6 are red, • 4 are yellow The ratio of red to white to yellow cars is 6 to 2 to 4. The ratio of yellow cars to white cars is 4 to 2 and may be expressed as 4:2 or 2:1. A ratio may be typically converted to a fraction when it is expressed as a ratio to the whole. In the above example, the ratio of yellow cars to the total cars in the lot is 4:12 or 1:3. We can convert these ratios to a fraction and say that 4/12 of the cars or 1/3 of the cars in the lot are yellow. Therefore, if a person randomly chose one car on the lot, then there is a one in three chance or probability that it would be yellow. ### Proper and improper common fractions Common fractions can be classified as either proper or improper. When the numerator and the denominator are both positive, the fraction is called proper if the numerator is less than the denominator, and improper otherwise.[4][5] In general, a common fraction is said to be a proper fraction if the absolute value of the fraction is strictly less than one—that is, if the fraction is between -1 and 1 (but not equal to -1 or 1).[6][7] It is said to be an improper fraction (U.S., British or Australian) or top-heavy fraction (British, occasionally North America) if the absolute value of the fraction is greater than or equal to 1. Examples of proper fractions are 2/3, -3/4, and 4/9; examples of improper fractions are 9/4, -4/3, and 8/3. ### Mixed numbers A mixed numeral (often called a mixed number, also called a mixed fraction) is the sum of a non-zero integer and a proper fraction. This sum is implied without the use of any visible operator such as "+". For example, in referring to two entire cakes and three quarters of another cake, the whole and fractional parts of the number are written next to each other: $2+\tfrac{3}{4}=2\tfrac{3}{4}$. This is not to be confused with the algebra rule of implied multiplication. When two algebraic expressions are written next to each other, the operation of multiplication is said to be "understood". In algebra, $a \tfrac{b}{c}$ for example is not a mixed number. Instead, multiplication is understood where $a \tfrac{b}{c} = a \times \tfrac{b}{c}$. To avoid confusion, the mutiplication is often explicitly expressed. So $a \tfrac{b}{c}$ may be written as $a \times \tfrac{b}{c}$, $a \cdot \tfrac{b}{c}$, or $a (\tfrac{b}{c})$. An improper fraction is another way to write a whole plus a part. A mixed number can be converted to an improper fraction as follows: 1. Write the mixed number $2\tfrac{3}{4}$ as a sum $2+\tfrac{3}{4}$. 2. Convert the whole number to an improper fraction with the same denominator as the fractional part, $2=\tfrac{8}{4}$. 3. Add the fractions. The resulting sum is the improper fraction. In the example, $2\tfrac{3}{4}=\tfrac{8}{4}+\tfrac{3}{4}=\tfrac{11}{4}$. Similarly, an improper fraction can be converted to a mixed number as follows: 1. Divide the numerator by the denominator. In the example, $\tfrac{11}{4}$, divide 11 by 4. 11 ÷ 4 = 2 with remainder 3. 2. The quotient (without the remainder) becomes the whole number part of the mixed number. The remainder becomes the numerator of the fractional part. In the example, 2 is the whole number part and 3 is the numerator of the fractional part. 3. The new denominator is the same as the denominator of the improper fraction. In the example, they are both 4. Thus $\tfrac{11}{4} =2\tfrac{3}{4}$. Mixed numbers can also be negative, as in $-2\tfrac{3}{4}$, which equals $-(2+\tfrac{3}{4}) = -2-\tfrac{3}{4}$. ### Reciprocals and the "invisible denominator" The reciprocal of a fraction is another fraction with the numerator and denominator reversed. The reciprocal of $\tfrac{3}{7}$, for instance, is $\tfrac{7}{3}$. The product of a fraction and its reciprocal is 1, hence the reciprocal is the multiplicative inverse of a fraction. Any integer can be written as a fraction with the number one as denominator. For example, 17 can be written as $\tfrac{17}{1}$, where 1 is sometimes referred to as the invisible denominator. Therefore, every fraction or integer except for zero has a reciprocal. The reciprocal of 17 is $\tfrac{1}{17}$. ### Complex fractions In a complex fraction, either the numerator, or the denominator, or both, is a fraction or a mixed number,[8][9] corresponding to division of fractions. For example, $\frac{\tfrac{1}{2}}{\tfrac{1}{3}}$ and $\frac{12\tfrac{3}{4}}{26}$ are complex fractions. To reduce a complex fraction to a simple fraction, treat the longest fraction line as representing division. For example: $\frac{\tfrac{1}{2}}{\tfrac{1}{3}}=\tfrac{1}{2}\times\tfrac{3}{1}=\tfrac{3}{2}=1\tfrac{1}{2}.$ $\frac{12\tfrac{3}{4}}{26} = 12\tfrac{3}{4} \cdot \tfrac{1}{26} = \tfrac{12 \cdot 4 + 3}{4} \cdot \tfrac{1}{26} = \tfrac{51}{4} \cdot \tfrac{1}{26} = \tfrac{51}{104}$ $\frac{\tfrac{3}{2}}5=\tfrac{3}{2}\times\tfrac{1}{5}=\tfrac{3}{10}.$ $\frac{8}{\tfrac{1}{3}}=8\times\tfrac{3}{1}=24.$ If, in a complex fraction, there is no clear way to tell which fraction line takes precedence, then the expression is improperly formed, and meaningless. ### Compound fractions A compound fraction is a fraction of a fraction, or any number of fractions connected with the word of,[8][9] corresponding to multiplication of fractions. To reduce a compound fraction to a simple fraction, just carry out the multiplication (see the section on multiplication). For example, $\tfrac{3}{4}$ of $\tfrac{5}{7}$ is a compound fraction, corresponding to $\tfrac{3}{4} \times \tfrac{5}{7} = \tfrac{15}{28}$. The terms compound fraction and complex fraction are closely related and sometimes one is used as a synonym for the other. ### Decimal fractions and percentages A decimal fraction is a fraction whose denominator is not given explicitly, but is understood to be an integer power of ten. Decimal fractions are commonly expressed using decimal notation in which the implied denominator is determined by the number of digits to the right of a decimal separator, the appearance of which (e.g., a period, a raised period (•), a comma) depends on the locale (for examples, see decimal separator). Thus for 0.75 the numerator is 75 and the implied denominator is 10 to the second power, viz. 100, because there are two digits to the right of the decimal separator. In decimal numbers greater than 1 (such as 3.75), the fractional part of the number is expressed by the digits to the right of the decimal (with a value of 0.75 in this case). 3.75 can be written either as an improper fraction, 375/100, or as a mixed number, $3\tfrac{75}{100}$. Decimal fractions can also be expressed using scientific notation with negative exponents, such as 6.023×10−7, which represents 0.0000006023. The 10−7 represents a denominator of 107. Dividing by 107 moves the decimal point 7 places to the left. Decimal fractions with infinitely many digits to the right of the decimal separator represent an infinite series. For example, 1/3 = 0.333... represents the infinite series 3/10 + 3/100 + 3/1000 + ... . Another kind of fraction is the percentage (Latin per centum meaning "per hundred", represented by the symbol %), in which the implied denominator is always 100. Thus, 51% means 51/100. Percentages greater than 100 or less than zero are treated in the same way, e.g. 311% equals 311/100, and -27% equals -27/100. The related concept of permille or parts per thousand has an implied denominator of 1000, while the more general parts-per notation, as in 75 parts per million, means that the proportion is 75/1,000,000. Whether common fractions or decimal fractions are used is often a matter of taste and context. Common fractions are used most often when the denominator is relatively small. By mental calculation, it is easier to multiply 16 by 3/16 than to do the same calculation using the fraction's decimal equivalent (0.1875). And it is more accurate to multiply 15 by 1/3, for example, than it is to multiply 15 by any decimal approximation of one third. Monetary values are commonly expressed as decimal fractions, for example \$3.75. However, as noted above, in pre-decimal British currency, shillings and pence were often given the form (but not the meaning) of a fraction, as, for example 3/6 (read "three and six") meaning 3 shillings and 6 pence, and having no relationship to the fraction 3/6. ### Special cases • A unit fraction is a vulgar fraction with a numerator of 1, e.g. $\tfrac{1}{7}$. Unit fractions can also be expressed using negative exponents, as in 2−1 which represents 1/2, and 2−2 which represents 1/(22) or 1/4. • An Egyptian fraction is the sum of distinct positive unit fractions, for example $\tfrac{1}{2}+\tfrac{1}{3}$. This definition derives from the fact that the ancient Egyptians expressed all fractions except $\tfrac{1}{2}$, $\tfrac{2}{3}$ and $\tfrac{3}{4}$ in this manner. Every positive rational number can be expanded as an Egyptian fraction. For example, $\tfrac{5}{7}$ can be written as $\tfrac{1}{2} + \tfrac{1}{6} + \tfrac{1}{21}.$ Any positive rational number can be written as a sum of unit fractions in infinitely many ways. Two ways to write $\tfrac{13}{17}$ are $\tfrac{1}{2}+\tfrac{1}{4}+\tfrac{1}{68}$ and $\tfrac{1}{3}+\tfrac{1}{4}+\tfrac{1}{6}+\tfrac{1}{68}$. • A dyadic fraction is a vulgar fraction in which the denominator is a power of two, e.g. $\tfrac{1}{8}$. ## Arithmetic with fractions Like whole numbers, fractions obey the commutative, associative, and distributive laws, and the rule against division by zero. ### Equivalent fractions Multiplying the numerator and denominator of a fraction by the same (non-zero) number results in a fraction that is equivalent to the original fraction. This is true because for any non-zero number $n$, the fraction $\tfrac{n}{n} = 1$. Therefore, multiplying by $\tfrac{n}{n}$ is equivalent to multiplying by one, and any number multiplied by one has the same value as the original number. By way of an example, start with the fraction $\tfrac{1}{2}$. When the numerator and denominator are both multiplied by 2, the result is $\tfrac{2}{4}$, which has the same value (0.5) as $\tfrac{1}{2}$. To picture this visually, imagine cutting a cake into four pieces; two of the pieces together ($\tfrac{2}{4}$) make up half the cake ($\tfrac{1}{2}$). Dividing the numerator and denominator of a fraction by the same non-zero number will also yield an equivalent fraction. This is called reducing or simplifying the fraction. A simple fraction in which the numerator and denominator are coprime (that is, the only positive integer that goes into both the numerator and denominator evenly is 1) is said to be irreducible, in lowest terms, or in simplest terms. For example, $\tfrac{3}{9}$ is not in lowest terms because both 3 and 9 can be exactly divided by 3. In contrast, $\tfrac{3}{8}$ is in lowest terms—the only positive integer that goes into both 3 and 8 evenly is 1. Using these rules, we can show that $\tfrac{5}{10}$ = $\tfrac{1}{2}$ = $\tfrac{10}{20}$ = $\tfrac{50}{100}$. A common fraction can be reduced to lowest terms by dividing both the numerator and denominator by their greatest common divisor. For example, as the greatest common divisor of 63 and 462 is 21, the fraction $\tfrac{63}{462}$ can be reduced to lowest terms by dividing the numerator and denominator by 21: $\tfrac{63}{462} = \tfrac{63 \div 21}{462 \div 21}= \tfrac{3}{22}$ The Euclidean algorithm gives a method for finding the greatest common divisor of any two positive integers. ### Comparing fractions Comparing fractions with the same denominator only requires comparing the numerators. $\tfrac{3}{4}>\tfrac{2}{4}$ because 3>2. If two positive fractions have the same numerator, then the fraction with the smaller denominator is the larger number. When a whole is divided into equal pieces, if fewer equal pieces are needed to make up the whole, then each piece must be larger. When two positive fractions have the same numerator, they represent the same number of parts, but in the fraction with the smaller denominator, the parts are larger. One way to compare fractions with different numerators and denominators is to find a common denominator. To compare $\tfrac{a}{b}$ and $\tfrac{c}{d}$, these are converted to $\tfrac{ad}{bd}$ and $\tfrac{bc}{bd}$. Then bd is a common denominator and the numerators ad and bc can be compared. $\tfrac{2}{3}$ ? $\tfrac{1}{2}$ gives $\tfrac{4}{6}>\tfrac{3}{6}$ It is not necessary to determine the value of the common denominator to compare fractions. This short cut is known as "cross multiplying" – you can just compare ad and bc, without computing the denominator. $\tfrac{5}{18}$ ? $\tfrac{4}{17}$ Multiply top and bottom of each fraction by the denominator of the other fraction, to get a common denominator: $\tfrac{5 \times 17}{18 \times 17}$ ? $\tfrac{4 \times 18}{17 \times 18}$ The denominators are now the same, but it is not necessary to calculate their value – only the numerators need to be compared. Since 5×17 (= 85) is greater than 4×18 (= 72), $\tfrac{5}{18}>\tfrac{4}{17}$. Also note that every negative number, including negative fractions, is less than zero, and every positive number, including positive fractions, is greater than zero, so every negative fraction is less than any positive fraction. ### Addition The first rule of addition is that only like quantities can be added; for example, various quantities of quarters. Unlike quantities, such as adding thirds to quarters, must first be converted to like quantities as described below: Imagine a pocket containing two quarters, and another pocket containing three quarters; in total, there are five quarters. Since four quarters is equivalent to one (dollar), this can be represented as follows: $\tfrac24+\tfrac34=\tfrac54=1\tfrac14$. If $\tfrac12$ of a cake is to be added to $\tfrac14$ of a cake, the pieces need to be converted into comparable quantities, such as cake-eighths or cake-quarters. #### Adding unlike quantities To add fractions containing unlike quantities (e.g. quarters and thirds), it is necessary to convert all amounts to like quantities. It is easy to work out the chosen type of fraction to convert to; simply multiply together the two denominators (bottom number) of each fraction. For adding quarters to thirds, both types of fraction are converted to twelfths, thus: $\tfrac14\ + \tfrac13=\tfrac{13}{43}\ + \tfrac{14}{34}=\tfrac3{12}\ + \tfrac4{12}=\tfrac7{12}$. Consider adding the following two quantities: $\tfrac35+\tfrac23$ First, convert $\tfrac35$ into fifteenths by multiplying both the numerator and denominator by three: $\tfrac35\times\tfrac33=\tfrac9{15}$. Since $\tfrac33$ equals 1, multiplication by $\tfrac33$ does not change the value of the fraction. Second, convert $\tfrac23$ into fifteenths by multiplying both the numerator and denominator by five: $\tfrac23\times\tfrac55=\tfrac{10}{15}$. Now it can be seen that: $\tfrac35+\tfrac23$ is equivalent to: $\tfrac9{15}+\tfrac{10}{15}=\tfrac{19}{15}=1\tfrac4{15}$ This method can be expressed algebraically: $\tfrac{a}{b} + \tfrac {c}{d} = \tfrac{ad+cb}{bd}$ And for expressions consisting of the addition of three fractions: $\tfrac{a}{b} + \tfrac {c}{d} + \tfrac{e}{f} = \tfrac{a(df)+c(bf)+e(bd)}{bdf}$ This method always works, but sometimes there is a smaller denominator that can be used (a least common denominator). For example, to add $\tfrac{3}{4}$ and $\tfrac{5}{12}$ the denominator 48 can be used (the product of 4 and 12), but the smaller denominator 12 may also be used, being the least common multiple of 4 and 12. $\tfrac34+\tfrac{5}{12}=\tfrac{9}{12}+\tfrac{5}{12}=\tfrac{14}{12}=\tfrac76=1\tfrac16$ ### Subtraction The process for subtracting fractions is, in essence, the same as that of adding them: find a common denominator, and change each fraction to an equivalent fraction with the chosen common denominator. The resulting fraction will have that denominator, and its numerator will be the result of subtracting the numerators of the original fractions. For instance, $\tfrac23-\tfrac12=\tfrac46-\tfrac36=\tfrac16$ ### Multiplication #### Multiplying a fraction by another fraction To multiply fractions, multiply the numerators and multiply the denominators. Thus: $\tfrac{2}{3} \times \tfrac{3}{4} = \tfrac{6}{12}$ Why does this work? First, consider one third of one quarter. Using the example of a cake, if three small slices of equal size make up a quarter, and four quarters make up a whole, twelve of these small, equal slices make up a whole. Therefore a third of a quarter is a twelfth. Now consider the numerators. The first fraction, two thirds, is twice as large as one third. Since one third of a quarter is one twelfth, two thirds of a quarter is two twelfth. The second fraction, three quarters, is three times as large as one quarter, so two thirds of three quarters is three times as large as two thirds of one quarter. Thus two thirds times three quarters is six twelfths. A short cut for multiplying fractions is called "cancellation". In effect, we reduce the answer to lowest terms during multiplication. For example: $\tfrac{2}{3} \times \tfrac{3}{4} = \tfrac{\cancel{2} ^{~1}}{\cancel{3} ^{~1}} \times \tfrac{\cancel{3} ^{~1}}{\cancel{4} ^{~2}} = \tfrac{1}{1} \times \tfrac{1}{2} = \tfrac{1}{2}$ A two is a common factor in both the numerator of the left fraction and the denominator of the right and is divided out of both. Three is a common factor of the left denominator and right numerator and is divided out of both. #### Multiplying a fraction by a whole number Place the whole number over one and multiply. $6 \times \tfrac{3}{4} = \tfrac{6}{1} \times \tfrac{3}{4} = \tfrac{18}{4}$ This method works because the fraction 6/1 means six equal parts, each one of which is a whole. #### Mixed numbers When multiplying mixed numbers, it's best to convert the mixed number into an improper fraction. For example: $3 \times 2\tfrac{3}{4} = 3 \times \left (\tfrac{8}{4} + \tfrac{3}{4} \right ) = 3 \times \tfrac{11}{4} = \tfrac{33}{4} = 8\tfrac{1}{4}$ In other words, $2\tfrac{3}{4}$ is the same as $\tfrac{8}{4} + \tfrac{3}{4}$, making 11 quarters in total (because 2 cakes, each split into quarters makes 8 quarters total) and 33 quarters is $8\tfrac{1}{4}$, since 8 cakes, each made of quarters, is 32 quarters in total. ### Division To divide a fraction by a whole number, you may either divide the numerator by the number, if it goes evenly into the numerator, or multiply the denominator by the number. For example, $\tfrac{10}{3} \div 5$ equals $\tfrac{2}{3}$ and also equals $\tfrac{10}{3 \cdot 5} = \tfrac{10}{15}$, which reduces to $\tfrac{2}{3}$. To divide a number by a fraction, multiply that number by the reciprocal of that fraction. Thus, $\tfrac{1}{2} \div \tfrac{3}{4} = \tfrac{1}{2} \times \tfrac{4}{3} = \tfrac{1 \cdot 4}{2 \cdot 3} = \tfrac{2}{3}$. ### Converting between decimals and fractions To change a common fraction to a decimal, divide the denominator into the numerator. Round the answer to the desired accuracy. For example, to change 1/4 to a decimal, divide 4 into 1.00, to obtain 0.25. To change 1/3 to a decimal, divide 3 into 1.0000..., and stop when the desired accuracy is obtained. Note that 1/4 can be written exactly with two decimal digits, while 1/3 cannot be written exactly with any finite number of decimal digits. To change a decimal to a fraction, write in the denominator a 1 followed by as many zeroes as there are digits to the right of the decimal point, and write in the numerator all the digits in the original decimal, omitting the decimal point. Thus 12.3456 = 123456/10000. #### Converting repeating decimals to fractions See also: Repeating decimal Decimal numbers, while arguably more useful to work with when performing calculations, sometimes lack the precision that common fractions have. Sometimes an infinite number of repeating decimals is required to convey the same kind of precision. Thus, it is often useful to convert repeating decimals into fractions. The preferred way to indicate a repeating decimal is to place a bar over the digits that repeat, for example 0.789 = 0.789789789… For repeating patterns where the repeating pattern begins immediately after the decimal point, a simple division of the pattern by the same number of nines as numbers it has will suffice. For example: 0.5 = 5/9 0.62 = 62/99 0.264 = 264/999 0.6291 = 6291/9999 In case leading zeros precede the pattern, the nines are suffixed by the same number of trailing zeros: 0.05 = 5/90 0.000392 = 392/999000 0.0012 = 12/9900 In case a non-repeating set of decimals precede the pattern (such as 0.1523987), we can write it as the sum of the non-repeating and repeating parts, respectively: 0.1523 + 0.0000987 Then, convert the repeating part to a fraction: 0.1523 + 987/9990000 ## Fractions in abstract mathematics In addition to being of great practical importance, fractions are also studied by mathematicians, who check that the rules for fractions given above are consistent and reliable. Mathematicians define a fraction as an ordered pair (a, b) of integers a and b ≠ 0, for which the operations addition, subtraction, multiplication, and division are defined as follows:[10] $(a,b) + (c,d) = (ad+bc,bd) \,$ $(a,b) - (c,d) = (ad-bc,bd) \,$ $(a,b) \cdot (c,d) = (ac,bd)$ $(a,b) \div (c,d) = (ad,bc)$ (when c ≠ 0) In addition, an equivalence relation is specified as follows: $(a, b)$ ~ $(c, d)$ if and only if $ad=bc$. These definitions agree in every case with the definitions given above; only the notation is different. More generally, a and b may be elements of any integral domain R, in which case a fraction is an element of the field of fractions of R. For example, when a and b are polynomials in one indeterminate, the field of fractions is the field of rational fractions (also known as the field of rational functions). When a and b are integers, the field of fractions is the field of rational numbers. ## Algebraic fractions Main article: Algebraic fraction An algebraic fraction is the indicated quotient of two algebraic expressions. Two examples of algebraic fractions are $\frac{3x}{x^2+2x-3}$ and $\frac{\sqrt{x+2}}{x^2-3}$. Algebraic fractions are subject to the same laws as arithmetic fractions. If the numerator and the denominator are polynomials, as in $\frac{3x}{x^2+2x-3}$, the algebraic fraction is called a rational fraction (or rational expression). An irrational fraction is one that contains the variable under a fractional exponent or root, as in $\frac{\sqrt{x+2}}{x^2-3}$. The terminology used to describe algebraic fractions is similar to that used for ordinary fractions. For example, an algebraic fraction is in lowest terms if the only factors common to the numerator and the denominator are 1 and –1. An algebraic fraction whose numerator or denominator, or both, contain a fraction, such as $\frac{1 + \tfrac{1}{x}}{1 - \tfrac{1}{x}}$, is called a complex fraction. Rational numbers are the quotient field of integers. Rational expressions are the quotient field of the polynomials (over some integral domain). Since a coefficient is a polynomial of degree zero, a radical expression such as √2/2 is a rational fraction. Another example (over the reals) is $\textstyle{\tfrac{\pi}{2}}$, the radian measure of a right angle. The term partial fraction is used when decomposing rational expressions into sums. The goal is to write the rational expression as the sum of other rational expressions with denominators of lesser degree. For example, the rational expression $\textstyle{2x \over x^2-1}$ can be rewritten as the sum of two fractions: $\textstyle{1 \over x+1}$ + $\textstyle{1 \over x-1}$. This is useful in many areas such as integral calculus and differential equations. ## Radical expressions Main articles: Nth root and Rationalization (mathematics) A fraction may also contain radicals in the numerator and/or the denominator. If the denominator contains radicals, it can be helpful to rationalize it (compare Simplified form of a radical expression), especially if further operations, such as adding or comparing that fraction to another, are to be carried out. It is also more convenient if division is to be done manually. When the denominator is a monomial square root, it can be rationalized by multiplying both the top and the bottom of the fraction by the denominator: $\frac{3}{\sqrt{7}} = \frac{3}{\sqrt{7}} \cdot \frac{\sqrt{7}}{\sqrt{7}} = \frac{3\sqrt{7}}{7}$ The process of rationalization of binomial denominators involves multiplying the top and the bottom of a fraction by the conjugate of the denominator so that the denominator becomes a rational number. For example: $\frac{3}{3-2\sqrt{5}} = \frac{3}{3-2\sqrt{5}} \cdot \frac{3+2\sqrt{5}}{3+2\sqrt{5}} = \frac{3(3+2\sqrt{5})}{{3}^2 - (2\sqrt{5})^2} = \frac{ 3 (3 + 2\sqrt{5} ) }{ 9 - 20 } = - \frac{ 9+6 \sqrt{5} }{11}$ $\frac{3}{3+2\sqrt{5}} = \frac{3}{3+2\sqrt{5}} \cdot \frac{3-2\sqrt{5}}{3-2\sqrt{5}} = \frac{3(3-2\sqrt{5})}{{3}^2 - (2\sqrt{5})^2} = \frac{ 3 (3 - 2\sqrt{5} ) }{ 9 - 20 } = - \frac{ 9-6 \sqrt{5} }{11}$ Even if this process results in the numerator being irrational, like in the examples above, the process may still facilitate subsequent manipulations by reducing the number of irrationals one has to work with in the denominator. ## Pronunciation and spelling This section . Please improve it by verifying the claims made and adding inline citations. Statements consisting only of original research may be removed. (March 2012) Further information: Ordinal number (linguistics) When reading fractions, it is customary in English to pronounce the denominator using ordinal nomenclature, as in "fifths" for fractions with a 5 in the denominator. Thus, for 3/5, we would say "three-fifths" and for 5/32, we would say "five thirty-seconds". This generally applies to whole number denominators greater than 2, though large denominators that are not powers of ten are often read using the cardinal number. Therefore, 1/123 might be read "one one hundred twenty-third" but is often read "one over one hundred twenty-three". In contrast, because one million is a power of ten, 1/1,000,000 is commonly read "one-millionth" or "one one-millionth". The denominators 1, 2, and 4 are special cases. The fraction 3/1 may be read "three wholes". The fraction 3/2 is usually read "three-halves", but never "three seconds". The fraction 3/4 may be read either "three fourths" or "three-quarters". Furthermore, since most fractions are used grammatically as adjectives of a noun, the fractional modifier is hyphenated. This is evident in standard prose in which one might write about "every two-tenths of a mile", "the quarter-mile run", or the Three-Fifths Compromise. When the fraction's numerator is one, then the word "one" may be omitted, such as "every tenth of a second" or "during the final quarter of the year". \| \| \| Denominator \| in words \| Denominator \| in words \| Denominator \| in words \| \|------------------\|---------------\|-------------------------\|---------------\|---------------\|--------------\|---------------\|------------------\| \| 2 \| 3 \| 4 \| 5 \| 6 \| 7 \| 8 \| 9 \| \| half (or halves) \| third(s) \| fourth(s) or quarter(s) \| fifth(s) \| sixth(s) \| seventh(s) \| eighth(s) \| ninth(s) \| \| \| \| \| \| \| \| \| \| \| 10 \| 11 \| 12 \| 13 \| 14 \| 15 \| 16 \| 22 \| \| tenth(s) \| eleventh(s) \| twelfth(s) \| thirteenth(s) \| fourteenth(s) \| fifteenth(s) \| sixteenth(s) \| twenty-second(s) \| \| \| \| \| \| \| \| \| \| \| 100 \| 1,000 \| 1,000,000 \| \| \| \| \| \| \| hundredth(s) \| thousandth(s) \| millionth(s) \| \| \| \| \| \| ## History The earliest fractions were reciprocals of integers: ancient symbols representing one part of two, one part of three, one part of four, and so on.[11] The Egyptians used Egyptian fractions ca. 1000 BC. About 4,000 years ago Egyptians divided with fractions using slightly different methods. They used least common multiples with unit fractions. Their methods gave the same answer as modern methods.[12] The Egyptians also had a different notation for dyadic fractions in the Akhmim Wooden Tablet and several Rhind Mathematical Papyrus problems. The Greeks used unit fractions and later continued fractions and followers of the Greek philosopher Pythagoras, ca. 530 BC, discovered that the square root of two cannot be expressed as a fraction. In 150 BC Jain mathematicians in India wrote the "Sthananga Sutra", which contains work on the theory of numbers, arithmetical operations, operations with fractions. The method of putting one number below the other and computing fractions first appeared in Aryabhatta's work around 499 CE.[citation needed] In Sanskrit literature, fractions, or rational numbers were always expressed by an integer followed by a fraction. When the integer is written on a line, the fraction is placed below it and is itself written on two lines, the numerator called amsa part on the first line, the denominator called cheda “divisor” on the second below. If the fraction is written without any particular additional sign, one understands that it is added to the integer above it. If it is marked by a small circle or a cross (the shape of the “plus” sign in the West) placed on its right, one understands that it is subtracted from the integer. For example, Bhaskara I writes[13] ६ १ २ १ १ १० ४ ५ ९ That is, 6 1 2 1 1 1० 4 5 9 to denote 6+1/4, 1+1/5, and 2–1/9 Al-Hassār, a Muslim mathematician from Fez, Morocco specializing in Islamic inheritance jurisprudence during the 12th century, first mentions the use of a fractional bar, where numerators and denominators are separated by a horizontal bar. In his discussion he writes, "... for example, if you are told to write three-fifths and a third of a fifth, write thus, $\frac{3 \quad 1}{5 \quad 3}$."[14] This same fractional notation appears soon after in the work of Leonardo Fibonacci in the 13th century.[15] In discussing the origins of decimal fractions, Dirk Jan Struik states:[16] "The introduction of decimal fractions as a common computational practice can be dated back to the Flemish pamphlet De Thiende, published at Leyden in 1585, together with a French translation, La Disme, by the Flemish mathematician Simon Stevin (1548-1620), then settled in the Northern Netherlands. It is true that decimal fractions were used by the Chinese many centuries before Stevin and that the Persian astronomer Al-Kāshī used both decimal and sexagesimal fractions with great ease in his Key to arithmetic (Samarkand, early fifteenth century).[17]" While the Persian mathematician Jamshīd al-Kāshī claimed to have discovered decimal fractions himself in the 15th century, J. Lennart Berggren notes that he was mistaken, as decimal fractions were first used five centuries before him by the Baghdadi mathematician Abu'l-Hasan al-Uqlidisi as early as the 10th century.[18][19] ## Pedagogical tools In primary schools, fractions have been demonstrated through Cuisenaire rods, fraction bars, fraction strips, fraction circles, paper (for folding or cutting), pattern blocks, pie-shaped pieces, plastic rectangles, grid paper, dot paper, geoboards, counters and computer software. ## References 1. H. Wu, The Mis-Education of Mathematics Teachers, Notices of the American Mathematical Society, Volume 58, Issue 03 (March 2011), page 374 2. Galen, Leslie Blackwell (March 2004), "Putting Fractions in Their Place", 111 (3) 3. ^ a b 4. ^ a b 5. "Fraction - Encyclopedia of Mathematics". Encyclopediaofmath.org. 2012-04-06. Retrieved 2012-08-15. 6. Eves, Howard ; with cultural connections by Jamie H. (1990). An introduction to the history of mathematics (6th ed. ed.). Philadelphia: Saunders College Pub. ISBN 0-03-029558-0. 7. Milo Gardner (December 19, 2005). "Math History". Retrieved 2006-01-18. See for examples and an explanation. 8. 9. A Source Book in Mathematics 1200-1800. New Jersey: Princeton University Press. 1986. ISBN 0-691-02397-2. 10. Die Rechenkunst bei Ğamšīd b. Mas'ūd al-Kāšī. Wiesbaden: Steiner. 1951. 11. Berggren, J. Lennart (2007). "Mathematics in Medieval Islam". The Mathematics of Egypt, Mesopotamia, China, India, and Islam: A Sourcebook. Princeton University Press. p. 518. ISBN 978-0-691-11485-9.	{"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": 128, "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": 0.9176041483879089, "perplexity_flag": "middle"}
http://mathoverflow.net/revisions/45577/list	## Return to Question 2 edited title; added 174 characters in body # CocompleteAnycocomplete category (s.t....)withadensesmallfullsubcategory is complete? Hello, I with to consider the following statement: If $C$ is a cocomplete category having a dense small full subcategory $D$, then $C$ is complete. (a full subcategory $D$ is dense in $C$ if every element of $C$ is canonical colimit of elements of $D$...) I think I know how to prove it (I give proof below), and I want someone to reassure me that this statement is true exactly as stated, as it seems a little bit surprising. Proof sketch: Consider the functor $Y : C \to psh(D)$, where $psh(D)$ denotes the category of presheaves on $D$ (Yoneda functor, i.e. $Y(X)(A) = Hom (A, X)$). $D$ being dense in $C$ is equivalent to this functor being fully faithful. Futhermore, we have a functor $F: psh(D) \to C$, namely, the one which extends the inclusion $D \to C$ by cocontinuity (as presheaf categories have the property of being free cocompletions: http://ncatlab.org/nlab/show/free+cocompletion). Then one can see that $Y$ is right adjoint to $F$. So this renders $C$ as reflective subcategory of $psh(D)$. psh(D)$( http://ncatlab.org/nlab/show/reflective+subcategory). Now,$psh(D)$is complete, and so every reflective subcategory of it. hence,$C\$ is complete. Thank you, Sasha 1 # Cocomplete category (s.t. ...) is complete Hello, I with to consider the following statement: If $C$ is a cocomplete category having a dense small full subcategory $D$, then $C$ is complete. (a full subcategory $D$ is dense in $C$ if every element of $C$ is canonical colimit of elements of $D$...) I think I know how to prove it (I give proof below), and I want someone to reassure me that this statement is true exactly as stated, as it seems a little bit surprising. Proof sketch: Consider the functor $Y : C \to psh(D)$, where $psh(D)$ denotes the category of presheaves on $D$ (Yoneda functor, i.e. $Y(X)(A) = Hom (A, X)$). $D$ being dense in $C$ is equivalent to this functor being fully faithful. Futhermore, we have a functor $F: psh(D) \to C$, namely, the one which extends the inclusion $D \to C$ by cocontinuity. Then one can see that $Y$ is right adjoint to $F$. So this renders $C$ as reflective subcategory of $psh(D)$. Now, $psh(D)$ is complete, and so every reflective subcategory of it. hence, $C$ is complete. Thank you, Sasha	{"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": 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": 0.9363947510719299, "perplexity_flag": "head"}
http://mathoverflow.net/questions/110338/octic-k3s-inside-cubic-4-folds/110341	## octic K3s inside cubic 4-folds ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) From the Thesis of B.Hassett I seem to understand that a smooth cubic 4-fold $X$ containing a $\mathbb{P}^2$ should contain also a octic K3, but I cannot see a natural way by which this K3 octic could appear inside $X$. How do you see that? One wild guess is: via linkage of a quartic 3-fold that contains a quadric surface contained in $X$... but this doesn't seem very consistent... - ## 1 Answer The $\mathbb{P}^{2}$ contained in your cubic fourfold $X$ is cut out by linear forms (say) $L_{1},L_{2},$ and $L_{3}.$ Since the homogeneous ideal of $X$ is contained in the homogeneous ideal generated by $L_{1},L_{2}$ and $L_{3},$ there exist quadrics $Q_{1},Q_{2},$ and $Q_{3}$ such that $X={L_{1}Q_{1}+L_{2}Q_{2}+L_{3}Q_{3}=0}.$ The octic K3 cut out by $Q_{1},Q_{2}$ and $Q_{3}$ is easily seen to lie in $X.$ - Cool. What's the relation between these quadrics and the quadric 3-folds obtained by projecting off $\mathbb{P}^2$? Are the 3-fold quadrics the restriction of the the 4-fold ones? – IMeasy Oct 22 at 15:37	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 18, "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": 0.9606870412826538, "perplexity_flag": "head"}
http://unapologetic.wordpress.com/2010/09/08/some-review/?like=1&_wpnonce=84af3d0b4c	# The Unapologetic Mathematician ## Some Review Before we push on into our new topic, let’s look back at some of the background that we’ve already covered. We’re talking about symmetric groups, which are, of course, groups. We have various ways of writing down an element of $S_n$, including the two-line notation and the cycle notation that are covered in our earlier description of the symmetric groups. As an example, the two-line notation $\displaystyle\left\lvert\begin{matrix}1&2&3&4\\2&1&4&3\end{matrix}\right\rvert$ and the cycle notation $(1\,2)(3\,4)$ both describe the permutation $\alpha\in S_4$ that sends $1$ to $2$, $2$ back to $1$, and similarly swaps $3$ and $4$. Similarly, the two-line notation the composition of $\displaystyle\left\lvert\begin{matrix}1&2&3&4\\4&2&1&3\end{matrix}\right\rvert$ and the cycle notation $(1\,4\,3)(2)$ or (equivalently) $(1\,4\,3)$ describe the permutation $\beta$ that cycles the elements $1$, $4$, and $3$ (in that order) and leaves $2$ untouched. We’re specifically concerned with complex representations of these groups. That is, we want to pick some complex vector space $V$, and for each permutation $\sigma\in S_n$ we want to come up with some linear transformation $\rho(\sigma):V\to V$ for which the composition of linear transformations and the composition of permutations are “the same” in the sense that given two permutations $\sigma$ and $\tau$, the transportation corresponding to the composite $\rho(\sigma\tau)$ is equal to the composite of the corresponding transformations $\rho(\sigma)\rho(\tau)$. We’re primarily interested in finite-dimensional representations. That is, ones for which $V$ is a finite-dimensional complex vector space. In this case, we know that we can always just assume that $V=\mathbb{C}^k$ — the space of $k$-tuples of complex numbers — and that linear transformations are described by matrices. Composition of transformations is reflected in matrix multiplication. That is, for every permutation $\sigma\in S_n$ we want to come up with an $k\times k$ matrix $\rho(\sigma)$ so that the matrix $\rho(\sigma\tau)$ corresponding to the composition of two permutations is the product $\rho(\sigma)\rho(\tau)$ of the matrices corresponding to the two permutations. I’ll be giving some more explicit examples soon. ## 5 Comments » 1. Minor quibble: I guess the ‘n’ in \$S_n\$ should be (possibly) different to the dimension of the representation? (Long-time reader: keep up the good work!) Comment by Doormat \| September 9, 2010 \| Reply • Good point. That could be confusing. Comment by \| September 9, 2010 \| Reply 2. [...] « Previous \| [...] Pingback by \| September 9, 2010 \| Reply 3. [...] Sample Representations As promised, we want to see some examples of matrix representations for those who might not have seen much of [...] Pingback by \| September 13, 2010 \| Reply 4. [...] –Por ejemplo ub blog que desconocia y que está hablando últimamente de la teoria de representaciones de grupos de permutaciones (que trata de como pasar del lenguaje de permutaciones al lenguaje de espacios vectoriales y transformaciones lineales (matrices). A cada permutación de un grupo dado se le asigna una transformación lineal de un espacio vectorial en si mismo): http://unapologetic.wordpress.com/2010/09/08/some-review/. [...] Pingback by \| September 29, 2010 \| Reply « Previous \| Next » ## About this weblog This is mainly an expository blath, with occasional high-level excursions, humorous observations, rants, and musings. The main-line exposition should be accessible to the “Generally Interested Lay Audience”, as long as you trace the links back towards the basics. Check the sidebar for specific topics (under “Categories”). I’m in the process of tweaking some aspects of the site to make it easier to refer back to older topics, so try to make the best of it for now. • ## Feedback Got something to say? Anonymous questions, comments, and suggestions at Formspring.me! %d bloggers like this:	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 33, "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": 0.8158494830131531, "perplexity_flag": "middle"}
http://mathoverflow.net/questions/67271/are-there-examples-of-nonconstructive-metaproofs/67279	## Are there examples of nonconstructive metaproofs? ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) This came up in a question on the xkcd forums. Is it possible to have a nonconstructive metaproof, i.e. a proof that there exists a proof in some formal system which does not construct said proof? Are there any known examples, preferably with some well-known formal system like PA? Conversely, is it possible to prove a meta-metatheorem saying that any metaproof can be used to find a proof? - 7 Sure. Metatheorem: any question I get on my homework has a proof. Metaproof: suppose otherwise. Then it would not be on my homework. – Qiaochu Yuan Jun 8 2011 at 16:01 Isn't there a story where a famous mathematician was assigned an open problem as homework by a devious professor, and ended up solving it? – David Diamondstone Jun 8 2011 at 16:03 5 @Qiaochu: you are very optimistic about the infallibility of your teacher. – Emil Jeřábek Jun 8 2011 at 16:03 Isn't there some question on how to show intuitionistically that Kripke models are complete for first-order intuitionistic (single-sorted) theories. Isn't the usual proof classical? – Andrej Bauer Jun 8 2011 at 16:34 1 Another example is completeness of Grothendieck toposes for higher-order intiotionistic logic. I think the proof is classical, and it's not clear how to make it intuitionistic. – Andrej Bauer Jun 8 2011 at 16:35 show 1 more comment ## 3 Answers In theory, David’s answer is correct. Nevertheless, in practice it is perfectly possible to prove the existence of a proof non-constructively (such as by manipulating models and then appealing to the completeness theorem) where no one has a clue how to actually find the proof. One example which springs to mind is Jacobson’s theorem: if $R$ is a ring such that for every $a\in R$ there exists an integer $n > 1$ such that $a=a^n$, then $R$ is commutative. By completeness of equational logic, this implies that for any $n > 1$, there exists an equational derivation of $xy=yx$ from the axioms of rings and $x^n=x$. Already finding such derivation for $n=3$ is a nontrivial exercise; explicit derivations are known for some $n$, but not in general. - Great answer. But I would still be interested in more examples, if anyone has some more. – David Diamondstone Jun 8 2011 at 19:19 2 Do you have your quantifiers in the right order in that statement? There are two closely related theorems of Jacobson that look like this and I'm not sure which one you mean. – Qiaochu Yuan Jun 8 2011 at 21:49 The quantifiers in the assumption of Jacobson’s theorem I have in mind are `$\forall a\in R\,\exists n > 1\,a^n=a$`. However, in order to express this in equational logic, the exponent has to be fixed, hence we only consider the corollary of the theorem with the stronger assumption `$\exists n\,\forall a\,a^n=a$` (more precisely, for each particular constant $n$, we consider the theorem with assumption `$\forall a\,a^n=a$`). – Emil Jeřábek Jun 9 2011 at 10:21 ### 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. If the proof system is recursively axiomatizable, this situation cannot occur. If there exists a proof of $\Theta$, there exists an algorithm to find that proof. Namely, search the recursively enumerable set of deductions until you find a proof of $\Theta$. This must terminate, as we have proved that $\Theta$ is provable. If the proof system is NOT recursive, then this may be possible. Consider the following set of axioms $\Sigma$ in the signature of arithmetic. Let $A$ be an infinite set which does not contain any infinite r.e. set. Define $$\Sigma = \{ (\bar k = \bar k) \wedge \sigma \mid k \in A, k > \ulcorner \sigma \urcorner, \mathfrak N \models \sigma \}$$ Now, note that any sentence provable in $Th(\cal N)$ is provable is in $\Sigma$. However, there is no algorithm to transform produce such proofs from $\Sigma$. To do so would require enumerating arbitrarily large elements of $A$, which is impossible - This can hardly be seen as a constructive proof, as the "constructive" part is entirely divorced from the "proof" part. – David Diamondstone Jun 8 2011 at 16:05 You are implicitly using Markov principle, I think. – Andrej Bauer Jun 8 2011 at 16:33 I don't know that this addresses the spirit of the question. This algorithm guarantees that it can find a proof but it does not guarantee that this will happen before I die or that the proof can be made short enough for me to finish reading it before I die, yet it is still possible that I can prove by other means that a proof exists in perhaps 10 minutes. – Qiaochu Yuan Jun 8 2011 at 16:47 1 There's a stronger form of this assertion. If the proof system is recursive, the statement "T proves $\phi$" is itself a $\Sigma_1$ statement, so if the non-constructive proof itself takes place in a "reasonable" system, it is possible to extract an explicit witness, which would be a direct proof. – Henry Towsner Jun 8 2011 at 17:58 What about examples from nonstandard analysis? By the transfer principle, given a proof of $\phi$ in nonstandard analysis, we know there exists a proof of $\phi$ using only standard techniques; but there is in general no nice way to extract the standard proof from the nonstandard proof, and if I recall correctly there are theorems which have a known nonstandard proof with no known standard proof. Would this count as a non-constructive metaproof? - 1 One can go from nonstandard proof to standard proof using ultrapowers. It is just that such a proof may not be as "elegant". – Gerald Edgar Jun 8 2011 at 16:28 If we're talking about proofs in formal systems of nonstandard analysis, there are general methods for extracting the standard proof from it. For instance, "Weak theories of nonstandard arithmetic and analysis" by Jeremy Avigad. – Henry Towsner Jun 8 2011 at 17:55	{"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": 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": 0.938130259513855, "perplexity_flag": "head"}
http://physics.stackexchange.com/questions/44238/fock-picture-of-bosonification-in-condensates	# Fock picture of bosonification in condensates I want to understand how bosonification in a condensate must be interpreted in the Fock states picture Say i have uncoupled fermions in a set of states $E_1$, $E_2$ ... over the vacuum $E_0$. They occupy all the levels up to the Fermi energy. Now, i introduce some coupling between them so, they become effectively boson pairs. Are the $E_i$ energy states out of the picture for these bosons? My random guess is that fermions would couple in states like $$\| E_1 \times E_2 \rangle - \| E_2 \times E_1 \rangle$$ $$\| E_1 \times E_2 \rangle + \| E_2 \times E_1 \rangle$$ $$\| E_3 \times E_4 \rangle - \| E_4 \times E_3 \rangle$$ $$\| E_3 \times E_4 \rangle + \| E_4 \times E_3 \rangle$$ And so on, but this is the aspect i'm not sure and i want clarification What are the states where the boson pairs are condensing into?? -	{"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": 4, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.919520914554596, "perplexity_flag": "middle"}
http://mathhelpforum.com/advanced-algebra/167180-orders-elements-isomorphisms.html	# Thread: 1. ## orders of elements and isomorphisms Show that if $\gamma: G_1 \rightarrow G_2$ is an isomorphism, then for each $g \in G_1, o(\gamma(g)) = o(g)$ Does this argument work? if $g \in G_1$ has order l $\Rightarrow g^l = e_1$ then $\gamma (g^l) = \gamma (e_1)$ then as the identity element of $G_1$ is mapped to the identity element $e_2 \in G_2$ $\gamma (g^l) = \gamma (e_1) = e_2$ then as $\gamma(x_1x_2)= \gamma(x_1)\gamma(x_2)$ must hold we have $\gamma(g^l) = \gamma(g) ^l = (g')^l = e_2$ where as bijection each $g' = \gamma(g)$ therefore $(g')^l = e_2$ in $\in G_2$ 2. Originally Posted by FGT12 Show that if $\gamma: G_1 \rightarrow G_2$ is an isomorphism, then for each $g \in G_1, o(\gamma(g)) = o(g)$ Does this argument work? if $g \in G_1$ has order l $\Rightarrow g^l = e_1$ then $\gamma (g^l) = \gamma (e_1)$ then as the identity element of $G_1$ is mapped to the identity element $e_2 \in G_2$ $\gamma (g^l) = \gamma (e_1) = e_2$ then as $\gamma(x_1x_2)= \gamma(x_1)\gamma(x_2)$ must hold we have $\gamma(g^l) = \gamma(g) ^l = (g')^l = e_2$ where as bijection each $g' = \gamma(g)$ therefore $(g')^l = e_2$ in $\in G_2$ This is only half the battle. You've shown that $\left\|\gamma(g)\right\|\mid\ell$, but you can use the same process in reverse (since all you used was that $\gamma$ is a homomorphism, but $\gamma^{-1}$ is as well!) to get that $\ell\mid\left\|\gamma(g)\right\|$.	{"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": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 30, "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": 0.9678863883018494, "perplexity_flag": "head"}
http://math.stackexchange.com/questions/89002/solve-in-terms-of-b-log-b-1-3x-3-log-b-x/89020	Solve in terms of $b$: $\log_b (1 - 3x) = 3 + \log_b x$ $$\log_b (1 - 3x) = 3 + \log_b x$$ If I use the properties of logs, I end up with $$\log_b \left(\frac{1 - 3x}{x}\right) = 3$$ From there, the example I have says to exponentiate both sides, however, they use a $\log_2(\text{equation}) = 2$. They then raise both sides to the $2$. I can't seem to do this here, does someone have an example that better fits my situation? - Raise $b$ to the powers involved. Note that $b^{\log_b(y)}=y$. So you should end up with $\frac{1-3x}{x}=b^3$. – André Nicolas Dec 6 '11 at 20:16 I've got it solved, thanks. If you answer this question with your comment, I'll mark this as the accepted answer. – erimar77 Dec 6 '11 at 20:31 It seems what is really missing here is an understanding of logarithms: Recall that $$\log_b(z) = y \quad\iff\quad b^y = z.$$ In other words, $\log_b(z)$ is the power to which you exponentiate $b$ to get $z$. In your case, $$\log_b\left( \frac{1-3x}{x} \right) = 3 \quad\iff\quad b^3 = \frac{1-3x}{x}.$$ – JavaMan Dec 6 '11 at 20:58 1 Answer Remember, any $\log$ to the base $b$, such as $\log_b(y)$, is in its heart an exponent. Let us start, then, from your expression $$\log_b\left(\frac{1-3x}{x}\right)=3.$$ Raise $b$ to the powers we see on each side. We get $$b^{\log_b\left(\frac{1-3x}{x}\right)}=b^3.$$ The left-hand side simplifies greatly. We get $$\frac{1-3x}{x}=b^3.$$ The rest is elementary algebra. The above equation is (for $x\ne 0$) equivalent to $$1-3x=b^3 x,$$ which is an easily solved linear equation. Comment: There was no need to do the preliminary manipulation. We are told that $$\log_b(1-3x)=3+\log_b x.$$ Raise $b$ to the power on the left-hand side, the right-hand side. We obtain $$b^{\log_b(1-3x)}=b^{3+\log_b x}.$$ By the "laws of logarithms" this yields $$1-3x=b^3 x.$$ -	{"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": 16, "mathjax_display_tex": 11, "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": 0.9631772041320801, "perplexity_flag": "head"}
http://crypto.stackexchange.com/questions/5166/what-is-the-complexity-of-the-square-attack-against-the-reduced-4-rounds-128-bit/5182	# What is the complexity of the Square attack against the reduced 4-rounds 128-bit Rijndael variant? I'm looking at a square attack against a reduced version of AES-128 with only 4 rounds (with block and key size of each 128 bit). I have a set of 256 plaintext-ciphertext block pairs. What is the complexity of the computations, in number of encryptions, partial encryptions, memory usage, and data requested? - – D.W. Oct 28 '12 at 4:09 Welcome to Cryptography Stack Exchange. Please note that the question titles should at least sound something like a question – what you put there would better be tags. I edited your question to use these tags, gave a better title and tried to edit the question to say what I understood you wanted to say. I hope I understood this right – feel free to edit this again (there is an edit button). Also, have a look at D.W.'s comment, and add the necessary information. – Paŭlo Ebermann♦ Oct 28 '12 at 14:02 ## 1 Answer There are two well-known distinguishers of reduced-rounds version of AES based on integral cryptanalysis. The first one is a 3-rounds distinguisher from Daemen and Rijmen while the other is a 4-rounds distinguisher from Gilbert and Minier. The 3-rounds distinguisher relies on the fact that one byte $y$ of the state in the third round is the xor of four one-to-one mappings $z_i:x\mapsto z_i(x)$ from a given byte $x$ in the state of the first round, and therefore, on the property that $\sum_{x\in GF(2^8)} \left[z_1(x)+z_2(x)+z_3(x)+z_4(x)\right]$. Hence, the distinguisher requires $2^8$ (well crafted) plaintexts and as many encryptions, but basically no memory. The 4-rounds distinguisher relies on the fact that the above-mentioned byte $z$ is actually a function of $x$ and three constants $c_1$, $c_2$, $c_3$ hereafter referred to as $c$ and the probability that there is an unusually high chance that the functions $z_{c'}$ and $z_{c''}$ are equal: this can be tested through the equality of the linear combination of four bytes resulting from the encryption of the two corresponding plaintext. The property is observable after browsing through $2^{16}$ values of $c$ (and enough of the possible values for $x$). Hence, the disinguisher requires about $2^{20}$ plaintexts and as many encryptions as well as less than $2^6$ bytes of memory. These distinguishers are meant to be turned into key recovery attacks, but these extend to at least 6-rounds AES. Since you mentionned 4-rounds attacks without further details, I assume you referred to one of the above. If you need a more precise answer, you should refine your question. Also, make it precise which version of Rijndael you're referring to as it may impact the underlying complexity. -	{"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": 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": 0.9390266537666321, "perplexity_flag": "middle"}
http://mathoverflow.net/questions/86143?sort=oldest	## Which trigonometric identities involve trigonometric functions? ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) Another question that's getting no answers on stackexchange: Once upon a time, when Wikipedia was only three-and-a-half years old and most people didn't know what it was, the article titled functional equation gave the identity $$\sin^2\theta+\cos^2\theta = 1$$ as an example of a functional equation. In this edit in July 2004, my summary said "I think the example recently put here is really lousy, because it's essentially just an algebraic equation in two variables." (Then some subsequent edits I did the same day brought the article to this state, and much further development of the article has happened since then.) The fact that it's really only an algebraic equation in two variables, $x^2+y^2=1$, makes it a lousy example of a functional equation. It doesn't really involve $x$ and $y$ as functions of $\theta$, since any other parametrization of the circle would have satisfied the same equation. In a sense, that explains why someone like Norman Wildberger can do all sorts of elaborate things with trigonometry without ever using trigonometric functions. But some trigonometric identities do involve trigonometric functions, e.g. $$\sin(\theta_1+\theta_2)=\sin\theta_1\cos\theta_2+\cos\theta_1\sin\theta_2$$ $$\sec(\theta_1+\cdots+\theta_n) = \frac{\sec\theta_1\cdots\sec\theta_n}{e_0-e_2+e_4-e_6+\cdots}$$ where $e_k$ is the $k$th-degree elementary symmetric polynomial in $\tan\theta_1,\ldots,\tan\theta_n$. These are good examples of satisfaction of functional equations. So at this point I wonder whether all trigonometric identities that do seem to depend on which parametrization of the circle is chosen involve adding or subtracting the arguments and no other operations. In some cases the addition or subtraction is written as a condition on which the identity depends, e.g. $$\text{If }x+y+z=\pi\text{ then }\tan x+\tan y+\tan z = \tan x\tan y\tan z.$$ QUESTION: Do all trigonometric identities that do involve trigonometric functions, in the sense that they are good examples of satisfaction of functional equations by trigonometric functions, get their non-triviality as such examples only from the addition or subtraction of arguments? Or is there some other kind? And if there is no other kind, can that be proved? In comments below the stackexchange posting, Gerry Myerson mentioned the identities $$\cos \frac x2=\sqrt{\frac{1+\cos x}{2}}$$ and $$\prod_{k=1}^\infty \cos\left(\frac{x}{2^k}\right)= \frac{\sin x}{x}$$ The latter is somewhat like the one involving tangents above: One can say that if $x_n = x_{n+1}+x_{n+1}$ for $n=1,2,3,\ldots$ then $$\prod_{k=1}^\infty \cos \left(\frac{x_k}{2^k}\right) = \frac{\sin x_1}{x_1}.$$ A similar but simpler thing applies to the half-angle formula. Postscript: Wikipedia's list of trigonometic identities is more interesting reading than you might think. It has not only the routine stuff that you learned in 10th grade, but also some exotic things that probably most mathematicians don't know about. It was initially created in September 2001 by Axel Boldt, who was for more than a year the principal author of nearly all of Wikipedia's mathematics articles---several hundred of them. - I'm a little confused: are you saying that $sin^2 θ + cos^2 θ = 1$ a bad example of a functional equation because $sin θ$ and $cos θ$ are not the only continuous solutions, in contrast to, say, $e^{x+y}=e^x e^y$? – Paul Siegel Jan 19 2012 at 22:05 @Paul Siegel : Let $f(t) = (t^2 - 1)/(t^2 + 1)$ and $g(t) = 2t/(t^2 + 1)$. The $f(t)^2 + g(t)^2 = 1$. That's what I meant when I wrote "any other parametrization of the circle would have satisfied the same equation." – Michael Hardy Jan 19 2012 at 22:41 In case anyone wants to see what Michael doesn't accept as an example, the m.se link is math.stackexchange.com/questions/99909/… – Gerry Myerson Jan 20 2012 at 4:47 Robert Israel has suggested the identity $\cos(t \sin(x)) = J_0(t) + 2 \sum_{k=1}^\infty J_{2k}(t) \cos(2kx)$. If we fix $t=1$ that becomes $\cos(\sin(x)) = J_0(1) + 2 \sum_{k=1}^\infty J_{2k}(1) \cos(2kx)$ and then it doesn't have anything like the function $t\mapsto J_{2k}(t)$, which is moderately exotic for something bearing a "trigonometry" tag, but it does have $k\mapsto J_{2k}(t)$, which is similarly moderately exotic. This does seem like a counterexample to the simplest interpretation of my guess. But one could ask whether the guess is still true if..... – Michael Hardy Jan 20 2012 at 19:26 ....one insists on having only finitely many terms, or if one insists on "elementary" functions instead of things like $k\mapsto J_{2k}(t)$. – Michael Hardy Jan 20 2012 at 19:27 show 1 more comment ## 2 Answers Unfortunately the category of smooth manifolds is not closed under finite limits. (Yes, to annoy my Wikipedian colleague Michael, I'm speaking Bourbaki.) This kind of undermines the idea of setting up an actual answer in terms of taking the usual map from the line to the circle, qua morphism, and generating some "category of relations" within which the question has a chance of an acceptable answer. But perhaps there is a germ of an idea here. Formulate the question via some equational logic that admits the more general parametrisations of the circle. - ### 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. Most of these identities have a different explanation, via holonomic functions, aka solutions of linear differential equations with polynomial coefficients. And, in the end, it is all a matter of linear algebra and bounding dimensions. This is very well explained in slides by Bruno Salvy (where trigonometrics are explicitly used for the simple examples, before moving on to the 'fun' stuff). To a certain extent, there is still a lot of geometry still going on in the above, but the geometry of the parameter space, rather than looking directly at the traces of the functions. Many details of the method can be found in the paper A non-holonomic systems approach to special function identities by Chyzak, Kauers and Salvy. -	{"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": 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": 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": 0.9423152208328247, "perplexity_flag": "middle"}
http://mathoverflow.net/questions/4411/quantitative-versions-of-ergodic-theorem/4416	## Quantitative versions of ergodic theorem ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) Are there any general theorems similar to Birkhoff's ergodic theorem, but giving quantitative estimates on the rate of convergence or average time of recurrence (perhaps with additional assumptions)? Take the example of an "irrational rotation" on the unit circle - are there any estimates on the average time it takes for a point to hit a certain interval? I know there are such theorems for very special systems (e.g. for Markov chains we have exponential convergence) - what can be said about a "generic" ergodic system? - ## 6 Answers The best effective estimate I know of in general is the very recent and impressive work of Einsiedler-Margulis-Venkatesh. They provide polynomial decay in the ergodic theorem (for unipotent and other flows) on a set of large (actually estimated) measure for quotients of most lie groups by arithmetic lattices. There were previously things like this: for closed hyperbolic surfaces, the error for the time average of the horocycle flow of length $T$ is something like $S(f) T^{-\epsilon}$ where $f$ is the smooth function you're averaging, $S(f)$ is a Sobolev norm, and $\epsilon$ depends on the spectrum of the surface. (In this case, every point is generic for the horocycle flow (by Ratner's theorem), and this effective estimate is independent of the point you use for the average!) See the nice survey "An introduction to effective equidistribution and property (tau)," by Einsiedler, found on his webpage. I would be very interested in any better estimates for flows on arithmetic hyperbolic $3$-manifolds given information on the trace field, quaternion algebra, et cetera. - The survey link is missing but thanks for the pointer! – Kaveh Khodjasteh Sep 20 2010 at 14:37 Sorry about the link rot. I edited to correct it. Let's hope it lasts a little longer this time. – Richard Kent Sep 20 2010 at 14:57 ### 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. For the specific case you mention of an irrational rotation of the circle, it depends on the rotation number $r$. You can get slow asymptotic convergence for something like $$r=\sum{10^{-n!}}$$ (and you can just keep adding factorial signs to make the convergence arbitrarily slow.) At the other end of the spectrum is rotation number equal to the golden ratio. In this case you know that for a Fibonacci number $q$, it will only take $q$ iterations to hit each interval of the form $[n/q, (n+1)/q]$. The behavior for an arbitrary irrational number is governed by its continued fraction expansion. Milnor's book Dynamics in one complex variable has a clear explanation. - The survey "The rate of convergence in ergodic theorems" by A. G. Kachurovskii (Russian Math. Surveys 1996) lists quite a number of results in this direction which might be of interest to you. It includes some negative results: for example, if I recall correctly, for any positive real sequence a`_n`=o(n) and any aperiodic measurable dynamical system, we can find a measurable function f taking only two distinct real values with the property that the ergodic sums of f are o(n) but not o(a`_n`). On the other hand, a number of sufficient conditions are presented for polynominal error estimates to hold in the ergodic theorem. - A. Leibman proved a quantitative lower bound for the averages $\frac{1}{N}\sum_{n=0}^{N-1} \mu(A\cap T^{-n}A)$ in terms of $\mu(A)$ (note: the sum begins at $n=0$). The bound is $$\frac{1}{N}\sum_{n=0}^{N-1} \mu(A\cap T^{-n}A) \geq \sqrt{\mu(A)^2+(1-\mu(A))^2} + \mu(A)-1$$ for all $N\geq 1$ when $T$ is a measure preserving transformation of a probability space, and this is the best possible such bound. - Dolpogyat has proved rates of convergence for Anosov systems (see #4 here.) Kac pioneered the study of recurrence times for stochastic processes (see here). - Usually such estimations require hyperbolicity or some particular kind of system (rotations, IET and similar....). For general systems an effective quantitative estimation on the rate of convergence is possible (altough not sharph), provided that the system is given effectively.(see J. Avigad, P. Gerhardy and H. Towsner, Local stability of ergodic averages, Transactions of the American Mathematical Society, 362 (2010), or 1 for a very shorth proof of a similar result). In your question it seems to me that you are mostly interested to the behavior of hitting times, and perhaps in rotations. As already remarked in another answer this depend on the arithmetical properties of the rotation. If the rotation has an angle which is well approximated by rationals you can have long hitting times for certain intervals (the time you need to wait for the interval to be hit is much more than the inverse of the lenght of the interval). Quantitative convergence results are given by the so called "discrepancy" estimations, in function of the Diofantine type of the angle (these are classical results you can find in many books). If you consider multidimensional rotations you can have even stronger pathological behaviors of the hitting time, but I do not know if you are interested in 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": 15, "mathjax_display_tex": 2, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9189023375511169, "perplexity_flag": "head"}
http://sciencehouse.wordpress.com/category/physics/kinetic-theory/	# Scientific Clearing House Carson C. Chow ## Archive for the ‘Kinetic Theory’ Category ### New paper on neural networks March 22, 2013 Michael Buice and I have just published a review paper of our work on how to go beyond mean field theory for systems of coupled neurons. The paper can be obtained here. Michael and I actually pursued two lines of thought on how to go beyond mean field theory and we show how the two are related in this review. The first line started in trying to understand how to create a dynamic statistical theory of a high dimensional fully deterministic system. We first applied the method to the Kuramoto system of coupled oscillators but the formalism could apply to any system. Our recent paper in PLoS Computational Biology was an application for a network of synaptically coupled spiking neurons. I’ve written about this work multiple times (e.g. here, here, and here). In this series of papers, we looked at how you can compute fluctuations around the infinite system size limit, which defines mean field theory for the system, when you have a finite number of neurons. We used the inverse number of neurons as a perturbative expansion parameter but the formalism could be generalized to expand in any small parameter, such as the inverse of a slow time scale. The second line of thought was with regards to the question of how to generalize the Wilson-Cowan equation, which is a phenomenological population activity equation for a set of neurons, which I summarized here. That paper built upon the work that Michael had started in his PhD thesis with Jack Cowan. The Wilson-Cowan equation is a mean field theory of some system but it does not specify what that system is. Michael considered the variable in the Wilson-Cowan equation to be the rate (stochastic intensity) of a Poisson process and prescribed a microscopic stochastic system, dubbed the spike model, that was consistent with the Wilson-Cowan equation. He then considered deviations away from pure Poisson statistics. The expansion parameter in this case was more obscure. Away from a bifurcation (i.e. critical point) the statistics of firing would be pure Poisson but they would deviate near the critical point, so the small parameter was the inverse distance to criticality. Michael, Jack and I then derived a set of self-consistent set of equations for the mean rate and rate correlations that generalized the Wilson-Cowan equation. The unifying theme of both approaches is that these systems can be described by either a hierarchy of moment equations or equivalently as a functional or path integral. This all boils down to the fact that any stochastic system is equivalently described by a distribution function or the moments of the distribution. Generally, it is impossible to explicitly calculate or compute these quantities but one can apply perturbation theory to extract meaningful quantities. For a path integral, this involves using Laplace’s method or the method of steepest descents to approximate an integral and in the moment hierarchy method it involves finding ways to truncate or close the system. These methods are also directly related to WKB expansion, but I’ll leave that connection to another post. Posted in Computational neuroscience, Kinetic Theory, Papers \| 6 Comments » ### New paper on finite size effects in spiking neural networks January 25, 2013 Michael Buice and I have finally published our paper entitled “Dynamic finite size effects in spiking neural networks” in PLoS Computational Biology (link here). Finishing this paper seemed like a Sisyphean ordeal and it is only the first of a series of papers that we hope to eventually publish. This paper outlines a systematic perturbative formalism to compute fluctuations and correlations in a coupled network of a finite but large number of spiking neurons. The formalism borrows heavily from the kinetic theory of plasmas and statistical field theory and is similar to what we used in our previous work on the Kuramoto model (see here and here) and the “Spike model” (see here). Our heuristic paper on path integral methods is here. Some recent talks and summaries can be found here and here. (more…) Posted in Computational neuroscience, Kinetic Theory, Papers, Physics, Probablity \| 8 Comments » ### Talk today at Johns Hopkins December 12, 2012 I’m giving a computational neuroscience lunch seminar today at Johns Hopkins. I will be talking about my work with Michael Buice, now at the Allen Institute, on how to go beyond mean field theory in neural networks. Technically, I will present our recent work on computing correlations in a network of coupled neurons systematically with a controlled perturbation expansion around the inverse network size. The method uses ideas from kinetic theory with a path integral construction borrowed and adapted by Michael from nonequilibrium statistical mechanics. The talk is similar to the one I gave at MBI in October. Our paper on this topic will appear soon in PLoS Computational Biology. The slides can be found here. Posted in Computational neuroscience, Kinetic Theory, Talks \| 3 Comments » ### Complexity is the narrowing of possibilities December 6, 2012 Complexity is often described as a situation where the whole is greater than the sum of its parts. While this description is true on the surface, it actually misses the whole point about complexity. Complexity is really about the whole being much less than the sum of its parts. Let me explain. Consider a television screen with 100 pixels that can be either black or white. The number of possible images the screen can show is $2^{100}$. That’s a really big number. Most of those images would look like random white noise. However, a small set of them would look like things you recognize, like dogs and trees and salmon tartare coronets. This narrowing of possibilities, or a reduction in entropy to be more technical, increases information content and complexity. However, too much reduction of entropy, such as restricting the screen to be entirely black or white, would also be considered to have low complexity. Hence, what we call complexity is when the possibilities are restricted but not completely restricted. Another way to think about it is to consider a very high dimensional system, like a billion particles moving around. A complex system would be if the attractor of this six billion dimensional system (3 for position and 3 for velocity of each particle), is a lower dimensional surface or manifold. The flow of the particles would then be constrained to this attractor. The important thing to understand about the system would then not be the individual motions of the particles but the shape and structure of the attractor. In fact, if I gave you a list of the positions and velocities of each particle as a function of time, you would be hard pressed to discover that there even was a low dimensional attractor. Suppose the particles lived in a box and they moved according to Newton’s laws and only interacted through brief elastic collisions. This is an ideal gas and what would happen is that the motions of the positions of the particles would be uniformly distributed throughout the box while the velocities would obey a Normal distribution, called a Maxwell-Boltzmann distribution in physics. The variance of this distribution is proportional to the temperature. The pressure, volume, particle number and temperature will be related by the ideal gas law, PV=NkT, with the Boltzmann constant set by Nature. An ideal gas at equilibrium would not be considered complex because the attractor is a simple fixed point. However, it would be really difficult to discover the ideal gas law or even the notion of temperature if one only focused on the individual particles. The ideal gas law and all of thermodynamics was discovered empirically and only later justified microscopically through statistical mechanics and kinetic theory. However, knowledge of thermodynamics is sufficient for most engineering applications like designing a refrigerator. If you make the interactions longer range you can turn the ideal gas into a liquid and if you start to stir the liquid then you can end up with turbulence, which is a paradigm of complexity in applied mathematics. However, the main difference between an ideal gas and turbulent flow is the dimension of the attractor. In both cases, the attractor dimension is still much smaller than the full range of possibilities. The crucial point is that focusing on the individual motions can make you miss the big picture. You will literally miss the forest for the trees. What is interesting and important about a complex system is not what the individual constituents are doing but how they are related to each other. The restriction to a lower dimensional attractor is manifested by the subtle correlations of the entire system. The dynamics on the attractor can also often be represented by an “effective theory”. Here the use of the word “effective” is not to mean that it works but rather that the underlying microscopic theory is superseded by a macroscopic one. Thermodynamics is an effective theory of the interaction of many particles. The recent trend in biology and economics had been to focus on the detailed microscopic interactions (there is push back in economics in what has been dubbed the macro-wars). As I will relate in future posts, it is sometimes much more effective (in the works better sense) to consider the effective (in the macroscopic sense) theory than a detailed microscopic theory. In other words, there is no “theory” per se of a given system but rather sets of effective theories that are to be selected based on the questions being asked. Posted in Computer Science, Kinetic Theory, Mathematics, Philosophy, Physics \| 4 Comments » ### Two talks December 8, 2011 Last week I gave a talk on obesity at Georgia State University in Atlanta, GA. Tomorrow, I will be giving a talk on the kinetic theory of coupled oscillators at George Mason University in Fairfax, VA. Both of these talks are variations of ones I have given before so instead of uploading my slides, I’ll just point to links to previous talks, papers, and posts on the topics. For obesity, see here and for kinetic theory, see here, here and here. ### New paper November 17, 2011 A new paper in Physical Review E is now available on line here. In this paper Michael Buice and I show how you can derive an effective stochastic differential (Langevin) equation for a single element (e.g. neuron) embedded in a network by averaging over the unknown dynamics of the other elements. This then implies that given measurements from a single neuron, one might be able to infer properties of the network that it lives in. We hope to show this in the future. In this paper, we perform the calculation explicitly for the Kuramoto model of coupled oscillators (e.g. see here) but it can be generalized to any network of coupled elements. The calculation relies on the path or functional integral formalism Michael developed in his thesis and generalized at the NIH. It is a nice application of what is called “effective field theory”, where new dynamics (i.e. action) are obtained by marginalizing or integrating out unwanted degrees of freedom. The path integral formalism gives a nice platform to perform this averaging. The resulting Langevin equation has a noise term that is nonwhite, non-Gaussian and multiplicative. It is probably not something you would have guessed a priori. Michael A. Buice1,2 and Carson C. Chow11Laboratory of Biological Modeling, NIDDK, NIH, Bethesda, Maryland 20892, USA 2Center for Learning and Memory, University of Texas at Austin, Austin, Texas, USA Received 25 July 2011; revised 12 September 2011; published 17 November 2011 Complex systems are generally analytically intractable and difficult to simulate. We introduce a method for deriving an effective stochastic equation for a high-dimensional deterministic dynamical system for which some portion of the configuration is not precisely specified. We use a response function path integral to construct an equivalent distribution for the stochastic dynamics from the distribution of the incomplete information. We apply this method to the Kuramoto model of coupled oscillators to derive an effective stochastic equation for a single oscillator interacting with a bath of oscillators and also outline the procedure for other systems. DOI: 10.1103/PhysRevE.84.051120 PACS: 05.40.-a, 05.45.Xt, 05.20.Dd, 05.70.Ln Posted in Computational neuroscience, Kinetic Theory, Papers, Physics \| 2 Comments » ### Talk in Marseille October 8, 2011 I just returned from an excellent meeting in Marseille. I was quite impressed by the quality of talks, both in content and exposition. My talk may have been the least effective in that it provoked no questions. Although I don’t think it was a bad talk per se, I did fail to connect with the audience. I kind of made the classic mistake of not knowing my audience. My talk was about how to extend a previous formalism that much of the audience was unfamiliar with. Hence, they had no idea why it was interesting or useful. The workshop was on mean field methods in neuroscience and my talk was on how to make finite size corrections to classical mean field results. The problem is that many of the participants of the workshop don’t use or know these methods. The field has basically moved on. In the classical view, the mean field limit is one where the discreteness of the system has been averaged away and thus there are no fluctuations or correlations. I have been struggling over the past decade trying to figure out how to estimate finite system size corrections to mean field. This led to my work on the Kuramoto model with Eric Hildebrand and particularly Michael Buice. Michael and I have now extended the method to synaptically coupled neuron models. However, to this audience, mean field pertains more to what is known as the “balanced state”. This is the idea put forth by Carl van Vreeswijk and Haim Sompolinsky to explain why the brain seems so noisy. In classical mean field theory, the interactions are scaled by the number of neurons N so in the limit of N going to infinity the effect of any single neuron on the population is zero. Thus, there are no fluctuations or correlations. However in the balanced state the interactions are scaled by the square root of the number of neurons so in the mean field limit the fluctuations do not disappear. The brilliant stroke of insight by Carl and Haim was that a self consistent solution to such a situation is where the excitatory and inhibitory neurons balance exactly so the net mean activity in the network is zero but the fluctuations are not. In some sense, this is the inverse of the classical notion. Maybe it should have been called “variance field theory”. The nice thing about the balanced state is that it is a stable fixed point and no further tuning of parameters is required. Of course the scaling choice is still a form of tuning but it is not detailed tuning. Hence, to the younger generation of theorists in the audience, mean field theory already has fluctuations. Finite size corrections don’t seem that important. It may actually indicate the success of the field because in the past most computational neuroscientists were trained in either physics or mathematics and mean field theory would have the meaning it has in statistical mechanics. The current generation has been completely trained in computational neuroscience with it’s own canon of common knowledge. I should say that my talk wasn’t a complete failure. It did seem to stir up interest in learning the field theory methods we have developed as people did recognize it provides a very useful tool to solve the problems they are interested in. Addendum 2011-11-11 Here are some links to previous posts that pertain to the comments above. http://sciencehouse.wordpress.com/2009/06/03/talk-at-njit/ http://sciencehouse.wordpress.com/2009/03/22/path-integral-methods-for-stochastic-equations/ http://sciencehouse.wordpress.com/2009/01/17/kinetic-theory-of-coupled-oscillators/ http://sciencehouse.wordpress.com/2010/09/30/path-integral-methods-for-sdes/ http://sciencehouse.wordpress.com/2010/02/03/paper-now-in-print/ http://sciencehouse.wordpress.com/2009/02/27/systematic-fluctuation-expansion-for-neural-networks/ Posted in Computational neuroscience, Conferences, Kinetic Theory, Talks \| 1 Comment » ### Marseille talk October 4, 2011 I am currently at CIRM in Marseille, France for a workshop on mean field methods in neuroscience. My slides are here. ### Snowbird meeting May 24, 2011 I’m currently at the biannual SIAM Dynamical Systems Meeting in Snowbird Utah. If a massive avalanche were to roll down the mountain and bury the hotel at the bottom, much of applied dynamical systems research in the world would cease to exist. The meeting has been growing steadily for the past thirty years and has now maxed out the capacity of Snowbird. The meeting will either eventually have to move to a new venue or restrict the number of speakers. My inclination is to move but I don’t think that is the most popular sentiment. Thus far, I have found the invited talks to be very interesting. Climate change seems to be the big theme this year. Chris Jones and Raymond Pierrehumbert both gave talks on that topic. I chaired the session by noted endocrinologist and neuroscientist Stafford Lightman who gave a very well received talk on the dynamics of hormone secretion. Chiara Daraio gave a very impressive talk on manipulating sound propagation with chains of ball bearings. She’s basically creating the equivalent of nonlinear optics and electronics in acoustics. My talk this afternoon is on finite size effects in spiking neural networks. It is similar but not identical to the one I gave in New Orleans in January (see here). The slides are here. ### 2011 JMM talk January 10, 2011 I’m on my way back from the 2011 Joint Mathematics Meeting. I gave a talk yesterday on finite size effects in neural networks. I gave a pedagogical talk on the strategy that Michael Buice and I have employed to analyze finite size networks in networks of coupled spiking neurons. My slides are here. We’ve adapted the formalism we used to analyze the finite size effects of the Kuramoto system (see here for summary) to a system of synaptically coupled phase oscillators. Posted in Computational neuroscience, Kinetic Theory, Talks \| 1 Comment » ### Talk at Pitt September 10, 2010 I visited the University of Pittsburgh today to give a colloquium. I was supposed to have come in February but my plane was cancelled because of a snow storm. This was not the really big snow storm that closed Washington, DC and Baltimore for a week but a smaller one that hit New England and not the DC area. My flight was on Southwest and I presume that they have such a tightly correlated flight system, where planes circulate around the country in a “just in time” fashion, that a disturbance in one part of the country affects the rest of the country. So while other airlines just had cancellations in New England, Southwest flights were cancelled for the day all across the US. It seems that there is a trade off between business efficiency and robustness. I drove this time. My talk was on the finite size effects in the Kuramoto model, which I’ve given several times already. However, I have revised the slides on pedagogical grounds and they can be found here. ### Slides for second SIAM talk July 21, 2010 Here are the slides for my SIAM talk on generalizing the Wilson-Cowan equations to include correlations. This talk was mostly on the paper with Michael Buice and Jack Cowan that I summarized here. However, I also contrasted our work with the recent work of Paul Bressloff who uses a system size expansion of the Markov process that Michael and Jack proposed as a microscopic model for Wilson-Cowan in their 2007 paper. The difference between the two approaches stems from the interpretation of what the Wilson-Cowan equation describes. In our interpretation, the Wilson-Cowan equation describes the firing rate or stochastic intensity of a Poisson process. A Poisson distribution is notable because all cumulants are equal to the mean. Our expansion is in terms of factorial cumulants (we called them normal ordered cumulants in the paper because we didn’t know there was a name for them), which are deviations from Poisson statistics. Bressloff, on the other hand, considers the Wilson -Cowan equation to be the average population firing rate of a large population of neurons. In the infinite size limit, there are no fluctuations. His expansion is in terms of regular cumulants and the inverse system size is the small parameter. In our formulation, the expansion parameter is related to the distance to a critical point where the expansion would break down. In essence, we use a Bogoliubov hierarchy of time scales expansion where the higher order factorial cumulants decay to steady state much faster than the lower order ones. ### Two talks at University of Toronto December 2, 2009 I’m currently at the University of Toronto to give two talks in a series that is jointly hosted by the Physics department and the Fields Institute. The Fields Institute is like the Canadian version of the Mathematical Sciences Research Institute in the US and is named in honour of Canadian mathematician J.C. Fields, who started the Fields Medal (considered to be the most prestigious prize for mathematics). The abstracts for my talks are here. The talk today was a variation on my kinetic theory of coupled oscillators talk. The slides are here. I tried to be more pedagogical in this version and because it was to be only 45 minutes long, I also shortened it quite a bit. However, in many ways I felt that this talk was much less successful than the previous versions. In simplifying the story, I left out much of the history behind the topic and thus the results probably seemed somewhat disembodied. I didn’t really get across why a kinetic theory of coupled oscillators is interesting and useful. Here is the post giving more of the backstory on the topic, which has a link to an older version of the talk as well. Tomorrow, I’ll talk about my obesity work. ### Talk at NJIT June 3, 2009 I was at the FACM ’09 conference held at the New Jersey Institute of Technology the past two days. I gave a talk on “Effective theories for neural networks”. The slides are here. This was an unsatisfying talk on two accounts. The first was that I didn’t internalize how soon this talk came after the Snowbird conference and so I didn’t have enough time to properly prepare. I thus ended up giving a talk that provided enough information to be confusing and hopefully thought provoking but not enough to be understood. The second problem was that there is a flaw in what I presented. I’ll give a brief backdrop to the talk for those unfamiliar with neuroscience. The brain is composed of interconnected neurons and as a proxy for understanding the brain, computational neuroscientists try to understand what a collection of coupled neurons will do. The state of a neuron is characterized by the voltage across its membrane and the state of its membrane ion channels. When a neuron is given enough input, there can be a massive change of voltage and flow of ions called an action potential. One of the ions that flows into the cell is calcium, which can trigger the release of neurotransmitter to influence other neurons. Thus, neuroscientists are highly focused on how and when action potentials or spikes occur. We can thus model a neural network at many levels. At the bottom level, there is what I will call a microscopic description where we write down equations for the dynamics of the voltage and ion channels for each neuron. These neuron models are sometimes called conductance-based neurons and the Hodgkin-Huxley neuron is the first and most famous of them. They usually consist of two to four differential equations and can easily be a lot more. On the other hand, if one is more interested in just the spiking rate, then there is a reduced description for that. In fact, much of the early progress in mathematically understanding neural networks used rate equations, examples being Wison and Cowan, Grossberg, Hopfield and Amari. The question that I have always had was what is the precise connection between a microscopic description and a spike rate or activity description. If I start with a network of conductance-based neurons can I derive the appropriate activity based description? (more…) Posted in Kinetic Theory, Neuroscience, Papers, Probablity, Talks \| 2 Comments » ### Snowbird conference May 20, 2009 I’m currently at the SIAM Dynamical Systems meeting in Snowbird, Utah. I gave a short version of my talk on calculating finite size effects of the Kuramotor coupled oscillator model using kinetic theory and path integral approaches. Here is the longer and more informative version of the talk. I summarized the papers on this talk here. ### Systematic fluctuation expansion for neural networks slides March 24, 2009 I gave a talk today at the Mathematical Neuroscience workshop on my recent work with Michael Buice and Jack Cowan on deriving generalized activity equations for neural networks. My slides are here. The talk is based on the paper we uploaded to the arXiv recently and I summarized here. Posted in Kinetic Theory, Mathematics, Neuroscience, Physics, Talks \| 1 Comment » ### Path integral methods for stochastic equations March 22, 2009 I’m currently in Edinburgh for a Mathematical Neuroscience workshop. I gave a tutorial today on using field theoretic methods to solve stochastic differential equations (SDE’s). The slides are here. The methods I presented have been around for decades but as far as I know they haven’t been collated together into a pedagogical review for nonexperts. Also, there is an entire community of theorists and mathematicians that are unaware of path integral methods. In particular, I apply the response function formalism stemming from the work of Martin Siggia and Rose. Field theory and diagrammatic methods are a nice way to organize perturbation expansions for nonlinear SDE’s. I plan to write a review paper on this topic in the next few months and will post it here. Addendum: Jan 20, 2011. The review paper can be found here. Posted in Kinetic Theory, Neuroscience, Physics, Talks \| 2 Comments » ### Kinetic theory of coupled oscillators January 17, 2009 Last week, I gave a physics colloquium at the Catholic University of America about recent work on using kinetic theory and field theory approaches to analyze finite-size corrections to networks of coupled oscillators. My slides are here although they are converted from Keynote so the movies don’t work. Coupled oscillators arise in contexts as diverse as the brain, synchronized flashing of fireflies, coupled Josephson junctions, or unstable modes of the Millennium bridge in London. Steve Strogatz’s book Sync gives a popular account of the field. My talk considers the Kuramoto model $\dot{\theta}_i = \omega_i+\frac{K}{N}\sum_j\sin(\theta_j-\theta_i)$ (1) where the frequencies $\omega$ are drawn from a fixed distribution $g(\omega)$. The model describes the dynamics of the phases $\theta$ of an all-to-all connected network of oscillators. It can be considered to be the weak coupling limit of a set of nonlinear oscillators with different natural frequencies and a synchronizing phase response curve. (more…) Posted in Kinetic Theory, Mathematics, Neuroscience, Papers, Physics, Talks \| 11 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": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 5, "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": 0.9347360730171204, "perplexity_flag": "middle"}
http://math.stackexchange.com/questions/183948/how-can-i-figure-out-the-plot-of-the-set-m-x-y-in-r-frac12-leq-x-la	# How can I figure out the plot of the set $M:=\{(x,y) \in R:\frac{1}{2}\leq x \land 1<y<\frac{1}{x}\}$ I have some trouble to figure out the plot/graph of a set like $$M:=\{(x,y) \in R:\frac{1}{2}\leq x \land 1<y<\frac{1}{x}\}$$ On wolframalpha I saw the plot. Is there a trick how to figure out such kind of graph? Greetings. - 3 Graph all the points $(x,y)$ such that $1/2 \leq x$, then do the same for the points with $1<y$ and $y<1/x$, then consider the intersection of these three regions. – Daniel Pietrobon Aug 18 '12 at 11:12 ## 2 Answers I suppose there's no such trick. Simply, you have to make as much effort as you can to get the more comprehensive view in the structure of the set you are asked to plot. In this case, I suggest you to start with the fixed range of variables, namely $x\ge \frac{1}{2}$ and $y\gt1$. You see that the plane was divided into the quarters. The one you are interested in is the up-right. Then you can move to plotting the more sophisticated part, i.e. $y\lt \frac{1}{x}$. In order to do that, sketch the $y=\frac{1}{x}$ graph and mark the underlying part of the plane. Once you've finished drawing all parts, you have to think what kind of conjugates were used to describe the set. Notice that in your example one has only "and" operators. Therefore, to obtain the result you have to take an intersection of all previously driven components. Don't forget to exclude points which were probably marked by drawing boundaries, as it happens often when sketching sets defined with strict inequalities. You could do that by drawing all "edges" with dashed line first and then bolding ones which reflect weak inequalities. In this case, $y=1$ and $y=\frac{1}{x}$ should be excluded. Edit: To obtain the graph of $y=\frac{1}{x}$, you can start with calculating coordinates of the points from the graph lying on the boundaries of the up-right quarter, namely two points of the form $(x,\frac{1}{x})$, where first has $x$ coordinate equal to $\frac{1}{2}$ and second $y$ one equal to $1$. It provides you with the points $(\frac{1}{2},2)$ and $(1,1)$. The remaining question is: What is the curve I am supposed to draw between the points? If you want the curve to be accurate, you have to be sure about such its features like monotonicity, concavity/convexity... All of them can be termined with tools known from calculus. Let me stop here as I'm not sure if it was exactly what you were asking about. - I just get the three equalities $y=1$, $y=1/x$, and $x=1/2$. Then I graph those lines and choose test points in each region formed to see if they satisfy the inequalities or not. Then I also check the boundaries to see if they should be included in the region or not. Or, you can consider the region described by $1/2\leq x$, then $1<y$, and then $y<1/x$. Once you combine these all together--they all have to hold so you take the intersection--you get your region. -	{"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": 25, "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": 0.9745147228240967, "perplexity_flag": "head"}
http://mathhelpforum.com/algebra/41074-how-do-you-figure-out-can-you-help-me-please.html	# Thread: 1. ## How do you figure this out, can you help me please?? I have real problems trying to figure this out, as I was not in class when er learned this. Does anyone know how to?? Ok so here is the question: 21dd+8d-26= 0 What are the steps that you have to take??? Thank you so much for helping me seetbinti 2. Originally Posted by sweetbinti I have real problems trying to figure this out, as I was not in class when er learned this. Does anyone know how to?? Ok so here is the question: 21dd+8d-26= 0 What are the steps that you have to take??? Thank you so much for helping me seetbinti $21d^2 + 8d - 26 = 0$ The quickest approach is the quadratic formula. -Dan 3. Originally Posted by sweetbinti I have real problems trying to figure this out, as I was not in class when er learned this. Does anyone know how to?? Ok so here is the question: $21dd+8d-26= 0$ What are the steps that you have to take??? Thank you so much for helping me seetbinti $21dd+8d-26= 0 is also written as <br /> 21d^2+8d-26= 0$ now you can factorize this equation or as the above guy said use the quadratic formula, but assuming you don't know it I'll show you the factorizing method !!! so , $21d^2 + 8d - 26 = 0$ we can't split middle term so we do it by completing the square or by converting it to the form (a+b)^2 = 0 so, we first transpose -26 to the RHS so we get $21d^2 + 8d = 26$ we no divide the whole equation by 21 to x^2 coefficient to a perfect square and 1 is a perfect square. so we get $d^2 + 8/21d = 26/21$ can be written as $d^2 + 2(1)(4/21) = 26/21$ no we add (4/21)^2 to both sides just to make the whole expression (a^2 + 2ab + b^2)ish so, $d^2 + 2+(1)(4/21) + (4/21)^2 = 26/21 + (4/21)^2$ factorizing LHS we get $(d+4/21)^2 = 26/21 + 16/441$ $(d+4/21)^2 = (546 + 16)/441$ $(d+4/21)^2 = 562/441<br />$ $d+4/21$ = PLUS OR MINUS $Sqrt(562)/21$ $d$ = PLUS OR MINUS $Sqrt(562)/21 - 4/21$ and thats the solution. Ice Sync 4. Originally Posted by topsquark $21d^2 + 8d - 26 = 0$ The quickest approach is the quadratic formula. -Dan What he said!! $d=\frac{-4\pm\sqrt{562}}{21}$	{"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": 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": 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": 0.953454315662384, "perplexity_flag": "middle"}
http://physics.stackexchange.com/questions/tagged/unit-conversion+electromagnetism	# Tagged Questions 1answer 58 views ### Magnetic Field and the Speed of Light Is it just a historical choice that both magnetic field and the Lorentz force equation include the speed of light? I figure that whoever wrote up the equations (in cgs!) could have put both factors ... 1answer 47 views ### CGS Units for Magnetism Why does the formula for magnetic field force include the speed of light in the denominator in cgs units? Where does the extra $c$ go in SI units? 1answer 167 views ### CGS Units to SI Units This may be a stupid question but I am having trouble getting the same result in CGS units as I would if I used SI units for a unitless calculation. I have to calculate \$E_c=\frac{m_e c\, ... 3answers 102 views ### What is the missing proportionality constant in the magnetic levitation formula? The formula for magnetic levitation is $$B \frac{dB}{dz} = \frac{ \rho g }{\chi}$$ but as always, I have a hard time figuring the units in SI. The left hand side is $\mathrm{T^2 /m}$, while $\chi$ ... 1answer 136 views ### Does the expression of the orbital magnetic dipole moment have $c$? The orbital magnetic dipole moment of a particle with mass $m$ and charge $q$ can be shown to be related to the orbital angular momentum through the equation \displaystyle ... 3answers 808 views ### Coulomb's Law: why is $k = \dfrac{1}{4\pi\epsilon_0}$ This was supposed to be a long question but something went wrong and everything I typed was lost. Here goes. Why is $k = \dfrac{1}{4\pi\epsilon_0}$ in Coulomb's law? Is this an experimental fact? ... 1answer 307 views ### Magnetic susceptibility in 1/eV In this paper the authors refer to transverse susceptibility $\chi_{ \perp}$ [meV $^{−1}$] I was taught that the magnetic susceptibility is dimensionless. How do I get $\chi$ in the above ...	{"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": 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": 0.9366883039474487, "perplexity_flag": "middle"}
http://mathdl.maa.org/mathDL/47/?pa=content&sa=viewDocument&nodeId=3781&bodyId=4055	Search ## Search Loci: Resources: Keyword Advanced Search # Loci: Resources Page 10 of 12 show printer friendly send to a friend # CalcPlot3D, an Exploration Environment for Multivariable Calculus by Paul Seeburger (Monroe Community College) ## Taylor Polynomials of a Function of Two Variables (1st and 2nd degree) The way the Taylor polynomials of a function of one variable progressively converge to the graph of the function like y = cos x is really quite impressive and is inherently interesting. We can extend this topic into three dimensions using CalcPlot3D. As an exercise, I require my students to generate the linear and quadratic Taylor polynomials of a function of two variables using the partial derivatives of the function evaluated at a particular point. $$\begin{eqnarray} f(x,y) &\approx L(x,y) = f(a,b) &+ f_x(a,b)(x-a) + f_y(a,b)(y-b) \qquad (1^{st}\text{-deg. Taylor poly or tangent plane})\\ f(x,y) &\approx Q(x,y) = f(a,b) &+ f_x(a,b)(x-a) + f_y(a,b)(y-b) \\ &+\frac{f_{xx}(a,b)}{2}(x-a)^2 &+ f_{xy}(a,b)(x-a)(y-b) + \frac{f_{yy}(a,b)}{2}(y-b)^2 \qquad (2^{nd}\text{-deg. Taylor poly})\end{eqnarray}$$ Exercise: Determine the 1st and 2nd degree Taylor polynomials in two variables for the given function.Simplify both polynomials. Show all work including all partial derivatives and using the formula clearly with functional notation in the first step. Please also provide a printout of the given surface along with each of the Taylor polynomials. (That’s 2 printouts all together.) Include the point on the surface where the polynomial is tangent to the surface. Use the Format Surfaces option on the View Settings menu so that the Taylor polynomial is reverse color and transparent so it’s possible to tell the two surfaces apart. If necessary, zoom out and the rotate to a view that shows the surfaces clearly. Then use the Print Graph option on the File menu of the applet to print the graph. $$f(x,y) = \sin(2x) + \cos y$$ for x,y near (0,0) \| \| \| \| \|----------\|------------------------------------\|------------------------------------\| \| Answers: \| 1st-degree Taylor Polynomial of f: \| 2nd-degree Taylor Polynomial of f: \| \| \| $$L(x,y)=1+2x$$ \| $$Q(x,y)=1+2x-\frac{1}{2}y^2$$ \| \| \| \| \| There is also a feature of the applet that will allow you to demonstrate higher-degree Taylor polynomials for a function of two variables. ### Example: 1. Graph the function, $$f(x,y)=\cos(x)\sin(y)$$. Then zoom out to -4 to 4 in the x and y-directions. 2. Now select the View Taylor Polynomials option from the Tools menu at the top of the applet. It will take a few seconds as the computer calculates the partial derivatives and creates the Taylor polynomials. This example is successfully calculated all the way up to the 15th degree polynomial. Once it is ready, the original function is graphed as a wireframe and the 1st degree Taylor polynomial (the tangent plane) is shown. A scrollbar appears along the bottom edge of the 3D plot. Use this scrollbar to scroll through the various Taylor polynomials of this function. Note that only odd degrees add new terms for this particular function. As you increase the degree of the Taylor polynomial notice how the polynomial of two variables fits the original surface better and better around the origin until it is a fairly good approximation of the whole visible surface at the 15th degree. 3. To better view the Taylor polynomial itself (shown in the text window just above the 3D plot), you can click and drag on the equation and view all terms, dragging the equation left and right. You can also use the Tools menu option Use Factorials in Taylor Polynomials to switch this property on or off. Using factorials makes the form of the terms of the higher order Taylor polynomials easier to see, and the terms also generally take up less horizontal space each. 4. You can also vary the center point for the Taylor expansion using the Tools menu option just below View Taylor Polynomials. The default center point is the origin. 5. Other nice functions to try centered about the origin include: • $$f(x,y)=\cos(x)-\sin(y)$$ • $$f(x,y)=\sin(2x)-\cos(y)$$ • $$f(x,y)=\sin(x^2+y^2)$$ • $$f(x,y)=xe^y+1$$ • $$f(x,y)=e^{x^2+2x-y}$$ • $$f(x,y)=\arctan(xy)$$ • $$f(x,y)=\arctan(x+y)$$ Click here to open a pdf file which contains the instructions for the activity. Next page >> Lagrange Multiplier Optimization Pages: \| 1 \| 2 \| 3 \| 4 \| 5 \| 6 \| 7 \| 8 \| 9 \| 10 \| 11 \| 12 \| Seeburger, Paul, "CalcPlot3D, an Exploration Environment for Multivariable Calculus," Loci (September 2011), DOI: 10.4169/loci003781	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 12, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 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": 0.8575844764709473, "perplexity_flag": "middle"}
http://www.physicsforums.com/showthread.php?s=41db4cffe5d04f7195077350e912f91f&p=4200772	Physics Forums ## Equation for transverse wave What is [an] equation for a transverse wave with no boundary conditions, as a function of x and t? I want to model a fluctuation string where neither of the ends are bound. 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 The wave equation and the boundary conditions are separate matters. The simplest transverse wave equation is $$\frac{{{\partial ^2}y}}{{\partial {x^2}}} = {c^2}\frac{{{\partial ^2}y}}{{\partial {t^2}}}$$ You need to apply boundary conditions to establish the two arbitrary functions that appear in the solution. That's what I'm confused about. I know that I can express the motion of a vibrating string using sin and cos terms, but I don't know what BC's to apply if the ends of the string are free to move. I want to express the shape of a string using sin and cos, but I am not sure what combination of terms is appropriate here. Would y=Acos(kx-wt) + Bcos(kx+wt) + Csin(kx-wt) + Dsin(kx+wt) fully describe the motion? ## Equation for transverse wave Let the string be ABCD with A, D the ends of the string. If both A and D are 'free to move' then how do you apply tension to the string? If you clamp at intermediated points, say B and C the AB and CD play no part in the wave motion. It does not really matter but I am going to put the c2 in its conventional place in what follows. Sorry that is what happens when you trust to an aging memory. $$\frac{{{\partial ^2}y}}{{\partial {x^2}}} = \frac{1}{{{c^2}}}\frac{{{\partial ^2}y}}{{\partial {t^2}}}$$ Now for fixed ends the boundary conditions are $$y(0,t) = y(l,t) = 0$$ If the ends are not 'free' but still participating in the wave then they can be attributed initial displacement and velocity conditions $$\begin{array}{l} y(x,0) = f(x) \\ {\left( {\frac{{\partial y}}{{\partial t}}} \right)_{t = 0}} = g(x) \\ \end{array}$$ The wave equation itself may be solved by the method of separating the variables $$y(x,t) = F(x)G(t)$$ F is a function of x only and G a function of t only. Substituting and dividing through by y=FG $$\frac{1}{F}\frac{{{d^2}F}}{{d{x^2}}} = \frac{1}{{G{c^2}}}\frac{{{d^2}G}}{{d{t^2}}}$$ Both sides of this equation can only be equal if they are constant. Convention has this constant as -λ2. Some algebra on the resultant pair of ordinary diffrential equations will lead to your required trigonometric solution ( not the one you offered ) where F has the form $${F_n}(x) = \sin \frac{{n\pi x}}{l}$$ and G has the form $${G_n}(t) = {A_n}\cos {\omega _n}t + {B_n}\sin {\omega _n}t$$ Thus $${y_n}(x,t) = {F_n}(x){G_n}(t) = \sin \frac{{n\pi x}}{l}\left[ {{A_n}\cos {\omega _n}t + {B_n}\sin {\omega _n}t} \right]$$ Where A and B are determined by the intial conditiions. In general the solution above will not be complete since it depends upon n. To obtain a complete solution you need to sum solutions over n from n=1 to ∞ Thus $$\begin{array}{l} y(x,0) = f(x) = \sum\limits_{n = 1}^\infty {{A_n}} \sin \frac{{n\pi x}}{l} \\ {\left( {\frac{{\partial y}}{{\partial t}}} \right)_{t = 0}} = g(x) = \sum\limits_{n = 1}^\infty {{B_n}} {\omega _n}\sin \frac{{n\pi x}}{l} \\ \end{array}$$ Thank you, that makes sense. But why does Fn(x) only consist of sin terms, instead of both sin and cos terms? What I originally had was y = Ʃ sin(nπx/L)[An cos($\omega$t) + Bn sin($\omega$t] + cos(nπx/L)[Cn cos($\omega$t) + Dn sin($\omega$t]. Is there a reason not to include the second term? ps, the reason why the ends are unbound is because there is not supposed to be any tension- this is a string in Brownian motion (so there are no real initial conditions that I can apply either, since it does not have to be in any specific configuration at t=0). I guess it would have been better to talk about it as an abstract wave than an actual string. Without tension there is no wave in the string. To derive the wave equation from first mechanical principles you require a restoring force. There is none in your situation. You string takes up the classic random walk line under the influence of brownian motion. This crosses and recrosses itself many times so takes up a but of an undulatory shape. But it is not a wave. Are you familiar with the mathematics of a 'random walk'? You are correct. It is definitely not a real wave in the classical sense. I am familiar with the random walk, and I agree that that is what this is. But since the string will have an internal resistance to bending, that rigidity brings it back to equilibrium (so the resistance to bending is the restoring force). It will fluctuate away from this equilibrium (and I am neglecting the fact that the entire string is diffusing, rotating or doing anything else more complicated) and the time scale is determined by drag forces, and amplitude of these fluctuations is determined the rigidity. But the dynamics of the fluctuation should be able to be modeled as a simple wave equation- I just wanted to check what that wave equation is, since most of the time (for obvious reasons) there actually are boundary conditions. I've modeled the shape at t = 0 as Ʃ Asin(nπx/L) +Bcos(nπx/L). Can I just multiply this by (cos(wt) + sin(wt)) to make it a function of time? Post#3 Would y=Acos(kx-wt) + Bcos(kx+wt) + Csin(kx-wt) + Dsin(kx+wt) Post#6 y = Ʃ sin(nπx/L)[An cos(ωt) + Bn sin(ωt] + cos(nπx/L)[Cn cos(ωt) + Dn sin(ωt]. Post#8 I've modeled the shape at t = 0 as Ʃ Asin(nπx/L) +Bcos(nπx/L). Can I just multiply this by (cos(wt) + sin(wt)) to make it a function of time? I'm not getting at you but that's three different versions of some formula you have posted so far. I obtained my expression by carrying out the algebra (separating the partial equation into two ordinary differential equations, one in x, one in t) as stated in post#5 Using $$\frac{1}{F}\frac{{{d^2}F}}{{d{x^2}}} = - {\lambda ^2}$$ and the boundary conditions F(0)=F(l)=0 leads to the particular solution $${F_n}(x) = \sin \frac{{n\pi x}}{l}$$ Where $${\lambda _n} = \frac{{n\pi }}{l}$$ If we took any value but zero for the cosine term constant in the general solution we could not satisfy the boundary conditions. So we are left with only a sine function in x. Similarly the initial ordinary differential equation in t leads to $$\frac{1}{{G{v^2}}}\frac{{{d^2}G}}{{d{t^2}}} = - {\lambda ^2}$$ with intitial conditions as stated. Since this is an initial value problem that does not require G to vanish we put in our values to determine A and B in the general solution. You cannot avoid putting up some values for these to determine A and B I thought post # 3 and post 6 were equivalent through some trig identity. I see why you applied the boundary conditions and set the coefficient on the cos(n$\pi$x/L) term to zero. But if I want my wave to look and act like the progressive wave shown here: http://www.mta.ca/faculty/science/ph...e/Twave01.html I thought that I can't apply that boundary condition (because the ends aren't actually pinned). I remember deriving the equation that you showed for a particle in the infinate square well and always throwing out the spacial cos term. But I thought that in this case we need to keep it. What I am saying to you is that you need to find some initial conditions to put into f(x) and g(t) in post#5. Alternatively you need to find some other points on your string. You cannot solve for the arbitrary functions/constants without this. The whole point (trick) of the exercise is to choose easy to solve positions on the string. For sure the ends of the string must start and be somewhere. Quote by LizardCobra What is [an] equation for a transverse wave with no boundary conditions, as a function of x and t? I want to model a fluctuation string where neither of the ends are bound. You can use "periodic boundary conditions". You can assume that the displacement and velocity are periodic in distance with a period of length, κL. Here l This can be expressed by the following two equations. f(x,t)=f(x+κL,t), and, ∂f(x,t)/∂t=∂f(x+κL,t)/∂t, for all x and all t. Here, κ is an arbitrary constant that can be any real number. L is a scale length. Thread Tools \| \| \| \| \|---------------------------------------------------\|-------------------------------\|---------\| \| Similar Threads for: Equation for transverse wave \| \| \| \| Thread \| Forum \| Replies \| \| \| Introductory Physics Homework \| 1 \| \| \| Introductory Physics Homework \| 1 \| \| \| Introductory Physics Homework \| 0 \| \| \| Introductory Physics Homework \| 3 \|	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 5, "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": 0.18, "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": 0.9299297332763672, "perplexity_flag": "middle"}
http://mathoverflow.net/questions/32083?sort=newest	## For a morphism f from a regular scheme, should there exist an open subscheme U of the target such that fibre of f at each point of U is regular. ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) For a finite type morphism $f:X\to S$, $X$ is a regular scheme, should there always exist an open (dense) subscheme $U\subset S$ such that the fibre of $f$ at each Zariski point of $U$ is regular? All schemes are excellent. If the anwser is 'yes', then: could one choose such an $U$ such that the preimage of any regular subscheme of $U$ is regular? Are these conditions on $U$ equivalent? - 8 Assume $X$ connected, hence irred. Assume $f$ dominant, or else the complement of closure of the image does the trick. Thus, $S$ is irred (and presumably you assume quasi-compact, so noetherian since excellent). By generic flatness, can assume $f$ flat by passing to dense open in $S$. Generic fiber is regular, so if generic point of $S$ is char. 0 then generic fiber is smooth. Then by flatness, $f$ is smooth over dense open (EGA IV$_4$, sec. 17 somewhere), so win without excellence. False without generic char. 0 (Speyer's example). Final equivalence unclear (I guess false); feels useless. – BCnrd Jul 16 2010 at 0:54 ## 1 Answer Over a field of characteristic zero, your result is true. This is Corollary III.10.7 in Hartshorne. In characteristic $p$ no. The simplest example is to take $k$ an algebraically closed field and map $\mathbb{A}^1_k$ to itself by $x \mapsto x^p$. For every $t \in k$, the fiber above $t$ is $\mathrm{Spec}\ k[x]/(x^p-t) \cong \mathrm{Spec} \ k[y]/y^p$ where the isomorphism is $x-t^{1/p} = y$. So every fiber is singular. There is a more interesting counter-example due to Serre: Let $k$ be an algebraically closed field of characteristic $2$. Consider the planar cubic `$$ F_{a,b,c}(x,y,z) :=a (y^2 z + y z^2) + b (x^2 z + x z^2) + c (x^2 y + x y^2) \quad (*)$$` We leave it to the reader to check that $F=0$ is singular at $(\sqrt{a}:\sqrt{b}:\sqrt{c})$. Generically, the singularity is a node cusp. Choose $(a_1, b_1, c_1)$ and $(a_2, b_2, c_2)$ in $k^3$ and try to map $\mathbb{P}^2_k \to \mathbb{P}^1_k$ by `$$(x:y:z) \mapsto (F_{(a_1,\ b_1,\ c_1)}(x,y,z) : F_{(a_2,\ b_2,\ c_2)}(x,y,z))$$` Then the fiber over $(t_1:t_2)$ is `$F_{(t_1 a_1+t_2 a_2,\ t_1 b_1+t_2 b_2,\ t_1 c_1+t_2 c_2)}=0$` which, as we just computed, is singular. More precisely, the above map is undefined at the $9$ points where $F_{(a_1,\ b_1,\ c_1)} = F_{(a_2,\ b_2,\ c_2)} =0$. But, if we take $X$ to be $\mathbb{P}^2$ blown up at those $9$ points, then we get a map from the regular $X$ to the regular $S$, where every fiber is a nodal cuspidal cubic or worse. Remark: Both of these counter-examples are still counter-examples if you replace "algebraically closed" by "perfect", but it would make my exposition messier. -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 40, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9092424511909485, "perplexity_flag": "head"}
http://en.wikipedia.org/wiki/Pauli_exclusion_principle	# Pauli exclusion principle Quantum mechanics Introduction Glossary · History Background Fundamental concepts Experiments Formulations Equations Interpretations Advanced topics Scientists The Pauli exclusion principle is the quantum mechanical principle that no two identical fermions (particles with half-integer spin) may occupy the same quantum state simultaneously. A more rigorous statement is that the total wave function for two identical fermions is anti-symmetric with respect to exchange of the particles. The principle was formulated by Austrian physicist Wolfgang Pauli in 1925. For example, no two electrons in a single atom can have the same four quantum numbers; if n, ℓ, and mℓ are the same, ms must be different such that the electrons have opposite spins, and so on. Integer spin particles, bosons, are not subject to the Pauli exclusion principle: any number of identical bosons can occupy the same quantum state, as with, for instance, photons produced by a laser and Bose–Einstein condensate. ## Overview The Pauli exclusion principle governs the behavior of all fermions (particles with "half-integer spin"), while bosons (particles with "integer spin") are not subject to it. Fermions include elementary particles such as quarks (the constituent particles of protons and neutrons), electrons and neutrinos. In addition, protons and neutrons (subatomic particles composed from three quarks) and some atoms are fermions, and are therefore subject to the Pauli exclusion principle as well. Atoms can have different overall "spin", which determines whether they are fermions or bosons — for example helium-3 has spin 1/2 and is therefore a fermion, in contrast to helium-4 which has spin 0 and is a boson. As such, the Pauli exclusion principle underpins many properties of everyday matter, from its large-scale stability, to the chemical behavior of atoms. "Half-integer spin" means that the intrinsic angular momentum value of fermions is $\hbar = h/2\pi$ (reduced Planck's constant) times a half-integer (1/2, 3/2, 5/2, etc.). In the theory of quantum mechanics fermions are described by antisymmetric states. In contrast, particles with integer spin (called bosons) have symmetric wave functions; unlike fermions they may share the same quantum states. Bosons include the photon, the Cooper pairs which are responsible for superconductivity, and the W and Z bosons. (Fermions take their name from the Fermi–Dirac statistical distribution that they obey, and bosons from their Bose–Einstein distribution). ## History In the early 20th century it became evident that atoms and molecules with even numbers of electrons are more chemically stable than those with odd numbers of electrons. In the famous 1916 article The Atom and the Molecule by Gilbert N. Lewis, for example, the third of his six postulates of chemical behavior states that the atom tends to hold an even number of electrons in the shell and especially to hold eight electrons which are normally arranged symmetrically at the eight corners of a cube (see: cubical atom). In 1919 chemist Irving Langmuir suggested that the periodic table could be explained if the electrons in an atom were connected or clustered in some manner. Groups of electrons were thought to occupy a set of electron shells about the nucleus.[1] In 1922, Niels Bohr updated his model of the atom by assuming that certain numbers of electrons (for example 2, 8 and 18) corresponded to stable "closed shells". Pauli looked for an explanation for these numbers, which were at first only empirical. At the same time he was trying to explain experimental results in the Zeeman effect in atomic spectroscopy and in ferromagnetism. He found an essential clue in a 1924 paper by Edmund C. Stoner which pointed out that for a given value of the principal quantum number (n), the number of energy levels of a single electron in the alkali metal spectra in an external magnetic field, where all degenerate energy levels are separated, is equal to the number of electrons in the closed shell of the noble gases for the same value of n. This led Pauli to realize that the complicated numbers of electrons in closed shells can be reduced to the simple rule of one electron per state, if the electron states are defined using four quantum numbers. For this purpose he introduced a new two-valued quantum number, identified by Samuel Goudsmit and George Uhlenbeck as electron spin. ## Connection to quantum state symmetry The Pauli exclusion principle with a single-valued many-particle wavefunction is equivalent to requiring the wavefunction to be antisymmetric. An antisymmetric two-particle state is represented as a sum of states in which one particle is in state $\scriptstyle \|x \rangle$ and the other in state $\scriptstyle \|y\rangle$: $\|\psi\rangle = \sum_{x,y} A(x,y) \|x,y\rangle$ and antisymmetry under exchange means that A(x,y) = −A(y,x). This implies that A(x,x) = 0, which is Pauli exclusion. It is true in any basis, since unitary changes of basis keep antisymmetric matrices antisymmetric, although strictly speaking, the quantity A(x,y) is not a matrix but an antisymmetric rank-two tensor. Conversely, if the diagonal quantities A(x,x) are zero in every basis, then the wavefunction component: $A(x,y)=\langle \psi\|x,y\rangle = \langle \psi \| ( \|x\rangle \otimes \|y\rangle )$ is necessarily antisymmetric. To prove it, consider the matrix element: $\langle\psi\| ((\|x\rangle + \|y\rangle)\otimes(\|x\rangle + \|y\rangle)) \,$ This is zero, because the two particles have zero probability to both be in the superposition state $\scriptstyle \|x\rangle + \|y\rangle$. But this is equal to $\langle \psi \|x,x\rangle + \langle \psi \|x,y\rangle + \langle \psi \|y,x\rangle + \langle \psi \| y,y \rangle \,$ The first and last terms on the right hand side are diagonal elements and are zero, and the whole sum is equal to zero. So the wavefunction matrix elements obey: $\langle \psi\|x,y\rangle + \langle\psi \|y,x\rangle = 0 \,$. or $A(x,y)=-A(y,x) \,$ ### Pauli principle in advanced quantum theory According to the spin-statistics theorem, particles with integer spin occupy symmetric quantum states, and particles with half-integer spin occupy antisymmetric states; furthermore, only integer or half-integer values of spin are allowed by the principles of quantum mechanics. In relativistic quantum field theory, the Pauli principle follows from applying a rotation operator in imaginary time to particles of half-integer spin. Since, nonrelativistically, particles can have any statistics and any spin, there is no way to prove a spin-statistics theorem in nonrelativistic quantum mechanics. In one dimension, bosons, as well as fermions, can obey the exclusion principle. A one-dimensional Bose gas with delta function repulsive interactions of infinite strength is equivalent to a gas of free fermions. The reason for this is that, in one dimension, exchange of particles requires that they pass through each other; for infinitely strong repulsion this cannot happen. This model is described by a quantum nonlinear Schrödinger equation. In momentum space the exclusion principle is valid also for finite repulsion in a Bose gas with delta function interactions,[2] as well as for interacting spins and Hubbard model in one dimension, and for other models solvable by Bethe ansatz. The ground state in models solvable by Bethe ansatz is a Fermi sphere. ## Consequences ### Atoms and the Pauli principle The Pauli exclusion principle helps explain a wide variety of physical phenomena. One particularly important consequence of the principle is the elaborate electron shell structure of atoms and the way atoms share electrons, explaining the variety of chemical elements and their chemical combinations. An electrically neutral atom contains bound electrons equal in number to the protons in the nucleus. Electrons, being fermions, cannot occupy the same quantum state, so electrons have to "stack" within an atom, i.e. have different spins while at the same place. An example is the neutral helium atom, which has two bound electrons, both of which can occupy the lowest-energy (1s) states by acquiring opposite spin; as spin is part of the quantum state of the electron, the two electrons are in different quantum states and do not violate the Pauli principle. However, the spin can take only two different values (eigenvalues). In a lithium atom, with three bound electrons, the third electron cannot reside in a 1s state, and must occupy one of the higher-energy 2s states instead. Similarly, successively larger elements must have shells of successively higher energy. The chemical properties of an element largely depend on the number of electrons in the outermost shell; atoms with different numbers of shells but the same number of electrons in the outermost shell have similar properties, which gives rise to the periodic table of the elements. ### Solid state properties and the Pauli principle In conductors and semi-conductors, free electrons have to share entire bulk space. Thus, their energy levels stack up, creating band structure out of each atomic energy level. In strong conductors (metals) electrons are so degenerate that they can not even contribute much to the thermal capacity of a metal. Many mechanical, electrical, magnetic, optical and chemical properties of solids are the direct consequence of Pauli exclusion. ### Stability of matter The stability of the electrons in an atom itself is not related to the exclusion principle, but is described by the quantum theory of the atom. The underlying idea is that close approach of an electron to the nucleus of the atom necessarily increases its kinetic energy, an application of the uncertainty principle of Heisenberg.[3] However, stability of large systems with many electrons and many nuclei is a different matter, and requires the Pauli exclusion principle.[4] It has been shown that the Pauli exclusion principle is responsible for the fact that ordinary bulk matter is stable and occupies volume. This suggestion was first made in 1931 by Paul Ehrenfest, who pointed out that the electrons of each atom cannot all fall into the lowest-energy orbital and must occupy successively larger shells. Atoms therefore occupy a volume and cannot be squeezed too closely together.[5] A more rigorous proof was provided in 1967 by Freeman Dyson and Andrew Lenard, who considered the balance of attractive (electron–nuclear) and repulsive (electron–electron and nuclear–nuclear) forces and showed that ordinary matter would collapse and occupy a much smaller volume without the Pauli principle.[6] The consequence of the Pauli principle here is that electrons of the same spin are kept apart by a repulsive exchange interaction, which is a short-range effect, acting simultaneously with the long-range electrostatic or coulombic force. This effect is partly responsible for the everyday observation in the macroscopic world that two solid objects cannot be in the same place at the same time. ### Astrophysics and the Pauli principle Dyson and Lenard did not consider the extreme magnetic or gravitational forces which occur in some astronomical objects. In 1995 Elliott Lieb and coworkers showed that the Pauli principle still leads to stability in intense magnetic fields such as in neutron stars, although at a much higher density than in ordinary matter.[7] It is a consequence of general relativity that, in sufficiently intense gravitational fields, matter collapses to form a black hole. Astronomy provides a spectacular demonstration of the effect of the Pauli principle, in the form of white dwarf and neutron stars. In both types of body, atomic structure is disrupted by large gravitational forces, leaving the constituents supported by "degeneracy pressure" alone. This exotic form of matter is known as degenerate matter. In white dwarfs atoms are held apart by electron degeneracy pressure. In neutron stars, subject to even stronger gravitational forces, electrons have merged with protons to form neutrons. Neutrons are capable of producing an even higher degeneracy pressure, albeit over a shorter range. This can stabilize neutron stars from further collapse, but at a smaller size and higher density than a white dwarf. Neutrons are the most "rigid" objects known; their Young modulus (or more accurately, bulk modulus) is 20 orders of magnitude larger than that of diamond. However, even this enormous rigidity can be overcome by the gravitational field of a massive star or by the pressure of a supernova, leading to the formation of a black hole. ## References 1. Langmuir, Irving (1919). "The Arrangement of Electrons in Atoms and Molecules" (– ). Journal of the American Chemical Society 41 (6): 868–934. doi:10.1021/ja02227a002. Retrieved 2008-09-01. [][] 2. Elliot J. Lieb The Stability of Matter and Quantum Electrodynamics 3. This realization is attributed by Lieb and by GL Sewell (2002). Quantum Mechanics and Its Emergent Macrophysics. Princeton University Press. ISBN 0-691-05832-6. to FJ Dyson and A Lenard: Stability of Matter, Parts I and II (J. Math. Phys., 8, 423–434 (1967); J. Math. Phys., 9, 698–711 (1968) ). 4. As described by FJ Dyson (J.Math.Phys. 8, 1538–1545 (1967) ), Ehrenfest made this suggestion in his address on the occasion of the award of the Lorentz Medal to Pauli. 5. FJ Dyson and A Lenard: Stability of Matter, Parts I and II (J. Math. Phys., 8, 423–434 (1967); J. Math. Phys., 9, 698–711 (1968) ); FJ Dyson: Ground-State Energy of a Finite System of Charged Particles (J.Math.Phys. 8, 1538–1545 (1967) ) 6. E.H. Lieb, M. Loss and J.P. Solovej, Phys. Rev. Letters, 75, 985–9 (1995) "Stability of Matter in Magnetic Fields" • Dill, Dan (2006). "Chapter 3.5, Many-electron atoms: Fermi holes and Fermi heaps". Notes on General Chemistry (2nd ed.). W. H. Freeman. ISBN 1-4292-0068-5. • Griffiths, David J. (2004). Introduction to Quantum Mechanics (2nd ed.). Prentice Hall. ISBN 0-13-805326-X. • Liboff, Richard L. (2002). Introductory Quantum Mechanics. Addison-Wesley. ISBN 0-8053-8714-5. • Massimi, Michela (2005). Pauli's Exclusion Principle. Cambridge University Press. ISBN 0-521-83911-4. • Tipler, Paul; Llewellyn, Ralph (2002). Modern Physics (4th ed.). W. H. Freeman. ISBN 0-7167-4345-0.	{"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": 10, "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": 0.8932684659957886, "perplexity_flag": "middle"}
http://math.stackexchange.com/questions/89710/matrices-for-which-mathbfa-1-mathbfa	# Matrices for which $\mathbf{A}^{-1}=-\mathbf{A}$ Consider matrices $\mathbf{A}\in\mathbb{C}^{n\times n}$ (or maybe $A\in\mathbb{R}^{n\times n}$) for which $\mathbf{A}^{-1}=-\mathbf{A}$. A conical example (and the only one I can come up with) would be $\mathbf{A} = \boldsymbol{i}\mathbf{I},\quad \boldsymbol i^2=-1$. 1. Are there more matrices than this example matrix (I guess yes) or can they even be generally constructed somehow? 2. Are there also real matrices for which this holds? 3. Now each matrix that is both skew-Hermitian and unitary fulfills this property. But does it also hold in the other direction, meaning is each matrix for which $\mathbf{A}^{-1}=-\mathbf{A}$ both skew-Hermitian and unitary (maybe this is simple to prove, but I don't know where to start at the moment, but of course I know if one holds the other has to hold, too)? 4. Do such matrices have any practical meaning? For example I know that Hermitian and unitary matrices are reflections (in a general sense), but what about skew-Hermitian and unitary (if 3 holds)? This is just for personal interrest without any practical application. I just stumbled accross this property by accident and want to know more about its implications and applications. - 4 $\left(\array{0 & 1 \\ -1 & 0}\right)$ – Alexander Thumm Dec 8 '11 at 20:41 @AlexanderThumm Of course, as simple as it can be, me stupid! But what about more general matrices? – Christian Rau Dec 8 '11 at 20:44 3 Easier to deal with $A^2=-I$, so you can say something about the eigenvalues of $A$. – Thomas Andrews Dec 8 '11 at 20:45 Considering Characteristic equation, your matrix $A$ must satisfy $$cA+dI=0$$ where $c,d\in \mathbf{C}$ and we know their expression in terms of sum of K-ordered minors of $A$. – Tapu Dec 8 '11 at 21:21 ## 4 Answers Note that the condition is invariant under conjugation, so any conjugate of a matrix satisfying this property also satisfies this property. It is simpler to write the condition as $A^2 = -I$. Note that taking determinants of both sides gives $(\det A)^2 = (-1)^n$, so if $n$ is odd then $A$ cannot be real. (Another way to see this is to note that, since $x^2 + 1$ is irreducible over $\mathbb{R}$, the characteristic polynomial of $A$ is necessarily $(x^2 + 1)^k$ for some $k$.) lhf's example shows that real examples always exist when $n$ is even. Because the polynomial $x^2 = -1$ has no repeated roots, $A$ is diagonalizable with eigenvalues $\pm i$ and the converse holds. It is bad practice to ask whether a matrix is skew-Hermitian or unitary. This is not really a property of a matrix. It is a property of a linear operator on a complex inner product space. It should not be hard to construct examples which are neither skew-Hermitian nor unitary by conjugating a diagonal matrix with entries $\pm i$ by a non-unitary matrix. - Thanks for your answer, but I have difficulties to understand the last sentence. What do you mean with "by conjugating by a non-unitary matrix". Now I understand that it is invariant under conjugation. But aren't the skew-Hermitian and unitary properties invariant under conjugation, too? I don't know how cojugation helps with constructing a counter-example. Or do you mean something different (I don't really understand what conjugation by a non-unitary matrix means)? – Christian Rau Dec 9 '11 at 1:21 1 @Christian: skew-Hermitian and unitary are only invariant under conjugation by a unitary matrix. "By" refers to the matrix $G$ doing the conjugating $D \mapsto GDG^{-1}$. – Qiaochu Yuan Dec 9 '11 at 1:22 I just cannot get what conjugation by a unitary matrix means. I guess with conjugation you mean not just the simple complex conjugate of the matrix elements? Sorry, but I'm not that deeply versed in linear algebra, especially if it gets too theoretical. – Christian Rau Dec 9 '11 at 1:24 Ah, ok. Now I understand what you mean by conjugation. And I understand why this keeps the mentioned property but destroys the unitarity property. Thanks again. – Christian Rau Dec 9 '11 at 1:26 Let $J=\left(\array{0 & 1 \\ -1 & 0}\right)$ as in Alexander's comment. Then $A=\operatorname{diag}(J,\dots,J)$ is a $2n \times 2n$ matrix with $A^2=-I$. - The equation $A^{-1} = -A$ is equivalent to $A^2 = -I$. A matrix (over the complex numbers) that satisfies this must be similar to a diagonal matrix with diagonal entries $i$ and $-i$. A real matrix that satisfies it must be similar to lhf's example. - It is clear that your condition is equivalent to $-I=A^2$ (Multiply both sides with A). Now, Any matrix B that satisfy $B^2 = I$, will give a matrix $A = \pm i B$ that satisfy your relation, since $(\pm iB)^2 = i^2 B^2 = -I.$ Conversely, If $A^2 = -I$ then $B = \pm i A$ yields a matrix $B$ with the property $B^2 = 1.$ Thus, you are essentially looking for matrices that satisfy $B^2=I$ These matrices, if I am not mistake, are essentially reflections of different types. - 1 The standard $\mathbb R^{2\times 2}$ example is not a reflection, but a rotation by 90°. In fact, reflections cannot be examples, as they always satisfy $B^2=I$ rather than $B^2=-I$ – Henning Makholm Dec 8 '11 at 21:09 @HenningMakholm He just means that a matrix with the searched property can be constructed from any reflection matrix. Though his last paragraph is a bit obsolete, but +1 for the first paragraph. – Christian Rau Dec 9 '11 at 1:56	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 48, "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": 0.9352269172668457, "perplexity_flag": "head"}
http://meetings.aps.org/Meeting/MAR10/SessionIndex3/?SessionEventID=125408	# Bulletin of the American Physical Society ## APS March Meeting 2010 Volume 55, Number 2 Monday–Friday, March 15–19, 2010; Portland, Oregon ### Session W26: Quantum Control and Resources for Quantum Computing Sponsoring Units: GQI Chair: Frank Wilhelm, University of Waterloo Room: D136 Thursday, March 18, 2010 11:15AM - 11:27AM W26.00001: Quantum Eraser and Phase-Matching for Exponential Spin-Squeezing via Coherent Optical Feedback Collin Trail , Ivan Deutsch , Poul Jessen , Leigh Norris A scheme for squeezing collective atomic spin states via coherent optical feedback was proposed by M. Takeuchi et. al., Phys. Rev. Lett. 94, 023003, 2005. In the first pass, the Faraday effect acts to entangle the light with the atoms. In a coherent second pass, this information is imprinted back onto the atoms, creating an effective nonlinear interaction and entanglement between atoms. However, the light is still entangled to the atoms when it escapes, leading to substantial decoherence, and moreover, the interaction slowly rotates the system out of sync with the squeezing axis, both of which result in suboptimal squeezing. We show how the addition of a quantum eraser and phase matching can lead to radically improved exponential scaling. We analyze this system in the presence of realistic imperfections such as photon scattering, optical pumping, losses in transmission and reflection, finite detector efficiency, and nonprojective measurements, and show that spin squeezing near 10 dB should be possible. Thursday, March 18, 2010 11:27AM - 11:39AM W26.00002: High fidelity gates with simple pulses Felix Motzoi , Jay Gambetta , Frank Wilhelm Fast oscillating terms in the Hamiltonian such as off-resonant excitation and counter- rotating terms can be a significant source of error in quantum information implementations. We expand on techniques first developed for controlling 3-level systems, Derivative Removal by Adiabatic Gate (DRAG) published in PRL 103 110501, which was experimentally tested in arXiv:0908.1955v1. Here we show how this technique can be applied specifically to 2-level systems, selective qubit addressing, and in general to multi-channel leakage systems to vastly improve fidelities. In all these cases, the error corresponds to a dragging of the adiabatic frame used for the coherent rotation, which is proportional to the derivative and can easily be corrected using a second quadrature control. The pulse shapes obtained are smooth and very short in duration, corresponding to very few cycles of the unwanted fast oscillation. Thursday, March 18, 2010 11:39AM - 11:51AM W26.00003: Generation of ion-photon entangled state with trapped barium ions Nathan Kurz , Gang Shu , Matthew Dietrich , Katherine Mitchell , Boris Blinov Trapped ions are an attractive qubit candidate, particularly for their long coherence times and natural coupling to photons. We trap single Barium-138 ions in a linear Paul trap and excite these ions with mode-locked pulses to generate a single photon per excitation. The Zeeman sublevels of the ground state can then be entangled with orthogonal polarization modes of the photons emitted upon decay from either the $6P_{3/2}$ or $6P_{1/2}$ state at 455 and 493 nm, respectively. Preliminary results have been obtained using weak continuous wave excitation to generate single photons at 493 nm, as confirmed by anti-coincidence measurements. Such work represents a crucial step toward the generation of multi-ion entangled states. Thursday, March 18, 2010 11:51AM - 12:03PM W26.00004: Qubit decoherence due to detector switching Frank Wilhelm , Ioana Serban We provide insight into the qubit measurement process involving a switching type of detector. We study the switching-induced decoherence during escape events. We present a simple method to obtain analytical results for the qubit dephasing and bit-flip errors, which can be easily adapted to various systems. Within this frame we investigate potential of switching detectors for a fast but only weakly invasive type of detection. We show that the mechanism that leads to strong dephasing, and thus fast measurement, inverts potential bit flip errors due to an intrinsic approximate time reversal symmetry. Based on arXiv:0905.3045 Thursday, March 18, 2010 12:03PM - 12:15PM W26.00005: CNOT gates with weakly coupled qubits: Dependence of fidelity on form of interaction Joydip Ghosh , Michael Geller An approach to the construction of a CNOT quantum logic gate for a 4-dimensional coupled-qubit model with weak but otherwise arbitrary coupling has been given recently (e-print arXiv0906.5209). How does the resulting fidelity depend on the form of qubit-qubit coupling? We calculate intrinsic fidelity curves (fidelity vs. total gate time) for a variety of qubit-qubit interactions, including the commonly occurring isotropic Heisenberg and XY models, as well as randomly generated ones. For interactions not too close to that of the Ising model, we find that the fidelity curves do not significantly depend on the form of the interaction, and we provide the fidelity curve for the non-Ising-like cases and a criterion for determining its applicability. Thursday, March 18, 2010 12:15PM - 12:27PM W26.00006: Optical generation of Fock states in a weakly nonlinear oscillator Botan Khani , Jay Gambetta , Felix Motzoi , Frank Wilhelm We apply optimal control theory to determine the shortest time in which an energy eigenstate of a weakly anharmonic oscillator can be created under the practical constraint of linear driving. We show that the optimal pulses are beatings of mostly the transition frequencies for the transitions up to the desired state and the next leakage level. The time of a shortest possible pulse for a given nonlinearity scale with the nonlinearity parameter as a power law. This power law is weaker than the one expected by a simple spectroscopic argument, emphasizing the benefits of quantum interference. Furthermore we confirm that even with realistic models for decoherence high fidelity energy eigenstates can be achieved. Thursday, March 18, 2010 12:27PM - 12:39PM W26.00007: Free-time and Fixed End-point Multi-target Optimal Control Theory: Application to Quantum Computing Kenji Mishima , Koichi Yamashita An extension of monotonically convergent free-time and fixed end-point optimal control theory (FRFP-OCT) to monotonically convergent free-time and fixed end-point \textit{multi-target} optimal control theory (FRFP-MTOCT) is presented. The features of our theory include optimization of the laser pulses with high transition probabilities, that of the temporal duration, the monotonic convergence, and the ability to optimize multiple-laser pulses simultaneously. The advantage of the theory and a comparison with conventional fixed-time and fixed end-point multi-target optimal control theory (FIFP-MTOCT) are presented by comparing data calculated using the present theory with those published previously [K. Mishima and K. Yamashita, Chem. Phys. \textbf{361}, 106 (2009)], where qubit system of interest consists of two polar NaCl molecules coupled by dipole-dipole interaction. Thursday, March 18, 2010 12:39PM - 12:51PM W26.00008: Feedback Control of the Two Components of a Schroedinger-Cat Justin Finn , Kurt Jacobs While quantum resonators can exist in mesoscopic superpositions of different locations in phase space (so called Schroedinger-cat states), to date feedback control protocols have been restricted to stabilizing such systems about a single point in phase space. This is due to the fact that measurements usually destroy phase-space superpositions. Here we show how it is possible to realize the feedback control of a Schroedinger-cat state of a mesoscopic resonator, by using a combination of linear and quadratic measurements of position. We show how these measurements can be realized experimentally, and present an explicit protocol for tracking and controlling the components of a cat-state in real-time. Thursday, March 18, 2010 12:51PM - 1:03PM W26.00009: Manipulation of the dynamics of many-body systems via quantum control methods Julie Dinerman , Lea Santos We investigate how dynamical decoupling methods may be used to manipulate the transport behavior of quantum many-body systems. These methods consist of sequences of unitary transformations designed to induce a desired dynamics. The systems considered for the analysis are one-dimensional spin-1/2 models, which, according to the parameters of the Hamiltonian, may be in the integrable or non-integrable limits, and in the gapped or gapless phases. Given a system in a certain regime, we develop control sequences that lead to an effective evolution typical of a system in the opposite regime, that is, a chaotic chain evolves as an integrable one and a system in the gapless phase behaves as a gapped one. Thursday, March 18, 2010 1:03PM - 1:15PM W26.00010: Magic recovery of spin coherence in an interacting quantum bath Nan Zhao , Jian-Liang Hu , Ren-Bao Liu The magic recovery of the spin coherence is a distinguished feature due to the quantum nature of the spin bath, which was first proposed from a pseudo-spin model. Here we proved without resorting to a specific bath model that this magic coherence recovery is general. Based on this proof and numerical simulations, we proposed the conditions under which the $\sqrt{2}\tau$ recovery would be experimentally observed. In particular, we explored the possibility of observing the phenomenon by measuring central nuclear spin coherence in solid state NMR experiments, where the coherence time $T_{2}$ and the inhomogeneous broadening time $T^{*}_{2}$ are of the same order. Thursday, March 18, 2010 1:15PM - 1:27PM W26.00011: Path integral representation of a two qubit system Justin Wilson , Victor Galitski In the path integral representation of a one qubit system, extra degrees of freedom are needed to pass from the Hamiltonian formulation to the path integral (Lagrangian) formulation. This leads to a topological term in the Lagrangian much like a Wess-Zumino term. Such a term is topological and is related to the Hopf fibration of $S^3$ by $S^1$ over $S^2$ (and indeed this term appears even when the Hamiltonian is zero). There is an analogous Hopf fibration for the two qubit state from $S^7$ by $S^3$ over $S^4$. We explore how this is related to the topological term in the path integral formulation for two qubit systems. Thursday, March 18, 2010 1:27PM - 1:39PM W26.00012: Fermionic Resources for Quantum Teleportation Adam D'Souza , David Feder The measurement-based quantum computing (MBQC) model requires the creation of a massively entangled resource state,'' on which computation proceeds via single-qubit measurements. Although 2D resource states are believed necessary for universal MBQC, 1D states can serve as resources for certain tasks as well, such as quantum teleportation. One possible route to a resource state is to cool a gapped, two-body system whose ground state encodes the resource. I will discuss our recent work in this area, in which we investigate candidate fermionic systems using the Density Matrix Renormalization Group method and the Matrix Product States description of highly entangled 1D states. Thursday, March 18, 2010 1:39PM - 1:51PM W26.00013: Quantum computation in the ground state of interacting fermions David Feder , Gora Shlyapnikov In measurement-based quantum computation (MBQC), an algorithm proceeds entirely by making projective measurements on successive qubits comprising some highly entangled `resource state.' While two-dimensional cluster states are known to be universal resources for MBQC, it has been proven that they cannot be the unique ground states of any two-body spin Hamiltonian. We show that a particular ground state of non-interacting fermions (equivalent to a many-body spin system) is formally equivalent to a cluster state, though only capable of simulating a limited set of quantum operations. In the presence of two-particle interactions, however, the ground state becomes a universal resource for MBQC. This result suggests that arbitrary quantum algorithms could be simulated fault-tolerantly simply by measuring a cold gas of interacting fermions, such as ultracold atoms in optical lattices. Thursday, March 18, 2010 1:51PM - 2:03PM W26.00014: How to implement a quantum algorithm on a large number of qubits by controlling one central qubit Alexander Zagoskin , Sahel Ashhab , J.R. Johansson , Franco Nori It is desirable to minimize the number of control parameters needed to perform a quantum algorithm. We show that, under certain conditions, an entire quantum algorithm can be efficiently implemented by controlling a single central qubit in a quantum computer. We also show that the different system parameters do not need to be designed accurately during fabrication. They can be determined through the response of the central qubit to external driving. Our proposal is well suited for hybrid architectures that combine microscopic and macroscopic qubits. More details can be found in: A.M. Zagoskin, S. Ashhab, J.R. Johansson, F. Nori, Quantum two-level systems in Josephson junctions as naturally formed qubits, Phys. Rev. Lett. 97, 077001 (2006); and S. Ashhab, J.R. Johansson, F. Nori, Rabi oscillations in a qubit coupled to a quantum two-level system, New J. Phys. 8, 103 (2006). Thursday, March 18, 2010 2:03PM - 2:15PM W26.00015: Self-organization in optical lattices studied within the positive-P representation Ray Ng , Erik S. Sorensen The positive-P representation is a commonly used quantum phase space method in quantum optics. It allows for the conversion of the master equation of a quantum mechanical system to a Fokker-Planck Equation, which can then be mapped on to a set of Stochastic Differential Equations. This makes it an ideal method when dealing with open systems and for studying real time dynamics. We use the positive-P representation to simulate ultra cold atoms trapped in an optical lattice within a cavity in the presence of a coupling to a resonant mode. It has been proposed that in this system the trapped atoms self-organize from a uniform starting configuration of equally occupied lattice sites to one where either only even or odd lattice sites are occupied.	{"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": 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": 0.8894544243812561, "perplexity_flag": "middle"}
http://math.stackexchange.com/questions/44004/why-do-books-say-of-course-its-never-that-simple-differential-equations	# Why do books say “of course” it's never that simple, differential equations? I'm reading about differential equations. It says that a first order ODE may be written: M(x,y)dx + N(x,y)dy = 0 Then if there is some function f(x,y) such that df/dx = M and df/dy = N you can write: ( df/dx )dx + ( df/dy ) dy = 0 Then it says "Of course, the only way that such an equation holds is if df/dx = 0 and df/dy = 0. And this entails that the function f be identically constant. f(x,y) =c." Here is the whole excerpt: I don't see why this is the case. First, I thought df/dy and df/dx were not zero but rather df/dx =M and df/dy =N... also, if two things add to zero they both don't need to be zero, one can be negative the other positive? The fact that it says "of course" is making me feel pretty dumb. This is from the section on "The Method of Exact Equations" Thanks. - 2 @tutorscomputer: Could you give the source and/or a complete quote? What would make more sense if it said that $dM/dy = dN/dx$. – Arturo Magidin Jun 8 '11 at 4:09 1 @tutorscomputer: you added material after the assertion. To provide context, you need to provide material leading up to the assertion. If you could quote the entire paragraph leading up to the assertion, that would help. – Arturo Magidin Jun 8 '11 at 4:20 1 Could it be that the $dx$ and $dy$ mean that there can be no "formal cancellation"? – Arturo Magidin Jun 8 '11 at 4:29 1 From the Schaum's Outline of Differential Equations maybe? Or the Schaum's Outline of Partial Differential Equations? – Doug Spoonwood Jun 8 '11 at 5:23 1 – Martin Sleziak Jun 8 '11 at 6:48 show 9 more comments ## 2 Answers The equation $$M(x,y) dx + N(x,y) dy = 0$$ means the following: that $x$ and $y$ are related in such a way that if we start at any point $(x,y)$, and then make an infinitesimal change $dx$ in $x$, then the corresponding change in $dy$ is such that the above expression equals zero. It might be easier to understand if you rewrite this as $$M\bigl(x,y(x)\bigr) + N\bigl(x,y(x)\bigr) \dfrac{dy}{dx} = 0.$$ This means that $y$ is a function of $x$, and the derivative $dy/dx$ satisfies the preceding equation. It has the same meaning as the preceding equation with $dx$ and $dy$ separated. Now suppose that $M(x,y) = \dfrac{\partial f}{\partial x}(x,y)$ and $N(x,y) = \dfrac{\partial f}{\partial y}(x,y).$ Then we may rewrite the preceding equation as $$\frac{\partial f}{\partial x}\bigl(x,y(x)\bigr) + \frac{\partial f}{\partial y}\bigl(x,y(x)\bigr) \frac{dy}{dx} = 0,$$ which by the chain rule is the same as saying that $$\frac{d f\bigl(x,y(x)\bigr)}{d x} = 0,$$ or equivalently, that $f\bigl(x,y(x)\bigr)$ is constant, as $x$ varies. Your textbook phrases things slightly differently, because it is using the first version of the differential equation: it writes instead $$\frac{\partial f}{\partial x} (x,y) d x + \frac{\partial f}{\partial y}(x,y) dy = 0.$$ Now this is the total change in $f(x,y)$ when $x$ changes by $dx$ and $y$ changes by $dy$. So what this equation says is that if $x$ and $y$ are related so that the changes $dx$ and $dy$ are related to each other by the original equation, then the quantity $f(x,y)$ will not change as $x$ (and hence $y$) changes. So it is not asserting that $f(x,y)$ is constant for any values of $x$ and $y$; rather, that $f(x,y)$ is constant as $x$ and $y$ change according to the constraints of the original equation. In short, $$f(x,y) =c$$ is a solution of the differential equation. [As far as I can tell, the reasoning, such as there is, in the text that follows the "Of course" is completely bogus; it is arguing as if $dx$ and $dy$ are independent quanities. If they were, i.e. if the change in $f(x,y)$ were zero no matter how we varied $x$ and $y$, then indeed $f(x,y)$ would be constant for all values of $x$ and $y$, but then the differential equation wouldn't be very interesting, since $M$ and $N$ themselves would be the zero functions. Rather, the d.e. asserts that the change in $f(x,y)$ is zero if we vary $x$ and $y$ in some related way, and the conclusion one wants to draw is the one I drew above, namely that $f(x,y)$ is constant when $x$ and $y$ vary according to the particular constraint of the d.e. Let me give an example: we could take $f(x,y) = xy$, just to choose one of the infinitely many different functions $f(x,y)$ of two variables that you could write down. Then $M = y$ and $N = x$, and the d.e. is $$y dx + x dy = 0,$$ or, in the second form that I wrote it, $$y + x \frac{dy}{dx} = 0.$$ Now the solutions is given by $x y = c$, i.e. by $$y = \frac{c}{x},$$ for an arbitrary constant $c$. (Check!) Note that is certainly not true that the partial derivatives of $f$, i.e. $M$ and $N$, vanish identically! So one (slightly sarcastic) answer to the question as to why an author writes "of course" is that this is mathematical shorthand for making an unjustified assertion that is quite possibly wrong!] - The book is being lazy with this derivation, and the "of course" is meant as a bluff to make you not think very hard about why the next step is true. It should read: Then if there is some function $f(x,y)$ such that $$\frac{\partial f}{\partial x} = M\qquad\text{and}\qquad\frac{\partial f}{\partial y} = N\text{,}$$ then we can rewrite the differential equation as $$\frac{\partial f}{\partial x}dx \,+\, \frac{\partial f}{\partial y}dy \;=\; 0.$$ From the multivariable chain rule, we can recognize the left side of the equation as the total differential $df$ of the function $f$. In particular, the equation states that $df = 0$, i.e. that $f$ is identically constant. In other words, $$f(x,y) \;\equiv\; c.$$ -	{"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": 66, "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": 0.960873007774353, "perplexity_flag": "head"}
http://mathoverflow.net/questions/9307?sort=newest	## Units in Ornstein-Uhlenbeck(OU) process ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) Take an OU process characterized by ````X(0) = x dX(t) = - a X(t) dt + b dW(t) ```` where a,b >0. The parameter a is usually interpreted a dissipative term, and b is a volatility term. My question is this: What are the units of a and b? Is it true that a is (time) -1 , and b is unitless? Then how can one make sense of the variance which approaches (b 2 /(2 a)) as t goes to infinity? - 1 You don't seem to have defined W. – Qiaochu Yuan Dec 18 2009 at 20:39 $W_t$ is the Wiener process. – Steve Huntsman Dec 18 2009 at 20:43 ## 2 Answers Say $X$ is a displacement and is measured in meters. Then $a$ indeed has units $1/s$ and $b$ has units $m/\sqrt{s}$; $dt$ as usual has units $s$ and $dW$ has units $1/\sqrt{s}$. This can be verified by looking at a physical model of the OU process, such as Hooke's law with damping and a noise term (see Wikipedia). Then $a = - k/\gamma$, $b^2 = 2 k_b T/\gamma$, where $k$ is Hooke's constant in kg/s^2, $\gamma$ the friction coefficient in kg/s, and $k_b T$ is in Joules (kg*m^2/s^2). - ### 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 your setup the Wiener term carries units. Think of the fact that $W_t - W_s \sim \mathcal{N}(0,t-s)$ for $s < t$. - And just to clarify: this means that $dW_t$ has units of (time)^{1/2}, so b has units of (time)^{-1/2}. Since as you believed a has units of (time)^{-1}, this means that the variance is dimensionless. – Steve Huntsman Dec 18 2009 at 20:42 Thinking of $dW_t$ as having units of (time)^{1/2} is a useful heuristic for checking if expressions in stochastic calculus are dimensionally consistent. – Michael Lugo Dec 18 2009 at 20:44 I always remember it as dW^2 = dt by Ito calculus, hence the 1/2. – Alex R. Jan 8 2011 at 3:23	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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": 0.9371208548545837, "perplexity_flag": "middle"}
http://mathoverflow.net/questions/39813/pointer-to-literature-on-double-enrichment-and-functors-among-enriching-categorie/39815	## Pointer to literature on double enrichment and functors among enriching categories? ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) I'm currently working with the following two situations: 1. $\mathbb A$ is a monoidal category, $\mathbb B$ is an $\mathbb A$-enriched monoidal category, and $\mathbb C$ is a $\mathbb B$-enriched category (and ${\mathbb A}\ncong {\mathbb B}$, ${\mathbb B}\ncong {\mathbb C}$, ${\mathbb A}\ncong {\mathbb C}$). 2. $\mathbb V$ and $\mathbb W$ are monoidal categories, $v$ is a $\mathbb V$-enriched category, $w$ is a $\mathbb W$-enriched category, and there is a monoidal functor $F:{\mathbb V}\to{\mathbb W}$. I haven't been able to find much in Kelly's Basic Concepts of Enriched Category Theory, but surely others have come across these two situations before; there are lots of easy results (like, the image of $v$ under $F$ is a $\mathbb W$-enriched category) which must have been known for ages. Are there terms for the situations above? That would help me search the literature. If not, even just one paper on either of these cases with a bibliography would be a good foothold. So far everything I've found on enriched category theory seems to focus on the case where there's only one enriching category in which all of the enriched categories are enriched (or the self-enrichment case where $\mathbb D$ is isomorphic to a $\mathbb D$-enriched category). Thank you! - ## 2 Answers One key phrase to look for is "change of base" (which of course means many other things, but category theorists have also used it to mean your #2). I don't know a canonical reference at a commonly used level of generality, but all these facts are certainly folklore in categorical circles. Your #1 might be called "iterated enrichment," and in good situations, it is a special case of #2. If A is a monoidal category and B is a monoidal A-category, then (the underlying ordinary category of) B is monoidal, and we have a (lax) monoidal functor B→A given by the A-valued $\underline{Hom}(I,-)$ where I is the unit object of B. In particular, every B-enriched category becomes an A-enriched category in a canonical way. If B is cocomplete, then this functor has a left adjoint (given by copowers of I) which is (automatically) colax monoidal, and strong monoidal if B is closed. Conversely, given a lax monoidal functor R:B→A where B is closed and hence a B-enriched category, then B becomes A-enriched according to #2. Edit: I remembered one reference that I like: Geoff Cruttwell's thesis. - Thanks Mike! Sadly the phrase "change of base" means (as you mention) so many things that it isn't really workable as a search phrase. I like the term "iterated enrichment" quite a bit, and will adopt it immediately. As you guessed, the two scenarios are related, but unfortunately I can't assume that $\mathbb B$ is self-enriched (closed). I voted your answer up, but I'm going to hold off on accepting an answer in order to see if somebody can come up with a citation of some sort. – Adam Sep 27 2010 at 4:28 I remembered one reference. – Mike Shulman Sep 27 2010 at 20:59 I think that Eilenberg, S. & Kelly, G.M. Closed categories in: Proceedings of the Conference on Categorical Algebra. (La Jolla, 1965) is the earliest reference treating change of base in the sense you mention. – Daniel Schäppi Mar 11 2012 at 3:16 ### 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. The thesis of Dominic Verity treats base change in a very abstract 2-categorical setting - looks like a lot of specialization is necessary until you arrive at the sort of situation sketched in the question. -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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": 0.3, "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": 0.9465652704238892, "perplexity_flag": "middle"}
http://mathhelpforum.com/advanced-statistics/85396-what-heck-standard-error.html	Thread: 1. What the heck is the standard error? I want to understand better what a standard error really is (I read conflicting things). Can anyone help me to get it? Is it correct to say a standard error is the standard deviation of the sampling distribution of a statistic? By sampling distribution I mean a statistic (a mean or a regression coefficient for example) is computed based on a sample and is thus a random variable. That random variable has a distribution. That distribution is called the sampling distribution. The standard deviation of that distribution is the standard error. Am I correct?? Is a standard error always used as a component in a confidence interval? If you are using statistical software and get a "standard error" listed with the point estimate for a regression coefficient lets say, can you multiply it by 1.96 and construct a confidence interval or does it depend on if there are better formulas for the CI? 2. 1.96 is a N(0,1) percentile that gives you 95 precent coverage in your intervals, i.e., P(-1.96<Z<1.96)=.95. So, that's for a two-sided interval with probability .95. Now the standard error is the estimate of the standard deviation. For example, the variance of the sample mean, for i.i.d. rvs is... $V(\bar X)={\sigma^2\over n}$. Hence the standard deviation of $\bar X$ is ${\sigma\over \sqrt n}$. BUT we usually don't know $\sigma$, so we estimate the st deviation with this standard error. Our estimate of $\sigma$ is S so the standard error of $\bar X$ is ${S\over \sqrt n}$ our standard error. 3. Originally Posted by B_Miner I want to understand better what a standard error really is (I read conflicting things). Can anyone help me to get it? Is it correct to say a standard error is the standard deviation of the sampling distribution of a statistic? By sampling distribution I mean a statistic (a mean or a regression coefficient for example) is computed based on a sample and is thus a random variable. That random variable has a distribution. That distribution is called the sampling distribution. The standard deviation of that distribution is the standard error. Am I correct?? Is a standard error always used as a component in a confidence interval? If you are using statistical software and get a "standard error" listed with the point estimate for a regression coefficient lets say, can you multiply it by 1.96 and construct a confidence interval or does it depend on if there are better formulas for the CI?	{"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": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.8923053741455078, "perplexity_flag": "head"}
http://math.stackexchange.com/questions/105613/solution-of-a-system-with-exponentials	# Solution of a system with exponentials I want to solve the following system: $$1 + (0.2\pi/W)^{2N} = (1/0.89125)^2$$ and $$1 + (0.3\pi/W)^{2N} = (1/0.17783)^2$$ but i can't see how i can do that without getting to many confusing calculations. Is there a 'smarter' way to solve this system? - 4 Smarter than subtracting $1$ from both sides and taking logs? – Peter Taylor Feb 4 '12 at 12:48 yes that's the obvious way but it has a substantial bit of calculations! – nikos Feb 4 '12 at 12:52 Not really... $\phantom{}$ – J. M. Feb 4 '12 at 12:53 ## 2 Answers Let's denote : $A =(1/0.89125)^2$ , and $B=(1/0.17783)^2$ . So: $(0.2\pi/W)^{2N}=A-1$ $(0.3\pi/W)^{2N}=B-1$ Now divide first equation by second : $\left(\frac{2}{3}\right)^{2N}=\frac{A-1}{B-1} \Rightarrow N = \frac{1}{2} \cdot \log_{2/3} \frac{A-1}{B-1}$ After you find variable $N$ substitute it into one of the starting equations . - The first equation can be rearranged like so: $$2N\log(0.2\pi/W) = \log((1/0.89125)^2-1)$$ and a similar thing can be done for the second equation. Perform a division to cancel out the $N$, leading to $$\frac{\log(0.2\pi/W)}{\log(0.3\pi/W)} = \frac{\log((1/0.89125)^2-1)}{\log((1/0.17783)^2-1)}$$ or $$\frac{\log(0.2\pi)-\log\,W}{\log(0.3\pi)-\log\,W} = \frac{\log((1/0.89125)^2-1)}{\log((1/0.17783)^2-1)}$$ Solving for $\log\,W$, and then $W$, should be a snap. Once you have the value of $W$, substitute into any of the two original equations, or the equation like the first one I gave in this answer to solve for $N$. -	{"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": 5, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9196072816848755, "perplexity_flag": "middle"}
http://mathoverflow.net/questions/19732?sort=votes	## Ackermann-related function ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) While doing some research, I came up with a problem of proving that $f(a,b,c)=\begin{cases}1 &\text{ if }A(a,b)=c\ \\ 0 &\text{ otherwise }\end{cases}$ is primitively recursive ($A$ is the Ackermann's function). Any references, ideas or proofs? (This may not be a good MO question, but since the participants in problem-solving sites listed in MO posting FAQ failed to solve it - I posted it there before - I was hoping for a solution here.) - ## 3 Answers Here's a sketch of an argument which I expect could be made into a proof. The key fact is that the Ackermann function fails to be primitive recursive only because it grows so quickly. More formally: Claim. There exists a Turing machine T and a primitive recursive function f(a, b, c) (which is an increasing function of c) such that on input (a, b), T computes A(a, b) in at most f(a, b, A(a, b)) steps. "Proof". Starting with the expression "A(a, b)", repeatedly expand terms of the form A(x, y) with the recursive definition, but do not simplify any of the resulting additions. The length of the string increases at every step, if we agree that the symbol "+" is "longer" than "A". The resulting string is a formal sum of positive integers, so its length is bounded by (a multiple of) A(a, b); hence the number of steps is also bounded above in terms of A(a, b), and we may perform each step in time polynomial in A(a, b). Now, we can simulate a given Turing machine for a fixed number of steps using a primitive recursive function. We may therefore compute the graph of the Ackermann function with a primitive recursive function as follows: Given a, b, c, • Compute f(a, b, c). • Simulate the Turing machine T on input (a, b) for f(a, b, c) steps. • If T has halted, then return whether c is equal to the output of T. If T has not halted, then c < A(a, b) so return false. - ### 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 key to do this is to realize the difference between computation and verification. Although computing value A(a,b) of the Ackermann function cannot be done primitive recursively, verifying whether a proposed number c is the correct value of A(a,b) can be done primitive recursively. (Note that computation and verification is also what distinguishes P and NP.) In this case, the fact that this can be done hinges on the strong monotonicity properties of Ackermann's function. Indeed, if A(a,b) = c then all the previous values of A needed to compute A(a,b) are bounded by c. Therefore, the search for a valid computation verifying that A(a,b) = c can be bounded by a primitive recursive function B(a,b,c). Knowing this function B we can take a proposed value c for A(a,b), compute the bound B(a,b,c) and search up to this bound for a valid computation verifying that c is indeed equal to A(a,b). If no such computation is found we return 1, otherwise we return 0. To make this a little more specific, let's say that computation for A(a,b) = c is a (coded) finite sequence of triples (ai,bi,ci) which ends with the triple (a,b,c). Each such triple codes the fact that A(ai,bi) = ci. The computation is valid when each such triple follows from previous triples and the rules for computing the Ackermann function. (Checking this is obviously primitive recursive.) For example a valid computation for A(1,1) = 3 is the sequence (0,1,2), (0,2,3), (1,0,2), (1,1,3). Using the rules for computing Ackermann function and our specific method for coding finite sequences, we can compute an explicit bound B(a,b,c) on a valid computation verifying A(a,b) = c. The verification that B is primitive recursive should be straightforward if our coding of finite sequences is reasonable. Therefore, verifying A(a,b) = c can be done by a primitive recursive computation. (The details of the last paragraph are best carried out in the privacy of one's office.) - Googling for "graph of the Ackermann function" gives this note by George Tourlakis, which proves both that the Ackermann function is not primitive recursive, and, at the end, that the graph is. -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.885282039642334, "perplexity_flag": "middle"}
http://physics.stackexchange.com/questions/tagged/higgs-mechanism?sort=unanswered&pagesize=15	# Tagged Questions The higgs-mechanism tag has no wiki summary. 2answers 81 views ### What is the process that gives mass to free relativitic particles? When a free particle move in space with a known momentum and energy then what is the physical process that gives mass to that free (relativistic) particle? What is role does the Higgs field in that ... 1answer 73 views ### How to find the Higgs coupling with a mixing matrix? It is known that the couplings to the Higgs are proportional to the mass for fermions; $$g_{hff}=\frac{M_f}{v}$$ where $v$ is the VEV of the Higgs field. I'm trying to figure out why this is true ... 0answers 109 views ### Relation among anomaly, unitarity bound and renormalizability There is something I'm not sure about that has come up in a comment to other question: Why do we not have spin greater than 2? It's a good question--- the violation of renormalizability is linked ... 0answers 66 views ### Could one theoretically build the Higgs equivalent of a Faraday cage? My understanding is, within quantum mechanics, in a pure vacuum, all known fields have a lowest energy state of zero. The Higgs field is the only exception -- it's lowest energy state is not zero. ... 0answers 127 views ### Field content and symmetry groups of Minimal Composite Higgs Models I'm trying to teach myself the Composite Higgs Model, both its theory and its LHC phenomenology (particularly the 4DCHM). Unfortunately, I'm struggling; the literature is contradictory and/or omits ... 0answers 113 views ### Do all the particles acquire mass in the Standard Model due to the Higgs mechanism only? I know that a mass term for an intermediate boson is not compatible with the gauge symmetry. But in principle a mass term for the electron field does not violate a gauge symmetry. However to build an ... 0answers 54 views ### In a composite Higgs, should the Z0 decay at the same rate that the neutral pion? I am sorry to ask a question that obviously is a sum of separate questions about a process: Should the decay rate of Z0 be related to the decay of the "higgs" composite field it is eating? And, should ... 0answers 55 views ### Higgs boson/field symmetries and local symmetries In the SM with gauge group U(1)xSU(2)xSU(3), those factors are associated to the gauge bosons associated with a local symmetry and the Higgs field provides masses to the elementary fermions AND the ... 0answers 147 views ### Describing the Higgs mechanism to non-particle physicists I'm sure I'm not the only person with this problem at the moment. I have been asked to give a public (not quite public, scientists, just not physicists) about 'this Higgs boson thing'. I am trying to ...	{"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": 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": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.93266361951828, "perplexity_flag": "middle"}
http://mathoverflow.net/questions/9984/what-does-faltings-theorem-look-like-over-function-fields	## What does Faltings' theorem look like over function fields? ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) Minhyong Kim's reply to a question John Baez once asked about the analogy between $\text{Spec } \mathbb{Z}$ and 3-manifolds contains the following snippet: Finally, regarding the field with one element. I'm all for general theory building, but I think this is one area where having some definite problems in mind might help to focus ideas better. From this perspective, there are two things to look for in the theory of $\mathbb{F}_1$. 1) A theory of differentiation with respect to the ground field. A well-known consequence of such a theory could include an array of effective theorems in Diophantine geometry, like an effective Mordell conjecture or the ABC conjecture. Over function fields, the ability to differentiate with respect to the field of constants is responsible for the considerably stronger theorems of Mordell conjecture type [emphasis mine], and makes the ABC conjecture trivial. What is the strongest such theorem? Does anyone have a reference? (A preliminary search led me to results that are too general for me to understand them. I'd prefer to just see effective bounds on the number of rational points on a curve of genus greater than 1 in the function field case.) - ## 2 Answers The "function field analogues" of Faltings' theorem were proved by Manin, Grauert and Samuel: see http://archive.numdam.org/ARCHIVE/PMIHES/PMIHES_1966__29_/PMIHES_1966__29__55_0/PMIHES_1966__29__55_0.pdf especially Theorem 4. (The quotation marks above are because all of this function field work came first: the above link is to Samuel's 1966 paper, whereas Faltings' theorem was proved circa 1982.) The statement is the same as the Mordell Conjecture, except that there is an extra hypothesis on "nonisotriviality", i.e., one does not want the curve have constant moduli. For some discussion on why this hypothesis is necessary, see e.g. p. 7 of http://math.uga.edu/~pete/hassebjornv2.pdf An effective height bound in the function field case is given in Corollaire 2, Section 8 of Szpiro, L.(F-PARIS6-G) Discriminant et conducteur des courbes elliptiques. (French) [Discriminant and conductor of elliptic curves] Séminaire sur les Pinceaux de Courbes Elliptiques (Paris, 1988). Astérisque No. 183 (1990), 7--18. Note that effectivity on the height is much better than effectivity on the number of rational points (Faltings' proof does give the latter). This is not to be confused with uniform bounds on the number of rational points, for which I believe there are only conditional results known in any case. - MR0222088 (36 #5140) Samuel, P. Lectures on old and new results on algebraic curves. Notes by S. Anantharaman. Tata Institute of Fundamental Research Lectures on Mathematics, No. 36 Tata Institute of Fundamental Research, Bombay 1966 ii+127+iii pp. – Chandan Singh Dalawat Dec 30 2009 at 3:33 ### 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 don't know enough arithmetic geometry to understand more than, like, half the words that I had to think about to find this reference, but if I'm understanding everything correctly, I think that there's an effective bound on the number of rational points on (nonisotrivial, genus greater than one) varieties over function fields given in this paper (p. 16). There's also a paper of Miyaoka, apparently, but I haven't found an open-access version. - Heier's paper -- which applies to function fields of curves over C -- gives a bound on the number of rational points which depends not only on the genus but also on the number of singular fibers. (My initial guess is that when the ground field is algebraically closed, uniformity in terms only of the genus might be false...) – Pete L. Clark Dec 28 2009 at 22:47 1 For effective bounds for the number of points for function fields in char. 0, see Buium's paper in Duke Math. J. 71 (1993),475--499. For char. p see Buium and myself in Compositio Math. 103 (1996), 1--6. These bounds depend on the genus and on the Mordell-Weil rank of the Jacobian. – Felipe Voloch Dec 29 2009 at 1:45	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9112664461135864, "perplexity_flag": "head"}
http://mathoverflow.net/questions/35462/dynamic-programming-and-combinatorics/35954	dynamic programming and combinatorics Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) Suppose there are $K$ buckets each can be filled upto $N-1$ balls. The gain on putting $i$ balls in the $k^{th}$ bucket is given by $\Delta l_{k,i}, \, i \in [1,N-1]$. The problem is to put $\lambda$ balls in those buckets to maximize the overall gain. How do we solve it? - 2 Answers You can use dynamic programming as you suggest in the title. Let $w_{ij}$ be the max gain you can get putting $j$ balls into first $i$ buckets. Then $w_{ij}$ has the following recursive relation: $$w_{i,j} = \max_{0 \le t \le \min(N-1, j)}(w_{i-1,j-t} + \Delta l_{i, t})$$ There $t$ is the number of balls you put into $i$-th bucket. Hence you can calculate $w_{K, \lambda}$ (your answer) in time $O(K \lambda N)$. - True. This is the right answer. However, it is far too complex. i have come up with a simpler one. which will select the bucket at every iteration (ball being added). Here is the recursion. $\Delta R(p+1)= \displaystyle \max_{i} \{ \Delta R(p-i)+ \displaystyle \max_{k \in [1,K]} \Delta l_{k,i}(p)\}$ Here, $R(p+1)$ is the optimal gain in putting p+1 balls. Can you comment if it is optimal? – pravesh Aug 13 2010 at 11:45 Is $R(p)$ the optimal gain in putting $p$ balls considering all the buckets? If so then I don't understand your formula. If you have put $i$ balls into $k$-th bucket you can't use this bucket any more. – falagar Aug 13 2010 at 11:50 And I don't know if there exists faster solution. – falagar Aug 13 2010 at 11:52 Yes if you have put i balls in the kth bucket, you can only fill the bucket with remaining N-i balls. Btw, you can also take them out if there is a better solution in the later stages. In this solution, I am filling one ball at a time by finding the right bucket for that ball. Now since each bucket can be filled with N-1 balls, I am also exploring that dependency by going back p−i times, where i <= N. Does it make sense? – pravesh Aug 13 2010 at 12:10 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 tried to work out the problem in a faster way. Here goes my answer. Let the set ${\chi_p}={t_i, 1 \leq i \leq l_p }$ collects indices of the buckets for optimally filling $p$ balls. $\bar{\chi_p}={1,\dots,K}/ \chi_p$ is the set of buckets not participating in the $p$ allocations. The DP we propose derives from the fact that while going inductively from $p$ to $p+1$ allocations, we do not add more than one element (bucket) from the set $\bar{\chi_p}$. Hence, at every stage of the DP, the goal is to identify bucket that needs to be added and the number of balls to be put in the selected bucket in order to maximize the resulting overall gain. We prove this property. \subsubsection{Proof} Let us assume that while going from the stage $p$ to $p+1$, we add $r$ new buckets from the set $\bar{\chi_p}$ into $\chi_{p+1}$ with indices $x_1,\dots,x_r$ containing $b_1,\dots,b_r$ balls. The corresponding gain accrued from the added buckets are $g_i, \, i \in {1,\dots,r}$. We also alter the allocation in $n$ buckets from $\chi_p$ with indices $y_1,\dots,y_n$ removing $d_1,\dots,d_n$ balls respectively. Similarly, let the total loss due to this alteration be $q$. Since we add one ball while going from $p$ to $p+1^{th}$ stage, following is true: $(b_1+b_2\dots +b_{r-1}+b_r)- (d_1+d_2 +\dots d_n)=1$ Let $\alpha = b_1+b_2\dots +b_{r-1}$, be the total number balls placed in all the added buckets except the $r^{th}$ bucket of index $x_r$. Similarly, let $\beta= (d_1+d_2 \dots +d_n)$ total number of balls removed from the set $\chi_p$ while transitioning from $p^{th}$ to $p+1^{th}$ allocation. By $\alpha+b_r=\beta+1$ Since $b_r \geq 1$, $\alpha \leq \beta$. Now, consider the partition of the total removed balls $\beta$ such that $\beta=\alpha + \epsilon, \, \epsilon \geq 0$. The loss $q$ can also be partitioned as $q = q_{\alpha}+ q_{\epsilon}$, such that $q_{\alpha}$ is maximum. The gain due to the allocation of $\alpha$ balls in $p+1^{th}$ step in new buckets is $\sum_{i=1}^{r-1} g_i$. It cannot be less then $q_{\alpha}$ else, it is not optimal, as there exists an allocation with higher gain. This implies $\sum_{i=1}^{r-1} g_i \geq q_{\alpha}$. Moreover, $q_{\alpha}$ was a part of optimal solution of $p$ allocations. Hence, using same argument, $q_{\alpha} \geq \sum_{i=1}^{r-1} g_i$. This means $\sum_{i=1}^{r-1} g_i = q_{\alpha}$. This is only possible when $x_1,x_2\dots, x_{r-1} \in \chi_p$. This implies that no more than one bucket is derived from $\bar{\chi_p}$ in the $p+1^{th}$ step. Note that it is also possible that the $p+1^{th}$ ball is allocated to $\chi_p$. The DP approach is based on the above property. Given that we know the optimal allocation for $p$ balls, and optimal gains for $1,2\dots,p$ allocations, we can approach the $p+1^{th}$ by exploiting these optimal substructures. Let us call the $K \times N-1$ matrix $\Delta {\bf L}$ whose elements are $\Delta l_{k,i}$ as gain matrix. Let the optimal gain and corresponding updated gain matrix after $p$ allocations be $\Delta R(p)$ and $\Delta {\bf L}(p)$ respectively. Since, at most one bucket is picked at every step, and this bucket has the capacity of $N-1$, hence every successive allocation is dependent only on previous $N-1$ decisions. It can be written as follows: $\Delta R(p+1)= \displaystyle \max_{i} { \Delta R(p-i)+ \displaystyle \max_{k \in [1,K]} \Delta l_{k,i}(p)}$ At every step, the recursion involves choosing the optimal bucket and number of balls to be put in it such that new total gain is maximized. -	{"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": 84, "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": 0.3, "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": 0.950600266456604, "perplexity_flag": "head"}
http://unapologetic.wordpress.com/2011/06/08/the-lie-algebra-of-a-lie-group/?like=1&_wpnonce=68d26d17fa	# The Unapologetic Mathematician ## The Lie Algebra of a Lie Group Since a Lie group $G$ is a smooth manifold we know that the collection of vector fields $\mathfrak{X}G$ form a Lie algebra. But this is a big, messy object because smoothness isn’t a very stringent requirement on a vector field. The value can’t vary too wildly from point to point, but it can still flop around a huge amount. What we really want is something more tightly controlled, and hopefully something related to the algebraic structure on $G$ to boot. To this end, we consider the “left-invariant” vector fields on $G$. A vector field $X\in\mathfrak{X}G$ is left-invariant if the diffeomorphism $L_h:G\to G$ of left-translation intertwines $X$ with itself for all $h\in G$. That is, $X$ must satisfy $L_{h}\circ X=X\circ L_h$; or to put it another way: $L_{h}\left(X(g)\right)=X(hg)$. This is a very strong condition indeed, for any left-invariant vector field is determined by its value at the identity $e\in G$. Just set $g=e$ and find that $X(h)=L_{h}\left(X(e)\right)$ The really essential thing for our purposes is that left-invariant vector fields form a Lie subalgebra. That is, if $X$ and $Y$ are left-invariant vector fields, then so is their sum $X+Y$, scalar multiples $cX$ — where $c$ is a constant and not a function varying as we move around $M$ — and their bracket $[X,Y]$. And indeed left-invariance of sums and scalar multiples are obvious, using the formula $X(h)=L_{h}\left(X(e)\right)$ and the fact that $L_{h}$ is linear on individual tangent spaces. As for brackets, this follows from the lemma we proved when we first discussed maps intertwining vector fields. So given a Lie group $G$ we get a Lie algebra we’ll write as $\mathfrak{g}$. In general Lie groups are written with capital Roman letters, while their corresponding Lie algebras are written with corresponding lowercase fraktur letters. When $G$ has dimension $n$, $\mathfrak{g}$ also has dimension $n$ — this time as a vector space — since each vector field in $\mathfrak{g}$ is uniquely determined by a single vector in $\mathcal{T}_eG$. We should keep in mind that while $\mathfrak{g}$ is canonically isomorphic to $\mathcal{T}_eG$ as a vector space, the Lie algebra structure comes not from that tangent space itself, but from the way left-invariant vector fields interact with each other. And of course there’s the glaring asymmetry that we’ve chosen left-invariant vector fields instead of right-invariant vector fields. Indeed, we could have set everything up in terms of right-invariant vector fields and the right-translation diffeomorphism $R_g:G\to G$. But it turns out that the inversion diffeomorphism $i:G\to G$ interchanges left- and right-invariant vector fields, and so we end up in the same place anyway. How does the inversion $i$ act on vector fields? We recognize that $i^{-1}=i$, and find that it sends the vector field $X$ to $i_\circ X\circ i$. Now if $X$ is left-invariant then $L_{h}\circ X=X\circ L_h$ for all $h\in G$. We can then calculate $\displaystyle\begin{aligned}R_{h}\circ\left(i_\circ X\circ i\right)&=\left(R_h\circ i\right)_\circ X\circ i\\&=\left(i\circ L_{h^{-1}}\right)_\circ X\circ i\\&=i_\circ L_{h^{-1}}\circ X\circ i\\&=i_\circ X\circ L_{h^{-1}}\circ i\\&=\left(i_\circ X\circ i\right)\circ R_h\end{aligned}$ where the identities $R_h\circ i=i\circ L_{h^{-1}}$ and $L_h^{-1}\circ i=i\circ R_h$ reflect the simple group equations $g^{-1}h=\left(h^{-1}g\right)^{-1}$ and $h^{-1}g^{-1}=\left(gh\right)^{-1}$, respectively. Thus we conclude that if $X$ is left-invariant then $i_\circ X\circ i$ is right-invariant. The proof of the converse is similar. The one thing that’s left is proving that if $X$ and $Y$ are left-invariant then their right-invariant images have the same bracket. This will follow from the fact that $i_*(X(e))=-X(e)$, but rather than prove this now we’ll just push ahead and use left-invariant vector fields. ## 7 Comments » 1. “we get a Lie algebra we’ll write as \mathfrak{G}” – this should be lowercase, given the context. Comment by Andrei \| June 8, 2011 \| Reply 2. Sorry, typo. Fixed. Comment by \| June 8, 2011 \| Reply 3. [...] In particular, we can identify it with the tangent space at the identity matrix , and thus with the Lie algebra of [...] Pingback by \| June 9, 2011 \| Reply 4. [...] a Vector Field, Vector fields on compact manifolds are complete, Maps Intertwining Vector Fields, The Lie algebra of a Lie group, General linear groups are Lie groups, The Lie algebra of a general linear [...] Pingback by \| June 12, 2011 \| Reply 5. [...] space at back to itself: . But we know that this tangent space is canonically isomorphic to the Lie algebra . That is, . So now we can define by . We call this the “adjoint representation” of [...] Pingback by \| June 13, 2011 \| Reply 6. [...] The Lie Algebra of a Lie Group (unapologetic.wordpress.com) [...] Pingback by \| June 15, 2011 \| Reply 7. [...] and Flows When we discussed the Lie algebra of a Lie group we discussed “left-invariant” vector fields. More generally than this if is a [...] Pingback by \| June 17, 2011 \| Reply « Previous \| Next » ## About this weblog This is mainly an expository blath, with occasional high-level excursions, humorous observations, rants, and musings. The main-line exposition should be accessible to the “Generally Interested Lay Audience”, as long as you trace the links back towards the basics. Check the sidebar for specific topics (under “Categories”). I’m in the process of tweaking some aspects of the site to make it easier to refer back to older topics, so try to make the best of it for now.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 52, "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": 0.9064945578575134, "perplexity_flag": "head"}
http://sbseminar.wordpress.com/category/papers/	## Man and machine thinking about SPC4June 29, 2009 Posted by Scott Morrison in crazy ideas, link homology, low-dimensional topology, papers, Uncategorized. 13 comments I’ve just uploaded a paper to the arXiv, Man and machine thinking about the smooth 4-dimensional Poincaré conjecture, joint with Michael Freedman, Robert Gompf, and Kevin Walker. The smooth 4-dimensional Poincaré conjecture (SPC4) is the “last man standing in geometric topology”: the last open problem immediately recognizable to a topologist from the 1950s. It says, of course: A smooth four dimensional manifold $\Sigma$ homeomorphic to the 4-sphere $S^4$ is actually diffeomorphic to it, $\Sigma = S^4$. We try to have it both ways in this paper, hoping to both prove and disprove the conjecture! Unsuprisingly we’re not particularly successful in either direction, but we think there are some interesting things to say regardless. When I say we “hope to prove the conjecture”, really I mean that we suggest a conjecture equivalent to SPC4, but perhaps friendlier looking to 3-manifold topologists. When I say we “hope to disprove the conjecture”, really I mean that we explain an potential computable obstruction, which might suffice to establish a counterexample. We also get to draw some amazingly complicated links: (more…) ## Have someone else write your bibliographyJune 14, 2009 Posted by Scott Morrison in papers, the arXiv, websites. 11 comments Whenever I’m finishing off a paper, at some point I have to sit down and clean up all the references, which generally look something like \cite{Popa?} or \cite{that paper by Marco and co}. Wouldn’t it be nice if someone else could do the rest? If you don’t already know about it, one great resource is mathscinet, which will produce nicely formatted BIBTEX entries for you (example). If you want to be even more efficient, you can wander around mathscinet, saving articles to your “clipboard”, and then ask mathscinet to give you the BIBTEX entries for everything at once. (After you have articles on the clipboard, follow the “clipboard” link in the top right of the page, then select BIBTEX from the drop-down box and click “SaveClip”.) If you’re even lazier, you could use the two command-line scripts that I use (download find-missing-bibitems and get-mathscinet-bibtex and put them on your path; you’ll need linux/OSX/cygwin to run). Now, when you cite items in LaTeX, cite them via their mathscinet identifiers, e.g. \cite{MR1278111} instead of \cite{Popa?}. Now, if you usually type latex article to compile, and bibtex article to generate the bibliography, you can also type find-missing-bibitems article, and all the missing BIBTEX entries will appear! For example, after adding \cite{MR1278111} somewhere in my text, the output of find-missing-bibitems article is ```@article {MR1278111, AUTHOR = {Popa, Sorin}, TITLE = {Classification of amenable subfactors of type {II}}, JOURNAL = {Acta Math.}, FJOURNAL = {Acta Mathematica}, VOLUME = {172}, YEAR = {1994}, NUMBER = {2}, PAGES = {163--255}, ISSN = {0001-5962}, CODEN = {ACMAA8}, MRCLASS = {46L37 (46L10 46L40)}, MRNUMBER = {MR1278111 (95f:46105)}, MRREVIEWER = {V. S. Sunder}, }``` If you’re brave, you could run something like `find-missing-bibitems article >> bibliography.bib` to automatically append any missing entries to your BIBTEX file. The really enthusiastic could incorporate this script into the standard latex-latex-bibtex-latex cycle. Really, I like to have more in my BIBTEX file: I generally use the note field to include a link to the mathscinet review, and a link to the DOI for the paper on the publisher’s webpage. If available, I want a link to the arxiv version of the paper too, for people without institutional access to the published version. Currently, the scripts can’t do this automatically, but it’s might not be much more work. Maybe next time. ## This paper was written for our blogAugust 21, 2008 Posted by David Speyer in Algebraic Geometry, mathematical physics, Number theory, Paper Advertisement, papers, representation theory. 8 comments I’ve recently been reading a paper which ties together a number of this blog’s themes: Canonical Quantization of Symplectic Vector Spaces over Finite Fields by Gurevich and Hadani. I’m going to try to write an introduction to this paper, in order to motivate you all to look at it. It really has something for everyone: symplectic vector spaces, analogies to physics, Fourier transforms, representation theory of finite groups, gauss sums, perverse sheaves and, yes, $\theta$ functions. In a later paper, together with Roger Howe, the authors use these methods to prove the law of quadratic reciprocity and to compute the sign of the Gauss sum. For the experts, Gurevich and Hadani’s result can be summarized as follows: they provide a conceptual explanation of why there is no analgoue of the metaplectic group over a finite field. Not an expert? Keep reading! (more…) ## Gale and Koszul duality, together at lastJuly 14, 2008 Posted by Ben Webster in category O, combinatorics, hyperplanes, mathematical physics, papers, the arXiv. 2 comments So, in past posts, I’ve attempted to explain a bit about Gale duality and about Koszul duality, so now I feel like I should try to explain what they have to do with each other, since I (and some other people) just posted a preprint called “Gale duality and Koszul duality” to the arXiv. The short version is this: we describe a way of getting a category $\mathcal{C}(\mathcal{V})$ (or equivalently, an algebra) from a linear program $\mathcal{V}$ (or as we call it, a polarized hyperplane arrangement). Before describing the construction of this category, let me tell you some of the properties that make it appealing. Theorem. $\mathcal{C}(\mathcal{V})$ is Koszul (that is, it can be given a grading for which the induced grading on the Ext-algebra of the simples matches the homological grading). In fact, this category satisfies a somewhat stronger property: it is standard Koszul (as defined by Ágoston, Dlab and Lukács. Those of you with Springer access can get the paper here). In short, the category has a special set of objects called “standard modules” (which you should think of as analogous to Verma modules) which make it a “highest weight category,” such that these modules are sent by Koszul duality to a set of standards for the Koszul dual. Of course, whenever confronted with a Koszul category, we immediately ask ourselves what its Koszul dual is. In our case, there is a rather nice answer. Theorem. The Koszul dual to $\mathcal{C}(\mathcal{V})$ is $\mathcal{C}(\mathcal{V}^\vee)$, the category associated to the Gale dual $\mathcal{V}^\vee$ of $\mathcal{V}$. Now, part of the data of a linear program is an “objective function” (which we’ll denote by $\xi$) and of bounds for the contraints (which will be encoded by a vector $\eta$). Stripping these way, we end up with a vector arrangement, simply a choice of a set of vectors in a vector space, which will specify the constraints. Theorem. If two linear programs have same underlying vector arrangment, the categories $\mathcal C(\mathcal V)$ may not be equivalent, but they will be derived equivalent, that is, their bounded derived categories will be equivalent. Interestingly, these equivalences are far from being canonical. In the course of their construction, one actually obtains a large group of auto-equivalences acting on the derived category of $\mathcal{C}(\mathcal{V})$, which we conjecture to include the fundamental group of the space of generic choices of objective function. (more…) ## New PhotographMarch 25, 2008 Posted by A.J. Tolland in low-dimensional topology, papers, quantum algebra, talks, tqft. 38 comments Last Friday, we had a seminar at Berkeley — or rather, at Noah’s house — featuring Mike Freedman and some quantity of beer. Mike spoke about some of the hurdles he had to overcome in writing his recent paper with Danny Calegari and Kevin Walker. One of the main results of this paper is that there is a “complexity function” c, which maps from the set of closed 3-manifolds to an ordered set, and that this function satisfies the “topological” Cauchy-Schwarz inequality. $c(A \cup_S B) \leq max \{c(A \cup_S A),c(B \cup_S B)\}$ Here, $A$ and $B$ are 3-manifolds with boundary $S$. [EDIT: and equality is only achieved if $A = B$] This inequality looks like the sort of things you might derive from topological field theory, using the fact that $Z(A \cup_S B) = \langle Z(A), Z(B) \rangle_{Z(S)}$. Unfortunately, it’s difficult to actually derive this sort of theorem from any well-understood TQFT, thanks to an old theorem of Vafa’s, which states roughly, that there’s always two 3-manifolds related by a Dehn twist that a given rational TQFT can’t distinguish. Mike speculated that non-rational TQFT might be able to do the trick, but what he and his collaborators actually did was an end run around the TQFT problem. They simply proved that that the function $c$ exists. I tell you all this, not because I’m about to explain what $c$ is, but to explain our new banner picture. We realized after the talk that there were a fair number of us Secret Blogging Seminarians in one place, and that we ought to take a photo. ## oldmathpapers.org, version 0March 6, 2008 Posted by Scott Morrison in evil journals, papers, the arXiv, Uncategorized. 9 comments A while ago we discussed the idea of “oldmathpapers.org”, a public repository for maths papers that aren’t readily available online. Many people quickly pointed out that this was a dangerous idea, getting very quickly into the deep waters of copyright violation. Nevertheless, here’s version 0, ready for your consumption! It neatly sidesteps the whole copyright issue by not keeping copies (or even looking at) actual versions of the paper — it’s simply intended to keep track of links to old maths papers, hosted elsewhere. That elsewhere, of course, is meant to be your and my web pages! Functionality is extremely limited; you can add a paper, you can list everything there so far, but there’s no searching, no sorting, no deleting, no correcting. On the other hand, I think that won’t be too too hard to add. The most important thing to note about the design of “oldmathpapers.org” is that it relies on MathSciNet identifiers to keep track of things. These exist for pretty much every published maths paper, and they’re a ready source of high quality metadata — and it’s this that will hopefully make the searching and sorting easy. Below, I’ll walk you through adding a paper: “Canonical bases in tensor products and graphical calculus for U_q(sl_2)”, by I. Frenkel and M. Khovanov. After that, please take a moment to contribute some old math papers! (more…) ## ProofreadingOctober 29, 2007 Posted by Ben Webster in papers. 13 comments It strikes me as a real structural problem of the mathematics community that math papers are on the whole simply not very well proof-read. Many papers, sometimes very important ones, are often full of errors, in large part because it is simply a very time-confusing and difficult task to check a math paper for errors. On a more general level, I think most math papers could use a good once over by a talented editor who also understands the mathematics. Too bad such people are in short supply, and generally have better things to do. I’ll confess that proofreading is sometimes a bit of a problem for me, probably more so than for some more detail-oriented people. While there’s been quite a range in the number of errors found by the referee, I’ve gotten a couple of reports on papers I’ve submitted that found quite a few errors, and in fact, I received just such a report (for my paper with Geordie Williamson) just yesterday. What’s most dispiriting about such a report is that it’s not as though I didn’t proofread the paper, more than once (as did Geordie, and at least one other mathematician). I just went right over a number of errors that become glaring when the referee pointed them out. And while it’s generally pretty easy to fix whatever the referee actually points out, one knows that there are yet more errors hiding in there (especially in this case, where the referee indicates that they lost patience, and didn’t carefully proofread the whole paper), and it seems hopeless to think you will catch all of them. Does anyone have recommendations other than Ritalin? Of course, it would be best if one could dispatch a horde of flying monkeys blog readers to proofread one’s papers for one, but that doesn’t seem like a sustainable plan (not that I wouldn’t appreciate any comments readers have on the paper. It’s on my webpage here. Don’t look at the arXived version. That’s out of date). ## Zotero!September 26, 2007 Posted by Ben Webster in papers, the arXiv. 5 comments Has anyone else tried Zotero yet? It’s a Firefox add-on intended to help you organize the papers you use in your research (organize notes, make bibliographies [it seems to have bibtex support], etc.). It still seems to have some kinks that need working out (for example, it seems to work better with the arXiv’s website than the Front’s), but it does look promising. And if it actually allows me to keep track of which arxiv articles I have downloaded on my computer, it will be invaluable for that alone (at the moment, it’s much easier for me to download the PDF from the web again rather than locate it in my downloads folder, a ridiculous state of affairs). ## Zeta function relations and linearly equivalent group actionsAugust 29, 2007 Posted by Ben Webster in Number theory, papers, representation theory. 23 comments Since we’ve already had one post on the relationship between group theory and algebraic number theory, I don’t see any reason to stop, so I thought I would write about some old research I did as an undergraduate (6 years ago! yeesh). Now, every number field has a zeta function (often called the Dedekind zeta function), generalizing the Riemann zeta function. Now, I hope you will all remember that the Riemann zeta function (which is the zeta function of the rationals) is defined by $\displaystyle{\zeta(z)=\sum_{n\in \mathbb{Z}_{>0}}n^{-z}=\prod_{p \text{ prime}}(1-p^{-z})^{-1}}.$ If you take a more general number field $K$, then the Dedekind zeta function of that field is a similar sum or product over the ideals $\mathcal{I}_K$ or primes $\mathcal{P}_K$ of $K$. $\displaystyle{\zeta_K(z)=\sum_{\mathfrak n\in \mathcal{I}_K}N(\mathfrak n)^{-z}=\prod_{\mathfrak p \in \mathcal{P}_K}(1-N(\mathfrak p)^{-z})^{-1}}.$ Here $N(\mathfrak n)$ is the norm of $\mathfrak n$. So how can we understand this function? (more…) ## Soergel bimodulesJuly 23, 2007 Posted by Ben Webster in category O, link homology, papers, Soergel bimodules. 2 comments The next ingredient in my paper with Geordie is understanding a bit about Soergel bimodules. Soergel bimodules (or “Soergelsche Bimoduln” auf Deutsch) are a remarkable category of bimodules over a polynomial ring. The main thing that’s remarkable about them is that they categorify the Hecke algebra of $S_n$ (for those of you who don’t know any other Hecke algebras, pretend I just said “Hecke algebra”). Theorem (Soergel). The split Grothendieck group of the category of Soergel bimodules for $S_n$ is the Hecke algebra $\mathcal{H}_n$ of $S_n$. We’ll unpack that a bit later. For the moment, bear with me. Of course, one of the most important things about Grothendieck groups is that they have a natural basis. If one uses a split Grothendieck group (like a normal Grothendieck group, except we only have a relation that says $[M]+[N]=[M\oplus N]$, not for non-trivial extensions), then this basis will be the classes of the indecomposable elements. Theorem (Soergel). The basis of indecomposable objects is the famous basis of Kazhdan and Lusztig. Now that I’ve told you why people like Soergel bimodules, I guess I had better define them. (more…)	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 36, "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": 0.9186320304870605, "perplexity_flag": "middle"}
http://www.physicsforums.com/showthread.php?t=267025	Physics Forums ## Induced Electric Field Hello all, In Faraday's law we have that the variation of the magnetic flux in time will generate an induced electric field that loops on itself. My question concerns this induced electric field. If we imagine a bar magnet moving through a circular surface and consider the electric field induced on the circle at the boundary of this surface. (Note: the circle may be imaginary or an actual circle made of wire) 1. What is the electric potential of this induced field (measured on the rest frame of the surface) for one complete circular loop? 2. Since it is a loop, if we placed an imaginary test charge on it, what would be the energy that this charge would acquire (or work done to move this charge around the loop)? You can also imagine a superconducting wire with no resistance going around the induced electric field. Basically I don't understand the ideal of an electric field looping on itself. In my mind when we move along an electric field line we must have a delta in the potential. So is this delta in potential uniquely defined? What if I do one lap or two or infinite laps? what would the delta potential be? 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 >the variation of the magnetic flux in time >will generate an induced electric field (...) >My question concerns this induced electric field. >What is the electric potential of this induced field (...) That's the point: There is NO electric potential for this induced field. If we had one, i.e., if we had E = - grad Phi, that would entail curl E = 0, instead of Faraday's curl E = - d_t B, where the r.h.s. is nonzero in the situation you describe. So when this electric field appears due to the d_t B if I place a test charge there and circulate the field I will do no work? "Bossavit" <bossavit@lgep.supelec.removethis.fr> wrote in message news:ge176p$ku0$1@fb07-hees.theo.physik.uni-giessen.de... > > >the variation of the magnetic flux in time > >will generate an induced electric field (...) > >My question concerns this induced electric field. > >What is the electric potential of this induced field (...) > > That's the point: There is NO electric potential for this induced > field. If we had one, i.e., if we had E = - grad Phi, that would > entail curl E = 0, instead of Faraday's curl E = - d_t B, where the > r.h.s. is nonzero in the situation you describe. ## Induced Electric Field ElectromagStudent schrieb: > Hello all, > > In Faraday's law we have that the variation of the magnetic flux in time > will generate an induced electric field that loops on itself. > > My question concerns this induced electric field. > > If we imagine a bar magnet moving through a circular surface and consider > the electric field induced on the circle at the boundary of this surface. > (Note: the circle may be imaginary or an actual circle made of wire) > > 1. What is the electric potential of this induced field (measured on the > rest frame of the surface) for one complete circular loop? > > 2. Since it is a loop, if we placed an imaginary test charge on it, what > would be the energy that this charge would acquire (or work done to move > this charge around the loop)? You can also imagine a superconducting wire > with no resistance going around the induced electric field. > > Basically I don't understand the ideal of an electric field looping on > itself. In my mind when we move along an electric field line we must have a > delta in the potential. So is this delta in potential uniquely defined? What > if I do one lap or two or infinite laps? what would the delta potential be? The Maxwell equations are a linear set, so for the fundamentals one is able to understand everything in a linear setting. Assume a constant magnetic field rising with linear in time t with constant rate b B = t (0,0,b). This fixes the induced part of the electric field by the law of induction (t = ctime) dB/dt + curl E = 0 Near to the origin of space and time you may choose eg a circular electric field around the origin E = 1/2 (-y, x, 0) Of course, electric fields around other origins or superpositions with constant fields yield the same curl E = 1/2 (-(y-y0) s , (x-x0)(1-s)), s,x0,y0 some real parameter. So the actual electric field in such a situation will be specified by boundary or symmetry conditions of the real situation. This may be a non constant magnetic field to fix a midpoint. Or you have boundary conditons of conducting surfaces - eg the superconducting ring with a slit to measure the electric line integral along the wire aka the induced voltage over the slit. Of course, there exists a 4-potential for this situation. A = (A0,A1,A2,A3) = 1/2 t (0 , -(y-y0) s, (x-x0)(1-s), 0) with B and E given by B = curl spacepart A = curl(A1,A2,A3) E = spacepart d/dt A Scalar potentials for B and E do not exist, since line integrals of (-y,x,0) depend on the contour of integration chosen. In a closed normal conducting wire the time changing magnetic field is transferring energy to the electron gas by induction heating up the wire. In a superconducting ring the rising magnetic field - assumed starting at zero strength - is inducing a surface current distribution. The resulting magnetic field is a topologically determined solution of the time-linear Maxwell boundary problem with the effect, that the magnetic flux through the superconducting torus up to its outermost contuour is exactly zero. The inner part of the torus is free of any fields. In the superconducting surface Maxwells equation are satisfied by an exponential decaying distribution of the surface current and the shielded outer magnetic field (Meissner effect). These topological implications of Maxwells equations cannot be understood locally in terms of fields and potentials. They are deeply connected to the quantum mechanical description of electric charges and currents and the aspect of electromagnetic fields as a locally gauge field for linear and angular momentum and of charged quantum particles. The kind of solutions of Maxwell equations with superconducting boundaries depend on the topological genus of surface. There is a correspondence of the vector space of current distributions on surfaces without poles and zeros on the one hand and the number of holes in the surface of the other hand. Any spherelike object has no holes and so there is no nonsingular current distribution to shield an external or internal magnetic field: Spheres, cylinders, cubes dont show super currents. Any topologically toruslike object has two linear independent current distibutions as solutions of Maxwells equations. The first solution - a flow along the rim around the axis without any carge winding through the hole - is the boundary (and the physical source) of external magnetic fields with internal zero field. The second - a flow through the hole without any charge rotating arund the axis - generates an internal magnetic field with zero field in the outer space like a closed ore infinite selenoid. Funny things and a branch of pure mathematics. It shows the way by which the many degrees of freedom, the Maxwell equations don't fix, can be used to model the charge and current field distibutions in ideal boundaries. -- Roland Franzius ElectromagStudent <test@hotmail.com> wrote: > Hello all, > > In Faraday's law we have that the variation of the magnetic flux in time > will generate an induced electric field that loops on itself. > > My question concerns this induced electric field. > > If we imagine a bar magnet moving through a circular surface and consider > the electric field induced on the circle at the boundary of this surface. > (Note: the circle may be imaginary or an actual circle made of wire) > > 1. What is the electric potential of this induced field (measured on the > rest frame of the surface) for one complete circular loop? > > 2. Since it is a loop, if we placed an imaginary test charge on it, what > would be the energy that this charge would acquire (or work done to move > this charge around the loop)? You can also imagine a superconducting wire > with no resistance going around the induced electric field. > > Basically I don't understand the ideal of an electric field looping on > itself. In my mind when we move along an electric field line we must have a > delta in the potential. So is this delta in potential uniquely defined? What > if I do one lap or two or infinite laps? what would the delta potential be? What you need to understand is that there is in general no potential. There is a simple undergraduate experiment to demonstrate this. Take a long thin coil, and feed it altenating current. Loop a wire (or a string of resistors) around it. Alternating current will be induced. What is the potential over a given resistor? When you try to measure it with an AC voltmeter you will find that the result depends on the way the leads are arranged. Hence no unique well defined potential difference. Jan ElectromagStudent <test@hotmail.com> wrote: > So when this electric field appears due to the d_t B if I place a test > charge there and circulate the field I will do no work? The E-field will do work. To get down to practicalities you might try to understand how a tokamak device works. Jan -- A: Because it messes up the order in which people normally read text. Q: Why is it such a bad thing? A: Top-posting. Q: What is the most annoying thing on usenet? > "Bossavit" <bossavit@lgep.supelec.removethis.fr> wrote in message > news:ge176p$ku0$1@fb07-hees.theo.physik.uni-giessen.de... > > > > >the variation of the magnetic flux in time > > >will generate an induced electric field (...) > > >My question concerns this induced electric field. > > >What is the electric potential of this induced field (...) > > > > That's the point: There is NO electric potential for this induced > > field. If we had one, i.e., if we had E = - grad Phi, that would > > entail curl E = 0, instead of Faraday's curl E = - d_t B, where the > > r.h.s. is nonzero in the situation you describe. >So when this electric field appears due to the d_t B if I place a test >charge there and circulate the field I will do no work? No, there is some work involved indeed. But it cannot be expressed in terms of some electric potential, which in the situation you describe does not exist. Consider this highly idealized setup: A ring, made of some magic material which contains only ONE electron and will not let it escape outside, but will let it go freely around the loop (no drag). At t = 0, B = 0 and E = 0 all over, and the electron is idle, somewhere. For t > 0, let B change, by the effect of some outside agency. (Your hand moving a magnet, for instance.) This creates an electric field E, which obeys the equation rot E = - dB/dt, among others. The charge is pushed by E, and accelerates. After the first lap, which takes some time T > 0 to complete, it has acquired kinetic energy, equal [give or take a minus sign that I'm too lazy to think about] to the flux of B, at time T, through the loop. This energy has been pumped out from the agency, whatever it is, that caused the built-up of B. [You may be able to devise an experiment in which you'd feel* the difference it makes, when you move the magnet around, whether some conductor is close to you or not.] The force on the electric charge has two parts, Coulomb's and Lorentz's, but only the former does work. You can't determine the force from Faraday's equation alone, extra information is needed about the divergence of E, whereas Faraday only tells you about its curl. But when the charge performs a closed loop, you still can compute the work performed by this force, which is that thing called "circulation" of E around the loop, and depends on the curl of E only, by the Stokes theorem. After Faraday's law, this circulation is the flux of dB/dt across the loop, which you know if you know how B evolves in time. 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http://www.physicsforums.com/showthread.php?t=206632	Physics Forums Thread Closed Page 1 of 2 1 2 > ## [SOLVED] Gravitation & Escape Speed - Stone leaving Earth and reached the Moon. Hello everyone, I have one question on gravitation and escape speed. In the earlier part of the question, the minimum speed for the stone to leave the Earth is calculated and its final destination is the Moon. In the next part, the question asks if the stone will hit the Moon with a speed greater than or smaller than the minimum speed calculated in the previous part. Can anyone help me with the above question by providing a suitable explanation as well? Many thanks in advance. 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 Have you tried the conservation of energy law? Quote by Irid Have you tried the conservation of energy law? To explain? Actually I have not. I don't even know if the speed will be greater than or less than the original escape speed that was calculated in the earlier part of the question. ## [SOLVED] Gravitation & Escape Speed - Stone leaving Earth and reached the Moon. Well, conservation of energy will enable to calculate the value of the speed, so you'll know if it is more or less smth else. Do you know the formula for gravitational potential energy and also for kinetic energy? Quote by Irid Well, conservation of energy will enable to calculate the value of the speed, so you'll know if it is more or less smth else. Do you know the formula for gravitational potential energy and also for kinetic energy? I believe that you have misunderstood my question. I have already calculated the escape speed for the first part. I just need the second pat solved which asks if the speed at which the stone will hit the Moon's surface will be greater than or smaller than that of the speed it was launched (ie. the escape speed which I had already calculated earlier). I apologize for the misunderstanding. Oh yes, I see, my bad. Maybe there's a shorter way, but I'd still calculate the velocity at the Moon's surface to obtain a figure, and then just compare the two. Quote by Irid Oh yes, I see, my bad. Maybe there's a shorter way, but I'd still calculate the velocity at the Moon's surface to obtain a figure, and then just compare the two. Alright, so what you are saying here is that I should calculate the Moon's escape speed and compare that to that of the Earth's? It's significant smaller and I don't see how this helps me tell what speed the stone will land on the Moon. I think you're still misunderstanding the question. I suggest you to calculate the speed of stone as it reaches the Moon's surface (assume a head-on collision for a greatest possible speed). Recognitions: Gold Member Homework Help Science Advisor You have now TWO gravitational potentials; that from the Earth, and that from the Moon Letting R be the distance the stone has from the Earth centre, r the distance it has from the Moon centre, conservation of mechanical energy means: $$\frac{1}{2}mv^{2}-\frac{Gm_{e}m}{R}-\frac{Gm_{m}m}{r}=C$$ where m is the stone's mass, me the Earth mass, mm the moon mass and G the universal gravitation constant (C being the constant value of the energy). See if you can use this. Quote by arildno You have now TWO gravitational potentials; that from the Earth, and that from the Moon Letting R be the distance the stone has from the Earth centre, r the distance it has from the Moon centre, conservation of mechanical energy means: $$\frac{1}{2}mv^{2}-\frac{Gm_{e}m}{R}-\frac{Gm_{m}m}{r}=C$$ where m is the stone's mass, me the Earth mass, mm the moon mass and G the universal gravitation constant (C being the constant value of the energy). See if you can use this. I do not actually need to solve and calculate a value and I am also unable to due to the lack of information provided in the question. I am, however, required to explain it. Can you help me phrase this in words? (making it a little clearer for me to understand as well ;-) ) Quote by Icetray Thank you for your reply. I do not actually need to solve and calculate a value and I am also unable to due to the lack of information provided in the question. I am, however, required to explain it. Can you help me phrase this in words? (making it a little clearer for me to understand as well ;-) ) How did you find the minimum speed for the stone? Didn't you consider a particular point during the stone' s path? Quote by Rainbow Child How did you find the minimum speed for the stone? Didn't you consider a particular point during the stone' s path? We were given a gravitational potential graph from the Moon to the Earth and we were provided with the values for the gravitational potential for the Earth and Moon at their surfaces. Since Escape speed is independent of its mass, I was able to calculate it using the Earth's gravitational potential. Yes but as arildno, pointed out you have TWO gravitational potentials. That means that the stone feels two forces one from the Earth and one from the Moon. The escape velocity depends from both. Quote by Rainbow Child Yes but as arildno, pointed out you have TWO gravitational potentials. That means that the stone feels two forces one from the Earth and one from the Moon. The escape velocity depends from both. Hmm...it seems that I have failed to consider that. I always thought that the stone just required to escape the pull of the Earth. Since it's heading to the moon anyway, how will the Moon's pull affect the escape speed? Wouldn't the stone just be required to escape Earth's field alone? Think aboout the forces that acting on the stone when you throw it away. The force from the Earth resists or helps the stone to move? What about the force from the Moon? And futhermore, what happens to the values of the two forces while the stone is moving? Quote by Rainbow Child Think aboout the forces that acting on the stone when you throw it away. The force from the Earth resists or helps the stone to move? What about the force from the Moon? And futhermore, what happens to the values of the two forces while the stone is moving? But the minimum velocity of 11.2 m/s helps the stone leave the Earth's atmosphere. After it has left the Earth's atmosphere, the stone will automatically head towards Earth and later get dragged in towards the Moon's field (if this is what you're trying to tell me). At this stage, why would I need bother about the Moon's gravitational potential? I don't see how it affects the minimum speed. I don't want the stone to escape the Moon's pull as well. I want the stone to land on the Moon. Sorry, but I am really lost right now. ): Recognitions: Gold Member Homework Help Science Advisor Here, I must disagree with your interpretation, Rainbow Child: I think you are to USE the escape velocity, as calculated in the first part, and then calculate the impact velocity on the moon. For simplicity, I'll regard the distance between the Moon and the Earth, D as a constant. Let Re be the Earth radius, Rm the moon radius, then the conservation of energy tells us: $$\frac{1}{2}mv_{i}^{2}-\frac{Gm_{e}m}{D+R_{e}}-\frac{Gm_{m}m}{R_{m}}=\frac{1}{2}mv_{e}^{2}-\frac{Gm_{e}m}{R_{e}}-\frac{Gm_{m}m}{D+R_{m}}$$ where ve is the calculated escape velocity, and vi the impact velocity you were to find. Now, rearrange terms and evaluate what terms are greater than others. Thread Closed Page 1 of 2 1 2 > Thread Tools \| \| \| \| \|----------------------------------------------------------------------------------------------------------\|-------------------------------\|---------\| \| Similar Threads for: [SOLVED] Gravitation & Escape Speed - Stone leaving Earth and reached the Moon. \| \| \| \| Thread \| Forum \| Replies \| \| \| General Discussion \| 70 \| \| \| Introductory Physics Homework \| 1 \| \| \| Introductory Physics Homework \| 1 \| \| \| General Astronomy \| 15 \| \| \| General Physics \| 7 \|	{"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": 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": 0.949656069278717, "perplexity_flag": "middle"}
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On the other ...	{"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": 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": 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": 0.9461004734039307, "perplexity_flag": "middle"}
http://mathoverflow.net/questions/89993/continuous-variation-from-solution-of-easy-problem-to-solution-of-hard-problem	Continuous variation from solution of easy problem to solution of hard problem Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) I asked this question a week ago over on math.stackexchange and got no reply, so I am asking here with slightly different wording. I am trying to prove that there exists a solution to a problem. I can't solve this problem in general, but I can always continuously vary the parameters of the general problem to turn it into a specific case which is easy to solve. I believe that the solution of the easy problem can then be varied continuously to be a solution to the hard problem. It seems to me that this would be a widely used technique, but I haven't seen it before, and I don't know how to prove that it works. To be specific, suppose that $g:\mathbb{R}^n \times \mathbb{R} \to \mathbb{R}^n$. I have a solution $x_0$ to $g(x_0, t=0)=0$. I desire to show the existence of a solution to $g(x_1, t=1)=0$. My strategy is to take the known solution $x_0$ and vary it from $t=0$ to $t=1$. What I need is a theorem of the following form: Suppose $g(x_0, t=0)=0$, and suppose that $\exists f$ such that \begin{equation} \frac{dx}{dt}=f(x,t) \implies \frac{dg(x,t)}{dt}=0, \end{equation} with $x$ a function of $t$. If $f$ and $g$ are "well behaved enough" in the vicinity $\|g(x,t)\| < \epsilon$ for some $\epsilon$, then $\exists x$ such that $g(x,1)=0$. $f$ and $g$ in my case are "well behaved enough" that I can probably prove just about any sort of continuity condition that is needed. What properties of $f$ and $g$ are needed, and what theorem will help me here? It looks like Picard-Lindelof may help, but it seems to only give the existence of a unique solution to the differential equation, and I need to show that that solution satisfies $g(x_1, t=1)=0$. Furthermore, $f$ is not well behaved when $g$ is far from zero, and so it seems I cannot use Picard-Lindelof without prior assumption that $g(x,t)$ stays small (which is kind of assumes the fact that I am trying to prove). - I don't see how to interpret the left-hand-side of the left part of your implication. $\hspace{1.7 in}$ $x$ does not appear to depend on $t$ (or anything, for that matter). $\;\;$ – Ricky Demer Mar 1 2012 at 23:10 4 I may be all wrong here, but it sounds very much like you're talking about homotopy continuation methods. ams.org/journals/tran/1978-242-00/… – Gilead Mar 1 2012 at 23:27 Not directly an answer to your question, but the idea of starting with a solution of an easy problem and continuously following its evolution as parameters are changed is also the idea behind the quantum adiabatic algorithm: arxiv.org/abs/quant-ph/0104129 – Yoav Kallus Mar 2 2012 at 7:45 @Ricky: I've edited to make clear that x is a function of t. @Gilead: Thanks. – Dan Stahlke Mar 2 2012 at 14:35 1 In the special case that your functions are continuously differentiable and the nice conditions: (1) Whenever $g(x,t) = 0$, then $D_x g(x,t)$ is an isomorphism; (2) Given $g(x_n,t_n) \equiv 0$ with $t_n \to t$, then the sequence $x_n$ has a convergent subsequence, then you can proceed directly with the implicit function theorem without appealing to degree theory (although its still there behind the scenes). – Aaron Hoffman Mar 2 2012 at 21:39 show 1 more comment 1 Answer The right tag for this question is topology and the answer is degree theory. You could start by reading, say, http://en.wikipedia.org/wiki/Degree_of_a_map or/and http://unapologetic.wordpress.com/2011/12/10/calculating-the-degree-of-a-proper-map/ for a quick introduction. Read also the book Differential forms in algebraic topology by Bott and Tu for in depth discussion. (Actually, read this book in any case!) Warning: Wikipedia article confuses local diffeomorphisms and covering maps, but you do not need to worry about this. It also unnecessarily restricts the discussion to the case of bounded domains, while all you need to assume is that the homotopy is proper, see below. Here is the upshot: For general continuous (or even smooth) functions $g$, the existence of solution for $t=0$ does not imply existence of solution for $t=1$. However, if you assume that $g(x, 0)$ has nonzero degree over its value $0$ and the family $g(\cdot , t)$ is a proper homotopy, then $g(x,1)$ also has nonzero degree over $0$, in particular, the equation $g(x,1)=0$ also has (at least one) solution. The key principles are: i. proper homotopy preserves the degree and ii. map $h$ has nonzero degree $\Rightarrow$ existence of solution of the equation $h(x)=0$ (the function $h$ is surjective). Here, every continuous map $g(x,t)$ defines a homotopy of the function $h_0=g(x,0)$ to the function $h_1=g(x,1)$. This homotopy is proper if the map $g: {\mathbb R}^n \times [0,1]\to {\mathbb R}^n$ is a proper map (inverse image of compact is compact). In calculus terms: $$\lim_{\|x\|\to\infty, t\to t_0} g(x,t)=\infty$$ Below are two examples to think about ($n=1$): 1. $g(x,t)=x^2 + t - \frac{1}{2}$. Then the equation $g(x,t)=0$ has solution for $t=0$ and all $t\le 1/2$ but no solutions for $t>1/2$. In this case, $g(x,t)$ (as a function of $x$) has zero degree at $0$ for every $t$. 2. $g(x,t)= (t-1)x +1$. In this case $g(x,t)=0$ again has a solution for all $t\ne 1$, but the equation $g(x,1)=0$ has no solutions. In this case, the map $g(x,t)$ (as a function of $x$) has nonzero degree for all $t\ne 1$, but the homotopy is not proper (the map $g(x,1)$ is not a proper map). In most places, you will read about degree of maps between compact manifolds, while you are interested in maps of ${\mathbb R}^n$. However, the 1-point compactification of ${\mathbb R}^n$ is the sphere $S^n$. Properness allows you to extend your homotopy $g(\cdot, t)$ to $S^n$, so now you can appeal to the usual degree theory. Interestingly, the degree theory generalizes (with some difficulty) to the case of maps of Banach spaces which, in turn, is very useful for proving existence of solutions of differential equations. (Google "Degree Theory" + "Banach Spaces".) - Thank you very much for such a good answer. I am not a mathematician so I have quite some reading ahead of me! – Dan Stahlke Mar 2 2012 at 14:25	{"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": 74, "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": 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": 0.9478201270103455, "perplexity_flag": "head"}
http://math.stackexchange.com/questions/3815/why-does-the-set-of-all-singleton-sets-not-exist	# Why does the set of all singleton sets not exist? Proposition: For a set $X$ and its power set $P(X)$, any function $f\colon P(X)\to X$ has at least two sets $A\neq B\subseteq X$ such that $f(A)=f(B)$. I can see how this would be true if $X$ is a finite set, since $\|P(X)\|\gt \|X\|$, so by the pidgeonhole principle, at least two of the elements in $P(X)$ would have to map to the same element. Does this proposition still hold for $X$ an infinite set? And if so, how does this show that the set of all singleton sets cannot exist? - Is this homework? – Aryabhata Sep 1 '10 at 20:40 ## 5 Answers If you had a function $f\colon P(X)\to X$ such that $f(A)=f(B)\Rightarrow A=B$, then $f$ would be one-to-one, which would imply $\|P(X)\|\leq \|X\|$; since $\|X\|\lt \|P(X)\|$ holds by Cantor's Theorem, that would give you $\|P(X)\|\lt\|P(X)\|$, which is impossible. This argument does not use finiteness of $X$, so it holds for $X$ infinite as well. Suppose that the collection of all singletons is a set $X$. Define a map $f\colon P(X)\to X$ by $f(A) = \{A\}$; this is well defined, since $\{A\}$ is a singleton. If $A\neq B$, then $\{A\}\neq\{B\}$, so $f$ would be one-to-one. Since, by the proposition, this is impossible, we conclude that our assumption that the collection of all singletons is a set must be false. Thus, there is no "set of all singletons." - Thank you, a very clear explanation. Much appreciated. – yunone Sep 1 '10 at 20:51 Don't forget to accept an answer if you are satisfied; doesn't have to be mine, but if you are essentially "done" here, you might want to pick one and accept it. – Arturo Magidin Sep 2 '10 at 18:41 If $x$ is a set containing all singletons, $\cup x$ is a set containing all sets, which leads to Russel's paradox. Thus, in $ZF$ there is no such $x$. - This, in my opinion, is the "correct" answer to the title question... no fancy functions or regularity needed. Just the union axiom and separation :) – Dylan Wilson Sep 2 '10 at 5:00 1 There are a lot of ways; but the poser specifically asked how the result quoted (no injective functions $P(X)\to X$) could be used to establish that there is no set of all singletons, so presumably a direct application of the result is what was sought here. – Arturo Magidin Sep 2 '10 at 18:35 Very nice! Thanks! – George S. Nov 14 '10 at 10:05 You would have to add a proof of $\forall x \exists y (\cup x = y)$ – PyRulez Jan 6 at 14:59 This is the union axiom. Of course I work in ZF (or some tiny fragment). – Martin Brandenburg Jan 26 at 22:09 If you accept that the collection of all sets (denoted by $V$ usually) is not a set (but in fact a proper class) then if you look at the function $f(x) = \{ x \}$ it is 1-1 from $V$ into the sets of all singletons, thus resulting that $V$ would be a set, in contradiction to the fact that it is in fact a proper class. But of course you'd have to accept that $V$ is a proper class as well, this is simpler and follows by the argument given in Arturo's answer (that the power-set of $V$ would be of the same cardinality). - Your presupposition is incorrect; there are a number of set theories (some of them provably equiconsistent with ZF) in which the set of all singletons exists; see Limiting set theory using symmetry, or Forster’s Oxford Logic Guide on the subject. (Disclaimer: Forster discusses my work in the book, so I’m hardly unbiased, but it is the standard work.) - So if they are equiconsistent, does that mean the set of all singletons does indeed exist in ZF? I was only referring to ZF in my question. – yunone Mar 9 '11 at 18:20 – Flash Sheridan Mar 15 '11 at 3:53 The question in the title is a consequence of the axiom of regularity. In fact If there was such a set $X$, then $\{X\}\in X$. But obviously we always have $X\in\{X\}$. Hence, we would be able to construct an infinite downward chain of sets which contradicts the axiom of regularity. - Note however that the question in the statement also holds in ZF-{Regularity}+{Aczel's Anti-Foundation Axiom}, so Regularity is not needed. – Arturo Magidin Sep 2 '10 at 4:46 You need to put two backslashes before a curly bracket rather than just one; otherwise, the curly bracket does not show up. – Arturo Magidin Sep 2 '10 at 4:53 @Arturo Magidin: This is really crazy :D I am certain I was able to TeX curly braises a couple of days ago...or wasn't I?? Thanks! – AD. Sep 2 '10 at 5:01	{"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": 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": 0.9487513303756714, "perplexity_flag": "head"}
http://math.stackexchange.com/questions/tagged/computational-complexity+calculus	# Tagged Questions 1answer 47 views ### Difficulty proving / disproving the following equalities relations ( Big Ω) I have left with some functions I can't find witenesses for proving/disproving Big Ω equalities relations. Here are the three relations: $\sum\limits_{i=1}^{n} (i^3 - i ^2) = \Omega(n^4)$ ... 0answers 22 views ### Evaluating a simple sum bound I'm trying to evaluate and prove a simple statement but It seems really raw/bad solution. I would like to advise with you if this is the right way because It is really getting more complicated than It ... 1answer 63 views ### Prove that the little-o definition doesn't hold for two function (f and g) I need your help with the following statement: Show there exist two function $f(n), g(n)$ such that meet the following definition: $g(n) = O(f(n))$ and $f(n) \ne O(g(n))$ But don't meet the ... 2answers 83 views ### Some Big-O complexity definition proofs I'm trying to prove (by definition) the following but to no avail: $n^{n/2} \ne O(3^{n/2})$ $n! \ne O(3^n)$ $(n-b)^a = \Theta(n^a)$ $a,b$ are both constants whereas $a > 0$ and $b$ ... 2answers 170 views ### Little-o proof by definition I'm trying to figure out how to prove the following but to no avail. Given the following functions : $f(n) = n^3 -4n$ $g(n) = 5n^2 + 3n$ I have to show that $g(n) = o(f(n))$ by definition, that ... 3answers 89 views ### Order of magnitudes comparasions I have a list of order of magnitudes I want to compare. My only idea is using calculus methods (limits , integral, etc...) to assert the functions relation. I need your help with the following. I ... 1answer 113 views ### How to continue this argument/proof? I was wondering to myself what the actual run time of Mergesort was, so I thought like this: We have the sort operation that takes time $s(2) = 1$ and $s(1) = 0$. Merging two sorted sequences with ...	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 15, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": 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": 0.9424025416374207, "perplexity_flag": "middle"}
http://crypto.stackexchange.com/questions/1800/does-rsa-padding-have-to-be-unpredictable-if-the-payload-is	# Does RSA padding have to be unpredictable if the payload is? I'm trying to understand the precise requirements on padding when using RSA for encryption. Suppose Alice uses RSA to encrypt a payload $M$ that cannot be guessed (say, a random nonce): Alice send $\mathrm{RSA}_{B_{pub}}(\mathrm{pad}(\mathtt{?}) \|\| M)$ to Bob, with a padding procedure to be determined. Assume reasonable sizes, say a 1024-bit RSA key and a 256-bit payload. Alice may send the same payload $M$ to multiple recipients, encrypted with each recipient's private key. Only a recipient must be able to recover $M$. But it's ok if an external observer can tell that Alice sent the same $M$ to Bob and Charlie (sent as $\mathrm{RSA}_{B_{pub}}(\mathrm{pad}(\mathtt{?}) \|\| M)$ and $\mathrm{RSA}_{C_{pub}}(\mathrm{pad}(\mathtt{?}) \|\| M)$ respectively)¹. With no padding, this is broken: the plaintext can be reconstructed from $e$ ciphertexts. In general, RSA padding must be random and not recoverable without knowing the private key, if only because otherwise the attacker could verify a guess of the payload; but here I assume the payload is unguessable. So what are the requirements on padding in this scenario? • Is it ok to use a deterministic padding function that depends only the recipient, say $\mathrm{pad}(B) = \mathrm{SHA256}(B_{pub})$? • Is it ok to use a predictable padding function, say by incrementing global counter for each recipient (assume Alice does have a reliable monotonic counter)? • Is it ok to use random padding that is leaked? In other words, in this scenario, is the mere fact that the attacker knows the padding a problem? ¹ What prompted this line of thought was a message $M$ that was a single-use symmetric encryption key; the long message encrypted with that symmetric key would be identical for all recipients anyway. - The safe method Thomas Pornin describes in the answer you linked to is basically RSA-KEM, at least once you add the fairly obvious step of using the derived single-use key $h(x)$ only to encrypt a second symmetric key, which may then be safely used to encrypt data for multiple recipients. – Ilmari Karonen Feb 10 '12 at 1:17 ## 2 Answers From what I understand, you consider a scenario in which a fixed, random-looking message $M$ is sent to multiple recipients, encrypted using RSA with some padding (this is called the "broadcast RSA" setting). Moreover, you consider affine paddings, in the sense that you apply the RSA function to an element of $\mathbb{Z}_N$ of the form $a_i M + b_i$ for some values $a_i, b_i$ depending on the recipient which may be known in part or in full by the adversary (in fact, we even have $a_i = 1$ in the cases you consider, but that doesn't really matter). In that case, expanding a bit on fgrieu's answer: • No, it is not OK to use a public deterministic padding that depends on the recipient, even if the padded message is as long as the modulus. This type of padding is broken in the broadcast setting by a generalization due to Håstad of the simple CRT attack alluded to by fgrieu. • For the same reason, a predictable padding function will not do, nor indeed any affine padding where the values $a_i, b_i$ are known by the adversary. • Even using such a padding where the $b_i$'s are randomized can be dangerous. For example, a padding of the form $R_i \\| M$ (with $R_i$ chosen at random for each recipient and long enough so that the padded message is as long as the modulus) can be attacked in the broadcast setting if $M$ is relatively long: see this paper (the attack is admittedly rather theoretical, but still). • This survey on RSA attacks by Boneh is over 10 years old already, but always good reading (you're especially interested in §4). Generally speaking, I wouldn't recommend using an affine or polynomial padding function for any new application except for interoperability reasons; and I would especially advise against using a custom affine padding that "looks strong enough" (unless you want to make cryptanalysts happy by providing them with a fresh new target :-). A better idea (which is essentially what Thomas Pornin suggested in the earlier discussion) is using RSA-KEM (RFC 5990): for each recipient, pick a random element $R_i$ in $\mathbb{Z}_N$, and send $C_i = R_i^e \mod N$ to the $i$-th recipient, together with the encryption of $M$ (with your favorite block cipher) under a key derived from $R_i$. Other options include paddings like RSA-OAEP (part of PKCS#1 v2.1). - The simplest and strongest security proofs for RSA-based cryptosystems are when the padded message is random-like on $Z_n$. None of the proposed scheme does that. At the very least, it is prudent and customary that the bit size of the padded message approach the bit size of the public modulus $n$. With 256-bit $M$, padding as $\mathrm{SHA256}(B_{pub})\|\|M$ gives 512 bits, and this is trivially breakable by cube root extraction for public exponent 3 and 1537-bit modulus or more. More sophisticated attacks work for a wider range of RSA keys. Thus, without provision for a minimum width, and with allowance for low public exponent, we can answer "no" to the first question. On the other hand, I know no attack when padding as $C\|\|\mathrm{SHA256}(B_{pub})\|\|M$ where $C$ is a public arbitrary constant of 513 bits less than the modulus (and not influencing the key generation process), and $M$ is a 256-bit random. My answer becomes "It might be secure, but we have no proof". I take the second question as replacing $\mathrm{SHA256}(B_{pub})$ with an incremental identifier of the receiver. My answer is basically the same, with slightly less confidence. Update: as pointed by Mehdi Tibouchi in another answer, the above schemes are insecure in the broadcast RSA setting considered in the question (at least, unless the public exponent exceeds the number of recipients), see this Håstad attack. Sorry for missing this one. Again, this shows the superiority of methods with a security proof. The safest is With a random padding, we expand $M$ as $Rnd\|\|M$ where $Rnd$ is secret, randomly generated for each message, and with 257 bits less than the modulus; we do NOT get a direct reduction to the basic RSA problem, because the rightmost portion $M$ is shared between multiple destinations. If we are short on randomness beside $M$, we can elect for a user-dependent deterministic random-like padding function expanding $M$ to the full modulus, e.g. generated by a Mask Generation Function as $\mathrm{MGF}(B_{pub}\|\|M)\|\|M$. We keep a tight security proof reduction to the above. Of course, there is the issue that identical $M$ to the same user cause identical cryptograms. In the third question, we are padding as $Rnd\|\|M$ where $Rnd$ is public, randomly generated for each message, with 257 bits less than the modulus. This one seems solid, though I see no security reduction to the usual statement of the RSA problem also succumbs to the attack pointed by Mehdi Tibouchi. -	{"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": 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": 0.919465959072113, "perplexity_flag": "middle"}
http://mathhelpforum.com/calculus/100161-i-can-t-seem-get-correct-answer.html	# Thread: 1. ## I can't seem to get the correct answer! differentiate: $y=cos(a^3+x^3)$ my answer: $y'=-sin(a^3+x^3)(3a^2+3x^2)$ correct answer: $y'=-3x^2sin(a^3+x^3)$ I tried this multiple times and get the same answer (my answer). Either I am doing it incorrectly or the back of the book is incorrect. 2. Originally Posted by yoman360 differentiate: $y=cos(a^3+x^3)$ my answer: $y'=-sin(a^3+x^3)(3a^2+3x^2)$ correct answer: $y'=-3x^2sin(a^3+x^3)$ I tried this multiple times and get the same answer (my answer). Either I am doing it incorrectly or the back of the book is incorrect. $a^3$ represents a constant ... $y = \cos(a^3 + x^3)$ $y' = -\sin(a^3 + x^3) \cdot 3x^2$	{"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": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 9, "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": 0.8884194493293762, "perplexity_flag": "middle"}
http://mathoverflow.net/revisions/36803/list	## Return to Answer 2 added 680 characters in body; added 6 characters in body There is a simple example in the case of the Hartman-Grobman theorem for maps in 3D. The example appears in the original article paper by Hartman, "A lemma in the theory of structural stability of differential equations", Proc. Amer. Math. Soc. 11, 1960. Let's consider the map $T: \mathbb R^3\to \mathbb R^3$ given by $$T(x,y,z)=(ax,\ ac(y+b xz),\ cz)),$$ where $a>1$, $b>0$, $0 < c<1$, $ac>1$. If $\varphi$ is a any linearizing map, then both $\varphi$ and $\varphi^{-1}$ are not of class $C^{1}$. In the 2D case, one can show that any map $T(X)=AX+F(X)$ of class $C^2$ such that $F$ and its gradient vanish at $X=0$ can be linearized in the neighborhood of $X=0$ with a $C^1$-diffeomorphism, provided that the matrix $A$ has no eigenvalue of absolute value of $0$ or $1$ (see another paper by Hartman, "On local homeomorphisms of Euclidean spaces", Bol. Soc. Mat. Mexicana (2) 5, 1960). 1 There is a simple example in the case of the Hartman-Grobman theorem for maps in 3D. The example appears in the original article by Hartman, "A lemma in the theory of structural stability of differential equations", Proc. Amer. Math. Soc. 11, 1960. Let's consider the map $T: \mathbb R^3\to \mathbb R^3$ given by $$T(x,y,z)=(ax,\ ac(y+b xz),\ cz)),$$ where $a>1$, $b>0$, $0 < c<1$, $ac>1$. If $\varphi$ is a linearizing map, then both $\varphi$ and $\varphi^{-1}$ are not of class $C^{1}$.	{"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": 27, "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": 0.8931705951690674, "perplexity_flag": "head"}
http://math.stackexchange.com/questions/199634/function-problem-vs-decision-problem	# Function problem vs. decision problem Take the set $FP$ of number-theoretic functions that are computable in polynomial time. Let us restrict to those functions with range in $\{0,1\}$, $FP_{0,1}$. Is there any correspondence with $FP_{0,1}$ and the class of decision problems solvable in polynomial time? It seems obvious that if we define $D_f=\{x\|f(x)=1\}$ with $f\in FP_{0,1}$, then $D_f$ is indeed a polynomial-time decidable language. What if we take arbitrary polynomial-time decidable language $L$. Is there always a function $g\in FP_{0,1}$ such that $D_g=L$? I'm struggling with something that seems to me as a contradiction, and I suspect it arises from some subtle difference between function and decision problems which I overlook. - ## 1 Answer Yes, if you take a polynomial-time decidable language, that language is always of the form $D_f$ for some polynomial-time computable $\{0,1\}$ valued function - because the definition of polynomial-time decidability is that the characteristic function of the language is polynomial-time computable. The main distinction between function and decision problems in computational complexity comes from functions that are not $\{0,1\}$ valued. A number-theoretic function $f$ is computable (with no time constraints) if and only if its graph $\{ \langle x,y\rangle : f(x) = y\}$ is computable. But the graph might be polynomial-time computable even though $f$ is not. For example, this happens when we take $f(x) = 2^x$. The problem is that when we want to decide whether a pair $\langle x,y\rangle$ is in the graph, we can work in time polynomial in $\|\langle x,y\rangle\|$, which is larger than both $\log(x)$ and $\log(y)$, but to try to compute $f(x)$ in polynomial time we have to bound ourselves to a time limit that is polynomial in $\log(x)$ alone. Now computing $y = 2^x$ in the naive way requires $x$ multiplications of numbers less than $y$, and $\log(y) = \log(2^x) = x$, and multiplication of two numbers less than $y$ requires time that is polynomial in $\log(y)$. So if we are allowed to take a polynomial amount of time relative to $\log(y)$ then we can verify whether $y = 2^x$, thus deciding the graph of $f(x) = 2^x$ in polynomial-time. But we cannot compute $f(x)$ from $x$ alone in polynomial time. - Thank you. It seems the contradiction I mentioned remains, I might post a question about it later (perhaps on MO). – rank Sep 21 '12 at 5:02 @rank: could you explain what contradiction you see? – Carl Mummert Sep 21 '12 at 11:19	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 34, "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": 0.938015878200531, "perplexity_flag": "head"}
http://math.stackexchange.com/questions/tagged/eigenvectors+geometry	# Tagged Questions 1answer 82 views ### Obtaining Least square adjusted single line by intersecting many 3D planes I am working with many 3D planes and looking for a Least square solution for below case. IF I am having many number of 3D planes knowing only one point and the normal vector (for eg. O1 and N1), ... 1answer 126 views ### ellipse equation from eigenvectors and eigenvalues I have a eigenvectors d1,d2 and eigenvalues v1,v2. The eigenvectors are axes of an ellipse that surrounds data points, with center u,v, and radii size of the eigenvalues. How can I find the ellipse ... 0answers 30 views ### Position on a path So the situation is: I have a path (which is represented as a two-dimensional array of GPS coordinates) and I have a percentage position on this path. I.e. I know that a person has walked 80% of the ... 6answers 952 views ### Find eigenvalues of a projection and explain what they mean Suppose B represents the matrix of orthogonal (perpendicular) projection of $\mathbb{R}^{3}$ onto the plane $x_{2} = x_{1}$. Compute the eigenvalues and eigenvectors of B and explain their geometric ...	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": 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": 0.9070329070091248, "perplexity_flag": "head"}
http://pediaview.com/openpedia/Control_theory	# Control theory The concept of the feedback loop to control the dynamic behavior of the system: this is negative feedback, because the sensed value is subtracted from the desired value to create the error signal, which is amplified by the controller. Control theory is an interdisciplinary branch of engineering and mathematics that deals with the behavior of dynamical systems with inputs. The external input of a system is called the reference. When one or more output variables of a system need to follow a certain reference over time, a controller manipulates the inputs to a system to obtain the desired effect on the output of the system. The usual objective of a control theory is to calculate solutions for the proper corrective action from the controller that result in system stability, that is, the system will hold the set point and not oscillate around it. The inputs and outputs of a continuous control system are generally related by differential equations. If these are linear with constant coefficients, a transfer function relating the input and output can be obtained by taking their Laplace transform. If the differential equations are nonlinear and have a known solution, it may be possible to linearize the nonlinear differential equations at that solution.[1] If the resulting linear differential equations have constant coefficients one can take their Laplace transform to obtain a transfer function. The transfer function is also known as the system function or network function. The transfer function is a mathematical representation, in terms of spatial or temporal frequency, of the relation between the input and output of a linear time-invariant solution of the nonlinear differential equations describing the system. Extensive use is usually made of a diagrammatic style known as the block diagram. ## Overview Smooth nonlinear trajectory planning with linear quadratic Gaussian feedback (LQR) control on a dual pendula system. Control theory is • a theory that deals with influencing the behavior of dynamical systems • an interdisciplinary subfield of science, which originated in engineering and mathematics, and evolved into use by the social sciences, like psychology, sociology, criminology and in the financial system. Control systems can be thought of as having four functions: Measure, Compare, Compute, and Correct. These four functions are completed by five elements: Detector, Transducer, Transmitter, Controller, and Final Control Element. The measuring function is completed by the detector, transducer and transmitter. In practical applications these three elements are typically contained in one unit. A standard example of a measuring unit is a Resistance thermometer. The compare and compute functions are completed within the controller which may be completed electronically through a Proportional control, PI Controller, PID Controller, Bistable, Hysteretic control or Programmable logic controller. Older controller units have been mechanical, as in a Centrifugal governor or a Carburetor. The correct function is completed with a final control element. The final control element changes an input or output in the control system which affect the manipulated or controlled variable. ### An example Consider a car's cruise control, which is a device designed to maintain vehicle speed at a constant desired or reference speed provided by the driver. The controller is the cruise control, the plant is the car, and the system is the car and the cruise control. The system output is the car's speed, and the control itself is the engine's throttle position which determines how much power the engine generates. A primitive way to implement cruise control is simply to lock the throttle position when the driver engages cruise control. However, if the cruise control is engaged on a stretch of flat road, then the car will travel slower going uphill and faster when going downhill. This type of controller is called an open-loop controller because no measurement of the system output (the car's speed) is used to alter the control (the throttle position.) As a result, the controller cannot compensate for changes acting on the car, like a change in the slope of the road. In a closed-loop control system, a sensor monitors the system output (the car's speed) and feeds the data to a controller which adjusts the control (the throttle position) as necessary to maintain the desired system output (match the car's speed to the reference speed.) Now when the car goes uphill the decrease in speed is measured, and the throttle position changed to increase engine power, speeding up the vehicle. Feedback from measuring the car's speed has allowed the controller to dynamically compensate for changes to the car's speed. It is from this feedback that the paradigm of the control loop arises: the control affects the system output, which in turn is measured and looped back to alter the control. ## History Centrifugal governor in a Boulton & Watt engine of 1788 Although control systems of various types date back to antiquity, a more formal analysis of the field began with a dynamics analysis of the centrifugal governor, conducted by the physicist James Clerk Maxwell in 1868 entitled On Governors.[2] This described and analyzed the phenomenon of "hunting", in which lags in the system can lead to overcompensation and unstable behavior. This generated a flurry of interest in the topic, during which Maxwell's classmate Edward John Routh generalized Maxwell's results for the general class of linear systems.[3] Independently, Adolf Hurwitz analyzed system stability using differential equations in 1877, resulting in what is now known as the Routh–Hurwitz theorem.[4][5] A notable application of dynamic control was in the area of manned flight. The Wright brothers made their first successful test flights on December 17, 1903 and were distinguished by their ability to control their flights for substantial periods (more so than the ability to produce lift from an airfoil, which was known). Continuous, reliable control of the airplane was necessary for flights lasting longer than a few seconds. By World War II, control theory was an important part of fire-control systems, guidance systems and electronics. Sometimes mechanical methods are used to improve the stability of systems. For example, ship stabilizers are fins mounted beneath the waterline and emerging laterally. In contemporary vessels, they may be gyroscopically controlled active fins, which have the capacity to change their angle of attack to counteract roll caused by wind or waves acting on the ship. The Sidewinder missile uses small control surfaces placed at the rear of the missile with spinning disks on their outer surface; these are known as rollerons. Airflow over the disk spins them to a high speed. If the missile starts to roll, the gyroscopic force of the disk drives the control surface into the airflow, cancelling the motion. Thus the Sidewinder team replaced a potentially complex control system with a simple mechanical solution. The Space Race also depended on accurate spacecraft control. However, control theory also saw an increasing use in fields such as economics. ## People in systems and control Main article: People in systems and control Many active and historical figures made significant contribution to control theory, including, for example: • Pierre-Simon Laplace (1749-1827) invented the Z-transform in his work on probability theory, now used to solve discrete-time control theory problems. The Z-transform is a discrete-time equivalent of the Laplace transform which is named after him. • Alexander Lyapunov (1857–1918) in the 1890s marks the beginning of stability theory. • Harold S. Black (1898–1983), invented the concept of negative feedback amplifiers in 1927. He managed to develop stable negative feedback amplifiers in the 1930s. • Harry Nyquist (1889–1976), developed the Nyquist stability criterion for feedback systems in the 1930s. • Richard Bellman (1920–1984), developed dynamic programming since the 1940s. • Andrey Kolmogorov (1903–1987) co-developed the Wiener–Kolmogorov filter (1941). • Norbert Wiener (1894–1964) co-developed the Wiener–Kolmogorov filter and coined the term cybernetics in the 1940s. • John R. Ragazzini (1912–1988) introduced digital control and the use of Z-transform in control theory (invented by Laplace) in the 1950s. • Lev Pontryagin (1908–1988) introduced the maximum principle and the bang-bang principle. ## Classical control theory To overcome the limitations of the open-loop controller, control theory introduces feedback. A closed-loop controller uses feedback to control states or outputs of a dynamical system. Its name comes from the information path in the system: process inputs (e.g., voltage applied to an electric motor) have an effect on the process outputs (e.g., speed or torque of the motor), which is measured with sensors and processed by the controller; the result (the control signal) is "fed back" as input to the process, closing the loop. Closed-loop controllers have the following advantages over open-loop controllers: • disturbance rejection (such as hills in the cruise control example above) • guaranteed performance even with model uncertainties, when the model structure does not match perfectly the real process and the model parameters are not exact • unstable processes can be stabilized • reduced sensitivity to parameter variations • improved reference tracking performance In some systems, closed-loop and open-loop control are used simultaneously. In such systems, the open-loop control is termed feedforward and serves to further improve reference tracking performance. A common closed-loop controller architecture is the PID controller. ### Closed-loop transfer function For more details on this topic, see Closed-loop transfer function. The output of the system y(t) is fed back through a sensor measurement F to the reference value r(t). The controller C then takes the error e (difference) between the reference and the output to change the inputs u to the system under control P. This is shown in the figure. This kind of controller is a closed-loop controller or feedback controller. This is called a single-input-single-output (SISO) control system; MIMO (i.e., Multi-Input-Multi-Output) systems, with more than one input/output, are common. In such cases variables are represented through vectors instead of simple scalar values. For some distributed parameter systems the vectors may be infinite-dimensional (typically functions). If we assume the controller C, the plant P, and the sensor F are linear and time-invariant (i.e., elements of their transfer function C(s), P(s), and F(s) do not depend on time), the systems above can be analysed using the Laplace transform on the variables. This gives the following relations: $Y(s) = P(s) U(s)\,\!$ $U(s) = C(s) E(s)\,\!$ $E(s) = R(s) - F(s)Y(s).\,\!$ Solving for Y(s) in terms of R(s) gives: $Y(s) = \left( \frac{P(s)C(s)}{1 + F(s)P(s)C(s)} \right) R(s) = H(s)R(s).$ The expression $H(s) = \frac{P(s)C(s)}{1 + F(s)P(s)C(s)}$ is referred to as the closed-loop transfer function of the system. The numerator is the forward (open-loop) gain from r to y, and the denominator is one plus the gain in going around the feedback loop, the so-called loop gain. If $\|P(s)C(s)\| \gg 1$, i.e., it has a large norm with each value of s, and if $\|F(s)\| \approx 1$, then Y(s) is approximately equal to R(s) and the output closely tracks the reference input. ### PID controller For more details on this topic, see PID controller. The PID controller is probably the most-used feedback control design. PID is an acronym for Proportional-Integral-Derivative, referring to the three terms operating on the error signal to produce a control signal. If u(t) is the control signal sent to the system, y(t) is the measured output and r(t) is the desired output, and tracking error $e(t)=r(t)- y(t)$, a PID controller has the general form $u(t) = K_P e(t) + K_I \int e(t)\text{d}t + K_D \frac{\text{d}}{\text{d}t}e(t).$ The desired closed loop dynamics is obtained by adjusting the three parameters $K_P$, $K_I$ and $K_D$, often iteratively by "tuning" and without specific knowledge of a plant model. Stability can often be ensured using only the proportional term. The integral term permits the rejection of a step disturbance (often a striking specification in process control). The derivative term is used to provide damping or shaping of the response. PID controllers are the most well established class of control systems: however, they cannot be used in several more complicated cases, especially if MIMO systems are considered. Applying Laplace transformation results in the transformed PID controller equation $u(s) = K_P e(s) + K_I \frac{1}{s} e(s) + K_D s e(s)$ $u(s) = \left(K_P + K_I \frac{1}{s} + K_D s\right) e(s)$ with the PID controller transfer function $C(s) = \left(K_P + K_I \frac{1}{s} + K_D s\right).$ It should be noted that for practical PID controllers a pure differentiator is neither physically realisable nor desirable due to amplification of noise and resonant modes in the system. Therefore a phase-lead compensator type approach is used instead, or a differentiator with low-pass roll-off. ## Modern control theory In contrast to the frequency domain analysis of the classical control theory, modern control theory utilizes the time-domain state space representation, a mathematical model of a physical system as a set of input, output and state variables related by first-order differential equations. To abstract from the number of inputs, outputs and states, the variables are expressed as vectors and the differential and algebraic equations are written in matrix form (the latter only being possible when the dynamical system is linear). The state space representation (also known as the "time-domain approach") provides a convenient and compact way to model and analyze systems with multiple inputs and outputs. With inputs and outputs, we would otherwise have to write down Laplace transforms to encode all the information about a system. Unlike the frequency domain approach, the use of the state space representation is not limited to systems with linear components and zero initial conditions. "State space" refers to the space whose axes are the state variables. The state of the system can be represented as a vector within that space.[6] ## Topics in control theory ### Stability The stability of a general dynamical system with no input can be described with Lyapunov stability criteria. A linear system that takes an input is called bounded-input bounded-output (BIBO) stable if its output will stay bounded for any bounded input. Stability for nonlinear systems that take an input is input-to-state stability (ISS), which combines Lyapunov stability and a notion similar to BIBO stability. For simplicity, the following descriptions focus on continuous-time and discrete-time linear systems. Mathematically, this means that for a causal linear system to be stable all of the poles of its transfer function must have negative-real values, i.e. the real part of all the poles are less than zero. Practically speaking, stability requires that the transfer function complex poles reside • in the open left half of the complex plane for continuous time, when the Laplace transform is used to obtain the transfer function. • inside the unit circle for discrete time, when the Z-transform is used. The difference between the two cases is simply due to the traditional method of plotting continuous time versus discrete time transfer functions. The continuous Laplace transform is in Cartesian coordinates where the $x$ axis is the real axis and the discrete Z-transform is in circular coordinates where the $\rho$ axis is the real axis. When the appropriate conditions above are satisfied a system is said to be asymptotically stable: the variables of an asymptotically stable control system always decrease from their initial value and do not show permanent oscillations. Permanent oscillations occur when a pole has a real part exactly equal to zero (in the continuous time case) or a modulus equal to one (in the discrete time case). If a simply stable system response neither decays nor grows over time, and has no oscillations, it is marginally stable: in this case the system transfer function has non-repeated poles at complex plane origin (i.e. their real and complex component is zero in the continuous time case). Oscillations are present when poles with real part equal to zero have an imaginary part not equal to zero. If a system in question has an impulse response of $\ x[n] = 0.5^n u[n]$ then the Z-transform (see this example), is given by $\ X(z) = \frac{1}{1 - 0.5z^{-1}}\$ which has a pole in $z = 0.5$ (zero imaginary part). This system is BIBO (asymptotically) stable since the pole is inside the unit circle. However, if the impulse response was $\ x[n] = 1.5^n u[n]$ then the Z-transform is $\ X(z) = \frac{1}{1 - 1.5z^{-1}}\$ which has a pole at $z = 1.5$ and is not BIBO stable since the pole has a modulus strictly greater than one. Numerous tools exist for the analysis of the poles of a system. These include graphical systems like the root locus, Bode plots or the Nyquist plots. Mechanical changes can make equipment (and control systems) more stable. Sailors add ballast to improve the stability of ships. Cruise ships use antiroll fins that extend transversely from the side of the ship for perhaps 30 feet (10 m) and are continuously rotated about their axes to develop forces that oppose the roll. ### Controllability and observability Main articles: Controllability and Observability Controllability and observability are main issues in the analysis of a system before deciding the best control strategy to be applied, or whether it is even possible to control or stabilize the system. Controllability is related to the possibility of forcing the system into a particular state by using an appropriate control signal. If a state is not controllable, then no signal will ever be able to control the state. If a state is not controllable, but its dynamics are stable, then the state is termed Stabilizable. Observability instead is related to the possibility of "observing", through output measurements, the state of a system. If a state is not observable, the controller will never be able to determine the behaviour of an unobservable state and hence cannot use it to stabilize the system. However, similar to the stabilizability condition above, if a state cannot be observed it might still be detectable. From a geometrical point of view, looking at the states of each variable of the system to be controlled, every "bad" state of these variables must be controllable and observable to ensure a good behaviour in the closed-loop system. That is, if one of the eigenvalues of the system is not both controllable and observable, this part of the dynamics will remain untouched in the closed-loop system. If such an eigenvalue is not stable, the dynamics of this eigenvalue will be present in the closed-loop system which therefore will be unstable. Unobservable poles are not present in the transfer function realization of a state-space representation, which is why sometimes the latter is preferred in dynamical systems analysis. Solutions to problems of uncontrollable or unobservable system include adding actuators and sensors. ### Control specification Several different control strategies have been devised in the past years. These vary from extremely general ones (PID controller), to others devoted to very particular classes of systems (especially robotics or aircraft cruise control). A control problem can have several specifications. Stability, of course, is always present: the controller must ensure that the closed-loop system is stable, regardless of the open-loop stability. A poor choice of controller can even worsen the stability of the open-loop system, which must normally be avoided. Sometimes it would be desired to obtain particular dynamics in the closed loop: i.e. that the poles have $Re[\lambda] < -\overline{\lambda}$, where $\overline{\lambda}$ is a fixed value strictly greater than zero, instead of simply asking that $Re[\lambda]<0$. Another typical specification is the rejection of a step disturbance; including an integrator in the open-loop chain (i.e. directly before the system under control) easily achieves this. Other classes of disturbances need different types of sub-systems to be included. Other "classical" control theory specifications regard the time-response of the closed-loop system: these include the rise time (the time needed by the control system to reach the desired value after a perturbation), peak overshoot (the highest value reached by the response before reaching the desired value) and others (settling time, quarter-decay). Frequency domain specifications are usually related to robustness (see after). Modern performance assessments use some variation of integrated tracking error (IAE,ISA,CQI). ### Model identification and robustness A control system must always have some robustness property. A robust controller is such that its properties do not change much if applied to a system slightly different from the mathematical one used for its synthesis. This specification is important: no real physical system truly behaves like the series of differential equations used to represent it mathematically. Typically a simpler mathematical model is chosen in order to simplify calculations, otherwise the true system dynamics can be so complicated that a complete model is impossible. System identification For more details on this topic, see System identification. The process of determining the equations that govern the model's dynamics is called system identification. This can be done off-line: for example, executing a series of measures from which to calculate an approximated mathematical model, typically its transfer function or matrix. Such identification from the output, however, cannot take account of unobservable dynamics. Sometimes the model is built directly starting from known physical equations: for example, in the case of a mass-spring-damper system we know that $m \ddot{{x}}(t) = - K x(t) - \Beta \dot{x}(t)$. Even assuming that a "complete" model is used in designing the controller, all the parameters included in these equations (called "nominal parameters") are never known with absolute precision; the control system will have to behave correctly even when connected to physical system with true parameter values away from nominal. Some advanced control techniques include an "on-line" identification process (see later). The parameters of the model are calculated ("identified") while the controller itself is running: in this way, if a drastic variation of the parameters ensues (for example, if the robot's arm releases a weight), the controller will adjust itself consequently in order to ensure the correct performance. Analysis Analysis of the robustness of a SISO (single input single output) control system can be performed in the frequency domain, considering the system's transfer function and using Nyquist and Bode diagrams. Topics include gain and phase margin and amplitude margin. For MIMO (multi input multi output) and, in general, more complicated control systems one must consider the theoretical results devised for each control technique (see next section): i.e., if particular robustness qualities are needed, the engineer must shift his attention to a control technique by including them in its properties. Constraints A particular robustness issue is the requirement for a control system to perform properly in the presence of input and state constraints. In the physical world every signal is limited. It could happen that a controller will send control signals that cannot be followed by the physical system: for example, trying to rotate a valve at excessive speed. This can produce undesired behavior of the closed-loop system, or even damage or break actuators or other subsystems. Specific control techniques are available to solve the problem: model predictive control (see later), and anti-wind up systems. The latter consists of an additional control block that ensures that the control signal never exceeds a given threshold. ## System classifications ### Linear systems control Main article: State space (controls) For MIMO systems, pole placement can be performed mathematically using a state space representation of the open-loop system and calculating a feedback matrix assigning poles in the desired positions. In complicated systems this can require computer-assisted calculation capabilities, and cannot always ensure robustness. Furthermore, all system states are not in general measured and so observers must be included and incorporated in pole placement design. ### Nonlinear systems control Main article: Nonlinear control Processes in industries like robotics and the aerospace industry typically have strong nonlinear dynamics. In control theory it is sometimes possible to linearize such classes of systems and apply linear techniques, but in many cases it can be necessary to devise from scratch theories permitting control of nonlinear systems. These, e.g., feedback linearization, backstepping, sliding mode control, trajectory linearization control normally take advantage of results based on Lyapunov's theory. Differential geometry has been widely used as a tool for generalizing well-known linear control concepts to the non-linear case, as well as showing the subtleties that make it a more challenging problem. ### Decentralized systems Main article: Distributed control system When the system is controlled by multiple controllers, the problem is one of decentralized control. Decentralization is helpful in many ways, for instance, it helps control systems operate over a larger geographical area. The agents in decentralized control systems can interact using communication channels and coordinate their actions. ## Main control strategies Every control system must guarantee first the stability of the closed-loop behavior. For linear systems, this can be obtained by directly placing the poles. Non-linear control systems use specific theories (normally based on Aleksandr Lyapunov's Theory) to ensure stability without regard to the inner dynamics of the system. The possibility to fulfill different specifications varies from the model considered and the control strategy chosen. Here a summary list of the main control techniques is shown: Adaptive control Adaptive control uses on-line identification of the process parameters, or modification of controller gains, thereby obtaining strong robustness properties. Adaptive controls were applied for the first time in the aerospace industry in the 1950s, and have found particular success in that field. Hierarchical control A Hierarchical control system is a type of Control system in which a set of devices and governing software is arranged in a hierarchical tree. When the links in the tree are implemented by a computer network, then that hierarchical control system is also a form of Networked control system. Intelligent control Intelligent control uses various AI computing approaches like neural networks, Bayesian probability, fuzzy logic,[7] machine learning, evolutionary computation and genetic algorithms to control a dynamic system. Optimal control Optimal control is a particular control technique in which the control signal optimizes a certain "cost index": for example, in the case of a satellite, the jet thrusts needed to bring it to desired trajectory that consume the least amount of fuel. Two optimal control design methods have been widely used in industrial applications, as it has been shown they can guarantee closed-loop stability. These are Model Predictive Control (MPC) and linear-quadratic-Gaussian control (LQG). The first can more explicitly take into account constraints on the signals in the system, which is an important feature in many industrial processes. However, the "optimal control" structure in MPC is only a means to achieve such a result, as it does not optimize a true performance index of the closed-loop control system. Together with PID controllers, MPC systems are the most widely used control technique in process control. Robust control Robust control deals explicitly with uncertainty in its approach to controller design. Controllers designed using robust control methods tend to be able to cope with small differences between the true system and the nominal model used for design. The early methods of Bode and others were fairly robust; the state-space methods invented in the 1960s and 1970s were sometimes found to lack robustness. A modern example of a robust control technique is H-infinity loop-shaping developed by Duncan McFarlane and Keith Glover of Cambridge University, United Kingdom. Robust methods aim to achieve robust performance and/or stability in the presence of small modeling errors. Stochastic control Stochastic control deals with control design with uncertainty in the model. In typical stochastic control problems, it is assumed that there exist random noise and disturbances in the model and the controller, and the control design must take into account these random deviations. Energy-shaping control Energy-shaping control view the plant and the controller as energy-transformation devices. The control strategy is formulated in terms of interconnection (in a power-preserving manner) in order to achieve a desired behavior. ## See also Examples of control systems Topics in control theory Other related topics ## References 1. Maxwell, J.C. (1868). "On Governors". Proceedings of the Royal Society of London 16: 270–283. doi:10.1098/rspl.1867.0055. JSTOR 112510. 2. Routh, E.J.; Fuller, A.T. (1975). Stability of motion. Taylor & Francis. 3. Routh, E.J. (1877). A Treatise on the Stability of a Given State of Motion, Particularly Steady Motion: Particularly Steady Motion. Macmillan and co. 4. Hurwitz, A. (1964). "On The Conditions Under Which An Equation Has Only Roots With Negative Real Parts". Selected Papers on Mathematical Trends in Control Theory. 5. Donald M Wiberg. State space & linear systems. Schaum's outline series. McGraw Hill. ISBN 0-07-070096-6 [Amazon-US \| Amazon-UK]. 6. Liu, Jie; Wang, Golnaraghi, Kubica (2010). "A novel fuzzy framework for nonlinear system control". Fuzzy Sets and Systems 161 (21): 2746–2759. ## Further reading • Levine, William S., ed. (1996). The Control Handbook. New York: CRC Press. ISBN 978-0-8493-8570-4 [Amazon-US \| Amazon-UK]. • Karl J. Åström and Richard M. Murray (2008). Feedback Systems: An Introduction for Scientists and Engineers.. Princeton University Press. ISBN 0-691-13576-2 [Amazon-US \| Amazon-UK]. • Christopher Kilian (2005). Modern Control Technology. Thompson Delmar Learning. ISBN 1-4018-5806-6 [Amazon-US \| Amazon-UK]. • Vannevar Bush (1929). Operational Circuit Analysis. John Wiley and Sons, Inc. • Robert F. Stengel (1994). Optimal Control and Estimation. Dover Publications. ISBN 0-486-68200-5 [Amazon-US \| Amazon-UK]. • Franklin et al. (2002). Feedback Control of Dynamic Systems (4 ed.). New Jersey: Prentice Hall. ISBN 0-13-032393-4 [Amazon-US \| Amazon-UK]. • Joseph L. Hellerstein, Dawn M. Tilbury, and Sujay Parekh (2004). Feedback Control of Computing Systems. John Wiley and Sons. ISBN 0-471-26637-X [Amazon-US \| Amazon-UK]. • Diederich Hinrichsen and Anthony J. Pritchard (2005). Mathematical Systems Theory I - Modelling, State Space Analysis, Stability and Robustness. Springer. ISBN 3-540-44125-5 [Amazon-US \| Amazon-UK]. • Andrei, Neculai (2005). Modern Control Theory - A historical Perspective. Retrieved 2007-10-10. • Sontag, Eduardo (1998). Mathematical Control Theory: Deterministic Finite Dimensional Systems. Second Edition. Springer. ISBN 0-387-98489-5 [Amazon-US \| Amazon-UK]. • Goodwin, Graham (2001). Control System Design. Prentice Hall. ISBN 0-13-958653-9 [Amazon-US \| Amazon-UK]. For Chemical Engineering • Luyben, William (1989). Process Modeling, Simulation, and Control for Chemical Engineers. Mc Graw Hill. ISBN 0-07-039159-9 [Amazon-US \| Amazon-UK]. ## Source Content is authored by an open community of volunteers and is not produced by or in any way affiliated with ore reviewed by PediaView.com. Licensed under the Creative Commons Attribution-ShareAlike 3.0 Unported License, using material from the Wikipedia article on "Control theory", which is available in its original form here: http://en.wikipedia.org/w/index.php?title=Control_theory • ## Finding More You are currently browsing the the PediaView.com open source encyclopedia. Please select from the menu above or use our search box at the top of the page. • ## Questions or Comments? If you have a question or comment about material in the open source 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. This open source encyclopedia supplement is brought to you by PediaView.com, the web's easiest resource for using Wikipedia content. All Wikipedia text is available under the terms of the Creative Commons Attribution-ShareAlike 3.0 Unported License. Wikipedia® itself is a registered trademark of the Wikimedia Foundation, Inc.	{"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": 27, "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": 0.8920259475708008, "perplexity_flag": "middle"}
http://mathhelpforum.com/pre-calculus/183997-cancelling-out-root-function-y.html	# Thread: 1. ## Cancelling out a root function of y Hi there, I've got an equation and I need to find the value of $y$. I've been able to get $y$ on it's own side of the equation but am stuck on the last steps. The left side of the equation is $y^{\frac{1}{4}}$ and I have then used the rule for powers to achieve $\sqrt[4]{y}$ but I don't know what I need to do now to both sides of the equation to get $y$ on it's own. Many thanks. 2. ## Re: Cancelling out a root function of y can you post the equation please? 3. ## Re: Cancelling out a root function of y I did think that the power of 4 would do it, but Mathcad isn't giving me $y$ on it's own. I'm starting to think I may have made an error in Mathcad. Would somebody please confirm if $(\sqrt[4]{y})^{4}=y$ Many thanks. 4. ## Re: Cancelling out a root function of y Originally Posted by BAdhi can you post the equation please? Hi, the original equation is $\frac{-2}{y^{\frac{1}{4}}}=-(2e^{x}-e^{-2x})^{-\frac{1}{2}}+\frac{1}{2}$ Thanks. 5. ## Re: Cancelling out a root function of y So far I've got $\frac{-2}{y^{\frac{1}{4}}}=-(2e^{x}-e^{-2x})^{-\frac{1}{2}}+\frac{1}{2}$ $-2=y^{\frac{1}{4}}[-(2e^{x}-e^{-2x})^{-\frac{1}{2}}+\frac{1}{2}]$ $y^{\frac{1}{4}}=\frac{-2}{-(2e^{x}-e^{-2x})^{-\frac{1}{2}}+\frac{1}{2}}$ 6. ## Re: Cancelling out a root function of y Would the next step be $y=[\frac{-2}{-(2e^{x}-e^{-2x})^{-\frac{1}{2}}+\frac{1}{2}}]^{4}$ Thanks. 7. ## Re: Cancelling out a root function of y $(\sqrt[4]{y})^{4}=y$ This is correct A quick and easy way to check is to give y a value and check. so let y = 1000 $(\sqrt[4]{1000})^{4}$ put this in your calculator to check, and you should get 1000. 8. ## Re: Cancelling out a root function of y Thanks for confirming, I've been able to complete the problem.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 14, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": 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": 0.9268106818199158, "perplexity_flag": "middle"}
http://physics.stackexchange.com/questions/1143/particle-wavefunction-and-gravity/1181	Particle wavefunction and gravity Suppose a particle has 50% probability of being at location $A$, and 50% probability being at location $B$ (see double slit experiment). According to QM the particle is at both $A$ and $B$ at the same time, so is there a force of gravity between the two particle superpositions? Is there self-gravity when a wave-function reaches over a finite distance? I cannot seem to wrap my head around this. Is the gravity a proportional fraction of the entire mass based on the probabilities. How do you combine a wavefunction with Gauss' law of gravity? I have being trying to think about self-gravity for a long time now. - 6 Answers There is some work by Roger Penrose on the subject. The papers title is, "On Gravity's Role in Quantum State Reduction", and it discuses how the interaction of two states that have different mass distributions with spacetime can cause the wavefunction to collapse in the one state or the other. There is also a following paper that discuses the same thing in Newtonian gravity, "Spherically-symmetric solutions of the Schrödinger-Newton equations" (and there is also this that you could have a look). There is one thing that I should point out that is also pointed out by David. In a situation as the one described in the question (double slit experiment), the particle is not at two different places at the same time and interacts with it self. It is the two states (wavefunctions) that interact to give you the interference. - +1 for the reference, although I would doubt (as most other physicists) that gravity have anything to do with the wavefunction collapse (Zurek's decoherence is way more plausible). – Piotr Migdal Nov 21 '10 at 10:53 Thanks for the links. I am a great admirer of Sir R. Penrose and his insights. – ja72 Nov 22 '10 at 3:14 I'm fairly sure it's not correct to say that the particle is at both A and B at the same time. If it interacts with something at A, then it's at A, not at B. And vice-versa. I believe the production of a gravitational field would be one such interaction (although perhaps we might need a quantum theory of gravity to be truly sure), so when you detect the gravitational field produced by the particle, it will appear to be "emanating" from either A or B, but not both. This would mean that the particle can't interact with itself, since if it exists at point A to be "emitting" the gravitational field, it can't also exist at point B to be reacting to the gravitational field. I believe the same question could apply to electromagnetic self-interaction of a charged particle. But for that case we do have a theory that should explain what happens, namely quantum electrodynamics. Perhaps someone else can explain that case in detail, or if I can figure out something, I'll edit it in here. - 1 The particle does interact with itself in the double slit experiment (not gravitationally) when in produces an interferance pattern on the back wall, even when the emision rate is slowed down to 1 particle at a time. If you detect the combined effect of gravity of the particle being at both location then you have not collapsed the wave function (still don't know where the particle is). – ja72 Nov 20 '10 at 5:49 If I remember correctly, the results drawn from the double split experiment are not conducted using a double split as depicted in textbooks. So I think that gravity is out of question in real life. – Robert Smith Nov 20 '10 at 5:54 @jalexious: How can you slow down the emission rate to 1 particle at a time? – Robert Smith Nov 20 '10 at 5:56 3 @jalexiou: that's not the particle interacting with itself, that's wavefunction interference. – David Zaslavsky♦ Nov 20 '10 at 6:05 2 @jalexious: No, measurement have nothing to do with virtual particles. I do not know what do you mean by 'particle represented by 2 wavefunctions'. However, there is a huge difference between: a) having particle at A and another at B b) having a one-paticle wavefunction with with non-zero values at A and B. – Piotr Migdal Nov 20 '10 at 9:44 show 4 more comments First a general comment - everything in the world is described by either classical or quantum fields. Point particles are a fiction, sometimes useful, sometimes not. Starting with classical field theories like Maxwell equations or general relativity, you find that you are forced to forget about point sources and repalce them by continuous charge or mass distribution, otherwise you get all kinds of nonsense (non-locality, acausality, etc. etc.). One of the reasons for that is the infinite self-force or self-energy problem that crops up already at the classical level. We can approximate a continuous distribution of matter by a "particle" if it obeys certain conditions, roughly speaking it has to be localized and weakly interacting. By "localized" I mean that all relevant observable quantities (expectation values of operators) are localized. This is not the situation you describe - the wavefunction is approximately localized (with two centres) but it is not observable. Relevant observable quantities like the expectation value of currents will not necessarily be localized. So, what you are asking in effect is the self-force for a particular distribution of mass (or charge). There is an answer for that, but since you are asking a question that had to do with short distance physics, the quantum mechanics of the gravitational (or electromagnetic) field comes into it. Probably not enough space-time to elaborate on this here. - This is similar to what my old boss called the "Lafyatis Problem," because it was first posed to him by Greg Lafyatis at Ohio State, which involved light emission from a particle in a "Schroedinger cat" state. The question is, if you have an atom in a superposition of two position states, and it emits light, should you expect to see an interference pattern in the emitted light due to the light being emitted by the two different possible locations of the atom? This got debated periodically, and as I recall the consensus answer that was settled on was that there wouldn't be any interference, but it was never fully settled, and kept being brought up from time to time. I know that Dave Wineland's group did experiments in which they put trapped ions into a "cat" state, a superposition of two positions in the trap (Science 272, 1131 (1996)). They didn't deal with this issue experimentally, but I think it's what prompted the discussion, and might be a place to start looking for information. - We don't know, really. Thinking in Quantum Mechanical terms, you should proceed like this: • The setup of the experiment is represented in quantised form in Schrödinger's equation (it gets more complex with QED and QFT, but the principle is pretty much the same). This includes the physical setup (screens, etc.) and all the fields you wish to account for. This is where you should include gravity. • The solution of the equation gives you a function which ultimately describes your experiment. You apply specific operators which correspond to measurements and obtain a series of possible outcomes and the probabilities of each. The thing of note here is that when you add gravity initially, its effects are taken into account globally (in Schrödinger's equation). So this would take into account all the effects you are talking about. What doesn't work in your line of thinking is that you account for gravity locally (i.e. with particle at point A or B or both), whereas, in a QM world you can't do that. Now, the problem is that obviously we don't have a good and valid representation of gravity in QM terms. - Let's first recognize that the gravitational force is 10^{-42} times weaker than the electrostatic force. Now pause to ingest that number. The earth is only about 10^{23} times more massive than we are. So the question is not a practical concern, but one of fundamental physical theory. We don't have a good theory in which we can analyze dynamical gravitational and quantum-mechanical processes. Hawking radiation is perhaps the best blend of the two, but it takes place in an extremely large gravitational field, where the discrepancy between the forces is much less. (Within string theory, which is a consistent quantum theory which includes gravity, the best quantum-gravity calculations regard the number of microstates of a black hole, i.e. its entropy.) To guess that an incoming state's gravitational self-interaction will affect its evolution -- and that this effect will be different for superpositions of localized states -- but I have no idea what would actually happen. I'm sure that people more connected with experiments will have a better feel, but a fundamental understanding is currently beyond our ken. - You've got a typo in the first sentence-- "weaker" should be "stronger." – Chad Orzel Nov 20 '10 at 19:24 Thanks -- fixed. – Eric Zaslow Nov 20 '10 at 20:57 Obviously this is a theoretical question to be asked. There is either an effect to be considered, or not. – ja72 Nov 21 '10 at 5:36	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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": 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": 0.957284152507782, "perplexity_flag": "middle"}
http://nrich.maths.org/417/clue	Amida Stage: 5 Challenge Level: Can you prove that for every Almida framework, no two paths ever end up at the foot of the same upright? We have to show that the system described is a permutation (re-arrangement) of the numbers $1$ to $n$ which occur at the top of the uprights. Imagine moving numbered counters down the paths at the same rate and every time a rung is encountered the two counters on adjacent uprights change places; this is called a transposition. This is a good way of recording the sequence transpositions $12345$ $21345$ $21354$ $23154$ $23514$ $32514$ $32541$ $35241$ $35214$ $35124$ $53124$ The NRICH Project aims to enrich the mathematical experiences of all learners. To support this aim, members of the NRICH team work in a wide range of capacities, including providing professional development for teachers wishing to embed rich mathematical tasks into everyday classroom practice. More information on many of our other activities can be found here.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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": 0.9315159916877747, "perplexity_flag": "middle"}
http://en.wikipedia.org/wiki/Spurious_relationship	# Spurious relationship It has been suggested that be merged into this article. (Discuss) Proposed since April 2011. In statistics, a spurious relationship (or, sometimes, spurious correlation) is a mathematical relationship in which two events or variables have no direct causal connection, yet it may be wrongly inferred that they do, due to either coincidence or the presence of a certain third, unseen factor (referred to as a "confounding factor" or "lurking variable"). Suppose there is found to be a correlation between A and B. Aside from coincidence, there are three possible relationships: A causes B, B causes A, OR C causes both A and B. In the last case there is a spurious correlation between A and B. In a regression model where A is regressed on B but C is actually the true causal factor for A, this misleading choice of independent variable (B instead of C) is called specification error. Because correlation can arise from the presence of a lurking variable rather than from direct causation, it is often said that "Correlation does not imply causation".[citation needed] A spurious relationship should not be confused with a spurious regression, which refers to a regression that shows significant results due to the presence of a unit root in both variables.[citation needed] ## General example An example of a spurious relationship can be illuminated examining a city's ice cream sales. These sales are highest when the rate of drownings in city swimming pools is highest. To allege that ice cream sales cause drowning, or vice-versa, would be to imply a spurious relationship between the two. In reality, a heat wave may have caused both. The heat wave is an example of a hidden or unseen variable, also known as a confounding variable. Another popular example is a series of Dutch statistics showing a positive correlation between the number of storks nesting in a series of springs and the number of human babies born at that time. Of course there was no causal connection; they were correlated with each other only because they were correlated with the weather nine months before the observations.[1] However Höfer et al [2] showed the correlation to be stronger than just weather variations as he could show in post reunification Germany that, while the number of clinical deliveries was not linked with the rise in stork population, out of hospital deliveries correlated with the stork population. ## Detecting spurious relationships The term "spurious relationship" is commonly used in statistics and in particular in experimental research techniques, both of which attempt to understand and predict direct causal relationships (X → Y). A non-causal correlation can be spuriously created by an antecedent which causes both (W → X and W → Y). Intervening variables (X → W → Y), if undetected, may make indirect causation look direct. Because of this, experimentally identified correlations do not represent causal relationships unless spurious relationships can be ruled out. ### Experiments In experiments, spurious relationships can often be identified by controlling for other factors, including those that have been theoretically identified as possible confounding factors. For example, consider a researcher trying to determine whether a new drug kills bacteria; when the researcher applies the drug to a bacterial culture, the bacteria die. But to help in ruling out the presence of a confounding variable, another culture is subjected to conditions that are as nearly identical as possible to those facing the first-mentioned culture, but the second culture is not subjected to the drug. If there is an unseen confounding factor in those conditions, this control culture will die as well, so that no conclusion of efficacy of the drug can be drawn from the results of the first culture. On the other hand, if the control culture does not die, then the researcher cannot reject the hypothesis that the drug is efficacious. ### Non-experimental statistical analyses Disciplines whose data are mostly non-experimental, such as economics, usually employ observational data to establish causal relationships. The body of statistical techniques used in economics is called econometrics. The main statistical method in econometrics is multivariate regression analysis. Typically a linear relationship such as $y = a_0 + a_1x_1 + a_2x_2 + \cdots + a_kx_k + e$ is hypothesized, in which $y$ is the dependent variable (hypothesized to be the caused variable), $x_j$ for j = 1, ..., k is the jth independent variable (hypothesized to be a causative variable), and $e$ is the error term (containing the combined effects of all other causative variables, which must be uncorrelated with the included independent variables). If there is reason to believe that none of the $x_j$s is caused by y, then estimates of the coefficients $a_j$ are obtained. If the null hypothesis that $a_j=0$ is rejected, then the alternative hypothesis that $a_{j} \ne 0$ and equivalently that $x_j$ causes y cannot be rejected. On the other hand, if the null hypothesis that $a_j=0$ cannot be rejected, then equivalently the hypothesis of no causal effect of $x_j$ on y cannot be rejected. Here the notion of causality is one of contributory causality: If the true value $a_j \ne 0$, then a change in $x_j$ will result in a change in y unless some other causative variable(s), either included in the regression or implicit in the error term, change in such a way as to exactly offset its effect; thus a change in $x_j$ is not sufficient to change y. Likewise, a change in $x_j$ is not necessary to change y, because a change in y could be caused by something implicit in the error term (or by some other causative explanatory variable included in the model). Regression analysis controls for other relevant variables by including them as regressors (explanatory variables). This helps to avoid mistaken inference of causality due to the presence of a third, underlying, variable that influences both the potentially causative variable and the potentially caused variable: its effect on the potentially caused variable is captured by directly including it in the regression, so that effect will not be picked up as a spurious effect of the potentially causative variable of interest. In addition, the use of multivariate regression helps to avoid wrongly inferring that an indirect effect of, say x1 (e.g., x1 → x2 → y) is a direct effect (x1 → y). Just as an experimenter must be careful to employ an experimental design that controls for every confounding factor, so also must the user of multiple regression be careful to control for all confounding factors by including them among the regressors. If a confounding factor is omitted from the regression, its effect is captured in the error term by default, and if the resulting error term is correlated with one (or more) of the included regressors, then the estimated regression may be biased or inconsistent (see omitted variable bias). ## Footnotes 1. Roger Sapsford, Victor Jupp, ed. (2006). Data Collection and Analysis. Sage. ISBN 0-7619-4362-5. 2. Höfer, Thomas; Hildegard Przyrembel and Silvia Verleger (2004). "New evidence for the Theory of the Stork". Paediatric and Perinatal Epidemiology 18: 18–22. Retrieved 19 November 2012. ## References • Pearl, Judea. Causality: Models, Reasoning and Inference, Cambridge University Press, 2000. • Yule, G.U, 1926, "Why do we sometimes get nonsense correlations between time series? A study in sampling and the nature of time series", Journal of the Royal Statistical Society 89, 1–64.	{"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": 15, "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": 0.9400689601898193, "perplexity_flag": "middle"}
http://math.stackexchange.com/questions/197330/another-cross-product	Another Cross Product So I understand most of the properties of cross products. However I ran into a small complication. I get that $i\times j = k$, $j\times k = i$. I also understand that $k \times j = -i$ and that $k\times k = j\times j = i\times i = 0$. But what happens when you have $-k\times k$? Does that also equal $0$, or something else? - 3 Well, it is $-0$. And $-0=0$. – Giuseppe Negro Sep 16 '12 at 3:15 1 Answer There is another algebraic property of the cross product such that $(ra) \times b = r(a \times b)$ where $a, b$ are vectors and $r$ is a constant. (http://en.wikipedia.org/wiki/Cross_product) Since you said that $k \times k = 0$, we can write $(-k) \times k = -(k \times k) = -0 = 0$ -	{"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": 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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9468488693237305, "perplexity_flag": "head"}
http://mathhelpforum.com/calculus/68254-integrals-approximating-arc-segment-lengths.html	# Thread: 1. ## integrals and approximating arc segment lengths Given a differentiable function y=f(x) on [a,b], and a regular partition a=x0<x1<...<xn-1<xn=b explain how approximating the arc segment length deltaLi by a straight line, then letting n -> infinity, leads to a definite integral for the total length L = integral from a to b sqrt(1 + (f'(x))^2) dx i was given this question just after being taught integration by parts ... and i cannot understand anything about it ... if someone could provide me with details on how to solve it ... not just the solution ... that would be great as this is due tomorrow 2. You could try to divide you circle in a pentagon, then hexagon, etc. With many sides, you get a better precision of the circumference. 3. If I understand this correctly, you are supposed to derive/demonstrate how the arc length formula works. Consider your function between partitions $x_1 , x_2$. Now consider the line between these two points as an approximation of your function. It might be way off, but as the difference between the points gets smaller it will be a better approx. So how can you represent this line in terms of the two points $(x_1 , y_1),(x_2 , y_2)$? Let the line be a hypotenuse of a right triangle and naturally the legs are $\Delta x , \Delta y$. Thus the line segment is $\sqrt{ (\Delta x)^2+ (\Delta y)^2}$ Now comes a trick. Factor out a delta x squared and you get $L= \Delta x \sqrt{1+\frac{(\Delta y)^2}{(\Delta x)^2}}$ Is this what you had in mind or am I completely off? #### 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": 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": 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": 0.9440222978591919, "perplexity_flag": "head"}
http://physics.stackexchange.com/questions/tagged/celestial-mechanics	# Tagged Questions The celestial-mechanics tag has no wiki summary. 3answers 72 views ### Parallax, obliquity, precession, and Orion? Today, the obliquity of the earth is about 23.4°. 6500 years ago, it was about 24.1° Imagine the blue square is the constellation of Orion, and the yellow star is the sun. Viewpoint B is you, on ... 1answer 100 views ### Does the possibility of large scale entanglement mean 2-Body Problems are also unsolvable? Experiments are showing that larger and larger objects can be entangled whereby proving that this quantum feature has no upper limit. Assuming this is true, does entangled celestial bodies mean even ... 0answers 25 views ### Planets motion around the sun [duplicate] Why planets of solar system move almost in the same plane? 1answer 49 views ### Relation between satellites' potential energy and quantum mechanical confinement energy? 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Except for Mercury, the planets in the Solar System have very small eccentricities. Is this property special to the Solar System? Wikipedia states: Most exoplanets with orbital periods of 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": 1, "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": 0.9290336966514587, "perplexity_flag": "middle"}
http://unapologetic.wordpress.com/2007/05/16/the-fundamental-involutory-quandle/	# The Unapologetic Mathematician ## The Fundamental Involutory Quandle As I discussed last time, coloring a knot with any abelian group is secretly using the dihedral quandle associated to that group. This is an involutory quandle with action $a\triangleright b=2a-b$. The reason knot coloring works out so nicely is that the axioms of (involutory) quandles line up with the Reidemeister moves. But for the moment we’re stuck with picking this or that involutory quandle and counting how many colorings it gives for a given knot. Different quandles give different coloring numbers, and we’d like to find a better way of thinking of them all at once. We’re going to construct a new involutory quandle from a knot that captures all of them. Take any diagram of the knot we’re interested in. Remember the knot table if you want to pick one out. Now each arc in the diagram has to get some color, no matter what quandle we’re using to color it. Instead of picking a color from a specific quandle, let’s just slap a label like $x$, $y$, or $z$ on each arc. Be sure to use a different label for each different arc. Now those labels will generate an involutory quandle. We can throw them together with the two quandle compositions to get “words” like $x\triangleright((y\triangleright z)\triangleright z)$. These words, of course, are subject to the normal quandle equivalences, but we need more relations for our purposes. At each crossing the values in a coloring have to satisfy a certain relation, so we’re going to build that right into our quandle. If the arcs labeled $x$ and $z$ meet under the crossing arc labeled $y$, then we must have $z=y\triangleright x$. This seems to depend on the choice of a diagram, though. Well, it sort of does, but any Reidemeister move gives an isomorphism of quandles relating the two sides. For example, performing the first one splits an arc into two pieces. Say label $x$ becomes $x_1$ and $x_2$. Then the relations we introduce say that $x_1\triangleright x_1=x_2$. But the axioms of quandles say that $x_1\triangleright x_1=x_1$, so $x_1=x_2$ and we can just drop one of these generators and the relation we’ve now “used up”. Try to find the isomorphisms for the other two moves. This justifies calling the quandle we’ve constructed (up to isomorphism) “the” fundamental involutory quandle $Q(K)$ of the knot $K$. So what’s a coloring? A coloring assigns an element of some quandle to each arc of the knot diagram. But arcs in the diagram are just generators of the fundamental quandle. That is, a coloring is a function that takes generators of the fundamental quandle to a selected target quandle. If it plays nicely with the relations between the generators, it will be a quandle homomorphism. In fact it does, precisely because we picked the relations between the generators to be exactly those required by colorings. A given relation comes from a crossing, and every coloring of a knot obeys the same restrictions at crosings. In the end we’ve found that the set of all colorings of $K$ by an involutory quandle $Q$ is the set of quandle homomorphisms $\hom_{\mathbf{Quan}}(Q(K),Q)$, so the number of $Q$-colorings is the cardinality of this set. If we have a good understanding of quandles and their homomorphisms, we can read off coloring numbers by involutory quandles from the fundamental involutory quandle. ### Like this: Posted by John Armstrong \| Knot theory, Quandles ## 1 Comment » 1. “The Involutory Quandle” would be a seriously good name for a blog. Comment by \| May 19, 2007 \| Reply « Previous \| Next » ## About this weblog This is mainly an expository blath, with occasional high-level excursions, humorous observations, rants, and musings. The main-line exposition should be accessible to the “Generally Interested Lay Audience”, as long as you trace the links back towards the basics. Check the sidebar for specific topics (under “Categories”). I’m in the process of tweaking some aspects of the site to make it easier to refer back to older topics, so try to make the best of it for now.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 21, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 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": 0.899776816368103, "perplexity_flag": "head"}
http://mathhelpforum.com/calculus/204618-multivariable-critical-points.html	# Thread: 1. ## Multivariable critical points Show that f(x,y) = $\sqrt{x^2+ y^2}$ has one critical point P and that f is nondifferentiable at P. Is P a saddle point, minimum, or maximum? So I found fx = $\frac {2x} {\sqrt {x^2 + y^2}}$ and fy = $\frac {2y} {\sqrt {x^2 + y^2}}$ and I tried some algebraic gymnastics to try and solve for critical points but unfortunately I'm just flat out stuck Thanks Anthony 2. ## Re: Multivariable critical points Originally Posted by cubejunkies Show that f(x,y) = $\sqrt{x^2+ y^2}$ has one critical point P and that f is nondifferentiable at P. Is P a saddle point, minimum, or maximum? So I found fx = $\frac {2x} {\sqrt {x^2 + y^2}}$ and fy = $\frac {2y} {\sqrt {x^2 + y^2}}$ and I tried some algebraic gymnastics to try and solve for critical points but unfortunately I'm just flat out stuck Thanks Anthony At critical points, all partial derivatives equal to 0. Also, for a function to be differentiable at a point, it need to be defined at that point, continuous at that point, and have continuous partial derivatives at that point. 3. ## Re: Multivariable critical points But how do you solve the partial derivatives when they're set equal to 0? I always get (0,0) but the first partial derivatives don't exist at (0,0) so I can't figure out whether (0,0) if it even is a critical point, is a max min or saddle point because the second partial derivatives will not exist at (0,0) either. 4. ## Re: Multivariable critical points Originally Posted by cubejunkies But how do you solve the partial derivatives when they're set equal to 0? I always get (0,0) but the first partial derivatives don't exist at (0,0)... ***************************** Compare that with how you initially posed the question: "Show that f(x,y) = has one critical point P and that f is nondifferentiable at P..." ***************************** Note that f is perfectly well defined at (0,0), even though its partial deriviatives aren't. To decide about (0,0), you'll need some technique other than the standard one (standard one = examining the eigenvalues of the Hessian). A good first step is to see if you can approximately "see" what the graph of the function looks like. To "see" the graph of the function, ask yourself about its "level curves". Think of it as graphed according to z = f(x, y). Then think of what points in the x-y plane correspond to z being a constant. Like z = r. This is like slicing the graph with a horizontal plane z = r. I don't know if that makes sense to you, but if it does, you should be able to "see" the graph, and then know the answer to your problem. Once you know the answer, it's pretty easy to "see" (and prove!) it algebraically too.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 6, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9416313767433167, "perplexity_flag": "head"}
http://mathoverflow.net/questions/51314/sums-of-inverses-of-odd-elements-modulo-p	## Sums of inverses of odd'' elements modulo $p$ ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) In a letter to Dirichlet, Gottold Eisenstein stated the congruence: $$q(u) \equiv u - \frac{u^2}{2}+ \frac{u^3}{3} - \frac{u^4}{4} + \cdots + \frac{u^{p-2}}{p-2}-\frac{u^{p-1}}{p-1} \pmod{p}$$ where $p$ in an odd prime number and $$q(u) = \frac{(1+u)^p -1 - u^p}{p}.$$ Putting $u=1$ and using $$1+\frac{1}{2}+\frac{1}{3}+ \cdots + \frac{1}{p-2}+\frac{1}{p-1} \equiv 0 \pmod{p}$$ we get $$1+\frac{1}{3}+\frac{1}{5}+ \cdots + \frac{1}{p-4}+\frac{1}{x} \equiv \frac{q(1)}{2} \pmod{p}$$ for $x=p-2.$ Question: There is a $t\quad \quad$ odd" solution of the following congruence: $$1+\frac{1}{3}+\frac{1}{5}+ \cdots + \frac{1}{p-4}+ \cdots + \frac{1}{t} \equiv q(1) \pmod{p}$$ i.e., $$t \in {p-2,p+2,p+4, \ldots}.$$ I just got $-2$ points on this post but I do not see why ? There not seem to exists negative votes ??? - Of course, such a $t$ exists provided $q(1)+\frac1{p-3}\not \equiv 0 (mod p)$ by the previous identity. Do you want $t$ as a function of $p$? – J.C. Ottem Jan 6 2011 at 19:15 @J.C.: thanks for comment. From the comment I detected a technical error in the formula. Hope new edit is OK and clearer. – Luis H Gallardo Jan 6 2011 at 21:38 It's still unclear. As far as I remember, the sum of reciprocals of odds $<p$ is treated in my joint paper arxiv.org/abs/1004.4337 (one of the lemmas). – Wadim Zudilin Jan 6 2011 at 22:07 1 17 doesn't have a solution. – Dror Speiser Jan 6 2011 at 22:10 @Wadim: Nice, thanks, you think to page 4, formula (21) of your paper ?. There are also very nice formulas for the fermat quotient , etc in an old paper of Emma Lemmer in Annals of Math 339, 2, 350--360, (1938). – Luis H Gallardo Jan 6 2011 at 22:27 show 6 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": 6, "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": 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": 0.8735095858573914, "perplexity_flag": "middle"}
http://mathoverflow.net/questions/78553?sort=votes	## Arithmetic Fuchsian group ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) Dear all, I have the following questions: Are all Fuchsian groups of signature $(0;2,2,2,\infty)$ arithmetic? What is known about the trace fields of these groups? Best, K. - ## 1 Answer The answer is NO. Any four times punctured sphere admits two involutions, the quotient by which is an orbifold of the signature you describe. Similarly, you can take a punctured torus, and the quotient by the elliptic involution is one of your surfaces. Conversely, you can cover one of them by a torus or a four-times-punctured sphere. Since very few tori are arithmetic, same is true of your class of groups. -	{"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": 0.9452193379402161, "perplexity_flag": "middle"}
http://mathoverflow.net/questions/99015/kernel-of-an-isogeny-and-coker-of-its-induced-map-on-the-tate-module/99016	## kernel of an isogeny and coker of its induced map on the Tate module ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) In a proof in Milne's note "Abelian Variety" (top on p.52), I saw an equality: $\mathrm{Ker}(\beta)(l) = \mathrm{Coker}(T_{l}(\beta))$, here $\beta$ is an (separable) isogeny of an abelian variety $A/k$, $l$ is a prime number different from the characteristic of the base field $k$, $(l)$ means the torsion points of order a power of $l$, $T_{l}(\beta)$ is the induced map of $\beta$ on the Tate module of $A$. I can't figure out why this equality holds. Do we have a natural map for these two groups? - ## 1 Answer Let $x$ be an $l^n$-torsion point in the kernel of the isogeny. Take any point $y$ such that $l^ny=x$. $l^n\beta(y)=\beta(l^ny)=\beta(x)=0$, so $\beta(y)$ is an $l^n$-torsion point. This is well-defined up to the image of an $l^n$-torsion point under the isogeny, since a division by $l^n$ is well-defined up to an $l^n$-torsion point. This gives a map from the $l^n$-torsion points of the kernel to the cokernel of the isogeny on $l^n$-torsion points. Take the limit of this chain of maps to get a map on Tate modules. - Thank you, that's clear. – unknown (google) Jun 7 at 8:34	{"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": 0.9064728617668152, "perplexity_flag": "head"}
http://mathoverflow.net/questions/868/etale-covers-of-the-affine-line/9935	## Etale covers of the affine line ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) In characteristic p there are nontrivial etale covers of the affine line, such as those obtained by adjoining solutions to x^2 + x + f(t) = 0 for f(t) in k[t]. Using an etale cohomology computation with the Artin-Schreier sequence I believe you can show that, at least, the abelianization of the absolute Galois group is terrible. What is known about the absolute Galois group of the affine line in characteristic p? In addition, can spaces which are not A^1 (or extensions of A^1 to a larger base field) occur as covers? - ## 7 Answers Indeed, you can get whatever genus you want even with a fixed Galois group G, so long as its order is divisible by p: this is a result of Pries: .pdf here. In fact, Pries has lots of papers about exactly what can happen; looking at her papers and the ones cited therein should give you a pretty thorough picture. We don't know the Galois group of the affine line, but we do know which finite groups occur as its quotients; this is a result of Harbater from 1994 ("Abhyankar's conjecture for Galois groups over curves.") Update: As a commenter pointed out, Harbater proved this fact for an arbitrary affine curve; the statement for the affine line was an earlier theorem of Raynaud. - 1 I interpreted your answer to mean that you can any attain finite group G as a Galois group as long as p divides #G. This is not true: a necessary condition (proved sufficient by Raynaud) for G to occur is that it be "quasi-p", i.e., generated by its Sylow p-subgroups. For instance an abelian group is quasi-p iff it's a p-group, not iff its order is divisible by p. I'm sure you know this, but others may not. – Pete L. Clark Dec 28 2009 at 6:17 ### 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. As an excuse to talk about one of my favorite results, I thought I'd put this out there (even though I've already mentioned this to Tyler privately). Abhyankar conjectured that the the collection of finite quotients of the étale fundamental group of the affine line in characteristic $p$ are exactly the quasi-$p$-groups. This was proved by Raynaud (as mentoned above). A slightly more complicated statement (for general curves) was quickly thereafter proved by Harbater. Here's an even more interesting (to my mind) result: Suppose $X$ a geometrically connected, projective variety of dimension over any field $K$ of positive characteristic. Suppose $L$ an ample line bundle on $X$, $D$ a closed subscheme of dimension less than $n$, and $S$ a $0$-dimensional subscheme of the regular locus of $X$ not meeting $D$. Then there exists a positive integer $r$ and an $(n+1)$-tuple of linearly independent sections of $L^{\otimes r}$ with no common zero such that the induced finite morphism $f : X \to P^n_K$ of $K$-schemes meets the following conditions. (1) If $H$ denotes the hyperplane at infinity, then $f$ is étale away from $H$. (2) The image $f(D)$ is contained in $H$. (3) The image $f(S)$ does not meet $H$. This was proved by Abhyankar in dimension $1$, and the general result is due to Kedlaya. The proof is just gorgeous; it's even simpler than his first paper on the subject, which only works for infinite fields $K$. This says something pretty remarkable: even though, in characteristic $0$, affine spaces are simply connected, in positive characteristic, every variety contains a Zariski open that is an étale cover of affine space! (Katz uses this kind of trick in his notes on Weil II.) - As stated by David Speyer and Clark Barwick, the awnser to the second question is the following: Any smooth projective curve $C$ defined over a field $k$ of positive characteristic $p$ can be realized as a finite cover of the projective line only ramified above one point. Here is a short constructive proof only based on Riemann-Roch theorem. It can be considered as an illustration of Kedlaya's proof, only dealing with curves. • First of all, there exists a generically étale finite cover $C\to\mathbf P^1$, induced by a rational function $f\in k(C)$ (in fact, any element of $k(C)-k(C)^p$ will do the job). • Denote by $R\in$ Div$(C)$ the (reduced) ramification divisor of the above cover (i.e. the ramified points are couted without multiplicity). From Riemann-Roch theorem, for large $n$, there exist a rational function $g\in k(C)$ having a pole of order $n$ at each point of the support of $R$. We may take $n$ strictly greater than $\frac{\deg(f)}p$. • Then, the rational function $h=f+g^p$ induces a cover $C\to\mathbf P^1$ only ramified above infinity. - In response to your second question, http://arxiv.org/abs/math/0207150 states "every curve over an infinite ﬁeld of characteristic p>0 admits a map to P^1 ramified over only one point!" The above paper proves a more general result, and cites the result about curves to a paper of Katz. - This is an amazing paper. Un/fortunately Kiran writes papers much faster than I can read them, but I will try to make an exception for this one. – Pete L. Clark Dec 28 2009 at 11:52 Here's a trick to generate a lot of examples over k = algebraic closure of a finite field (used in the Weil conjectures). Take any smooth curve C and a generically etale map f:C -> P^1. Then there is some open U in A^1 such that V = f^{-1}(U) -> U is finite etale. The reduced scheme underlying A^1 - U is a finite set of closed points which generate a finite subgroup G of the additive group A^1(k). Composing with the quotient A^1 -> A^1/G gives a map V -> A^1 - 0 that is finite etale. Composing with the Artin-Schreier isogeny A^1 - 0 -> A^1 sending x to x^p + 1/x gives a finite etale map V -> A^1. - I see that this question is from a while back, but I figured I add this little morsel: Manish Kumar proved for his thesis that the commutator subgroup of the algebraic fundamental group of A^1 (and, in fact, ANY smooth affine curve) over a countable algebraically closed field of characteristic p>0 is profinite free. He later (to people's astonishment, as the original statement was surprising to begin with) continued to prove this for any algebraically closed field of characteristic p>0 in this arxiv pre-print: http://arxiv.org/PS_cache/arxiv/pdf/0903/0903.4472v2.pdf - The statement about which finite groups occur as Galois groups of etale covers of the affine line -- any group which does not have a non-trivial quotient of order prime to p -- is due to Raynaud, not Harbater. (Harbater extended Raynaud's theorem to arbitrary affine curves.) -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 43, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9434999823570251, "perplexity_flag": "head"}
http://www.birs.ca/events/2012/2-day-workshops/12w2151	# Differential Schemes and Differential Cohomology (12w2151) Arriving Friday, June 22 and departing Sunday June 24, 2012 ## Organizers Richard Churchill (Hunter College, City University of New York (CUNY), Graduate Center, CUNY, and University of Calgary) Yang Zhang (The University of Manitoba) ## Objectives There are two main objectives. The first is to introduce very recent work on the modernization'' of differential algebra to algebraic geometers/number theorists having little or no acquaintance with the field. The basic properties of the differential spectrum (DiffSpec'') of a differential ring have now been established in a manner which workers in those areas can quickly grasp, but those ideas need a proper venue for communication. The informal nature of BIRS' workshops, which are familiar to both applicants, would be ideal. The second objective is to illustrate how algebraic problems which can be solved ''locally'' in terms of differential equations may admit global solutions in terms of differential cohomology. Specifically, very recent work by Ray Hoobler enables one to understand Kolchin's constrained cohomology in terms of the $,Delta$-flat topology, and that should offer new methods to algebraic geometers and number theorists. And of course the interaction of people in those areas with differential algebraists should result is new methods for use by the latter group.	{"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": 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": 0.3, "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": 0.9366123080253601, "perplexity_flag": "middle"}
http://math.stackexchange.com/questions/99295/geodesic-on-a-plane/99330	# Geodesic on a plane I guess that each geodesic on a plane is a straight line. Is it right? What can I use to prove it? I guess I have to use somehow Levi-Civita connection. - ## 3 Answers This just adds a few words and references, but maybe it will help you organize your thoughts. If you use the standard coordinates for $\mathbf R^2$, then the Euclidean metric has constant matrix $$(g_{ij}) = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}.$$ Therefore, the Christoffel symbols of the Levi-Civita connection are all zero. If you trace through the definition of the covariant derivative along a curve $\gamma(t) = (\gamma_1(t), \gamma_2(t))$ applied to the tangent vector field of $\gamma$, then you end up with the geodesic equation, which in this case requires that $$\frac{\partial^2\gamma_1}{\partial t^2} = \frac{\partial^2\gamma_2}{\partial t^2} = 0.$$ - I'm very willing to expand upon the proofs, but I didn't want to write a chapter of Riemannian geometry out. Let me know! – Dylan Moreland Jan 15 '12 at 18:37 I think it is quite simple, it is a classical example of calculus of variations. You find a function that minimizes a Lagrangian that represents the function's length. You use Euler-Lagrange equation to find that the function's second derivative is zero, concluding that it must be a straight line. For a full proof, see here. - Similar to what Dylan said, let $\mathbf{u}$ be a tangent vector to a path $\gamma$, then $\nabla_{u}u=0$ describes geodesic motion. Then noting that the Christoffel symbols vanish, we have $\partial_{A}u^{B}=0$. If we parameterize in $\tau$ we have $$\frac{d u^{x}(\tau)}{d\tau}=0$$ and $$\frac{du^y(\tau)}{d\tau}=0$$ which imples $u^x=v_{0}^x$ and $u^y=v_{0}^y$ and $x(\tau)=v_{0}^x\tau+x_0$ and $y(\tau)=v_{0}^y\tau +y_0$. These are lines. I hope this helps. -	{"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": 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": 0.3, "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": 0.9272562265396118, "perplexity_flag": "head"}
http://physics.stackexchange.com/questions/29719/finding-acceleration-for-a-car-after-finding-torque	# Finding acceleration for a car after finding torque I am trying to calculate the acceleration of a vehicle after finding the torque $\tau$. Assuming my Horse Power is 130 and my RPM is 1000, I calculated torque as: $$\tau = \frac{130 \times 33,000}{2\pi \times 1000} \approx 682.77$$ Assuming that the effective radius of my car tire is 12 inches, I calculated the force $F$: $$F = \frac{\tau}{12} \approx 56.89$$ Now I will "use the Force" to find the acceleration, using the formula $F = ma$. Rewriting in terms of $a$, and assuming the mass of my vehicle is 2712 lbs I get: $$a = \frac{F}{2712} \approx 0.0209$$ (I am not sure what the units should be here - the above torque equation I believe is using US customary units and the tire radius is in inches... so maybe the acceleration is in $ft/s^2$?) However, for larger RPMs, the acceleration seems to decrease, according to these operations. This seems to be the exact opposite of what I expected and what should happen (right?). - Hi Dylan to Physics.SE! You have to keep in mind that engine RPM is not wheel RPM. Other than that your conclusion is correct, higher RPM means less acceleration. What happens in a real engine though is that the power output also increases with RPM. About the units: sticking to SI units can reduce a lot of conversion problems, I would not recommend using HP, lbs, ft and inch in the same equation. – Alexander Jun 8 '12 at 7:59 Does the above torque equation refer to engine RPM or wheel RPM? Also, if power output increases with RPM, does this account for increase in instantaneous speed? How would one measure instantaneous speed at a time interval, with kinematic equations? (e.g. $s = s_{0} + at$) – Dylan Jun 8 '12 at 16:21 1) both, depends what you do with it. 2) The increase of output power with RPM is a general gasoline engine parameter, independent of any actual car speed. 3) $v = \Delta s/\Delta t$. – Alexander Jun 8 '12 at 16:30 With respect to 3), I am trying to use acceleration to determine speed. It seems you have given me an equation to find velocity using the change in speed over the change in time... Should I find the velocity with the equation $v = v_{0} + at$ and then use the equation you have and solve with respect to $s$? – Dylan Jun 8 '12 at 16:35 Not really. The equation I gave is for "measure instantaneous speed" as you asked for. To calculate it for a given initial speed $v_0$ and time dependent acceleration $a(t)$ the speed at time $t$ is $v(t) = v_0 + \int_0^t a(t) dt$. This site is not really made for discussions though, so please open a new question if it is still not clear how to approach the problem. – Alexander Jun 8 '12 at 22:34 ## 2 Answers I'd go along with Alexander's comment that working in SI units makes life a lot easier. However, assuming you have a good reason for sticking to US units ... The torque you've calculated is in foot pounds. The easy way to see this is that the 33,000 conversion factor you've used converts horsepower to foot pounds min$^{-1}$, and the rpm is in units of min$^{-1}$. The min$^{-1}$ on the top and bottom of the fraction cancel leaving the units as foot pounds. In the second equation you've put in the distance as inches, which makes life harder than it need be. If you take the wheel radius to be one foot rather than 12 inches you get the force equal to 682.77 pounds. For the last step you need to be aware that the "pound" is being used as a unit of force here i.e. it's the force exerted by an object weighing one pound in Earth's gravity. In units of feet per second the acceleration due to gravity is about 32.18 feel/sec$^2$, so the acceleration of an object weighing 2712 pounds will be: $$a = \frac{682.77}{2712} \times 32.18 = 8.09$$ and that's in units of feet/sec$^2$. Maybe I'm just used to SI units, but I repeated the calculation using SI units and got the same result a lot quicker! Apart from anything else it makes the distinction between mass and force a lot clearer so you wouldn't have forgotten you need to multiply by the acceleration due to gravity to get the acceleration. You're quite correct that if power is constant then torque is inversely proportional to rpm. However for most engines the power is roughly proportional to engine speed over a reasonable range, so the torque is approximately constant. If you look at the torque curves for family cars you'll see the torque is roughly constant in the middle of the engine speed range but tails off at high engine speeds. Very highly tuned engines have a torque that peask at higher engine speeds because they're tuned to develop a lot of power at high engine speeds. - How can I use this information to calculate the speed in MPH? Using an online unit conversion calculator, converting this to $mph$ yields a very large value... – Dylan Jun 8 '12 at 16:32 You've calculated the acceleration not the speed. The acceleration is 8.09 feet/sec$^2$, so after one second you're doing 8.09 feet/sec, after two seconds you're doing 16.18 feet/sec and so on. There are 5280 feet in a mile, so 8.09 feet/sec is 5.52 mph. So after one second you're doing 5.52 mph, after two seconds 11.04 mph and so on. – John Rennie Jun 8 '12 at 17:03 I am sorry, i meant to say that calculating $mph^2$ yields a large number. I thought I needed to convert acceleration to $mph^2$ units – Dylan Jun 8 '12 at 17:37 If you use mph$^2$ the number you'd get is the speed the car would be doing after accelerating steadily for an hour and this would indeed be a large number. I make it 19,872 miles per hour, which is probably faster than most cars could manage :-) I doubt anyone would find it useful to give acceleration in mph$^2$. Feet per sec$^2$ is more likely to be useful, or for us Europeans meters per sec$^2$. – John Rennie Jun 8 '12 at 17:45 Ok. Thank you. The math is easy, but the relationships are kind of hard to envision for a non-Physicist like me. :S – Dylan Jun 8 '12 at 17:52 show 5 more comments You should really include units. I believe the conversion factor horsepower is 1 horsepower $=33,000$ ft*lbf/min (not seconds) It is also in feet not inches. -	{"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": 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": 0.9532670974731445, "perplexity_flag": "middle"}
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First, let me say that this is merely something I have always wondered about, and can never seem to find a good reference for. I simply want to know... the geek in me. Why was $e$ (Euler's Number) ... 1answer 32 views ### Manipulating derivatives after substitution: $\xi=\gamma x$ I am following a quantum mechanics text book which uses a simple looking substitution in a derivative. The substitution is $$\xi=\gamma x\tag1$$ It then says that ...	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 76, "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": 0.9021442532539368, "perplexity_flag": "middle"}
http://math.stackexchange.com/questions/142575/does-there-exist-a-vector-space-with-30-elements?answertab=active	# Does there exist a vector space with 30 elements? Does there exist a vector space with 30 elements? How to determine whether there exist any vector space of particular cardinality? - 1 – anon May 8 '12 at 10:30 very useful thanks – srijan May 8 '12 at 10:50 ## 1 Answer Any finite field has order $p^k$ for some prime $p$ and $k\geq1$. Then a vector space of dimension $n$ over such a field has $(p^k)^n=p^{kn}$ elements, so in particular, the number of elements of a vector space over a finite field must be a prime power. So there is no vector space with $30$ elements. So for a general integer $x$, if $x\ne p^k$ for some $k\geq1$ and prime $p$ then there is no vector space of order $x$, and if $x=p^k$ for such $p$ and $k$, there is one vector space up to isomorphism for every (ordered) factorization of $k$ into two integers; if $k=k_1k_2$ then the $k_2$ dimensional vector space over the field of $p^{k_1}$ elements has order $x$. - i got my answer.thank you sir – srijan May 8 '12 at 10:28 Dear sir what about its sub spaces? can it have subspaces having dimension lying between 0 to 30? – srijan May 8 '12 at 10:44 2 Well, a subspace is also a vector space, so there are the same conditions on possible cardinalities (I assume you meant cardinality, not dimension here). The $k_2$ dimensional vector space over the field of $p^{k_1}$ elements has subspaces of cardinality $p^{k_1t}$ for $0\leq t\leq k_2$. – Matt Pressland May 8 '12 at 10:47 Thanks sir everything is cleared now – srijan May 8 '12 at 10:53	{"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": 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": 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": 0.9091271758079529, "perplexity_flag": "head"}
http://planetmath.org/basicalgebra	# basic algebra Let $A$ be a finite dimensional, unital algebra over a field $k$. By Krull-Schmidt Theorem $A$ can be decomposed as a (right) $A$-module as follows: $A\simeq P_{1}\oplus\cdots\oplus P_{k}$ where each $P_{i}$ is an indecomposable module and this decomposition is unique. Definition. The algebra $A$ is called (right) basic if $P_{i}$ is not isomorphic to $P_{j}$ when $i\neq j$. Of course we may easily define what does it mean for algebra to be left basic. Fortunetly these properties coincide. Let as state some known facts (originally can be found in [1]): Proposition. 1. 1. A finite algebra $A$ over a field $k$ is basic if and only if the algebra $A/\mathrm{rad}A$ is isomorphic to a product of fields $k\times\cdots\times k$. Thus $A$ is right basic iff it is left basic; 2. 2. Every simple module over a basic algebra is one-dimensional; 3. 3. For any finite-dimensional, unital algebra $A$ over $k$ there exists finite-dimensional, unital, basic algebra $B$ over $k$ such that the category of finite-dimensional modules over $A$ is $k$-linear equivalent to the category of finite-dimensional modules over $B$; 4. 4. Let $A$ be a finite-dimensional, basic and connected (i.e. cannot be written as a product of nontrivial algebras) algebra over a field $k$. Then there exists a bound quiver $(Q,I)$ such that $A\simeq kQ/I$; 5. 5. If $(Q,I)$ is a bound quiver over a field $k$, then both $kQ$ and $kQ/I$ are basic algebras. # References • 1 I. Assem, D. Simson, A. Skowronski, Elements of the Representation Theory of Associative Algebras, vol 1., Cambridge University Press 2006, 2007 Major Section: Reference Type of Math Object: Definition Parent: ## Mathematics Subject Classification 13B99 None of the above, but in MSC2010 section 13Bxx 20C99 None of the above, but in MSC2010 section 20Cxx 16S99 None of the above, but in MSC2010 section 16Sxx ## Recent Activity May 17 new image: sinx_approx.png by jeremyboden new image: approximation_to_sinx by jeremyboden new image: approximation_to_sinx by jeremyboden new question: Solving the word problem for isomorphic groups by mairiwalker new image: LineDiagrams.jpg by m759 new image: ProjPoints.jpg by m759 new image: AbstrExample3.jpg by m759 new image: four-diamond_figure.jpg by m759 May 16 new problem: Curve fitting using the Exchange Algorithm. by jeremyboden new question: Undirected graphs and their Chromatic Number by Serchinnho ## Info Owner: joking Added: 2011-02-22 - 04:52 Author(s): joking ## Versions (v5) by joking 2013-03-22	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 30, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 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": 0.8276761174201965, "perplexity_flag": "middle"}
http://mathoverflow.net/questions/19997/if-the-fourier-transformed-of-f-is-odd-is-f-odd/20000	## If the fourier transformed of f is odd, is f odd? ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) Let $f\in L^1(R)$ such that $F(f)$ is odd, where $F$ is the Fourier transform. Can I then say that $f$ is odd? If $F(f)$ is odd, then $\int \cos(x\xi) f(x) dx = 0 \:\:\forall \xi\in R$ Can I deduce from it that $f$ is odd? - ## 3 Answers Write $f^-(x)=f(-x)$ etc. Then $F(f^-)=F(f)^-$. If $F(f)$ is odd then $$0=F(f)+F(f)^-=F(f+f^-).$$ The only $L^1$ function with zero Fourier transform is $0$ so that $f+f^-=0$, that is, $f$ is odd. - Great! Thanks... – Nicolò Apr 1 2010 at 9:21 ### 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. Yes. Using tempered distributions it's immediately obvious, since $f = C \mathcal{F}^3 (\mathcal{F} f)$ for a constant $C$, and $\mathcal{F}$ maps odd distributions into odd distributions. The missing details of the proof are just simple exercises in distributions. Distribution theory is very useful for Fourier transform questions like this, since Fourier inversion works perfectly. - Look at $L^2$ first: In $L^2$ the FT is diagonalizable. The space of odd functions $\in L^2$ is the direct sum `$Eig(\mathcal{F},+i)\oplus Eig(\mathcal{F},-i)$ of the eigenspaces of $\mathcal{F}$ with respect to the eigenvalues $+i$ and $-i$. Because eigenspace are mapped into themselves, $\mathcal{F}$ maps odd functions to odd functions. In particular this is true for all $f\in L^1\cap L^2$ and by continuity it is true for all $f\in L^1$. (In fact an analogue statement is true for all tempered distributions.) EDIT: Oh, I just saw that you asked for the other direction. Using the same argument you can show that $\mathcal{F}f$ odd $\implies f=\mathcal{F}^3(\mathcal{F}f)$ odd is true for all $f\in L^2$. This time I'm not quite sure if it is possible to extend this from $L^1\cap L^2$ to $L^1$, but maybe the result for $L^1\cap L^2$ is useful for you too. -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 33, "mathjax_display_tex": 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": 0.9204843044281006, "perplexity_flag": "head"}
http://mathhelpforum.com/algebra/147812-sequences-series-natural-log.html	# Thread: 1. ## Sequences and Series + Natural Log I've spent so long trying to figure this out and I just CAN'T. I can see the relationship but I don't know how to actually calculate this through Sn = u1(r^n-1) / 4-1 or Sn = n/2(2u1 + (n-1)d) I tried calculating them both as geometric and as arithmetic but I can't figure out a consistent d or r? Thank you. 2. Originally Posted by positiveion I've spent so long trying to figure this out and I just CAN'T. I can see the relationship but I don't know how to actually calculate this through Sn = u1(r^n-1) / 4-1 or Sn = n/2(2u1 + (n-1)d) I tried calculating them both as geometric and as arithmetic but I can't figure out a consistent d or r? Thank you. For this sequence to be arithmetic each term must be separated by a common difference. In other words $U_3 - U_2 = U_2 - U_1$ Since $\left(\frac{[\ln(2)]^3}{3!} - \frac{[\ln(2)]^2}{2!}\right) \neq \left(\frac{[\ln(2)]^2}{2!} - \frac{[\ln(2)]}{1!}\right)$ then the series is not arithmetic More it to be geometric then the ratio of each successive term must be equal: $\frac{U_3}{U_2} = \frac{U_2}{U_1}$ $\left(\frac{(\ln2) ^3}{3!} \times \frac{2!}{(\ln2) ^2}\right) \neq \left(\frac{(\ln2) ^2}{2!} \times \frac{1!}{\ln2}\right)$ so it is not geometric either. From what I can gather this is of the form $\frac{(\ln2)^n}{n!}$ but I do not know where to go from there but hopefully someone else may be able to take it further edit: are you told how x and a relate to the sequence? 3. I recognize this as the Maclaurin series for $f(x) = a^x$ with $a = 2$ and $x =1$, but I'm not sure how this might help you to solve. 4. ahgh i feel so confused i print screened more of the worksheet does this additional info make any difference?	{"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": 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": 0.9648837447166443, "perplexity_flag": "head"}
http://mathhelpforum.com/advanced-algebra/189547-jordan-block-question.html	# Thread: 1. ## jordan block question what is the difference between the jordan form of matrices A and diagonal block matrices similar to A ? 2. ## Re: jordan block question Originally Posted by transgalactic what is the difference between the jordan form of matrices A and diagonal block matrices similar to A? Difference, in what terms? For example $A\sim J=\begin{bmatrix}{\lambda}&{1}&{0}\\{0}&{\lambda}& {1}\\{0}&{0}&{\lambda}\end{bmatrix}\Rightarrow A\sim K=\begin{bmatrix}{\lambda}&{2}&{0}\\{0}&{\lambda}& {3}\\{0}&{0}&{\lambda}\end{bmatrix}$ $J$ is the Jordan form of $A$, $K$ is a diagonal block matrix similar to $A$ but it is not its Jordan form. 3. ## Re: jordan block question because of the 2 ,3 on the smaller diagonal ? 4. ## Re: jordan block question Originally Posted by transgalactic because of the 2 ,3 on the smaller diagonal? It is irrelevant, only an example. The important thing is that $\dim \ker (K-\lambda I)=1$ so, the Jordan form of $K$ is $J$ . #### 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": 8, "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": 0.8998500108718872, "perplexity_flag": "middle"}
http://math.stackexchange.com/questions/185884/explain-the-arithmetics-of-this-trigonometrical-expression	# Explain the arithmetics of this trigonometrical expression $$(\frac{d_1}{2} + \frac{d_1}{2} \cdot \cos(2x))+(\frac{d_2}{2} \cdot \sin(2x)) = \frac{d_1}{2}.$$ Can someone explain how the arithmetics works here? (My teacher has only noted that "filtration removes $\cos(2x)$ and $\sin(2x)$") - So $d_1\cos(2x) + d_2 \sin(2x) = 0$? What are $d_1, d_2$? Just constants? – user2468 Aug 23 '12 at 14:50 If you are writing this equation then there must be some relationship b/w d1 and d2 – Rahul Taneja Aug 23 '12 at 15:23 Are you looking to solve for $d_1$ and $d_2$? Because, in particular, for $x=0$, the equation is false unless $d_1=0$, in which case $d_2=0$ can be seen by taking $x=\pi/4$, for instance. – Michael Boratko Aug 23 '12 at 15:31 Jennifer - Yes $d_1$ and $d_2$ are just constants. Rahul - The relationship between d1 and d2 are that they are 2 different carriers transmitted over the same channel and through the same signal (see my dropbox images for further mathematical explaination. Michael - No, I'm not looking to solve $d_1$ and $d_2$, I just want to know how you simplify the LHS expression into the simplified RHS expression, what is the arithmetics behind it? – user38541 Aug 23 '12 at 16:05 Images: modulation - dl.dropbox.com/u/16952797/temp_stuff/KT/QAM/20120823_174940.jpg demodulation - dl.dropbox.com/u/16952797/temp_stuff/KT/QAM/20120823_175218.jpg – user38541 Aug 23 '12 at 16:08 ## 1 Answer As has been pointed out in comments, for this equation to hold for all $x$, we need to have $d_1=d_2=0$. The reference to being removed by filtration doesn't refer to an arithmetic operation inherent in this equation, but to the application of a low-pass filter indicated in the second image linked to in a comment. - Alright, that makes sense. Thank you! – user38541 Aug 23 '12 at 17:23	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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": 0.9203037619590759, "perplexity_flag": "middle"}
http://mathoverflow.net/questions/90401?sort=votes	## Variance of exponential random variable ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) For a random variable $\xi$, what bounds can be achieved for Var $e^{\xi}$ in terms of E$\xi$ and Var $\xi$? - 2 Unfortunately, this question is a poor fit for this site since it is not research-level. You'd be better off asking this at math.stackexchange.com. At any rate, you might ponder the following two examples first: Let $X_1 = c$ almost surely for some $c \in \mathbb R$ and, let $X_2 = \exp(Y)$ where $Y$ has a Pareto distribution, i.e., with pdf $f(y) = \alpha y^{-\alpha-1}$ for all $y \geq 1$ and some $\alpha > 0$. – cardinal Mar 6 2012 at 23:35 ## 2 Answers None. Even if $X$ takes only two values, one of which with very small probability, Var$(e^X)$ can be made arbitrarily small or large. - ### 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. Here's a different way to think about your question, using the cumulant-generating function `$c(\theta) \equiv \log \operatorname{\mathbb{E}} e^{\theta \xi}$`. To visualize this function, note that CGFs (i) always go through the origin (obviously) and (ii) are always convex (by the Cauchy-Schwarz inequality). What you're asking for is a bound on var $e^{\xi}$, or $e^{c(2)} - e^{2 c(1)}$ in CGF notation. Properties (i) and (ii) alone tell us the (unsurprising!) fact that $c(2)$ is at least $2c(1)$, and is strictly greater than $2c(1)$ if $\xi$ has nonzero variance. So the question is what you can say about the relationship between $c(2)$ and $2c(1)$ based on the extra information you're given about the first cumulant (mean) and second cumulant (variance) of $\xi$. Translating into CGF terms, this extra information corresponds to the first and second derivatives of $c(\cdot)$ at the origin, `$c'(0)=\operatorname{\mathbb{E}} \xi$` and `$c''(0)=\operatorname{var} \xi$`. The problem that cardinal and Omer are pointing out is that if there's weird stuff going on in the higher cumulants (or equivalently moments), then there's a lot of room for flexibility in the higher derivatives at zero, $c'''(0)$, $c''''(0)$, and so on. This means you can get a lot of convexity in $c(\cdot)$ away from the origin, and hence there's basically nothing you can say, without further assumptions, about the gap between $c(2)$ and $2c(1)$. Here's a picture. The slope of the CGF at the origin is negative, reflecting the fact that `$\operatorname{\mathbb{E}} \xi$` is negative in this example. The curvature at the origin measures variance. Higher cumulants, reflected in higher order derivatives at the origin, make their presence felt via the "extra" (beyond quadratic) convexity that starts to become visible around $\theta=1$. The picture is drawn for an example in which var $e^{\xi}$ is finite, i.e. $c(2)$ is finite, which need not be the case in general. Perhaps thinking in these terms will help you figure out what extra constraints you might need to put on your problem to be able to put bounds on var $e^{\xi}$? One final remark: your terminology ("exponential random variable") is rather misleading. -	{"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": 32, "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": 0.9426395297050476, "perplexity_flag": "head"}
http://mathoverflow.net/questions/25337/lifting-varieties-to-characteristic-zero	## Lifting varieties to characteristic zero. ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) If you want to compute crystalline cohomology of a smooth proper variety $X$ over a perfect field $k$ of characteristic $p$, the first thing you might want to try is to lift $X$ to the Witt ring $W_k$ of $k$. If that succeeds, compute de Rham cohomology of the lift over $W_k$ instead, which in general will be much easier to do. Neglecting torsion, this de Rham cohomology is the same as the crystalline cohomology of $X$. I would like to have an example at hand where this approach fails: Can you give an example for A smooth proper variety $X$ over the finite field with $p$ elements, such that there is no smooth proper scheme of finite type over $\mathbb Z_p$ whose special fibre is $X$. The reason why such examples have to exist is metamathematical: If there werent any, the pain one undergoes constructing crystalline cohomology would be unnecessary. - 1 Even if such lifting always existed, it need not be functorial, which is a big deal for a cohomology theory. – Victor Protsak May 20 2010 at 11:14 I think very similar questions have come up on MO before. You might try searching for them, especially under the tag characteristic-p. – Pete L. Clark May 20 2010 at 11:30 Here is a link: mathoverflow.net/questions/423/… – Ravi Vakil Mar 13 at 6:15 ## 1 Answer This paper of Serre gives an example (I've justed pasted I. Barsotti's math-sci review). (The paper can be found in Serre's "Collected Works vol. II 1960-1971) Serre, Jean-Pierre Exemples de variétés projectives en caractéristique $p$ non relevables en caractéristique zéro. (French) Proc. Nat. Acad. Sci. U.S.A. 47 1961 108--109. An example of a non-singular projective variety $X_0$, over an algebraically closed field $k$ of characteristic $p$, which is not the image, $\text{mod}\,p$, of any variety $X$ over a complete local ring of characteristic 0 with $k$ as residue field. The variety $X_0$ is obtained by selecting, in a 5-dimensional projective space $S$, and for $p>5$, a non-singular variety $Y_0$ which has no fixed point for an abelian finite subgroup $G$ with at least 5 generators of period $p$, of the group $\Pi(k)$ of projective transformations of $S$, but which is transformed into itself by $G$; then $X_0=Y_0/G$. The reason for the impossibility is that $\Pi(K)$, for a $K$ of characteristic 0, does not contain a subgroup isomorphic to $G$. {Misprint: on the last line on p. 108 one should read $s(\sigma)=\exp(h(\sigma)N)$.} -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 33, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": 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": 0.894325852394104, "perplexity_flag": "head"}
http://mathhelpforum.com/differential-geometry/176374-closed-open-int-adh-boundary.html	# Thread: 1. ## Closed, open, int, adh and boundary Verify if the following sets are closed, open and find $\text{int}\, A,\overline A,\partial A$ and the set of accumulation points. (Does the last set have a name?) a) $A=\{(x,y)\in\mathbb R^2\|y>x\}$ b) If $a\in\mathbb R^n,$ $A=\{x\in\mathbb R^n\|x\cdot a=1\}$ c) $A=\displaystyle\overline{B}\left( \left( \frac{3}{4},0 \right),\frac{1}{4} \right)\cup \bigcup\limits_{k=2}^{\infty }{B\left( \left( \frac{3}{{{2}^{k+1}}},0 \right),\frac{1}{{{2}^{k+1}}} \right)}.$ I have the definitions but I'm not sure how to proceed. 2. There is a link between continuity of maps and open and closed sets. 3. I believe that the set of limit points of $A$ can be called the derived set of $A$, and is sometimes denoted by $A'$. The boundary of the first set is $y=x$. Note that none of the boundary points are in the set. What does this tell you about the set being open or closed?	{"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": 9, "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": 0.9607239961624146, "perplexity_flag": "head"}
http://math.stackexchange.com/questions/133222/name-of-probability-distribution	Name of probability distribution Does this distribution have a name: $f(x) = yx^{y-1}$ for $0 < x<1$ and $y>0$? It looks like an exponential distribution. Or is it a nameless distribution? - 1 Aside: When $y \in \mathbb N$, this is the density function of the maximum of a sample of size $y$ of iid $\mathcal U(0,1)$ random variables. – cardinal Apr 18 '12 at 0:17 5 – Byron Schmuland Apr 18 '12 at 0:25 1 Answer As Byron Schmuland says, this is a beta distribution with the second parameter $\beta=1$. It is sometimes called a "standard power-function distribution". -	{"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": 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": 0.7927303314208984, "perplexity_flag": "middle"}
http://blogs.agu.org/martianchronicles/2009/07/10/solar-system-creator/	Home - Astrobiology - Solar System Creator 10 July 2009 ## Solar System Creator Posted by Ryan As I mentioned last month, on top of research and grad school duties, I’m in the process of planning out a sci-fi novel. It began with the month-long outlining challenge “Midsommer Madness” over at the Liberty Hall writing site, and I am continuing with it in my spare time. I am trying to make my novel grounded in reality whenever possible. It is set in a known star system, 55 Cancri. The 55 Cancri system has 5 known planets, but I also took some artistic license and added moons and small planets that observations would likely have missed. Then, once I had planets and moons, I needed to figure out which ones would be habitable! I happen to know a thing or two about planets, so I put together a handy spreadsheet to use to calculate things like surface temperature, surface gravity and orbital period given things like how bright the star is, how far away the planet is, etc. Once I had the spreadsheet made, I realized that there are likely other people out there who might find it useful. So, whether you are a writer trying to come up with a plausible setting for your bestselling sci-fi epic, or a student learning about the solar system, or just plain curious about planets, please feel free to use and modify this spreadsheet. Right now it is set up for our solar system to give you an idea of what reasonable values are for the different variables, and to show that the results are generally pretty good despite the simplicity of the calculations. If you find this useful or have any questions, feel free to contact me by leaving a comment on this post. I described how I calculated the surface temperature below. You don’t need to read the explanation to use the spreadsheet: it should just work if you enter numbers, but I encourage you to try to follow the derivation. Even if you don’t follow the algebra, I tried to explain everything in words to give some conceptual understanding of the ideas behind the math, and the ideas are what matter. Note for Students: The spreadsheet is free for you to use, but be sure you cite this page as a source. Also show your work for all calculations! Copying values from the spreadsheet without showing your work probably won’t earn you any points, and may be considered plagiarism, which is grounds for failure and/or expulsion at most schools. And really, it’s not that hard to do the calculations, especially since the rest of this post is spent walking you through them! You might even learn something! Ok, so how does it work? Well, the calculation of a planet’s surface temperature is based on the very simple idea that if its average temperature is not changing, then the amount of energy the planet absorbs must match the amount that it emits. Pretty much common sense! If the amount in and out were different, then the temperature would change until they balanced! First, the absorption. The energy source is the star, which has a certain luminosity $L$ (given in watts). This says how much energy the star puts out in all directions per second. We want to know how much energy per square meter hits the planet, so we take the luminosity and spread it out evenly over the surface of a sphere with a radius $R$ equal to the distance from the planet to the star. The surface area of a sphere is $A=4\pi R^2$, so the amount of energy from the star hitting each square meter of the planet’s cross section is: $L/A=\dfrac{L}{4\pi R^2}$ The planet’s cross section is just the area of a circle with the planet’s radius: $\pi r^2$. Note that we’re using the area of a circle and not a sphere! That’s because the starlight doesn’t hit the whole planet, it just hits the part of the planet that is visible. Can you see all sides of a sphere at once? Neither can I, and neither can the star. What we see is the 2 dimensional cross section: a circle. So now we have an equation for how much energy the planet absorbs per second: $Energy Absorbed Per Second= \dfrac{L \pi r^2}{4\pi R^2}$ But that is assuming that the planet absorbs every bit of light that hits it, which we know isn’t true: we see planets in the night sky by their reflected light! So we can add a correction called albedo. Albedo, $A$, is the fraction of starlight that the planet reflects back out into space, and $(1-A)$ is the fraction of starlight a planet absorbs. So with that correction, our equation becomes: $Energy Absorbed Per Second= \dfrac{(1-A) L \pi r^2}{4\pi R^2}$ Now we have to figure out an expression for the energy that the planet emits. Here we have to make an assumption to simplify things: we assume that the energy absorbed by the planet is immediately redistributed evenly over the whole planet. Obviously this isn’t right, it is much warmer on the day side than the night side, but this assumption makes our lifes much easier. We just have to remember that the value we get is going to be an average of day and night temperatures. We also assume that the planet radiates away its energy like a blackbody. A blackbody is something that absorbs and emits all radiation perfectly. We’re not going to worry too much about this assumption. For our purposes, planets are close enough to being blackbodies that it doesn’t matter much. I know, I know, we just made an adjustment for albedo two paragraphs ago, implying that the planet is not a perfect blackbody! Just calm down. It works pretty well, and that’s all we need. Anyway, if we assume the planet is a uniform temperature blackbody, then we can use the handy equation for blackbody emission: $Energy Emitted Per Square Meter Per Second = \sigma T^4$. Sigma is called the Stefan-Boltzmann constant, and is given by $\sigma = 5.67\times 10^{-8} W m^{-2} K^{-4}$ To get rid of that pesky “per square meter” part of the equation, we just multiply by the surface area of the thing doing the emitting: in this case, the planet. Here we do use the surface area of a sphere, remember our assumption that the energy absorbed gets spread out over the whole surface? This is why we did that. The result is: $Energy Emitted Per Second = 4 \pi r^2 \sigma T^4$ Now that’s a fine equation if your planet emits every bit of energy that it receives straight back to space. But that’s not how it works for planets with atmospheres. There’s this effect where the atmosphere traps energy in the system for a longer time, resulting in a warmer planet… you may have heard of it: the Greenhouse Effect! It would be nice if we could add that to our model! If we don’t, we’ll never get the surface temperature right for a planet like Venus, where the greenhouse effect dominates. To actually do a proper simulation of the greenhouse effect is very difficult and complicated, so instead we are going to use a fudge factor. The bottom line is that the greenhouse effect $GE$ reduces the amount of energy radiated from the surface that escapes to space. So we can do something very similar to our albedo adjustment: $GE$ gives the amount of energy that the atmosphere absorbs, and $1-GE$ gives the amount of energy that actually escapes to space. For the Earth $GE \approx 0.4$ and for Venus $GE \approx 0.99$. Our modified equation is now: $Energy Emitted Per Second = (1-GE) 4 \pi r^2 \sigma T^4$ Now, remember why we were doing all of this? We want to find the planet’s average surface temperature $T$. To get this, we have to set our two equations equal to each other and solve: $Energy Emitted Per Second = Energy Absorbed Per Second$ $(1-GE) 4 \pi r^2 \sigma T^4 = \dfrac{(1-A) L \pi r^2}{4\pi R^2}$ Look! The planet’s radius appears on both sides of the equation! That means it cancels out, and that a planet’s radius has no effect on its surface temperature! Ok, don’t get too excited, we still need to solve for T. $T^4 = \dfrac{L (1-A)}{16\sigma\pi R^{2}(1-GE)}$ $T = \left(\dfrac{L (1-A)}{16\sigma\pi R^{2}(1-GE)}\right)^{1/4}$ Voila! There is the expression for the equilibrium surface temperature of a planet, taking into account the planet’s reflectivity and the greenhouse effect. I hope this sheds a little light into how to think about the energy budget of a planet, and how my spreadsheet works. Again, if you have any questions, post them in the comments and I’ll answer them! Posted in: Astrobiology, Earth, exoplanets, Fun Stuff, Planets in General, Science Fiction ## Comments Solar System Creator ### 8 Comments 1. Vagueofgodalming said on 10 July 2009 Cool! No doubt you’ve already thought of this but if you search the Arxiv for ‘Cancri’ you get a number of papers on dynamical stability in the gaps in that system. Depending on the degree of realism you want. • Ryan said on 13 July 2009 Awesome, thanks for the tip! It turns out that my fictional terrestrial planet is pretty close to the inner edge of one of the stable zones, so I just nudged it 0.1 AU outward and am calling that plausible. (Afer all, I want it habitable even if it is a bit cold…) 2. Sharon E. Dreyer said on 10 July 2009 Wow! This is a great article. While my brother is the rocket scientist (aerospace engineer) of the family, I’m the writer. Don’t have a clue whether your math is real or, but I know how difficult it can be to work on a novel and go to school, or work for a living at the same time. Check out my first and recently released novel, Long Journey to Rneadal. This exciting tale is a romantic action adventure in space and is more about the characters than the technology. • Ryan said on 13 July 2009 Congrats on the novel, but be sure to check with sites like Writer Beware and SFWA to be sure you’re not being cheated by your publisher: http://www.sfwa.org/BEWARE/general.html#Literary 3. jshoer said on 13 July 2009 Oh, come on, no accounting for tidal heating? I just have to put in a plug for the as-yet-undiscovered Europan squid and Titanian anaerobic bacterium. 4. Ryan said on 13 July 2009 Joe: Nope, no tidal heating. I suppose I could probably add that to the energy balance, along with radioactive decay, but I suspect it’s not really worth the effort… 5. I recently came across your blog and have been reading along.I don’t know what to say except that I have enjoyed reading. Nice blog. I will keep visiting this blog very often. • Ryan said on 28 November 2009 Thanks for the kind words! If there’s anything you’d like to hear more about on the blog, let me know! Solar System Creator ### 2 Trackbacks • ##### Solar System Creator \| Liberty Hall Writers on 10 July 2009 • ##### 21st Century Waves » Welcome to the Carnival of Space #111 — The Apollo 11 Launch Anniversary Edition on 13 July 2009	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 23, "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": 0.9364000558853149, "perplexity_flag": "middle"}
http://www.physicsforums.com/showthread.php?p=3973256	Physics Forums Analytical solution to integro-differential equation Hi all, I'm trying to find an analytical solution to the following integro-differential equation: [itex] a f'(x)\int_0^x f(x)dx + b f'(x) + a [f(x)]^2 - a f(x) = 0 [/itex] with initial condition: [itex] f(0)=1 [/itex] This is a simplified problem for which I know the solution: $f(x)=1$. I'm trying to find a general method to solve this equation that I can use for more complex problems. The main difficulty is the product of the differential and the integral. Can anyone point me in the right direction? Integral transforms (e.g. Laplace) seem to be the general way to tackle integro-differential equations but I'm not sure how to apply those here. Many thanks in advance, Maarten 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 I'm not sure that this can be solved analytically. One thing you can try is dividing with f'(x) (assuming it's not zero) and then differentiating (the other case, f'(x)=0, makes your job very easy). That will get rid of the integral, but although the expression you get is not really easy to work with, it's reduced to a second order differential equation. Hi ! There are two obvious solutions f(x)=0 and f(x)=1. In order to show that there are no other solution, let : $F(x)=\int_0^x f(x)dx$ Then, bring back F(x), f(x)=F'(x) and f'(x)=F''(x) into the ODE This leads to a second order ODE, rather easy to solve, thanks to the conditions F'(0)=f(0)=1 and F(0)=0. Recognitions: Homework Help Analytical solution to integro-differential equation Quote by mbraakhekke Integral transforms (e.g. Laplace) seem to be the general way to tackle integro-differential equations but I'm not sure how to apply those here. Integral transforms are typically only useful for linear equations. This equation is nonlinear, so the typical integral transforms are not likely to work at all. (The laplace transform won't). Quote by JJacquelin Hi ! There are two obvious solutions f(x)=0 and f(x)=1. In order to show that there are no other solution, let : $F(x)=\int_0^x f(x)dx$ Then, bring back F(x), f(x)=F'(x) and f'(x)=F''(x) into the ODE This leads to a second order ODE, rather easy to solve, thanks to the conditions F'(0)=f(0)=1 and F(0)=0. This is what I would do. So, I suppose that you did it. Attached Thumbnails Thanks to all for your suggestions. @JJacquelin: your derivation is very helpful. The remaining equation without the boundary conditions: $(F'(x)-1)(F(x)+k)=C$ should be also solvable for more complex situations. Mathematica delivers a general solution in terms of the Lambert W (product log) function. Anyway, this is something I can work with. Thanks again, Maarten Hello ! I agree. Without the boundary conditions, the general solution f(x) involves the Lambert-W function : Attached Thumbnails Thread Tools \| \| \| \| \|---------------------------------------------------------------------------\|----------------------------\|---------\| \| Similar Threads for: Analytical solution to integro-differential equation \| \| \| \| Thread \| Forum \| Replies \| \| \| Calculus & Beyond Homework \| 7 \| \| \| Math & Science Software \| 1 \| \| \| Classical Physics \| 0 \| \| \| Differential Equations \| 1 \| \| \| Calculus & Beyond Homework \| 7 \|	{"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": 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": 0.9133636951446533, "perplexity_flag": "middle"}
http://physics.stackexchange.com/questions/38106/does-nantenna-nano-antenna-violates-2nd-law-of-thermodynamics/48437	# Does Nantenna (nano antenna) violates 2nd Law of Thermodynamics? Does Nantenna (nano antenna) violates 2nd Law of Thermodynamics ? Nantennas absorb infrared heat and convert it in direct current. Quote from Wikipedia: He did not discuss whether or not this would violate the second law of thermodynamics. Some people said that that Nantennas and Solar Cells are not thermodynamics. I do not understand that thesis. That is why I have re-phrased question: Are Nantennas a perpetual motion machine of Second Kind ? As I know: 2nd law XOR perpetual motion of second kind PS: I can't agree that answers are prooving that that second law is not violated by nantennas and MIM diods. If someone has to say anything about the subject then I would be happy to read that. Thank You all for responses! - Please, comment if you minus. I quote wikipedia, It is mentions that my question make sense. If someone says it does not violates then I have to argue that it does when it make sense. – Max Sep 30 '12 at 17:34 I am the one who put those words into the wikipedia article. What I wanted to say was: "Obviously this application would violate the second law of thermodynamics." But I ended up writing that more understated and diplomatic sentence which you quote, because of wikipedia's referencing rules. Indeed, if you have a room at temperature T, and a nantenna device at the same temperature T, it cannot charge a battery. This is a textbook example of the Second Law. – Steve B Oct 5 '12 at 21:11 Inventor of nantennas Dr. Steven Novack claims applications for nantennas that violate 2nd law in his interview on radio station. But absolutely most science community believe he is wrong and such applications are impossible as they violate 2nd law? Is that correct? – Max Oct 5 '12 at 21:44 I would bet a large sum of money that the overwhelming majority of professional physicists would agree with my assessment that the device cannot possibly work as advertised in the application in question, because if it worked then it would violate the Second Law. However, I have not actually done a survey! – Steve B Oct 5 '12 at 23:08 Thank You! I find it strange that someone invented nantennas, got nanotechnology award for the invention and claimed wrong or revolutional application for the invention and it is shyly ignored by science community for 4 years already. – Max Oct 5 '12 at 23:42 ## 3 Answers First of all, nantennas in general don't violate the second law of thermodynamics, so they are not perpetual motion machines of second kind. As long as the total entropy goes up, the second law is obeyed. In other variables, it really means that a part of the incoming heat has to heat the nantenna up but there may still be a lot of energy left for energy production, much like in any other heat engine. The Wikipedia suggestion that natennas could violate the second law only referred to a particular application hypothesized by Mr Novack. If he could be cooling the room while getting energy out of it, and if the gadget to cool the room were not connected to any cooler heat bath, then it would indeed be a perpetual motion machine of second kind and it would be impossible. The reason why Nature makes it impossible is kind of trivial. If the room has temperature $T$, then the nantenna or "power plant" may only be kept at the same temperature $T$ if there's equilibrium. But if that's the case, the nantenna emits thermal radiation, too. So even if it absorbs some incoming radiation, it still radiates its own. They're balanced and the energy gain is zero. Solar cells and "legitimate applications" of nantennas can only create energy because they work with incoming light whose "own" temperature is higher than the temperature of the solar cell or nantenna itself. For example, solar radiation has the temperature comparable to 5,500 Celsius degrees. The solar cells are effectively heat engines operating between this high temperature and a much lower temperature of the ground. The same is really true about life on Earth, too. The energy from the Sun may be converted and is often converted to useful energy or work because the high-energy photons from the Sun – which correspond to a high temperature and therefore a low entropy per unit energy ($E\sim TS$) – are processed on Earth and the energy is finally emitted in much lower-temperature "infrared" thermal photons – which carry a higher entropy. So the entropy can go up even if a part of the incoming energy is converted to useful work. The temperature inequality between the solar surface (and the solar radiation) on one hand and the cool temperature of the outer space is necessary for the Sun to play this often praised beneficial role. - But they have MIM diodes connected to Nantennas. Such diods operated at up to 150 THz. Diods are rectifying currents. That is why it will not radiate back. It covers most IR spectrum in the room. When news about nantennas arrived in 2008 I read that it has cooling effect. But I can't find it now. Same rectifying effect at lower frequencies are used in wireless chargers for mobile phones. – Max Sep 23 '12 at 17:50 4 – mmc Sep 23 '12 at 22:29 There are random alternating currents that diodes rectify. That is why nantenna becomes colder and energy goes away as direct rectified current. – Max Sep 24 '12 at 15:58 @Max there is no perfect rectifier. Read and understand the Brownian ratchet article. – user2963 Sep 26 '12 at 14:55 If rectifier efficiency would be 1% then it still will take energy out from system as direct current. MIM diodes operating at 150Thz have 40% efficiency. 150Thz is most IR spectrum. Every body radiate at IR spectrum. – Max Sep 26 '12 at 15:28 This is not the unique instance where heat is transformed in electricity. Thermocouple works on this principle and is perfectly described in classical thermodynamics. http://en.wikipedia.org/wiki/Thermocouple - Does thermocouple have a cooling effect? I think it requires difference of temperatures to produce electricity. – Max Sep 23 '12 at 18:57 Yes they do have a cooling effect which is even used in some aquarium tank cooling device. And you are right, they require a temperature difference to produce electricity. – Shaktyai Sep 24 '12 at 12:08 – Max Sep 24 '12 at 12:56 It is just a thought : on the quantum level is that still true. I explain, it is possible that on quantum level, energy is absorbed and emitted or converted into other energy. For example : the room is cooled by a quantum-pump. This pump generates electrical power which turns a ventilator. The net entropy is lower at the end. Which is impossible. But on the quantum level this is different. Hence the choice of my pump; a quantum pump. Imagine it uses quantum-fluctuations of the energy of the vacuum created at the planck-length to say something (which may or may not be possible). This fluctuations are the same in all directions. Imagine we can control its directions.This control-mechanism woulc cost the same energy as the net result of the quantum-pump. But now : imagine that the control-mechanisme uses information as its source, not energy. So the fact that the quantum-pump is organised in a way, would control the net-energy of the pump.So,the information (the way you organise this pump) put into the pump, would result in a lower entropy and make turn the ventilator. - What kind of information you need to make ventilator to rotate? Do you need to know location and direction of fluctuation? – Max Jan 5 at 22:40 ## protected by Qmechanic♦Jan 5 at 22:19 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": 3, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9396747350692749, "perplexity_flag": "middle"}
http://math.stackexchange.com/questions/90567/intersection-of-a-number-field-with-a-cyclotomic-field	# Intersection of a number field with a cyclotomic field Let $K$ be a number field and $N$ a positive integer. Prove that if the absolute discriminant of $K$ is coprime to $N$, then $K \cap \mathbb{Q}[\zeta_{N}]=\mathbb{Q}.$ This is something that the Childress book on CFT leaves kind of vague when proving Artin's Lemma. I would like to see a proof if possible. Thanks in advance! - 3 Prove the more general fact that if $K, L$ are number fields with coprime discriminants, then $K \cap L = \mathbb{Q}$. (To do this, recall the even more general fact that if $K \subset L$, then $D_K \| D_L$.) – Qiaochu Yuan Dec 11 '11 at 20:33 4 Another proof: Recall that there are no unramified extensions of $\mathbb{Q}$ and that the primes that ramify are precisely those dividing the discriminant. If the discriminants in your example are coprime, then no primes in the intersection can ramify, so the intersection has to be $\mathbb{Q}$. – Fredrik Meyer Dec 11 '11 at 21:56 @Fredrik +1 for making explicit why the intersection is $\mathbb{Q}$. In particular, it is not always true in the relative setting ($L$ and $K$ extensions of a number field $F$). – M Turgeon Dec 14 '11 at 1:20 ## 1 Answer If $L$ is an intermediate extension of $E$ over $F$, and $\Delta_{N/M}$ denotes the discriminant of $N$ over $M$, then $$\Delta_{E/F} = N^{L}_F(\Delta_{E/L})\Delta_{L/F}^{[E:L]},$$ where $N^L_F(\cdot)$ is the norm map from $L$ to $F$. In particular, the discriminant of $L/F$ divides the discriminant of $E/F$. Let $L = K\cap\mathbb{Q}(\zeta_N)$. Then the discriminant of $L$ over $\mathbb{Q}$ has to divide the discriminant of $K$, and also has to divide the discriminant of $\mathbb{Q}(\zeta_N)$. The discriminant of $\mathbb{Q}(\zeta_N)$ is $$(-1)^{\varphi(N)/2}\left(\frac{N^{\varphi(N)}}{\prod\limits_{p\|N}p^{\varphi(N)/(p-1)}}\right).$$ In particular, it can only be divisible by primes that divide $N$. - Thanks for the clear answer! – Anna Dec 11 '11 at 20:55	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 33, "mathjax_display_tex": 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": 0.8882017135620117, "perplexity_flag": "head"}
http://mathoverflow.net/questions/616/what-is-an-example-of-a-presheaf-p-where-p-is-not-a-sheaf-only-a-separated-pre/3514	## What is an example of a presheaf P where P^+ is not a sheaf, only a separated presheaf? ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) There is a standard way to construct the sheafification of a presheaf on a Grothendieck topology which involves matching families. Details may be found here: http://ncatlab.org/nlab/show/matching+family In short, there is a functor + sending presheaves to separated presheaves and then separated presheaves to sheaves. So P^++ is always a sheaf. Gelfand/Manin's Methods of Homological Algebra has a wrong proof that P^+ is a sheaf, and I have seen in several places a proof that P^++ is a sheaf. However, it seems that for any presheaf P I run into, P^+ is already a sheaf. Does anyone know an example of a presheaf P where P^+ is not a sheaf i.e. where you actually need to apply the functor + twice to get a sheaf? - Just to stimulate interest, the sheaf P^++ is indeed the sheafification of P. So this expresses the sheafification endofunctor as the square of the separation endofunctor; pretty cool! – Andrew Critch Oct 15 2009 at 17:44 1 I'm under the impression that if you want an example not of the type in Anton's response, you need a Grothendieck topology which doesn't come from a topological space. – Reid Barton Oct 15 2009 at 20:01 1 Do you think so? I was hoping that there is a way to build a presheaf on a space that stays nonseparated as you keep restricting to finer and finer open covers, and that this will cause us to miss some sections when we apply +. – AH Oct 15 2009 at 20:14 It's not so much that I believe it to be true as that I seem to recall reading it somewhere. But I don't remember where, sorry... – Reid Barton Oct 15 2009 at 23:05 ## 4 Answers I think this works: Consider a topological space consisting of 4 points $A$, $B$, $C$, $D$, where the topology is given by open sets $ABC$, $BCD$, $B$, $C$, $ABCD$, $\emptyset$. Then let the presheaf $\mathcal{F}$ be given by: $$\mathcal{F}(ABC)=\mathbb{Z}$$ $$\mathcal{F}(BCD)=\mathbb{Z}$$ $$\mathcal{F}(BC)=\mathbb{Z}$$ $$\mathcal{F}(ABCD)=\mathbb{Z}$$ $$\mathcal{F}(B)=\mathbb{Z}/2\mathbb{Z}$$ $$\mathcal{F}(C)=\mathbb{Z}/2\mathbb{Z}$$ $$\mathcal{F}(\emptyset)=0$$ where all restrictions are what you expect (identity in the case of $\mathbb{Z} \to \mathbb{Z}$ and canonical surjection in the case $\mathbb{Z} \to \mathbb{Z}/2 \mathbb{Z}$). Then if we we get $\mathcal{F}^+$ is given by: $$\mathcal{F}^+(ABC)=\mathbb{Z}$$ $$\mathcal{F}^+ (BCD)=\mathbb{Z}$$ $$\mathcal{F}^+ (BC)= \mathbb{Z}/2\mathbb{Z} \oplus \mathbb{Z}/2\mathbb{Z}$$ $$\mathcal{F}^+ (ABCD)=\mathbb{Z}$$ $$\mathcal{F}^+ (B)= \mathbb{Z}/2\mathbb{Z}$$ $$\mathcal{F}^+ (C)=\mathbb{Z}/2\mathbb{Z}$$ $$\mathcal{F}^+ (\emptyset)=0$$ where the map from $\mathcal{F}^+ (BCD)$ to $\mathcal{F}^+ (BC)$ is given by taking the canonical surjection on both copies, and other restrictions are obvious. Then note that if we take 1 over $BCD$ and 3 over $ABC$, these two are compatible over $BC$ but they do not patch. The key point is that being compatible over a refinement is not the same thing as being compatible. That is, the way the plus construction works is by taking $F^+$ of a space to be some direct limit over open covers of guys on the covers which are compatible on intersections. If we had said instead take direct limit over open covers of guys on the covers which compatible on some refinement of the intersection, then applying just once probably works. So in our example, 1 and 3, over $ABC$ and $BCD$, in our original presheaf were compatible on a refinement of $BC$ but not on $BC$. - ### 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 haven't checked carefully that I mean the same thing by + as you do, but I think the following example works. Take X={p,q} with the discrete topology, let S be any set with \|S\|>1, and let F be the constant presheaf which returns S for any open subset of X (and all restrictions are identity maps). In particular, F(∅)=S. Then it is easy to see that F++(X)=SxS, but I claim that F+(X)=S. To see this, suppose you have two sections s∈S=F({p}) and s'∈S=F({q}). These section "agree on intersections" only if their restrictions in F({p}∩{q})=F(∅)=S agree (i.e. only if s=s'). Note that F+(∅) is a one point set because ∅ is covered by the empty cover (a covering by no sets at all, not even ∅), and any two sections of F(∅) agree on this cover, so when you take F++, this problem doesn't happen again. - 1 This seems to work, but it wasn't really what I had in mind. I was hoping for something like a presheaf which stays nonseparated as we restrict to arbitrarily small open sets. Suppose we revise the question and assume that P(∅) = one point. Can we find an example then? – AH Oct 15 2009 at 19:39 Is this example why some texts (such as Hartshorne) require a presheaf to send ∅ to the final object? That always bugged me. But I guess it simplifies the construction of the sheafification (when your Grothendieck topology comes from a topological space). – Reid Barton Oct 15 2009 at 19:44 I think the key point behind the + functor is that it allows us to do sheafification without referring to stalks, which is something we may not have for a general Grothendieck topology. As I recall, Hartshorne's construction of the sheafification is different and makes use of stalks in an essential way. In any case, I don't think that concerns about P(∅) tell the full story. I'm not satisfied with the above example because I suspect that there is another reason why P^+ can fail to be a sheaf which has nothing to do with P(∅) not being a point. – AH Oct 15 2009 at 19:55 I don't know an example that doesn't use ∅ in this sneaky way. I think it's silly to assume that a presheaf sends ∅ to the final object; I prefer to just say a presheaf is a functor from the topology (thought of as a category) to some other category. For a sheaf, the sheaf axiom already implies that ∅ goes to the final object. – Anton Geraschenko♦ Oct 15 2009 at 19:57 I believe if you translate Hartshorne's construction of his + functor, by replacing elements of stalks with equivalence classes of representing pairs (U,s), you get everyone else's + functor. Taking H^0 and passing to the limit of refinements is the same as taking compatible functions to the union of stalks. The compatibility for these functions becomes interesting exactly when you have a cover by disjoint open sets. (Note also the bizarre statement of Hartshorne Chapter 2, exercise 1.1.) – S. Carnahan♦ Oct 16 2009 at 2:17 show 1 more comment Anton's example can be modified to avoid the "empty set". Here's one way: let X be the category associated to the partially ordered set a ≤ b ≤ c ≤ e b ≤ d ≤ e Give this the minimal topology in which c and e cover d and a covers b. Let F be the presheaf with F(a) = 1, F(b) = F(c) = F(d) = S, F(e) = ∅. Then F+(b) = 1, F+(c) = F+(d) = F+(e) = S. This is not a sheaf, since F++(e) = S x S. Of course, all I have done is to introduce an object a to play the role of the empty cover in Anton's example. It's probably worth remarking, however, that what makes the empty set empty (from the point of view of sheaves) is that it is covered by the empty cover. In fact, the usual topology on the category of open subsets of {p,q} can be modified so that the empty subset is not covered by the empty cover. With respect to this topology, the presheaf F that Anton defined is already a sheaf. - An example is given in MacLane's "Sheaves in Geometry and Logic." Consider the constant presheaf on a space $X$ with $P(U) = S$ where $S$ is a set with more than one element and restriction maps are identities. The plus construction doesn't change anything except that $P(0) = 0$, and one can show easily that this is not a sheaf. In general it is true that the plus construction turns separated presheaves into sheaves and any presheaf into a separated presheaf; hence ++ = sheafification. -	{"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": 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": 0.18, "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": 0.9475953578948975, "perplexity_flag": "head"}
http://math.stackexchange.com/questions/253266/what-is-the-value-of-int-gamma-barz-dz?answertab=active	# What is the value of $\int_{\gamma} \bar{z} dz$? I could use some help in calculating $$\int_{\gamma} \bar{z} \; dz,$$ where $\gamma$ may or may not be a closed curve. Of course, if $\gamma$ is known then this process can be done quite directly (eg. Evaluate $\int \bar z dz$), though that is not the case here. For instance, if $\gamma$ is indeed a closed curve then I can show the above integral is purely imaginary, but still don't know how to explicitly calculate it. Thanks! - ## 4 Answers Let $\gamma$ has parametric representation $$\gamma=\{(x, \ y)\in\mathbb{R^2}\colon\quad x=\varphi(t), \, y=\psi(t),\; t\in[a, \, b] \}.$$ Then for $z=x+iy \in \gamma$ \begin{gather} \int\limits_{\gamma}{\bar{z} \ dz}=\int\limits_{\gamma}{(x-iy) \ (dx+i\:dy)}=\int\limits_{\gamma}{(x\ dx +y\ dy)}+i\int\limits_{\gamma}{(x \ dy - y \ dx )}= \\ =\int\limits_{a}^{b}{\left(\varphi(t) \varphi'(t)+\psi(t) \psi'(t)\right)dt}+i \int\limits_{a}^{b}{\left(\varphi(t) \psi'(t)-\psi(t) \varphi'(t)\right)dt}= \\ =\dfrac{1}{2}\int\limits_{a}^{b}{d\left( \varphi^2(t)+\psi^2(t) \right)dt}+i \int\limits_{a}^{b}{\left(\varphi(t) \psi'(t)-\psi(t) \varphi'(t)\right)dt}= \\ =\varphi^2(b)-\varphi^2(a)+\psi^2(b)-\psi^2(a)+i \int\limits_{a}^{b}{\left(\varphi(t) \psi'(t)-\psi(t) \varphi'(t)\right)dt}. \end{gather} - You have $$\int_{\gamma} \bar{z} dz = \int_{\gamma} (x - iy) (dx + idy) = \int_{\gamma} xdx + ydy + i \int_{\gamma} x dy - ydx.$$ The integrand of the real part is an exact differential, and so it depends only on the end points of $\gamma$ and vanishes if $\gamma$ is closed. You can write an explicit formula for this part. The imaginary part is not exact, and so in general, the integral depends on the path. Like WimC mentions above, if $\gamma$ is a simple closed curve, you can use Green's theorem to relate this to the enclosed area. If $\gamma$ is not closed, I don't think you can expect any more information. - For a closed curve it is $2 i$ times the oriented area of the region enclosed by $\gamma$. Here "oriented area" means that the winding number around a point determines how that point contributes to the area. For example, for any circle of radius $r$ traversed in counter clockwise direction the integral equals $2i\, \pi r^2$. To see the general formula note that if $\gamma: [0,1] \to \mathbb{C}$, $\gamma(t) = x(t) + iy(t)$ then $$\operatorname{Im} \int_{\gamma} \overline{z} dz = \int_0^1 \det \begin{pmatrix} x(t) & x'(t) \\ y(t) & y'(t) \end{pmatrix} dt$$ and this is indeed twice the area of the region enclosed by $\gamma$. You already know that the real part is zero. - As the integrand is not holomorphic, the integral will depend on the whole path $\gamma$ and not only on the endpoints. In that sense your expression is already the most compact expression one can write without knowledge of the path $\gamma$. What kind of final (closed form) expression do you have in mind? -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 20, "mathjax_display_tex": 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": 0.8337447047233582, "perplexity_flag": "head"}
http://conservapedia.com/Special_relativity	Special theory of relativity From Conservapedia (Redirected from Special relativity) Special Relativity (SRT) is a generalization of classical mechanics. In contrast to General Relativity, Special Relativity deals with processes observed in so called inertial frames (frames of observation without the influence of acceleration or gravity). It is based on two main observations from different experiments. 1. The speed of light is constant for all (inertial) observers, regardless of their velocities relative to each other. 2. The laws of physics are identical in all inertial reference frames. It is not difficult to demonstrate that it is impossible to reconcile these conditions with a Newtonian mechanics, in which the coordinates are formed in a three-dimensional space. In SRT, each object moving in its own inertial frame has its own time, which constitutes a fourth coordinate describing its state. The three most prominent SRT effects are time dilation, length contraction and the equivalence of mass and energy. Mathematics The central equation of special relativity is[1]: γ=$\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}$, where: • γ is the factor relating relativistic time, mass, and momentum to non-relativistic time, mass, and momentum. • v is the speed of the object in question • c is the speed of light Classical mechanics When v, the speed of the object in question, is low relative to c, γ approaches $\frac{1}{\sqrt{1-0}} = 1$, causing the Einstein-Lorentz relativity equations to be equivalent to Newton's equations. This is why classical mechanics, governed by Isaac Newton's laws of motion, works for particles at low mass and low speed. However, at higher velocities, γ diverges, causing relativity to be essential. Electromagnetism, including for light and gamma radiation, where the quanta (photons) travel at light speed have no rest mass, is always relativistic. Universal speed limit At the other extreme, when v approaches c, γ approaches $\frac{1}{\sqrt{1-1}} = \frac{1}{0} = \infty$. Since infinite γ means infinite mass, no object can ever reach the speed of light. (Some theorists have postulated hypothetical tachyons, which would always travel faster than the speed of light[2]. No evidence has been found for them. If they existed, special relativity would mean that they cannot ever travel slower than the speed of light.) Experimental Proofs 1. Michelson-Moreley experiment 2. Blackbody radiation spectrum 3. GPS clocks (general relativity) 4. Lifetimes of fast traveling particles History In the beginning of the last century the two assumption mentioned before where found experimentally in the form of the Maxwell equations, which describe electromagnetic waves. The general idea at that point was that any wave is carried by some medium, called the luminiferous aether. However, as the earth is moving through space, there should be a difference in the interferometrically determined light wavelength, when measured at noon, at evening and at night at the same geographic location. However the Michelson-Morley experiment did reject the aether hypothesis. With this rejection it became an inevitable fact that equations descibing physics, which where fitting to a three dimensional space with a single timeframe for all observers exist. The mathematical framework and physical significance was developed by Henri Poincaré and Hendrik Lorentz. Albert Einstein gave an alternate derivation in terms of postulates. Many other scientists contributed modifications of this theory. In particular, the Irish Physicist Fitzgerald proposed that the failure of the Michelson Morley experiment was as a result of a length contraction effect. This idea was taken up by Hendrik Lorentz and shown by others to be a useful mechanism by which theory could be forced into conformance with experimental results. However, in 2005, Michael Strauss a computer engineer invalidated much of Special Relativity theory by showing clear contradictions in the theory. Special Relativity Spin as a relativistic effect The relativistic extension of Quantum Mechanics, described by the Dirac Equation allows, due to the symmetry of the equation in 4-space, an additional quantum number to exist, called spin. electron spin was known from chemistry before relativity arose. Interpretation and paradoxes Some consequences of the SRT are: 1. It is impossible ever to transmit information faster than the speed of light.[3] 2. The laws of physics are identical, without any variation, in every location throughout the universe. 3. The laws of physics are identical, without any variation, no matter how fast something is traveling (in an inertial reference frame). Special relativity alters Isaac Newton's laws of motion by assuming that the speed of light will be the same for all observers, despite their relative velocities and the source of the light. (Therefore, if A sends a beam of light to B, and both measure the speed, it will be the same for both, no matter what the relative velocity of A and B. In Newtonian/Galilean mechanics, If A sends a physical object at a particular velocity towards B, and nothing slows it, the velocity of the object relative to B depends on the velocities of the object and of B relative to A.) In the framework of special relativity, several thought experiments can be constructed, which lead to apparent paradoxes. The most striking one is the twin paradox. If you take twins, one on earth, and one in traveling to the next star with high speed and back, their biological age will not be the same, even though you could redefine the system of the traveling twin to be resting. This paradox is resolved because the second twin is not in an inertial frame - he has accelerated, most significantly at his turn-around point. This points out, that while neglected in special relativity, acceleration has a non-trivial role. This is considered in the General theory of relativity. References 1. ↑ http://www2.slac.stanford.edu/vvc/theory/relativity.html 2. ↑ http://www.math.ucr.edu/home/baez/physics/ParticleAndNuclear/tachyons.html 3. ↑ This assumption is commonly restated in this manner. For example, a discussion of hypothetical tachyons talks "about using tachyons to transmit information faster than the speed of light, in violation of Special Relativity."[1] However, there is some question whether the Theory of Special Relativity really restricts faster-than-light communication of information. Related Links Mass Energy Equivalence Example Time Dilation Example	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 3, "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": 0.9256912469863892, "perplexity_flag": "middle"}
http://quant.stackexchange.com/questions/4672/why-are-regressors-squared-and-not-1-5-or-2-2-or-2-5/4694	# Why are regressors squared and not ^1.5 or ^2.2 or ^2.5? When a researcher in economics or finance wants to apply a linear regression model but suspects a non-linear relationship between one of the regressors and the dependent variable, it is typical to also include the square of that regressor, then (maybe, but usually not) do something like a Ramsey RESET test afterwards. My question is; why square it? Why not take the exponent (call it $x$) to be $x \in [1.5,2.5]$, for example? Getting the right "shape" of the line is impotant to make the assumption $E[\epsilon_i \| \mathbf{X}]=0$ hold; surely $x$ exactly equals $2$ does not achieve this optimally? Here, of course, I'm talking about variables that assume strictly positive values so we don't get complex outcomes. An example is $Age$ in education$\to$income studies. - 1 The main reason is parsimony. – John Dec 4 '12 at 15:03 @John I don't understand. Do you mean that it's like this because $2$ looks prettier than $x\in[1.5,2.5]$? Surely parsimonious should be defined statistically instead of based on how pretty something looks from a qualitative perspective. – Jase Dec 4 '12 at 22:50 Instead of performing a non-linear least squares routine, the researcher has effectively imposed constraints on the coefficient. They want to handle non-lineraities without too many extra variables. So they just square it. Parsimony. – John Dec 4 '12 at 23:12 @John What are these extra variables? – Jase Dec 4 '12 at 23:25 2 Because a negative number raised to a non-integer exponent is complex. Of what use would regressing physical economic values have here? – pat Dec 5 '12 at 5:11 show 1 more comment ## 1 Answer I basically agree with @John, let me expand: We want to model $y$ using a simple linear model, the most basic setup is $$y = c + \mathbf{X}\beta$$ with $y$ the $N$ observations, $c$ a constant, $\mathbf{X}$ the $N \times M$ matrix of regressors and $\beta$ a $M$-dimensional vector of coefficients. This model has $M$ parameters, the elements of $\beta$. The above model is estimated and the Ramsey RESET test finds that the model to be misspecified and the researcher wants to fix this. As you propose the above model is easily extended $$y = c + \mathbf{X}\beta + \mathbf{X}'\gamma$$ where $\mathbf{X}'_{i, j} = \mathbf{X}_{i, j}^{e_i}$, $\mathbf{e}$ is a $M$-dimensional vector and $\gamma$ a $M$-dimensional vector of coefficients. This model has $3M$ parameters, the elements of $\beta$, $\gamma$ and $e$ and much harder to estimate because of the nonlinearity. This can be easily solved by fixing all $e_i$ a priori. This yields another question: to which value do we fix it? As @pat notes, raising to a non-integer is a bad idea in the general case. But, as you note, one could use the absolute of the regressor raised to a rational exponent since $f(q) = \|a^q\|$ is continuous and real for all real $q \in \mathbb{Q}$. So why the insistence on integer valued exponents? One simple reason is laziness: it is much simpler to compute $x^2$ than $x^{1.95}$, a second reason is convention. A third reason is that small changes in the exponent have a small impact on the model. These arguments do not apply to the case where a rational exponent would yield a significant improvement. Unfortunately this has severe methodological problems: as argued above, making the exponent parameters makes estimation much harder and, perhaps more importantly, reduces parsimony. The last option of fixing the exponent is possible. However it would require a strong economic argument to defend this particular choice. If your application is such that it is absolutely clear that exponentiation with $q \in \mathbb{Q}$ is justified then you're free to do that. There are no methodological problems that I know of. But prepared for your critics who will notice and wil require justification of your particular choice for $q$. Another reason to choose $e_i = 2$ is the symmetry with taking cross products of the regressors, from this perspective is a square is a cross product with itself. - Very nice response. Could you address my reply to @pat in the comments section? Then I will accept the answer. – Jase Dec 6 '12 at 5:22	{"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": 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": 0.9295276999473572, "perplexity_flag": "middle"}
http://mathhelpforum.com/algebra/197304-can-somebody-explain-why-you-take-derivative-curve-find-its-length.html	1Thanks • 1 Post By Prove It # Thread: 1. ## Can somebody explain why you take the derivative of a curve to find its length? Given a curve: $G(t)=\binom{t}{y(t)}$ It goes between the two points: $A=(t_1,y_1) B=(t_2,y_2)$ To find its length: $Length=\int_{t_1}^{t_2}\sqrt{\left(\frac{d}{dt}G(t )\right)^2}\, dt$ What I understand is that you use the integral going from the values $t_1$ and $t_2$ to find the entire length between these two values. The squareroot and the exponential comes from the Pythagorean theorem to find the hypotenuse. What I don't understand is why we take the derivative of the curve $G(t)$. 2. ## Re: Can somebody explain why you take the derivative of a curve to find its length? Arc length - Wikipedia, the free encyclopedia Go to "Finding arc lengths by integrating".	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 6, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.929429292678833, "perplexity_flag": "head"}
http://math.stackexchange.com/questions/43571/example-of-an-infinite-balanced-nonabelian-group?answertab=oldest	# Example of an infinite balanced nonabelian group A group is called a balanced group provided that, for all $a,b \in G$, either $ab=ba$ or $a^2=b^2$. Example of (possibly infinite) balanced groups are abelian groups. An example of a finite balanced nonabelian group is the Quaternion group. Does it exist an infinite balanced nonabelian group? - 3 Yes, $Q_8 \times \mathbb{Z}_2^\omega$ is balanced, where $Q_8$ is the quaternion group of order $8$ and $\mathbb{Z}_2^\omega$ is the direct sum of infinitely many copies of $\mathbb{Z}_2$. – Jim Belk Jun 6 '11 at 8:16 @Jim: Oh yes you're right! Why didn't I think about such a simple example? Thanks. – Thomas Connor Jun 6 '11 at 8:28 @Jim: I think you should post this as an answer so that the question can be closed by having an accepted answer. – t.b. Jun 6 '11 at 8:47 @Theo: OK, I'd be happy to. – Jim Belk Jun 6 '11 at 8:58 @Jim: Thanks! ${}$ – t.b. Jun 6 '11 at 9:01 ## 2 Answers Yes, $Q_8 \times \mathbb{Z}_2^\omega$ is balanced, where $Q_8$ is the quaternion group of order $8$ and $\mathbb{Z}_2^\omega$ is the direct sum of infinitely many copies of $\mathbb{Z}_2$. - As Jim Belk has quite rightly pointed out, there do exist infinite non-abelian examples. However, there do not exist any infinite finitely generated examples of non-abelian balanced groups. Proof: Suppose $G$ is non-abelian, balanced, infinite, and finitely generated. Group elements are either central or non-central (obviously), and the non-central ones square to some fixed element, $u$. As groups do not contain non-trivial idempotents (elements such that $e^2=e$), one has that $u \in Z(G)$. Noting that for $x$ some non-central generator, $xu\not\in Z(G)$ so $u=(xu)^2=u^3$ and so $u^2=1$. Further, $uw^2=u$ for all $w\in Z(G)$, by the same logic. Thus, $W^2=u$ for all $W\in G\setminus Z(G)$ while $W^2=1$ for all $W \in Z(G)$. Therefore, every element has order either 2 or 4. One can either plow on ahead (I don't think it is too hard, but it is a tad tedious) to prove the problem outright, or one can apply an early result about Burnsides Problem: a finitely generated group in which each element has order a divisor of 4 is finite. Ta-da! -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 28, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9378979206085205, "perplexity_flag": "head"}
http://nrich.maths.org/2532/solution?nomenu=1	Your answer will be a multiple of $7$. Let the number thought of be $x$. Then the final number is $$4(2x + 3) - 5- x = 7x + 7 = 7(x+ 1)$$	{"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": 2, "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": 0.8428065776824951, "perplexity_flag": "head"}
http://mathhelpforum.com/advanced-applied-math/144957-forming-de-f-ma-print.html	# Forming a DE from F=ma Printable View • May 16th 2010, 02:15 AM craig Forming a DE from F=ma A particle of mass $m$ is projected with speed $u$ along a straight horizontal track. The first section of the track has length $d$. On this section of the track the motion is resisted by a constant force of magnitude $mk$, where $k$ is a positive constant. The particle does not come to rest on this first section of the track. Show that the speed $V$ of the particle at the end of the first section of the track is given by: $V = \sqrt{u^2 - 2kd}$ Using the equation $ma = F$, I got the following equation. If we let the speed of the particle = $v$ $m\frac{dv}{dt} = v-mk$, however in the solutions they just have $m\frac{dv}{dt} = -mk$, what's happened to the $v$, I would have thought that the forces acting on the paticle would be $v$ and $-mk$? Thanks for the help • May 16th 2010, 05:36 PM roninpro If you think about your units for a second, $v$ cannot possibly be a force. • May 17th 2010, 02:04 AM craig Ohh yeh, actually that's rather obvious isn't it :S Think I must be used to dealing with a constant force produced by the engine. Thanks for spotting that obvious mistake! All times are GMT -8. The time now is 03:44 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": 15, "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": 0.9599109292030334, "perplexity_flag": "head"}
http://scicomp.stackexchange.com/questions/tagged/lapack+linear-algebra	# Tagged Questions 1answer 54 views ### $\mathbf{UDU}^\top$ decomposition routines in LAPACK/Eigen? I would like to compute the decomposition of a real symmetric positive definite matrix $\mathbf{A} = \mathbf{UDU}^\top$. LINPACK seems to have it as DSIFA, but I ... 1answer 151 views ### Magma vs. Plasma I'm having a difficult time understanding the difference between the linear algebra packages MAGMA and PLASMA from just a quick glance. It looks like MAGMA is oriented towards GPU's and vector ... 2answers 259 views ### What are the fastest available implementations of BLAS/LAPACK or other linear algebra routines on GPU systems? nVidia, for example, has CUBLAS, which promises 7-14x speedup. Naively, this is nowhere near the theoretical throughput of any of nVidia's GPU cards. What are the challenges in speeding up linear ... 1answer 306 views ### BLAS/LAPACK subroutine to add two matrices with different offsets and leading dimensions I currently searching for a subroutine from BLAS or LAPACK which realizes the following operation A = alphaA + beta B where A and B have different leading ... 1answer 61 views ### Efficient computation of the extension of a linear basis to completion when the basis is almost complete (ideally using LAPACK routines) I have a $p \times n$ matrix $B$ (where $n < p$) with orthonormal columns and would like to find a numerically efficient way to extend this matrix to get a complete $p$-dimensional orthonormal ... 3answers 1k views ### Understanding how Numpy does SVD I have been using different methods to calculate both the rank of a matrix and the solution of a matrix system of equations. I came across the function linalg.svd. Comparing this to my own effort of ... 3answers 465 views ### Matrix exponential of a real asymmetric matrix with Fortran 95 and LAPACK I recently asked a question along the same lines for skew-Hermitian matrices. Inspired by the success of that question, and after banging my head against a wall for a couple of hours, I'm looking at ...	{"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": 0.9231175184249878, "perplexity_flag": "middle"}
http://mathoverflow.net/questions/117486/direct-product-of-rings	## Direct product of rings ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) Is there an infinite family $\lbrace R_\alpha\rbrace_\alpha$ of rings (with identity $1\neq 0$) such that their direct product is a (semi) hereditary ring ? I think the answer must be negative but i have no proof or counterexample yet. - I have no time right now, but if I did have time the first place I'd look is T.Y. Lam's Lectures on Modules and Rings. I think there are a number of pages available through Google Books. The answer might be there – David White Dec 29 at 19:37 @David White: Exactly, its always true that there are a number of pages available through Google Books ! – chatish Dec 29 at 21:33 ## 1 Answer Products of fields are semihereditary. This follows from the facts that products of fields are von Neumann regular and that von Neumann regular rings are semihereditary. A proof can be found (as suggested by David White) in Lam's Lectures on modules and rings, Example 2.32 d). (In the above, fields and rings are not necessarily commutative.) - @Fred Rohrer: But an infinite direct product of non-zero rings can not be countable so there is a gap in your argument for the hereditary case. – chatish Dec 29 at 21:52 1 But that von Neumann regular rings are semi-hereditary seems not to be in doubt, so that answers one question. – Todd Trimble Dec 29 at 21:54 Oh dear, let me edit my silly mistake. – Fred Rohrer Dec 29 at 21:57	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9238049983978271, "perplexity_flag": "middle"}
http://crypto.stackexchange.com/questions/1727/helper-data-authentication-in-pufs?answertab=votes	Helper data authentication in PUFs As I understand, PUFs work by using two procedures: generation and reproduction. Generation reads a value $w$ from a fuzzy source and generates a key $R$ and helper data $P$. Then, in the reproduction procedure, it reads $w'$ from the fuzzy source and, using $P$, is able to recover $R$. My question is: this seems to assume that $P$ can be stored safely and can be authenticated, since if a attacker changes the value $P$ that is stored then he can force the system to use a different key. How can this be done, if you don't have a key in the first place? Or am I missing something? (I have followed the notation from this paper) - 2 Answers It is very similar to how we authenticate ourselves to a website. During registration, the website must store enough information to, at some time in the future, convince itself that the person trying to authenticate now is the same person who registered at some time in the past. For online services, this typically involves storing some function of the password. The online service also wants to make sure that the stored information is stored safely (i.e., unauthorized entities cannot get it) and authenticated (i.e., no unauthorized parties can change it). Otherwise authentication would fail. With a PUF, it is almost the same. You register the PUF with the system by extracting $R$ and $P$ from $w$. $R$ and $P$ would then be stored in, for example, a database. When the PUF is used again to authenticate, the service would read $w'$, pull $P$ from the database and run the reproduction procedure to get $R'$. It could then pull $R$ from the database and see if $R==R'$. If so, authentication is successful and access is granted. How you protect $R$ and $P$ from the generation phase is up to you. It also depends on how paranoid you are, how valuable the service is, your threat model, etc. For some, simply storing them in a database that (hopefully) only they have read access to is fine. Others might want to encrypt and sign the values. It should be noted that $P$ is a public value. It does not need to be kept secret for the system to be secure. As you noted, however, if someone tampers with $P$, they could trick the system into authenticating the wrong party. Equivalently, if I change the hash stored in a database with login credentials, I can now become that user. Active Attacker Update Things become tricky when you allow your adversary to modify $P$. AFAIK, there are no guarantees that the attacker cannot modify $P$ in a devistating way (e.g., to publically reveal $R$). And I am not aware of any PUF research to mitigate the problem. Fortunately PUFs and biometrics are very similar. 1. Both are a noisy source 2. Both require fuzzy extraction to be used to handle the noise Given that, hopefully the following will at least help to solve the problem. In "Robust and Reusable Fuzzy Extractors" by Boyen in the book "Security with Noisy Data", Boyen tackles a similar problem. From the chapter: Unfortunately, ordinary fuzzy extractors do not address the issue of an active adversary that can modify $P$ maliciously, either on the storage server or while in transit to the user. I don't fully understand the details of the work yet, so I don't feel it wise to try and describe the algorithms and protocols. NOTE: It looks like much of that chapter was taken from a paper by Boyen and some of his colleagues titled "Secure Remote Authentication using Biometric Data". I have not read the paper to confirm that all the detail you might be looking for is there, however. - I see, it makes sense if the PUF is used for authentication. But what if the PUF is used to generate a key for a device? – Conrado PLG Jan 24 '12 at 15:27 @ConradoPLG, for key generation, you use R from the generation stage (or some function of it) to encrypt the data. Then you can store P and throw away R. Upon regeneration, if someone has modified P, they won't be able to decrypt. You shouldn't even need to store R for key generation. And, P gives an attacker negligible information, so it can remain public. Protect it against modification if necessary using standard methods. – mikeazo♦ Jan 24 '12 at 17:15 Thanks for the clarification, but I still don't understand something: how can I protect it against modification? For that I would need a key, but I'm using the PUF to get a key in the first place. Is there any kind of guarantee that an attacker can't change $P$ in such a way to force the system to derive a key $R'$ of his choice? – Conrado PLG Jan 24 '12 at 17:49 @ConradoPLG, I know of no such guarantee. Why can't you use a different key to make sure $P$ isn't changed? Generate a GPG public/private key pair. Sign the value $P$ with the private key, then anyone with the public key can verify that $P$ has not been modified. Or, use an HMAC with a password of some sort. The key to protect $P$ is completely different from $R$. – mikeazo♦ Jan 24 '12 at 19:23 1 I guess my question wasn't clear enough, sorry for that! Suppose that the PUF is used by some device to derive a key that is used to encrypt and authenticate its hard drive, for example. There is no user interaction. If the device generates and stores a key to sign $P$, then an attacker can change both the key and $P$ (that is, assume that there is no secure storage) and the security would be compromised. If you assume there is a safe place to store the second key and $P$, then why use a PUF in the first place if you can generate a key and store it there? – Conrado PLG Jan 26 '12 at 10:42 show 1 more comment You can also think about PUFs in terms of challenges/responses. If I want to authenticate you with a PUF, I need to be in possession of it first. I make a list of challenges and determine, for each, the reading from the PUF for that challenge. The reading won't necessarily be a compact, high entropy string. So to distil the reading down to a high entropy value, I use an extractor. Extractors are fine if the readings are consistent. However with most PUFs, the readings will be noisy and moderately inconsistent. The $P$ value is sometimes called a secure sketch ($P$). All it does is help do error correction. The sketch is not specific to using PUFs; it is part of a "fuzzy extraction" process that can be done after getting a noisy reading from any source. For each challenge, I record the value extracted from the reading as the correct response. I then send you the PUF. When you want to authenticate, I send a challenge that I haven't used before, you get a reading off the PUF, use the profile to extract the response, and send the response. If it matches, I scratch the challenge & response pair off the list and authenticate you for that session. The sketches can be made public. They leak no useful information to an adversary trying to guess the correct response. The adversary will have just as much luck guessing the response with or without the sketch. There is no threat of the adversary changes the sketch server-side: I've already used the sketch to generate the responses and I'm done with it. An adversary might change the sketch client-side. This would cause a denial-of-service: a legitimate user will no longer be able to authenticate. That is an accepted limitation of the threat model. -	{"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": 38, "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": 0.9418573975563049, "perplexity_flag": "middle"}
http://crypto.stackexchange.com/questions/5487/is-there-an-advantage-to-storing-keys-split-between-several-hashes	# Is there an advantage to storing keys split between several hashes? I have a question about the way to store a key or password that was used for encryption, so that the application can check if the user put in the right key for decryption. If I make a mistake, please advise me and I will try to avoid such a mistake in the future. Normally, the key that was used for encryption is stored hashed on the system, so that the application can check if the user used the correct key. Would it be better to store not the whole key hashed, but split the key into several parts and hash each of these parts? This should make cracking more difficult, because the attacker has to crack several hashes, with different salts (if used). Am I totally wrong with my line of thought? If yes a little hint would be nice, so that I can investigate in that direction and avoid this mistake in future. - 3 "but split the key into several parts and hash each of these parts" that'd be much easier to crack. Read a bit about LANMAN hashes. – CodesInChaos Nov 26 '12 at 20:55 ## 2 Answers Lets take your idea to the extreme to see its weakness. For simplicity I'll scale things down. Let's assume an 8 bit key chosen randomly. Call this key $k$. If I break into your database and get $d=H(k,s)$ along with $s$ (where $H$ is a hash function and $s$ is the salt), it would take on average $2^{8-1}=2^7=128$ computations (or calls to $H$) to find $k$. Now, using your idea, lets say the server instead splits $k$ into $(b_1,b_2,\cdots,b_8)$ (i.e., bits) and then computes $d_i=H(k_i,s_i)$ (so there is a different salt for each bit of the key). Now, the attacker who breaks into your database and steals the pairs $(d_i, s_i)$ needs to crack all $8$ hashes. More difficult? No. The reason is, how much work does it take to break $d_1$? Well, the attacker only need compute $H(1,s_1)$ and $H(0,s_1)$ to find $b_1$. That is only two operations (one on average though since half of the time the attacker will get the right value on the first try). Do this for all $8$ bits and the attacker has to make a maximum of $16$ hash function calls to break the key $k$. The average number of hash function calls would be $12$. This is much, much less than in the normal case above. Asymptotically speaking, the first system (which hashes the entire password) is something like $O(2^n)$ where $n$ is the number of bits in the key and the second system (your proposal) is $O(n)$. - Thanks for that clear answer. – Simon Rühle Nov 26 '12 at 21:21 Instead of storing a value based on the key, look at the problem you're trying to solve. You're not actually interested in the value of the key, you just need to ensure decryption works. If you feel you must test the user's input immediately, might I suggest instead storing a value that was encrypted with the key, and testing decryption directly? The advantage is that security remains equal to the security of the encryption algorithm itself, as opposed to the less-understood (and therefore much more risky) security of a home-brewed hash mechanism. The disadvantage is that your software would be holding both the challenge and response data, which might help enable a known plaintext attack. However, rather than storing any test value whatsoever, I advise you take a step back and determine the overall risk to your system of attempting to decrypt with the "wrong" key. Your application should know as soon as any encrypted data arrives whether or not the key was correct, as the decrypted data will either be legible or it will be useless. What is the impact to your system if the entered key was wrong? Does it mean the user has to wait for encrypted data to arrive before he knows the inputted key was correct? Is that truly an intolerable situation for your system? - The problem with the wrong key is, that I copy the encrpyted file and decrypt it in the original file, so that if the key was wrong, the data was lost, but your idea with testing if encryptions works with a short value. Is there any good suggestion what can be taken for such a thing (I mean what value, a dynamic and how to create it)? – Simon Rühle Dec 1 '12 at 18:08 It can't be dynamic, because you need to refer to it both at the time of encryption as well as decryption. Like I mentioned, a static value of "ABC123" provides a crib to an attacker: try a brute force approach and if any decrypts to the known value, you lose. I don't know of any asymmetric approach you could take here that wouldn't be easily bypassed. I think you'll just have to take the risk of the known plaintext attack. It's much safer than storing a hash of the key. – John Deters Dec 14 '12 at 14:35	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9626429080963135, "perplexity_flag": "middle"}
http://mathoverflow.net/questions/35288/undergraduate-roadmap-to-algebraic-geometry	## Undergraduate roadmap to algebraic geometry? ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) Hello, I'm sorry if this question isn't posted correctly. I hope that it is (since other questions regarding roadmaps have been allowed). Now to my question: From what I've heard from professors and such, algebraic geometry seems like an interesting branch of mathematics. I'd like to learn some basic results and maybe do some kind of thesis in a few years on the subject. So, what I'm curious about is you have any tips on what books to read? Say that one has read Artin's Algebra and Herstein's Topics in Algebra, and also has the basic courses in real analysis and topology, complex variables etc. down, where should one go to learn? What books? I'm also curious if algebraic geometry (at an "easy level") requires deep knowledge about other fields of mathematics too, so that one might have to read books that at first seems to have no relevance to algebraic geometry? Best regards. - This question has already been asked: mathoverflow.net/questions/1291/… – danseetea Aug 11 2010 at 23:11 See also: mathoverflow.net/questions/2446/… – danseetea Aug 11 2010 at 23:14 2 danseetea: Sorry, but I thought that this differed somewhat since this is aimed at an undergraduate level. – Dedalus Aug 11 2010 at 23:17 1 @danseetea: Both of those were recommendations for more advance students. – David Corwin Aug 11 2010 at 23:18 1 I've made this question community wiki. If you want to encourage more input, you might want to unmark Amitesh Datta's answer. It is unusual for roadmap questions to have answers accepted. – S. Carnahan♦ Aug 11 2010 at 23:46 show 3 more comments ## 8 Answers I believe there have been similar questions, but not one exactly of this flavor. To answer your last question, it is true that you need to know many different areas of mathematics in order to delve deeply into algebraic geometry. On the other hand, to get a basic grounding in the field, one need only have a basic understanding of abstract algebra. That being said, I will give my recommendations. If you have already done complex variables, and I'm not sure that every student in your position will have completed this, I recommend Algebraic Curves and Riemann Surfaces by Rick Miranda. Although this book also develops a complex analytic point of view, it also develops the basics of the theory of algebraic curves, as well as eventually reaching the theory of sheaf cohomology. Multiple graduate students have informed me that this book helped them greatly when reading Hartshorne later on. If you want a very elementary book, you should go with Miles Reid's Undergraduate Algebraic Geometry. This book, as its title indicates, has very few prerequisites and develops the necessary commutative algebra as it goes along. More advanced students may complain that this book does not get very far, but I think it may very well satisfy what you are looking for. Another book you might want to check out is the book Algebraic Curves by William Fulton, which you can thankfully find online for free. If you would not mind a computational approach, and furthermore a book which requires even fewer algebraic prerequisites than you seem to have, you might want to check out Ideals, Varieties, and Algorithms by Cox and O'Shea. Thierry Zell's suggestion is also supposed to be good. That being said, if you decide that you like algebraic geometry and decide to go more deeply into the subject, I highly recommend that you learn some commutative algebra (such as through Commutative Algebra by Atiyah and Macdonald). But for the moment, I think the above recommendations will suit you well. - +1 on Rick Miranda's book. It is a beautiful introduction to so many extremely important ideas throughout complex analysis and algebraic geometry. It also really helps bridge the gap between the probably-more-familiar analytic points of view and the much more abstract algebro-geometric points of view. – Jack Huizenga Aug 12 2010 at 0:19 +1 on Ried and +1 on Fulton's book. Both give students a very good idea of what algebraic geometry is about before tacking the hardcore accounts. – Andrew L Aug 12 2010 at 2:46 I don't think undergrad algebraic geometry can have commutative algebra as a prerequisite, even though it's essential if you want to get to a level where you can do research in the field. And learning a little bit of algebraic geometry should help immensely if you later take commutative algebra, because without algebraic geometry, many of the definitions and theorems are completely unmotivated. – Peter Shor Aug 12 2010 at 17:48 I agree, and none of my book recommendations have commutative algebra as a prerequisite. – David Corwin Aug 13 2010 at 14:44 ### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you. Try Shafarevich's Basic Algebraic Geometry I. It includes a good amount of examples from geometry in the exposition, as well as the necessary commutative algebra. I also like that differentials, divisors, and other gadgets are introduced without schemes (of course, you should eventually learn about schemes). The book moves pretty quickly, but if you have gone through Artin and Herstein then you should find it approachable. - I second that suggestion. It has a lot of good topics and very good exposition. – Thierry Zell Aug 11 2010 at 23:59 2 I like this book and it's second volume immensely as well.Many experts in AG-like Micheal Artin-strongly advise a stepwise learning of AG from classical machinery like algebraic curves to the "intermediate" formulation by varieties to the modern formulation by schemes.The modern machinery is so abstract,it's very hard to motivate it without this kind of "psuedohistorical" approach.I tend to agree,although I'm far from an expert. – Andrew L Aug 12 2010 at 2:44 I agree that it's very nice, though as far as I can tell, it requires some commutative algebra background (see, for example, the appendix to it). – David Corwin Aug 13 2010 at 14:45 If you can get hold of the first edition, or get the original Russian Math Surveys article it was based on, most of the needed commutative algebra is included in the text. – roy smith May 12 2011 at 21:16 I had the opportunity to teach an undergraduate course in algebraic geometry, and let me tell you that finding a book at that level was not easy. First, be aware that most books will not be self-contained and will farm out the necessary commutative algebra to outside references. For a first taste, I would go with An Invitation to Algebraic Geometry by Karen Smith et al. (Springer) It's not really designed for undergrads, and I'm not sure how it would work as a textbook for a course; rather, it was written for people with a certain mathematical maturity who don't know anything about the subject and want to learn the basics. I recommend it because it's short, recent, good for self-study, contains a lot of the classical stuff, and it can give you an idea of the flavor of the subject. There are tons of algebraic geometry books out there, so I'm sure other good recommendations are forthcoming. - Sara Lapan, a graduate student here, has notes on her website from Karen's introductory course: www-personal.umich.edu/~swlapan/math/… The room is always packed for these lectures. – Dylan Moreland Aug 12 2010 at 7:57 Thanks for the link. I noticed there are also notes for a second course on schemes, that might be of interest too. www-personal.umich.edu/~swlapan/math/… – Thierry Zell Aug 12 2010 at 11:37 @Dylan,Thierry: Wow,great find-thanks a lot,both of you! – Andrew L Aug 12 2010 at 22:52 Broadly speaking, algebraic geometry is the geometric study of solutions to polynomial equations. To begin with, you would start by working with solutions in affine space $\mathbb{A}_k^n= k^n$, where $k$ is an algebraically closed field (e.g. $\mathbb{C})$. Eventually, it becomes advantageous to add points at infinity by working in projective space $\mathbb{P}^n_k=\mathbb{A}^n_k\cup (\text{hyperplane at }\infty)$. After a while, you may move beyond even this. All of this is spelled out in the basic books on algebraic geometry. Although there are quite a number of these, very few are explicitly at the undergraduate level. If I had to come up with a list, it would probably be the similar to Davidac897's. Out of the lot, I'd recommend Reid's "Undergraduate Algebraic Geometry" as a good starting point: it is short, to the point, with minimal prerequisites. Once you've gotten through it, you'll have enough of a foundation to tackle something more ambitious if you're still so inclined. One other thought. I have a sense that you are planning to read this on your own. Things are easier if you find a group of fellow students to "share the pain". Have fun. - Interesting that you didn't recommend you own extensive notes on the subject at your website,Donu. I find them very interesting,stimulating and not too advanced.They're certainly a lot easier to break into then Harris or Hartshorne-but probably too difficult for any but exceptional undergraduates.I agree for them,Reid is a better choice. – Andrew L Aug 12 2010 at 22:48 Thank you, but they are too unpolished at this point to seriously recommend as a primary source. – Donu Arapura Aug 12 2010 at 23:00 There is an excellent book on algebraic geometry entitled Algebraic Geometry: A First Course by Joe Harris. This book, however, emphasizes the classical roots of the subject but if you have not yet seen too much of algebraic geometry, it is worthwhile getting this book and reading a few lectures. (The book is split into "lectures" rather than "chapters".) There are many beautiful constructions in classical algebraic geometry that can be understood without too much background (and which lay the foundations for some aspects of modern algebraic geometry) and this can perhaps give you a rough indication of the geometric intuitions in algebraic geometry. And in my opinion, the book does an excellent job of conveying the beauty and elegance of algebraic geometry. The prerequisites for reading this book (according to Harris) are: linear algebra, multilinear algebra and modern algebra. However, since this is a "Graduate Texts in Mathematics" book, there are some places where it is very helpful (but not essential to the point that you cannot read the book otherwise) to have a basic knowledge of commutative algebra, complex analysis and point-set topology. (E.g., basic facts about topological spaces, local rings, basic constructions in commutative algebra, holomorphic functions etc.) Atiyah and Macdonald's an An Introduction to Commutative Algebra should furnish more than enough preparation. (You can also concurrently read commutative algebra if that is your preference.) Since you are an undergraduate student, you should not worry too much about learning "background material" just yet before at least seeing what classical algebraic geometry is about. If at some point you decide to specialize in the subject, you will need to learn the "modern tools" such as, for example, schemes, sheaves and sheaf cohomology. The "classic book" for this is Robin Hartshorne's Algebraic Geometry but since that does require a solid background in commutative algebra (or at least the mathematical maturity to accept facts without proofs), you might want to try other books. (But this is, I hasten to add, an excellent book if you do have the background to understand it.) As Bcnrd (on MathOverflow) recommended to me, Qing Liu's Algebraic Geometry and Arithmetic Curves seems to be an excellent book on the subject. Most of the background material in commutative algebra is developed from scratch, and the first six chapters furnish a good introduction to the "modern tools". The last three chapters focus more on the arithmetic side of algebraic geometry, but you can always omit that if you so desire. (But if you are interested in number theory, definitely take a look at that!) Succinctly, I recommend: Take a look at Atiyah and Macdonald and at least read the first few chapters. (The book is roughly 120 pages so covering the first few chapters is not too hard. Though be warned: Some people say that Atiyah and Macdonald is "dense", but I personally found it a very readable book and I think the majority find that so as well.) Then you should have the right background to read Harris and I hope that that will show you how fascinating the subject of algebraic geometry is. Good luck! - Excellent and inspiring answer. – Dedalus Aug 11 2010 at 23:36 2 Harris's book is what I used for my undergraduate course, and it is not suitable for a first foray into algebraic geometry, especially for an undergraduate. I love the book, but one of its strength, the number of examples, can get really overwhelming for the total beginner. There is just not enough on the affine case, and the informal, almost conversational style is good for grad students, but not for undergrads. Harris is the ideal second book in Algebraic Geometry (or possibly Eisenbud-Harris is you want to get into schemes.) – Thierry Zell Aug 12 2010 at 11:47 Check out Hulek's «Elementary algebraic geometry», which is part of AMS's Student Mathematical Library. It's a great introduction to the subject for undergraduates that requires few prerequisites. - After having picked up some basic ideas - from sources mentioned above or maybe even from Kenji Ueno's Introduction to Algebraic Geometry, which according to the introduction is aimed at "scientists", not just mathematicians, and thus has almost zero prerequisites - you should not be afraid to take the next step and study schemes. For this I want to recommend warmly the 3 little, undergraduate readable, books by Kenji Ueno, Algebraic Geometry 1-3, which appeared in the AMS series "Translations of Mathematical Monographs" as nr. 185, 197 and 218. Each of these tomes has less than 200 pages - and the pages are small. Yet the author manages to cover all the basic topics of scheme theory, painlessly and in a very friendly style. To read the book you do not need to have studied Commutative Algebra before - instead you can have a copy of Matsumura ready while reading Ueno; for all the results he uses he gives precise references in H. Matsumura, Commutative Algebra, 2nd edition, Benjamin 1980. It is no accident that the AMS decided to have this translated this from Japanese, I find the text an extraordinary combination of good content and friendliness - I think that the Ueno trilogy (or anything involving schemes, for what matters) is too advanced for a first introduction as un undergraduate. Moreover, I doubt that one does not need to study commutative algebra before reading those books. Deep result will be quoted, but I cannot see how one can read a book about schemes without a firm grasp say of Noetherian rings. – Andrea Ferretti Aug 12 2010 at 14:23 I think the best way is to read a book on commutative algebra (Atiyah & MacDonald) and then, you can start reading Hartshorne's. Chapter 1 will give you a fairly concrete idea of what classical alg. geo. is about. You should also try to do all the exercises even though that will be time consuming. I think this is the best way one can do it. - 2 I seriously doubt that this is the best way to proceed. I mean, Hartshorne for a first approach to the subject, and at the undergraduate level? – Adrián Barquero Mar 3 2011 at 3:38 Yes. I'm myself doing it (undergrad). After reading Atiyah & MacDonald carefully (doing most exercises), Hartshorne is kind of a revelation of what all these commutative algebra is for. I can't believe I got -2 for this! – Brian Mar 3 2011 at 14:53	{"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": 4, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9548889994621277, "perplexity_flag": "middle"}

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