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http://math.stackexchange.com/questions/81145/why-this-limit-cant-be-computed-like-this | # Why this limit can't be computed like this?
This is the limit: $$\lim_{x \to 2} \frac{3x^2-x-10}{x^2-4}$$
The solution manual says it is: $\frac{11}{4}$
I've tried to solve it like a polynomial like this:
$$\frac{(\frac{3x^2}{3x^2}-\frac{x}{3x^2}-\frac{10}{3x^2})*3x^2}{(\frac{x^2}{x^2}-\frac{4}{x^2})*x^2}=$$
$$= \frac{(1-0-0)*3x^2}{(1+0)*x^2}=3$$
Can you please tell me where am I doing wrong? Or why these kind of operation doesn't work here? Thank you
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Take a closer look at the zeros in (1-0-0) and (1+0). Note that $x$ to going to 2, not infinity. – Byron Schmuland Nov 11 '11 at 16:18
$x\rightarrow 2$ not to $\infty$ , you have to factor both numerator and denominator.. – pedja Nov 11 '11 at 16:19
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Andrew, dividing by the leading term is best when evaluating "end behavior" (i.e. limits at infinity) – The Chaz 2.0 Nov 11 '11 at 16:24
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Like Byron said: dividing with the leading term is the strategy for the limit when $x\to\infty$, and doesn't work as well when $x\to 2$. – Willie Wong♦ Nov 11 '11 at 16:25
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@Andrew: no, $\frac{x}{3x^2}$ doesn't go to zero as $x \to 2$. It goes to $\frac{2}{12}$ – Ross Millikan Nov 11 '11 at 16:26
show 3 more comments
## 2 Answers
For this problem, you'll want to factor the numerator and the denominator,
(which both tend to $0$ as $x \to 2$), cancel a factor of $x-2$, and then try direct substitution again.
$\frac{3x^2 -x - 10}{x^2-4} = \frac{(3x +5)(x - 2)}{(x + 2)(x - 2)} = \frac{3x + 5}{x + 2}$ (when $x \neq 2$)
Now letting $x \to 2$, we get $\frac{3 \cdot 2 + 5}{2 + 4} = \frac{11}{4}$, as desired.
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What if we can't factorize? Should we assume then that the limit DNE? or is 0? – Andrew Nov 11 '11 at 16:33
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There are, in fact, at least 8 techniques (not counting L'Hospital's rule) for evaluating 0/0 limits. I say 8, since my AP Calc teacher gave a handout listing them. Sure wish I had that handout! Some of the techniques are: multiplying by radical conjugate, factoring, trig results (sinx/x $\to$ 1 as $x \to 0$, compound fraction/lcd manipulation, ... – The Chaz 2.0 Nov 11 '11 at 16:36
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I guess direct substitution should be on the list, since that's a good place to start! – The Chaz 2.0 Nov 11 '11 at 16:40
The original poster asked in a comment: "What if we can't factorize? Should we assume then that the limit DNE? or is 0?"
You had $$\lim_{x \to 2} \frac{3x^2-x-10}{x^2-4}$$ and as $x\to2$, the numerator and denominator both approach $0$. If the numerator and denominator are polynomials, then:
• If the numerator approaches a non-zero number and the denominator approaches $0$, then the limit does not exist (in some cases it's $\infty$ or $-\infty$, in which case the result is often phrased as "the limit does not exist").
• If the denominator is a non-zero number then just plug in the number that $x$ is approaching (in this case $2$) and that's the limit.
• (The really important case) If they both approach $0$, then use the fact from algebra described below.
Algebra tells us that if you plug a number into a polynomial and get $0$, that tells you something about how to factor it. If you plug $2$ into $3x^2-x-10$ and get $0$, that means $x-2$ is one of the factors. Similarly, you plug $2$ into $x^2-4$ and get $0$, and that tells you $x-2$ is one of the factors. So $$\frac{3x^2-x-10}{x^2-4} = \frac{(x-2)(\cdots\cdots)}{(x-2)(\cdots\cdots)}.$$ Then you need to figure out what goes in place of $(\cdots\cdots)$ in each case. You can do that by long division, dividing $3x^2-x-10$ by $x-2$, and similarly dividing $x^2-4$ by $x-2$. That will work even in cases that would be difficult to factor if you didn't have this way of knowing that $x-2$ is one of the factors. And you don't need to factor completely; you only need to pull out any factors that are $0$ when $x=\text{(in this case) }2$.
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+1 Assuming you were taught polynomial division, which I've found is rarer and rarer in high schools... – user7530 Nov 11 '11 at 20:01 | {"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": 43, "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": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9542601704597473, "perplexity_flag": "middle"} |
http://mathoverflow.net/questions/9218/probabilistic-proofs-of-analytic-facts/10312 | ## Probabilistic Proofs of Analytic Facts
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
What are some interesting examples of probabilistic reasoning to establish results that would traditionally be considered analysis? What I mean by "probabilistic reasoning" is that the approach should be motivated by the sort of intuition one gains from a study of probability, e.g. games, information, behavior of random walks and other processes. This is very vague, but hopefully some of you will know what I mean (and perhaps have a better description for what this intuition is).
I'll give one example that comes to mind, which I found quite inspiring when I worked through the details. Every Lipschitz function (in this case, $[0,1] \to \mathbb{R}$) is absolutely continuous, and thus is differentiable almost everywhere. We can use a probabilistic argument to actually construct a version of its derivative. One begins by considering the standard dyadic decompositions of [0,1), which gives us for each natural n a partition of [0,1) into $2^{n-1}$ half-open intervals of width $1/{2^{n-1}}$. We define a filtration by letting $\mathcal{F}_n$ be the sigma-algebra generated by the disjoint sets in our nth dyadic decomposition. So e.g. $\mathcal{F}_2$ is generated by ${[0,1/2), [1/2,1)}$. We can then define a sequence of random variables $Y_n(x) = 2^n (f(r_n(x)) - f(l_n(x))$ where $l_n(x)$ and $r_n(x)$ are defined to be the left and right endpoints of whatever interval contains x in our nth dyadic decomposition (for $x \in [0,1)$). So basically we are approximating the derivative. The sequence $Y_n$ is in fact a martingale with respect to $\mathcal{F}_n$, and the Lipschitz condition on $f$ makes this a bounded martingale. So the martingale convergence theorem applies and we have that $Y_n$ converges almost everywhere to some $Y$. Straightforward computations yield that we indeed have $f(b) - f(a) = \int_a^b Y$.
What I really like about this is that once you get the idea, the rest sort of works itself out. When I came across the result it was the first time I had thought of dyadic decompositions as generating a filtration, but it seems like a really natural idea. It seems much more structured than just the vague idea of "approximation", since e.g. the martingale condition controls the sort of refinement the next approximating term must yield over its predecessor. And although we could have achieved the same result easily by a traditional argument, I find it interesting to see from multiple points of view. So that's really my goal here.
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One example that comes to mind is the relationship between Brownian motion and harmonic functions. Is that the kind of thing you're thinking of? – Qiaochu Yuan Dec 18 2009 at 1:08
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The book The Probabilistic Method, by Alon and Spencer, includes probability-inspired proofs of results that don't belong to probability in sections called "The Probabilistic Lens", which are inserted between the various chapters. I don't have my copy at hand right now, and the only analytic one I remember being there is Bernstein's proof of the Weierstrass approximation theorem, which Harald Hanche-Olsen has already mentioned. – Michael Lugo Dec 18 2009 at 1:53
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I don't mean to be overly critical (really!), but isn't it immediate from the definition that a Lipshitz function is absolutely continuous? It's just a replay of the argument that a linear function with slope M is continuous: take delta = epsilon/M. I think HHO's example below is so good that it should be the exemplar of the question, perhaps. – Pete L. Clark Dec 18 2009 at 2:51
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@Yemon: Absolute continuity does not mention differentiability. It turns out, but is comparatively much deeper, that an AC function is differentiable almost everywhere. I still think that Lipschitz implies AC is done just by taking delta = epsilon/(Lipschitz constant). – Pete L. Clark Dec 18 2009 at 7:33
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Shouldn't this be community wiki? – Harry Gindi Dec 27 2009 at 0:40
show 5 more comments
## 19 Answers
One nice example is Bernstein's proof of the Weierstrass theorem. This proof analyses a simple game: Let $f$ be a continuous function on $[0,1]$, and run $n$ independent yes/no experiments in which the “yes” probability is $x$. Pay the gambler $f(m/n)$ if the answer “yes” comes up $m$ times. The gambler's expected gain from this is, of course, `$$p_n(x)=\sum_{k=0}^n f(k/n)\binom{n}{k}x^k(1-x)^{1-k}$$` (known as the Bernstein polynomial). The analysis shows that $p_n(x)\to f(x)$ uniformly.
S. N. Bernstein, A demonstration of the Weierstrass theorem based on the theory of probability, first published (in French) in 1912. It has been reprinted in Math. Scientist 29 (2004) 127–128 (MR2102260).
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This is freaking amazing!! – Pietro KC May 2 2010 at 20:06
### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you.
Question: Given $n$ points in Euclidean space (which we might as well take to be $\ell_2^n$), what is the smallest $k=k(n)$ so that these points can be moved into $k$-dimensional Euclidean space via a transformation which expands or contracts all pairwise distances by a factor of at most $1+\epsilon$?
Answer: $k(n)\le C \ {\log (n+1) \over {\epsilon^2}}$.
Proof: A (suitably normalized) random rank $k(n)$ orthogonal projection works.
Nowadays this is called the Johnson-Lindenstrauss Lemma. All known proofs in a form this strong use random linear operators.
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While I've forgotten most of the necessary technical details (ah for the days when I knew more about probability and less about homological algebra), one striking example is the exploitation of conformal invariance of planar Brownian motion to reprove results in complex analysis. See
Burgess Davis. Brownian Motion and Analytic Functions, Ann. Probab. Volume 7, Number 6 (1979), 913-932.
which in particular has a probabilistic proof of the little Picard theorem.
(I first learned of Davis' proof from a sketch in Körner's wonderful book Fourier Analysis, which I'd recommend for students as an antidote to the inevitable tedium and occasional narrowness of a first & second course in analysis.)
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This looks excellent! Thanks! – Erik Davis Dec 18 2009 at 10:42
## 1) perimeter of planar sets with constant width
I like the probabilistic proof that every set of constant width 1 has perimeter pi using Buffon's needle problem. See also the wikipedia article on Buffon's noodle problem.
Another beautiful analytic (of a sort) theorem where probability plays an important role is regarding the overhang problem. The description of the problem and the solution is taken from the abstract of the paper "maximum overhang" by Mike Paterson, Yuval Peres, Mikkel Thorup, Peter Winkler and Uri Zwick:
## 2) Maximum overhang
How far can a stack of $n$ identical blocks be made to hang over the edge of a table? The question dates back to at least the middle of the 19th century and the answer to it was widely believed to be of order $\log n$. Recently, Paterson and Zwick constructed $n$-block stacks with overhangs of order $n^{1/3}$, exponentially better than previously thought possible. We show here that order $n^{1/3}$ is indeed best possible, resolving the long-standing overhang problem up to a constant factor.
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There are probabilistic proofs of Atiyah-Singer or most anything else that can be done with a heat kernel.
(Rogers & Williams is rife with probabilistic proofs of analytic facts [as well as the fundemental theorem of algebra], and more generally just about all of potential theory can be recast in terms of martingales a la Doob, as Qiaochu points out; surely there are many more examples.)
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This is obviously a very old answer, but I see that you're still active on MO. Would you happen to have a reference for the probabilistic proof of Atiyah-Singer that you mention? Is it substantially different from the usual heat kernel approach? – Paul Siegel Feb 7 2012 at 19:44
amazon.com/… – Steve Huntsman Feb 8 2012 at 0:02
I believe that Krylov and Safonov's original proof of the Harnack inequality for solutions of elliptic equations in nondivergence form was a probabilistic one. PDE people wouldn't have the slightest idea just from glancing at the title of their paper that this is what they proved (or at least this PDE person). This was not an isolated incident. Much of Krylov's pioneering work in elliptic equations was originally written up in the language of Markov processes, etc, with analytic proofs appearing later.
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The Radon-Nikodym Theorem and the Lebesgue differentiation theorem can be proved by Martingale theory (see "Probabilty Theory" by Heinz Bauer, pp. 173-5).
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This paper (Prime Numbers and Brownian Motion, by Patrick Billingsley) is perhaps more about proving number theoretic facts than analytical, but at least to me they have a very analytical flavor anyway, and was the first thing to come into my mind when I read your question. I think you would find it interesting.
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Shannon's theorem giving the capacity of noisy channel is proved using random coding. (Efficiently-computable codes are not known.)
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How is this an analytic fact? – Marcin Kotowski Feb 8 2012 at 1:23
There are a number of probabilistic inequalities that are quite frequently used in harmonic analysis. For example, Khintchine's inequality (http://en.wikipedia.org/wiki/Khintchine_inequality).
The same idea of using random signs and taking expectations is rather common. One specific inequality proved in this manner which I've come across comes is the Rademacher-Menshov Theorem (for almost orthogonal functions). The theorem gives a way to control the $L^2$ norm of partial sums of a sequence of N "almost orthogonal" functions by the sum of the $L^2$ norms of each function modulo a logarithmic loss in N. A precise statement and proof of this inequality can be found on page 43 of this article by Ciprian Demeter, Terence Tao, and Christoph Thiele: http://arxiv.org/abs/math/0510581.
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Here's something that's pretty neat: find a measurable subset $A$ of $[0,1]$ such that for any subinterval $I$ of $[0,1]$, the Lebesgue measure $\mu(A\cap I)$ has $0 < \mu(A\cap I) < \mu(I)$. There's an explicit construction of such a set in Rudin, who describes such sets as "well-distributed". Balint Virag (and maybe others) found a very slick probabilistic construction.
Let $X_1, X_2, \ldots$ be i.i.d. coin flips, i.e. $X_1$ is $1$ with probability $1/2$ and $-1$ with probability $1/2$. Consider the (random) series
$$S:=\sum_{n=1}^\infty X_n/n.$$
By the Kolmogorov three-series theorem, it converges almost surely. However, it's a simple exercise to see that for any $a$, the event ${S > a}$ has non-trivial measure: for $a>0$, there's a positive chance of the first $e^a$ terms of the series being positive, so the $e^a$-th partial sum is positive, and the tail is independent and positive or negative with equal probability, due to symmetry. For $a\leq 0$, it's trivial, again because of symmetry.
A common way of realizing i.i.d. coin flips on the unit interval is as Rademacher functions: for $x\in[0,1]$, let ${b_n}$ be its binary expansion, and $X_n(x) = (-1)^{b_n}$. Realized this way, the random sum $S$ becomes an almost everywhere finite measurable function from $[0,1]$ to $\R$. It only takes a bit more work to see that the set ${S>a}$ is exactly a well-distributed set.
Alex Bloemendal has written this up in a short note, but I'm not sure if he's published it anywhere.
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You shouldn't use $x$ for both the variable in $[0,1]$ and the number that you want $S$ greater than. – Robert Israel Feb 7 2012 at 20:47
D'oh. Thanks... – Andrew Stewart Feb 8 2012 at 0:20
I'm not sure how kosher it is for me to answer my question, but since there had been several comments about my original post I did not want to make any major edits to it. I've posed this question to my probability professor and he mentioned his favorite, from the paper "Triple points: from non-Brownian filtrations to harmonic measures." by Tsirelson. It's pretty far over my head, but it claims to have a probabilistic proof of (I'm quoting the description)
A conjecture by C. Bishop (1991) about harmonic measures for three arbitrary (not just regular) non-intersecting domains in Rn. Roughly speaking, trilateral contact is always rare harmonically (though not topologically).
This seems like it goes hand in hand with some of the above comments, where basically knowledge of things like hitting probabilities of brownian motion and similar things for other processes can assist in understanding the fine properties of various domains, useful to people in PDE and harmonic analysis.
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Davie's construction of subspaces of $c_0$ and $\ell_p$ ($p\in (2, \infty)$) without the approximation property, as outlined in Section 2.d of Lindenstrauss and Tzafriri's book Classical Banach Spaces I, uses a probabilistic lemma (Lemma 2.d.4, p.87-88). I do not know Davie's proof all that intimately, having been through it only once - courtesy of a fellow grad student who took a couple of hours to go over it in a research group seminar... I remember that it looked like magic at the time.
(Edited once for a typo)
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One of the basic constructions in the theory of singular integral operators is the Calderón-Zygmund decomposition, which follows from a simple stopping time argument. This result has numerous important applications in harmonic analysis; for instance, it plays a role in proving the $L^p$-convergence of Fourier series ($1 < p < \infty$).
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Since probability theory is usefully formalized as a special case of quantum probability, a related question is what examples are there of quantum proofs for classical (non-quantum) results. There are now sufficiently many examples to merit a survey by Drucker and deWolf “Quantum Proofs for Classical Theorems.” I blogged about two such examples on FXPAL's blog.
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None of the results in the (excellent) paper you reference are ven remotely analytic. – Marcin Kotowski Feb 8 2012 at 1:23
Some examples can be found in the book "Statistical independence in probability, analysis and number theory" by Mark Kac.
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How about the Convolution Theorem, which can be seen as a consequence of $${\rm E}[e^{iu(X+Y)}]={\rm E}[e^{iuX}]{\rm E}[e^{iuY}],\;\; u \in \mathbb{R},$$ where $X$ and $Y$ are independent random variables.
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Doeblin's proof of the fundamental limit theorem for regular Markov chains: (450 p. in Introduction to probability, available online.) The proof uses coupling.
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I realize this isn't an analytic fact – Yoo Dec 24 2009 at 5:37 | {"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": 72, "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": 0.9402767419815063, "perplexity_flag": "head"} |
http://math.stackexchange.com/questions/147166/does-my-definition-of-double-complex-noncommutative-numbers-make-any-sense/147170 | # Does my definition of double complex noncommutative numbers make any sense?
I wanted to factorize $a^2+b^2+c^2$ into two factors in a similar way to $$a^2+b^2 = (a+ib)(a-ib)$$ This doesn't seem to be possible using real or complex numbers. However I came up with the following idea $$(a + ib + jc) (a -ic -jc) = a^2+b^2+c^2$$ if we define $$i^2=j^2=-1$$ and $$ij = -ji.$$ So instead of using complex numbers I had to define a kind of double complex number obeying a non-commutative multiplication rule to solve my factorization problem. Does my definition of this numbers make sense or is it somehow inconsistent? Do these numbers already have a name in mathematics and are they used? Is there any literature about them?
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– Thomas Andrews May 19 '12 at 21:20
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Congratulations on rediscovering the quaternions! – Qiaochu Yuan May 20 '12 at 1:20
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I agree, you have had a very nice insight there. Hamilton searched for a long time -- it was a running joke among his kids every morning -- for a way to generalize complex numbers into 3D space. He "got it" while walking with his wife over a bridge, realizing that he needed three $i$-like quantities, not just two, to make everything work. Despite quaternions falling out of favor later on, they were instrumental both the the development of vectors and concepts such as dot and cross products, all of which are quaternions subsets. Maxwell's original equations use quaternions, not vectors. – Terry Bollinger May 20 '12 at 4:01
@QiaochuYuan Thank you. "One cannot invent useful things, one can only rediscover them." ;-) – asmaier May 20 '12 at 13:00
I should add on @Terry's comment that the bridge mentioned has an inscription about this realization. One can still see this in Dublin (if one is not too hazy from drinking). – Asaf Karagila May 22 '12 at 22:15
## 3 Answers
You have come across the quaternions. They are numbers of the form $$a+bi+cj+dk$$ where $a,b,c,d\in\mathbb{R}$ and $i$, $j$, and $k$ are symbols satisfying $$i^2=j^2=k^2=ijk=-1$$ $$ij=k,\quad jk=i,\quad ki=j$$ $$ji=-k,\quad kj=-i, \quad ik=-j$$ Multiplication of quaternions is non-commutative in general, but it is still associative.
The conjugate of a quaternion $q=a+bi+cj+dk$ is defined to be $$q^*=a-bi-cj-dk,$$ and the norm of a quaternion is defined to be. $$\|q\|=\sqrt{qq^*}.$$ After expanding everything out, one can see that $$\|a+bi+cj+dk\|=\sqrt{a^2+b^2+c^2+d^2}.$$ So, to factor the expression $a^2+b^2+c^2$, you were just using quaternions for which the coefficient of $k$ is equal to $0$: $$(a+bi+cj+0k)(a-bi-cj-0k)=a^2+b^2+c^2+0^2=a^2+b^2+c^2.$$ There are other extensions of the complex numbers that are possible: the other one usually mentioned are the octonions, which are $8$-dimensional over the real numbers.
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Is there a special name for quaternions which have the coefficient of $k$ set to zero? And can't there also be quintions which could then factorize $a^2+b^2+c^2+d^2+e^2$ into two factors ? – asmaier May 20 '12 at 12:55
No, there are no quintions. The next number system is octonions, which are not associative. – Stefan Smith May 20 '12 at 14:37
@asmaier Quaternions with their $k$ component zero have no special name because they aren't closed under multiplication; I've added an answer that explains the details on this. – Steven Stadnicki May 22 '12 at 22:32
It's worth pointing out that you need to go up to the quaternions - a four-dimensional space - because a factorization of the sort you're describing doesn't 'work' in just three dimensions. Suppose you had $z=a+bi+cj$ (and so $\bar{z}=a-bi-cj$) and $w=d+ei+fj$ (with $\bar{w}=d-ei-fj$). Then $x=zw$ would be of the form $r+si+tj$, with $r$, $s$ and $t$ each expressions in $a\ldots f$, and $\bar{x}$ would be $r-si-tj$; taking the norms (or in other words looking at $|x|^2 = x\bar{x} = |z|^2|w|^2$) would give you an identity of the form $(a^2+b^2+c^2)\cdot(d^2+e^2+f^2) = r^2+s^2+t^2$. But consider $(1^2+1^2+1^2)\cdot(4^2+2^2+1^2) = 3\cdot 21 = 63$; this number is of the form $8n+7$ and so by the Sum Of Three Squares theorem it can't be expressed as $r^2+s^2+t^2$ for any values of $r,s,t$. This means the three-squares identity (or in other words, the three-dimensional product you're looking for) can't exist.
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Going from complex numbers to the quaternions you lose commutativity. Going from quaternions to octonions you lose associativity. What’s next? There is very nice and elegant mathematics around these hypercomplex numbers. Adolf Hurwitz proved it in 1898 that every normed division algebra with an identity is isomorphic to one of the following four algebras: $\mathbb{R}$, $\mathbb{C}$, $\mathbb{H}$ and $\mathbb{O}$, that is the real numbers, the complex numbers, the quaternions and the octonions. And that’s it. No more, no less.
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http://mathoverflow.net/questions/45351/does-pi-1-have-a-right-adjoint/45375 | ## Does $\pi_1$ have a right adjoint?
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
Eilenberg and Mac Lane showed that given a group $G$ there exists a pointed topological space $X_G$ such that $\pi(X_G,\bullet)\cong G$. It is obviously a way to "invert direction" to the functor $\pi_1\colon \mathbf{Top}^\bullet\to \mathbf{Grp}$ to a functor $\mathcal K\colon \mathbf{Grp}\to \mathbf{Top}^\bullet$ such that $\pi_1(\mathcal K(G),\bullet)\cong G$ (almost by definition). This is equivalent to say that there exists a natural transformation (equivalence, in this case) between $\pi_1\circ\mathcal K$ and $\mathbf{1}_{\mathbf{Grp}}$, which turns out to resemble some sort of counity.
It would be wonderful if I could define an adjunction between the two categories in exam, given by the two functors. Everytime I try to think about some sort of unity to this hypotetical adjunction I poorly fail: considering the vast literature in the field of algebraic topology, I believe in only two possible cases. The first, nothing interesting arises from this adjunction. The second, there is no sort of adjunction.
The key point, quite trivial, to answer is: it is well known that an adjunction is uniquely determined by one among unity and counity, provided the one is universal. But $\boldsymbol\varepsilon\colon \pi_1\circ\mathcal K\to \mathbf{1}_{\mathbf{Grp}}$ is an equivalence: can I conclude that it is universal?
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## 2 Answers
In the more general setting the answer is no. Left adjoints preserve colimits and it is not true that $\pi_{1}(X\vee Y)\cong \pi_{1}(X)\ast\pi_{1}(Y)$ for all spaces $X,Y$ (even compact metric spaces). For instance, if $(\mathbb{HE},x)$ is the usual Hawaiian earring, let $X=Y=C\mathbb{HE}=\mathbb{HE}\times I/\mathbb{HE}\times{1}$ be the cone on the Hawaiian earring with basepoint the image of $(x,0)$ in the quotient. It is a theorem of Griffiths that $\pi_{1}(C\mathbb{HE}\vee C\mathbb{HE})$ is uncountable and not the free product of trivial groups.
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Clear, clear. Thanks. :) – tetrapharmakon Nov 9 2010 at 0:05
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Let $\mathcal{C}$ be the homotopy category of connected pointed CW complexes. Then the functor $\pi_1:\mathcal{C}\to\mathbf{Grp}$ does indeed have a right adjoint. The cleanest way to define it is to use the simplicial classifying space functor $B:\mathbf{Grp}\to\mathcal{C}$. For the unit of the adjunction you need a natural homotopy class of maps $X\to B(\pi_1(X))$. This can be done by obstruction theory, by induction up the skeleta of $X$. Alternatively, one can define a more canonical map from $|SX|$ (the geometric realisation of the singular complex) to $B(\pi_1(X))$ by simplicial methods, and then use the fact that there is a natural map $|SX|\to X$ which is a homotopy equivalence when $X\in\mathcal{C}$. The first place I would look for more details would be Peter May's old book on simplicial objects in algebraic topology.
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Goerss-Jardine is probably going to be more useful than Peter May's book, which is really outdated. – Harry Gindi Nov 8 2010 at 21:21
That copy of the book is copyright-infringing (not that I care), and it's probably better if you can avoid posting it on MO. – Harry Gindi Nov 9 2010 at 0:58 | {"extraction_info": {"found_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": 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": 0.93374103307724, "perplexity_flag": "head"} |
http://mathoverflow.net/questions/80430?sort=oldest | ## General form of Schwarz reflection principle
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Hello all,
It is easy to find results on reflecting holomorphic functions over circles and lines, but I am wondering what there is for reflecting over smooth curves, or analytic arcs, etc. In particular, I am interested in the conformal map f from the upper half-plane to $\{x+yi : y>1/(1+x^2)\}$ which maps $0$ to $i$ and fixes infinity (so, say, maps $i$ to $2i$). It seems to me that I should be able to extend $f$ to be analytic in a neighborhood of infinity, but I cannot find a reference. Any help will be appreciated.
Greg
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## 2 Answers
For reflection across analytic arcs see Caratheodory's book Conformal Representation pages 87-90 or his book Theory of Functions vol 2 pages 101-104 .
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Wonderful! Thank you. – Greg Markowsky Nov 15 2011 at 6:29
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For your specific question, note that the domain you describe $\mathbb{D}$, consisting of those $z = x+iy$ for which $y > 1/(1+x^2)$, when regarded as a domain in the extended complex plane, $\mathbb{CP}^1= \mathbb{C}\cup\infty$, is a disk with a smooth, real-analytic boundary in $\mathbb{CP}^1$, so the mapping you are describing does, in fact, extend analytically across the boundary, everywhere along the boundary. In particular, it extends analytically (meromorphically, actually) to a neighborhood of $z = \infty$.
- | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 12, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.918129563331604, "perplexity_flag": "head"} |
http://mathoverflow.net/questions/7508?sort=oldest | ## A hypersurface with many points
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Ok, it's time for me to ask my first question on MO.
Consider the affine curve $Y+Y^q=X^{q+1}$ over the finite field `$\mathbf{F}_q$`. It's interesting because it has the largest number of points over `$\mathbf{F}_{q^2}$` possible relative to its genus, which is $q(q-1)/2$. In other words, this curve realizes the Weil bound over `$\mathbf{F}_{q^2}$`. This seems to be well-known in the literature.
Now consider the following hypersurface $\mathcal{X}$ over the finite field `$\mathbf{F}_q$`:
```$$Z+Z^q+Z^{q^2}=\det\left(\begin{matrix}
0 & X & Y \\ Y^q & 1 & X^q \\ X^{q^2} & 0 & 1\end{matrix}\right)$$```
Empirical observation seems to point to the following: The compactly supported cohomology of $\mathcal{X}$ is only nonzero in degrees 2 and 4. In degree 2, the dimension of $H^2(\mathcal{X})$ is $q^2-1$ and the $q^3$-power frobenius acts as the scalar $q^3$. Which is all a fancy way of saying that for all $n$,
`$$\#\mathcal{X}(\mathbf{F}_{q^{3n}})=q^{6n}+q^{3n}(q^2-1).$$`
Thus $\mathcal{X}$ has the largest number of $\mathbf{F}_{q^3}$-points among any hypersurface with the same compactly supported Betti numbers.
Can anyone help me prove the above formula? (I can do $n=1$ alright...) Bonus points if you can also compute the automorphism group of $\mathcal{X}$. Many more bonus points if you can formulate the generalization to hypersurfaces of higher dimension!
The above hypersurface arises in the study of the bad reduction of Shimura varieties, if anyone cares to know.
EDIT: Admittedly, this is a narrow problem about a very particular surface. Therefore I'm going to accept an answer to the following question: Is there an algorithm to compute the zeta function of a hypersurface of this sort, that's quicker than counting points?
(I've already noticed that it's enough to recur over X and Y, and to test for each pair that the expression on the right lies in the image of the linear map defined by the expression on the left. But I can't think of anything faster than this.)
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If you're serious about bonus points, you should add a bounty =p. – Harry Gindi Dec 2 2009 at 3:55
How do I do that? – Jared Weinstein Dec 2 2009 at 7:41
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Jared -- when you view this question, I think you should see a little grey link labeled "start a bounty", right above "add a comment". Click that if you want to offer a bounty. See the FAQ for more details. (In my opinion, though, an elegant question like this will draw people regardless of whether you offer a bounty.) – David Speyer Dec 3 2009 at 5:45
## 3 Answers
Here is a complete solution to the main question when $n$ and $q$ are both odd, and a partial solution for the other parities. The partial solution includes a reduction to the case $n=2$.
Let `$\text{Tr}_k$` denote the trace map from `$\mathbb{F}_{q^n}$` to `$\mathbb{F}_{q^k}$`, assuming that $k|n$.
The equation is $$z+z^q + z^{q^2} = x^{q^2+q+1} - xy^q - x^{q^2}y.$$ First, make the change of variables $y \leftarrow -yx^{q^+1},$ so that the equation becomes $$z+z^q+z^{q^2} = x^{q^2+q+1}(y^q+y+1).$$ Second, when $q$ is odd, we can clarify matters a little with the change of variables $y \leftarrow y - \frac12$ to get rid of the constant. The image of the $q$-linear map `$z \mapsto z + z^q + z^{q^2}$`, acting on `$\mathbb{F}_{q^{3n}}$`, consists of those elements $Z'$ such that `$\text{Tr}_3(z') = \text{Tr}_1(z')$`. Whenever such a $z'$ is reached by the right side, there are $q^2$ solutions for $Z$. On the right side, $y \mapsto y' = y^q+y$ is an $q$-linear isomorphism when $n$ is odd, while when $n$ is even its image is the locus of `$\text{Tr}_2(y') = \text{Tr}_1(y')$`. Meanwhile $y' \mapsto x^{q^2+q+1}y'$ is a $q$-linear isomorphism unless $x=0$.
Thus when $n$ and $q$ are both odd, there are $(q^{3n}-q^2)(q^{3n}-1)$ solutions with $x,y \ne 0$. There are $q^2(2q^{3n}-1)$ more solutions when one of them is zero.
When $q$ is odd and $n$ is even, then in principle different $x' = x^{q^2 + q + 1}$ could behave differently in the equation $z' = x'y'$. For a fixed $x'$, the set of possible $x'y'$ is a certain $q$-linear hyperplane with codimension $1$, while the set of possible $z'$ is a certain $q$-linear hyperplane with codimension $2$. In the special case that $\{x'y'\}$ contains $\{z'\}$, then there are $q^{3n+1}$ solutions for that value of $x'$. For generic non-zero values of $x'$, there are $q^{3n}$ solutions. Dualize the hyperplanes with respect to `$\text{Tr}_1$`. The dual of $\{y'\}$ is the line $L$ of trace 0 elements in `$\mathbb{F}_{q^2}$`, while the dual of $\{z'\}$ is the plane $P$ of trace 0 elements in `$\mathbb{F}_{q^3}$`. Meanwhile $\{x'\}$ is the subgroup $G$ of $\mathbb{F}_{q^{3n}}$ of index $q^2+q+1$. A special value of $x'$ in this subgroup is one such that $x'L \subset P$. A priori I am not sure that it never happens. What I can say is that if $x'$ is special, then it must lie in `$\mathbb{F}_{q^6}$` because both $L$ and $P$ do. So you can reduce the counting problem to the case that $3n = 6$ or $n = 2$.
If $q$ is even, then the equation is $$z' = x'(y'+1),$$ where as before $z' = z+z^q+z^{q^2}$ and $y' = y + y^q$. In this case $y'$ is any element with zero trace, and the dual line $L$ is just $\mathbb{F}_q$ itself. I have not worked out exactly how it looks, but I suppose that it reduces to the case $n=1$ for similar reasons as above.
Afterthought: I don't feel like changing all of the equations, but I'm wondering now whether there a dual change of variables to put the right side in the form $y''(x^{q^2}+x)$. I think that the map $x \mapsto x^{q^2}+x$ is always non-singular when $q$ is odd.
A remark about where the trace conditions come from. If $a$ is an irreducible element of `$\mathbb{F}_{q^n}$`, then the map $x \mapsto x^q$ is a cyclic permutation matrix in the basis of conjugates of $a$. A map such as $z \mapsto z+z^q+z^{q^2}$ is then a sum of disjoint permutation matrices and it easy to compute its image and cokernel.
Some remarks about Jared's second, more general question: C.f. the answer to this other mathoverflow question about counting points on varieties. For fixed $q$, the equation of a hypersurface is equivalent to a general Boolean expression, and there may not be much that you can do other than count one by one. There are several strategies that work in the presence of special structure: You can use zeta function information, if you have it, to extrapolate to large values of $q$. You can count the points on a variety if you happen to know that it's linear, or maybe quadratic, or the coset space of a group. And you can use standard combinatorial counting tricks, which in algebraic geometry form amount to looking at fibrations, blowups, inclusion-exclusion for constructible sets, etc.
This particular variety decomposes a lot because it can be made jointly linear in $Y$ and $Z$, and $X$ only enters in a multiplicative form.
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Excellent work! I can now finish the argument when $n=2$, using your criterion. (The claim is false for $q=3$, by the way.) – Jared Weinstein Dec 5 2009 at 5:16
Aha! Now I know why I got stuck at the end. – Greg Kuperberg Dec 5 2009 at 6:26
I think people like Alan Lauder in Oxford know much better ways of computing zeta functions than the naive approach. I think the idea is that they deform to characteristic zero and compute Monsky-Washnitzer or overconvergent p-adic cohomology and then compute char polys of Frobenius. Sounds like using a sledgehammer to crack a nut but my understanding is that these fancy cohomology theories are actually very concrete in terms of cocycles over coboundaries and produce computationally effective point-counting methods. – Kevin Buzzard Dec 5 2009 at 8:14
Maybe you're right. – Greg Kuperberg Dec 5 2009 at 18:43
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I'll put up a quick proof that the curve has the claimed number of points, in case it inspires anyone. For every $X \in \mathbb{F}_{q^2}$, the left hand side $X^q+X$ is the trace of $X$, and is thus in $\mathbb{F}_q$. The trace is an $\mathbb{F}_q$ linear map, so each fiber has size $q$. For those $X$ where $Tr(X)=0$, there is one root $Y=0$. For the other $X$'s, there are $q+1$ roots of $Y^{q+1} = Tr(X)$ in $\mathbb{F}_{q^2}$. So there are $q+(q^2-q)(q+1) = q^3$ points on the affine curve.
I have no idea how to deal with that determinant though. I hope someone will have a better idea!
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The equation of the surface, for fixed $x$, is $F_q$-linear in $y,z$. I'd try to use that.
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I did notice that! I did some experiments: Fix $n\geq 1$, let $x\in\mathbf{F}_{q^{3n}}^*$, and let $N(x)$ be the number of values of $y\in\mathbf{F}_{q^{3n}}$ for which the above equation has a solution in $W\in \mathbf{F}_{q^{3n}}$. It seems that $N(x)=q^{3n-2}$ for all $x$! (Of course it's a power of $q$, because it's the zero locus of an additive polynomial in $Y$.) Establishing this would imply my guess for the zeta function. It's pretty miraculous, I think. – Jared Weinstein Dec 4 2009 at 1:00 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 116, "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": 0.9436866044998169, "perplexity_flag": "head"} |
http://math.stackexchange.com/questions/265264/prove-f-n1f-n-1-f-n2-1n-without-induction | # Prove $F_{n+1}F_{n-1}-(F_{n})^2=(-1)^n$ without induction
I am asked to pove the statement about fibonacci sequence. The task is from the passage about series and sequences. But the proof seems to need induction way, doesn't it?
Prove the statement $$F_{n+1}F_{n-1}-(F_{n})^2=(-1)^n$$ for all $n\ge 1$.
How can I prove this by thinking about limit and convergence?
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Without induction? – Nameless Dec 26 '12 at 10:13
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Then you seem to have defined $F_n$ without recursion? – Hagen von Eitzen Dec 26 '12 at 10:19
I cannot comprehend, why anybody would want to prove this kind of facts without induction? Induction and recursive definition (such as with Fibonacci sequence) go together like love and marriage. Ok, since you asked, you can probably also do this by using Binet's formula for Fibonacci numbers, but that is a lot messier, and you cannot prove Binet's formula without induction anyway :-) Another point: I don't see any limits or converging sequences here, do you? – Jyrki Lahtonen Dec 26 '12 at 10:19
this is the reason why i asked you guys. $F_{n}$ cannot be defined without recursion, so it is impossible to prove the above statement without induction? – doniyor Dec 26 '12 at 10:22
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– Mohan Dec 26 '12 at 10:37
show 5 more comments
## 2 Answers
I think this depends on what is meant by "mathematical induction". Recall that $F_n$ can be solved in the following way. As \begin{equation} \begin{pmatrix}F_{n+1}\\F_n\end{pmatrix} =\begin{pmatrix}1 & 1\\ 1&0\end{pmatrix} \begin{pmatrix}F_n\\F_{n-1}\end{pmatrix},\tag{1} \end{equation} we have \begin{equation} \begin{pmatrix}F_{n+1}\\F_n\end{pmatrix} =\begin{pmatrix}1 & 1\\ 1&0\end{pmatrix}^n \begin{pmatrix}F_1\\F_0\end{pmatrix}.\tag{2} \end{equation} In a broad sense, the derivation of $(2)$ from $(1)$ is mathematical induction, but in normal context, I think this is seldom regarded as such. Now, if this is not considered as mathematical induction, we can solve $(2)$ directly and hence we may and verify your inequality simply by plugging in the solution for $F_n$. (Edit: Or better, as Qiaochu Yuan suggests, one may simply compute the determinant of the matrix $n$-th power in $(2)$.)
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Alternately, you can just compute the determinant of that matrix you're using. – Qiaochu Yuan Dec 26 '12 at 10:53
Ha ha, of course! – user1551 Dec 26 '12 at 11:07
As, $F_{n+2}=F_{n+1}+F_n,$ the Characteristic equation of the recurrence relation will be $t^2-t-1=0$
If $a,b$ are the roots of the equation, $a+b=1,ab=-1$ and $F_n=Aa^n+Bb^n$ where $A,B$ are arbitrary constants.
So, $$F_{n+1}F_{n-1}-(F_{n})^2=(Aa^{n+1}+Bb^{n+1})(Aa^{n-1}+Bb^{n-1})-(Aa^n+Bb^n)^2$$ $$=AB\{a^{n+1}b^{n-1}+b^{n+1}a^{n-1}-2(ab)^n\}$$ $$=AB(ab)^n\{\frac{a^2+b^2}{ab}-2\}$$ $$=-5AB(-1)^n$$ as $\frac{a^2+b^2}{ab}-2=\frac{(a+b)^2}{ab}-4=-1-4=-5$
Now, $0=F_0=Aa^0+Bb^0\implies B=-A$
and $1=F_1=Aa+Bb=A(a-b)\implies A=\frac1{a-b}$
So, $AB=-A^2=\frac{-1}{(a-b)^2}=\frac{-1}{(a+b)^2-4ab}=-\frac15$
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thanks lab, great – doniyor Dec 26 '12 at 11:51
@doniyor, welcome – lab bhattacharjee Dec 26 '12 at 12:33 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 21, "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": 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": 0.9111589789390564, "perplexity_flag": "middle"} |
http://www.math.uni-bielefeld.de/sfb701/projects/view/22 | Faculty of Mathematics
Collaborative Research Centre 701
Spectral Structures and Topological Methods in Mathematics
# Project C8
## Finiteness Properties of Infinite Discrete Groups
Principal Investigator(s) Other Investigators
## Summary:
The main goal is to establish finiteness properties (finite generation, finte presentability, etc.) of arithmetic groups in positive characteristic, e.g., $SL_N(\mathbf{F}_q[t; t^{-1}])$. In particular, one aim is a proof of the {\em Rank Conjecture} which would give the exact finiteness length for reductive groups. Another point on the agenda is the question whether $SL_2(\mathbf{Z}[t; t^{-1}])$ is finitely generated. Finally the finiteness properties of Torelli subgroups (a) in Out($F_{N}$) and (b) in mappingclass groups of closed oriented surfaces are to be determined.
## Recent Preprints:
| | | |
|-------|----------------------------------------------------------------------------------------------------------|-------------|
| 13030 | Polytopes and groups | PDF | PS.GZ |
| 12146 | The braided Thompson’s groups are of type $F_\infty$ | PDF | PS.GZ |
| 12145 | Higher generation for pure braid groups | PDF | PS.GZ |
| 12070 | The Brin-Thompson groups $sV$ are of type $F_\infty$ | PDF | PS.GZ |
| 12069 | An Eilenberg–Ganea phenomenon for actions with virtually cyclic stabilisers | PDF | PS.GZ |
| 12060 | Higher Finiteness Properties of Reductive Arithmetic Groups in Positive Characteristic: The Rank Theorem | PDF | PS.GZ |
| 12055 | Rational homological stability for groups of partially symmetric automorphisms of free groups | PDF | PS.GZ |
| 12054 | A combinatorial proof of the degree theorem in Auter space | PDF | PS.GZ |
| 12053 | Brown's criterion in Bredon homology | PDF | PS.GZ |
| 12004 | On the classifying space for the family of virtually cyclic subgroups for elementary amenable groups | PDF | PS.GZ |
## Announced Talks interacting with the project:
May 21, 2013 16:15U2-135 !!! CANCELLED !!!Endlichkeitseigenschaften arithmetischer Gruppen über Funktionenkörpern (Teil 2)Kai-Uwe Bux
May 22, 2013 14:15T2-228 Classifying spaces and cohomology rings for Thompson's groups F and braided FMatthew Zaremsky
May 23, 2013 12:15V4-116 Ping-pong on a euclidean buildingStefan Witzel | {"extraction_info": {"found_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": 0.7214673161506653, "perplexity_flag": "middle"} |
http://math.stackexchange.com/questions/138958/can-i-apply-rolles-theorem-in-reverse | # Can I apply Rolle's theorem in reverse?
I want to prove the following statement:
Let $f$ and $g$ be two real functions continuous in some interval $[a, b]$ and differentiable in $(a, b)$. If $f' = g'$, then $f(x) = g(x) + c$, where $c$ is a constant.
I thought I could argue like this: Consider $h(x) = f(x) - g(x)$. Knowing that $h' = 0$, we want to prove that $h$ is constant. Pick any $k \in (a,b)$, then $h'(k) = 0$. Therefore (and this is the part I'm not sure about), by Rolle's theorem we can find $d, e \in (a, b)$ such that $d < k < e$ and $h(d) = h(e)$. Since this must be true for any $d, e, k$, $h$ is a constant.
Is this a correct application of Rolle's theorem?
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No, that's not what Rolle's theorem says. What you want here is the mean value theorem. – Chris Eagle Apr 30 '12 at 16:36
Your hypothesis are the wrong way: you want continuity in the closed interval, differentiability in the open interval. – Arturo Magidin Apr 30 '12 at 16:40
Hint. Let $r\in (a,b]$. By the Mean Value Theorem, there exists $c$ in $(a,r)$ such that $f'(c) = (f(r)-f(a))/(r-a)$. Therefore... – Arturo Magidin Apr 30 '12 at 16:41
@ArturoMagidin: I fixed the intervals. – Javier Badia Apr 30 '12 at 17:19
## 2 Answers
No, this is not correct. That is, the converse of Rolle's theorem does not hold (if this is what you're asking). For instance, let $f(x)=x^3$ on $[-1,1]$. Then $f'(0)=0$ but there are not two points $c$ and $d$ in $[-1,1]$ with $c\ne d$ and $f(c)=f(d)$.
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As Chris Eagle said. If $x, y \in (a,b)$ by mean value theorem, we have $$0 = h'(c) = \dfrac{h(x) - f(y)}{x-y}.$$ Hence, $h(x) = h(y)$ and $h$ is a constant.
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Thank you for your answer. While it helped me with the problem, David's answered the point of the question, so I accepted his. Thank you anyway. – Javier Badia Apr 30 '12 at 17:20 | {"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": 32, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9278144836425781, "perplexity_flag": "head"} |
http://mathoverflow.net/questions/68875?sort=oldest | ## How to invert the matrix [n choose 2j - i] ?
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In a certain model of a stat-physics type, one encounters a matrix $$A_n:=\left[\binom{n}{2j-i}\right]_{i,j=1}^{n-1}.$$ The determinant of this matrix (equal to $2^{\binom n2}$) counts the number of all possible configurations, and our understanding of the model would greatly increase if we would know the inverse of this matrix. So the question is,
is there a closed-form expression for the inverse of the matrix $A_n$?
A little more information about the matrix: its eigenvalues are $2^i\colon i=1,\dots,n-1$. The eigenvectors are not orthogonal to each other (and $A_n$ is not a symmetric matrix, for sure), but the vectors corresponding to even $i$'s are orthogonal to the ones which correspond to the odd $i$'s (i.e., the whole space orthogonally decomposes into the "even" and the "odd" parts).
PS I would appreciate any help or reference. I've looked through Krattenthaler's seminal "Advanced determinant calculus", but didn't find such a matrix there.
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1
Did you try using the formula $A^{-1}=\frac{1}{\det(A)}(\text{adjoint of } A)$ for $n=2,3$ (to start with) just to get an idea of what the inverse looks like? – Johan Öinert Jun 26 2011 at 20:32
If you clear the denominators, it looks like the even-odd pattern of the rows is the same as the even-odd pattern of Pascal's triangle. For example, the first column of $2^{19}A_{12}^{-1}$ is $88179, -533860, 1556475, -2842320, 3564470, -3182088, 2034942, -915824, 276471, -50388, 4199$ which has the same pattern $\mod 2$ as $1,10,45,120,210,252,210,120,45,10,1.$ The exponents of $2$ in the prime factorizations are symmetric, but not the same as in Pascal's triangle. – Douglas Zare Jun 26 2011 at 22:37
## 1 Answer
This is more an idea to explore than a complete answer.
You may interpret the binomial coefficient $\binom{n}{k}$ as the elementary symmetric function $e_k$ of $1,1,\ldots,1$ ($n$ variables evaluated at $1$). The coefficients of the adjoint matrix of $A_n$ become skew Schur functions of $1,1,\ldots,1$. Then there may be some further simplifications.
(By the way, this approach gives a nice proof for the value of the determinant of $A_n$: it is the value of the staircase Schur function $s_{(n-1,n-2,\ldots,1,0)}$ evaluated at $1,1,\ldots,1$. Note that the staircase Schur function at $x_1,x_2,\ldots,x_n$ is equal to $\prod_{i \lt j} (x_i+x_j)$).
EDIT: I find that the coefficient $(i,j)$ of the inverse is $(-1)^{i+j} s_{[j]'/(n-i)}(1,1,\ldots,1)/2^{\binom{n}{2}}$, where $[j]$ stands for the partition obtained from $(n-1,n-2,\ldots,1)$ by removing $j$, and $[j]'$ is its conjugate. At this point there is some hope to find a nice formula. First by expressing the skew Schur function as a sum a Schur functions by means of dual Pieri rule: $$s_{\lambda/(k)}=\sum s_{\nu}$$ where the sum is carried over all partitions $\nu$ obtained from $\lambda$ by removing a horizontal strip with $k$ boxes. After that by using the following formula found in Macdonald, I.3. Ex. 4: $$s_{\lambda'}(1,1,\ldots,1)=\prod_{x \in \lambda} \frac{n-c(x)}{h(x)}$$ where the evaluation is at $(1,1,\ldots,1)$ with $n$ ones, $\lambda'$ is the conjugate of $\lambda$ and $h(x)$ and $c(x)$ are the hook length and content respectively of the box $x$ in the diagram of $\lambda$.
Hopefully the formulas simplify.
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5
There is more on this sort of thing in: Symmetric Polynomials, Pascal Matrices, and Stirling Matrices Michael Z. Spivey1, Andrew M. Zimmer (full text seems to be readily available). – Igor Rivin Jun 27 2011 at 1:56
Thank you very much for this idea, I think it could really lead to something. – Leonid Petrov Jun 27 2011 at 6:12
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@Igor: Full text is available via my web site here: math.pugetsound.edu/~mspivey/Symmetric.pdf. – Mike Spivey Jun 27 2011 at 19:57
@Mike: Many thanks for the link! So I have managed to write the inverse in terms of the skew Schur functions as is suggested in the answer. Does your paper give any other, better way of writing the inverse? – Leonid Petrov Jun 28 2011 at 0:47
@Leonid: My apologies for responding so late to your question; somehow I didn't get pinged by your use of "@Mike." All of the matrices in our paper are lower triangular, so the methods may not directly apply. However, one of the core ideas is that elementary and complete symmetric polynomials are inverse to each other, in some sense, and maybe you can adapt that idea to your situation. See, for example, p. 296 of Richard Stanley's Enumerative Combinatorics, Vol. II. – Mike Spivey Oct 31 2011 at 4:18 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 42, "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": 0.918766438961029, "perplexity_flag": "head"} |
http://mathhelpforum.com/calculus/67727-gradient-directional-derivative-way-represent-vector-equations.html | Thread:
1. Gradient and directional derivative way of represent the vector equations
I have question about presentation of directional derivative.
Sometimes I see directional derivates and gradient to be represented as a function but I think that it is convenient to use "vector presentation" instead.
My question is that is there some cases and/or good reasons why should use functional presentation instead of vector?
To make my point clear I wrote an example what I mean about "presentations".
Simply I mean the presentation of vector, using comma $(2,1)$ versus using i and j components $2i + j$
There is one example twice, for "function presentation" and for "vector presentation".
Let we have a function: $f(x,y) = x^2 - y^2$ and we calculate gradient and directional derivative of function $f$.
Function presentation:
Gradient (I include all phases here):
$\nabla f(x,y) = \frac{\partial f}{\partial x} i+ \frac{\partial f}{\partial y}j$
$= \frac{\partial}{\partial x} (x^2 - y^2)i+ \frac{\partial}{\partial y}(x^2 - y^2)j$
$= (2x+0)i - (0-2y)j$
$= \underline{\underline{2xi - 2yj}}$
Next we calculate directional derivative of function f at this point (2,-1) to the direction of $\overline{\imath} - 3\overline{\jmath}$
The vector $\overline{\imath} - 3\overline{\jmath}$ has magnitude $\sqrt{1^2 + (-3)^2} = \sqrt{10}$. The unit vector in the direction $\overline{\imath} - 3\overline{\jmath}$ is thus $\hat{u} = \frac{1}{\sqrt{10}} (\overline{\imath} - 3\overline{\jmath})$
Directional derivative is then
$D_{\hat{u}} f(a,b) = \hat{u} \cdot \nabla f = \frac{1}{\sqrt{10}} (i -3j) \cdot (2xi-2yj)$
$=\frac{1}{\sqrt{10}} [1 \cdot 2x + (-3) \cdot (-2y)] = \frac{1}{\sqrt{10}} (2x + 6y)$
and plugging the point (2,-1) we get
$=\frac{1}{\sqrt{10}} (4-6) = \underline{\underline{\frac{-2}{\sqrt{10}}}}$
Vector presentation
$\nabla f(x,y) = \frac{\partial f}{\partial x} i , \frac{\partial f}{\partial y}j$
$= \frac{\partial}{\partial x} (x^2 - y^2)i , \frac{\partial}{\partial y}(x^2 - y^2)j$
$= (2x,2y)$
Next we calculate directional derivative of function f at this point (2,-1) to the direction of $(1,- 3)$
Let's plugin the point to the Gradient: $\underline{\underline{\nabla f(2,-1) = (4,2)}}$
The vector $(1,- 3)$ has magnitude $\sqrt{1^2 + (-3)^2} = \sqrt{10}$. The unit vector in the direction $\overline{\imath} - 3\overline{\jmath}$ is thus $\hat{u} = \frac{1}{\sqrt{10}} (1, -3)$
Directional derivative is then
$D_{\hat{u}} f(a,b) = \hat{u} \cdot \nabla f = \frac{1}{\sqrt{10}} (1,-3) \cdot (4,2)$
$<br /> =\frac{1}{\sqrt{10}} (4-6) = \underline{\underline{\frac{-2}{\sqrt{10}}}}<br />$
Calculations above are the same no matter when we plugin the point (2,-1).
I think that representing the gradient as a vector is more clear than function.
Now, I will repeat my question:
Is there some cases and/or good reasons why should use functional presentation (i,j components) instead of vector presentation in points (x,y)? | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 27, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": 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": 0.8978284597396851, "perplexity_flag": "middle"} |
http://math.stackexchange.com/questions/94645/expected-smallest-prime-factor | # Expected smallest prime factor
For a random integer $x$ chosen uniformly between 2 and $n$, what is the expected value of the smallest prime factor of $x$ as a function of $n$?
What is the behavior of the function as $n$ tends to infinity?
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## 3 Answers
A quick and dirty answer... (I began before Will answered...)
I first address the following question: what is the probability $\pi_n$ that a number has the n-th prime $p_n$ as smallest prime factor.
• A random number is even with proba ${1\over 2}$, so the smallest prime factor will be 2 with probability ${1\over 2}$.
• An odd number is a multiple of 3 with proba ${1\over 3}$, so the smallest prime factor will be $3$ with proba ${1\over 2\cdot 3}$.
• If it is not dividable by 2 or 3, which happens with probability $1 - {1\over 2} - {1\over 6} = {1\over 3}$, this will be $5$ with proba ${1\over 5}\times{1\over 3} = {2 \over 2\cdot 3 \cdot 5}$.
• Then it will be 7 with proba ${1\over 7} \times (1 - {1\over 2} - {1\over 6} - {1\over 15}) = {1\over 7} \times {4 \over 15} = {8 \over 2\cdot 3 \cdot 5\cdot 7}$.
Denoting this proba by $\pi_n$ and by $p_n$ the sequence of primes, we have $\pi_1 = {1\over 2}$ and $\pi_{n} = {1 \over p_n} \left( 1 - \sum_{i=1}^{n-1} \pi_i\right)$.
Edit I take some time to see the connection between this answer and Will’s. I compute the totient function :
$\phi(2) = 1$, $\phi(2\cdot3) = 1\cdot 2$, $\phi(2\cdot 3\cdot 5) = 1 \cdot 2 \cdot 4$. Denoting $p_n\# = \prod_{i\le n} p_i$, it appears that in the first few terms I get the following : $$\pi_n = {\phi(p_{n-1}\#) \over p_n\#},$$ which is slightly different – but Will is computing an expectation, and he is right cf edit 6 below.
Edit 2 This is logical from the definition of totient function : $\phi(p_{n-1}\#)\over p_{n-1}\#$ is the proportion numbers which are not dividable by $p_1, \dots, p_{n-1}$ ; multiply by $1\over p_n$ to get the proportion of numbers which are not dividable by $p_1, \dots, p_{n-1}$ but dividable by $p_n$.
If one manage to prove by recurrence that the above defined $\pi_n$ coincide with this, the fact that the sum is 1 should be clear.
Edit 3 It is not difficult to complete. If we prove that for all $n$, $$1 - \sum_{i\le n} \pi_i = {\phi(p_n\#) \over p_n\#},$$ we are done. This is true for $n=1, 2$. The induction step is: $$\begin{array}{rcl} 1 - \sum_{i\le n+1} \pi_i &=& \left(1 - \sum_{i\le n} \pi_i\right) - \pi_{n+1} \\ &=& \left(1 - \sum_{i\le n} \pi_i\right)\left( 1 - {1\over p_{n+1}}\right)\\ &=& \left({\phi(p_n\#) \over p_n\#} \right) \left( {p_{n+1} - 1\over p_{n+1}}\right)\\ &=& { \phi(p_n\#) \times (p_{n+1} - 1) \over p_n\# \times p_{n+1} }\\ &=& {\phi(p_{n+1}\#) \over p_{n+1}\#} \end{array},$$ so we are done.
Edit 4 Using Euler’s trick, we have easily that $$1 - \sum_{i=1}^\infty \pi_i = \prod_p \left(1-{1\over p}\right) = { 1 \over \sum_{n=1}^\infty {1\over n}} = 0,$$ which can surely be rewritten respecting the modern standards... I am not familiar with analytic number theory, but I am sure this product is a classic.
Edit 5 Yes, it is a classic, cf Merten’s third theorem, which tells that $\prod_{p\le n} \left(1-{1\over p}\right) \sim {e^{-\gamma}\over \log n}$. Using $p_n \sim n \log(n)$ we get that
$$1 - \sum_{i=1}^n \pi_i = \prod_{p\le p_n} \left(1-{1\over p}\right) \sim {e^{-\gamma}\over \log n + \log\log n} \sim {e^{-\gamma}\over \log n }$$ and $$\pi_n \sim {e^{-\gamma}\over n \log^2 n},$$ which gives you the asymptotic behaviour of this sequence.
Edit 6 In fact I didn’t adress the original question but this is possible now. The smallest prime factor of a number taken uniformly between 1 and $p_n-1$ is $p_k$ with probability $\simeq {\pi_k \over \sum_{\ell<n}\pi_\ell}$. Its expectation is $${ \sum_{\ell < n} p_\ell\pi_\ell \over \sum_{\ell < n} \pi_\ell} \sim \sum_{\ell < n} p_\ell\pi_\ell = \sum_{\ell < n} {\phi(p_\ell\#) \over p_\ell\#},$$ as Will first stated (oh my God, why did that took me so long?). The above equivalent shows that this goes to infinity as $n\rightarrow\infty$. Comparing to an integral leads to a $O\left( {p_n \over \log p_n} \right)$.
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The induction is easy: If $\pi_n=\frac{\phi(p_{n-1}\#)}{p_n\#}$, then $\pi_{n+1}=\frac1{p_{n+1}}\left(1-\sum_{i=1}^n\pi_i\right)=$ $\frac1{p_{n+1}}\left(1-\sum_{i=1}^{n-1}\pi_i-\frac{\phi(p_{n-1}\#)}{p_n\#}\right)=\frac1{p_{n+1}}\left(p_n\pi_n-\frac{\phi(p_{n-1}\#)}{p_n\#}\right)=\frac{(p_n-1)\phi(p_{n-1}\#)}{p_{n+1}\#}=\frac{\phi(p_n\#)}{p_{n+1}\#}$. – Brian M. Scott Dec 28 '11 at 10:57
Well, this is an other way to write it :) – Elvis Dec 28 '11 at 11:59
I posted that before I saw your third edit, or I’d not have bothered. – Brian M. Scott Dec 28 '11 at 12:13
No problem, I kept editing this for a while, it’s my fault. – Elvis Dec 28 '11 at 12:29
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Taking the primorials $$P_0 = 1, \; P_1 = 2, \; P_2 = 6, \; P_3 = 30, \; P_4 = 210,$$ I get your expected value as $n$ increases to $\infty$ as $$E = \sum_{k = 0}^\infty \; \frac{\phi(P_k)}{P_k},$$ wher $\phi$ is Euler's totient function. I'm not sure yet whether this is finite.
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are you saying $E(n) = \sum_k^n \phi(P_k)/P_k$? – user20266 Dec 28 '11 at 8:54
Intuitively, I would expect this to be of the order of $\dfrac{n}{\log_e n}$ on the grounds that it is slightly more than the sum of the primes less than or equal to $n$ divided by $n-1$, and that is slightly less than $n$ times the frequency of primes near $n$.
Empirically this looks reasonable: for $n$ between 32000 and 64000, something like $1.9\dfrac{n}{\log_e n}$ looks fairly close.
- | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 53, "mathjax_display_tex": 9, "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": 0.951667070388794, "perplexity_flag": "head"} |
http://mathhelpforum.com/advanced-algebra/39099-matrix-inverse.html | Thread:
1. matrix inverse
Find the inverse of the given matrix
[2 4 3]
[3 -4 -4]
[5 0 -1]
Can you please teach me the steps of sloving the inverse of this matrix? Thank you very much.
2. Which methods have you learned in class? You can perform row operations on your matrix until you get it into row echelon form and perform the exact same operations on the identity matrix to get your inverse.
Or you can utilize determinants ... Any method in particular?
3. Hi o_O,
I am just in the beginning of linear algebra.
Here is the method that we used :
1) form the augmented matrix [A l I sub n] and Use elementary row operations.
I saw the answer of this question in the book is " does not exist" . I don't know how to reach this answer from the given question.
4. $\left[ \begin{array}{ccc}2 & 4 & 3 \\ 3 & -4 & -4 \\ 5 & 0 & -1 \end{array} \Bigg| \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array} \right]$
Now, I presume you understand that whatever operations you do to reduce A to the identity matrix, you will get A-inverse if you perform those exact same operations to the identity matrix.
In order for an inverse to not exist, a row of 0's must occur in your augmented matrix after performing some row operations as this will prevent any possibilities of achieving your identity matrix. For example, [0 0 0] -> [0 0 1] you can't achieve this via elementary row operations.
If we just focus on trying to reduce A to the identity matrix, we will wind up with exactly that - a row of zeroes:
$\left[ \begin{array}{ccc}2 & 4 & 3 \\ 3 & -4 & -4 \\ 5 & 0 & -1 \end{array}\right]$ ${\color{white}.} \quad \stackrel{R_{1} \iff R_{3}}{\longrightarrow}$ ${\color{white}.} \quad \left[ \begin{array}{ccc}5 & 0 & -1 \\ 3 & -4 & -4 \\ 2 & 4 & 3 \end{array}\right]$ ${\color{white}.} \quad \stackrel{R_{3}' = 3R_{3} -2R_{2}}{\longrightarrow}$ ${\color{white}.} \quad \left[ \begin{array}{ccc}5 & 0 & -1 \\ 3 & -4 & -4 \\ 0 & 20 & 17 \end{array}\right]$
$\stackrel{R_{2}' = 5R_{2} - 3R_{1}}{\longrightarrow} \left[ \begin{array}{ccc}5 & 0 & -1 \\ 0 & -20 & -17 \\ 0 & 20 & 17 \end{array}\right]$
Now whatever you do with this last matrix, you will wind up with a row of 0's and will thus have no inverse. You can see more here (scroll down to Example 3 for a similar example)
5. Hello,
Or you can just say that the determinant is 0, therefore the matrix has no inverse.
6. Of course but the OP was asked to determine the matrix's invertibility by means of elementary row operations. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 7, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9312524199485779, "perplexity_flag": "head"} |
http://mathoverflow.net/questions/73553?sort=newest | ## When may Function (meromorphic) be expanded as power series with coefficients of integers
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Let F be meromorphic function,with what properties may it expanded as power series with coefficients of integers in such a form:
$$F=\sum_0^{\infty}a_i x^i,a_i\in \mathcal{N} \bigcup 0,\exists M \space a_i \leq M^i$$.
and when the coefficients consist of a sequence of computably enumerable relation.
If the question is ambiguous ,please tell me but please do not downvote it.
When may Function (meromorphic) be expanded as power series with coefficients of integers
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Without specifying what kind of properties you are asking for, this seems too open-ended for me. How about "F has the property that it may be expanded as a power series with integer coefficients"? Perfectly well-defined property. – Yemon Choi Aug 24 2011 at 10:08
I assume you mean on the whole complex plane. The power series coefficients come from C, so there are very few meromorphic functions with integer coefficients. Certainly 1/z^k * f(z) for f(z) = e(z), sin(z), cos(z), but it's hard to say something specific. Can you expound on your question? – Robert K Aug 24 2011 at 10:12
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@Robert: A power series with integer coefficients can never converge on the whole complex plane, unless it is a polynomial. Indeed, the Cauchy–Hadamard theorem implies that a series with integer coefficients, infinitely many of which are nonzero, has radius of convergence at most $1$. – Emil Jeřábek Aug 24 2011 at 10:31
3
A sufficient condition is: Write the meromorphic function as quotient $f/g$ of homolomorphic functions. If the powerseries of $f$ and $g$ (around the origin) have integral coefficients and $g(0) = 1$ holds, than the powerseries of the meromorphic function will also have integral coefficients. – Ralph Aug 24 2011 at 10:52
1
@Robert: Can you please clarify what $1/z^k*f(z) ...$ means.For instance, $z^{-1}\cdot sin(z)$ doesn't epand with integral coefficients. – Ralph Aug 24 2011 at 11:04
show 2 more comments
## 1 Answer
This was a large research subject in 1930-s. The key authors are G. Polya, Ch. Pisot and Raphael Robinson. The book of Bieberbach, Analytische Fortsetzung (in German, there is a Russian translation) contains a chapter with a survey of these results.
The general spirit of these results is the following: if you have a Taylor series with integer coefficients which has an analytic or meromorphic continuation in sufficiently large region, then the function must be rational, and in certain cases all such functions can be explicitly described. But there are too any results to mention them here.
By the way, the question is equivalent, via Borel-Laplace transform to a question about entire functions which take integer values at positive integers. So "Integer-values entire functions" is just another name of the same topic.
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@Alexandre Thank you,but I don't know German nor do I know Russian.The results are so interesting. – XL Aug 6 at 5:24
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Pisot's papers are in French and Robinson in English. – Alexandre Eremenko Aug 6 at 8:02
@Alexandre，Thank you ,I highly appreciate your answer – XL Aug 6 at 15:05 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 7, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": 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": 0.9142035841941833, "perplexity_flag": "middle"} |
http://harrisonbrown.wordpress.com/tag/math-nt/ | # Portrait of the Mathematician
Mathematics, statistics, philosophy, and way too much pop culture
## Posts Tagged ‘math.NT’
### The importance of choosing the right model
December 13, 2009
I’ve had the idea for this post bouncing around in my head for several months now, but now that I don’t have classes to worry about I can finally get around to writing it. I want to talk about some “pathological examples” in computer science, in particular in complexity and computability theory.
I wanted to post a picture of William H. Mills with this post, in the spirit of Dick Lipton’s blog. Unfortunately I can’t find one! Mills was a student of Emil Artin in the ’40s; he finished his thesis in 1949 and promptly (as far as I can tell) disappeared until the ’70s, when we find some work by a William H. Mills on combinatorial designs. After this, there’s nothing of note until Dr. Mills passed away several years ago at the age of 85.
While his work (assuming it’s his) on combinatorial designs looks interesting, W.H. Mills’ small place in history is assured by a short and unassuming paper from 1947, published when he was still a student. In this paper he showed that there’s a constant, which he called A, such that $[A^{3^n}]$ is prime for all integers $n \geq 1$. Nowadays it’s called Mills’ constant.
(more…)
Tags:cs.CC, graph theory, math.CO, math.LO, math.NT, prime numbers, wild speculation
Posted in expository, open questions | 3 Comments »
### GILA 4: The pseudorandom case
September 19, 2009
So in the last post, we defined the discrete Fourier transform and gave some of its basic properties. At the end, we claimed that it gives us a simple notion of pseudorandomness that allows us to make rigorous the intuition that “pseudorandom subsets of $\mathbb{Z}/N\mathbb{Z}$ should have many arithmetic progressions.” Today we’re going to justify this notion of pseudorandomness, and work through this — the easy case — of the proof of Roth’s theorem for $\mathbb{Z}/N\mathbb{Z}$. At the end, we’ll sketch how to modify the method for the regular, finitary version of the theorem.
(more…)
Tags:DFT, GILA, math.CO, math.NT, probabilistic method, pseudorandom, randomness, roth's theorem
### GILA 3: The Fourier transform, and why it works
September 3, 2009
At the end of the last post, we outlined an approach to proving Roth’s theorem via a density-increment method. I’ll re-post it here, a little more fleshed out, for convenience.
Sketch. The idea is to deal with “structured sets” (like sets that are themselves long arithmetic progressions) and “pseudo-random sets” in different ways; we saw last time that the density-increment argument looks promising for structured sets, but “random” sets won’t succumb to it.
First, we need to make rigorous our intuition that “random enough” sets of positive density will have length-3 APs. The obvious way to do this is by showing the contrapositive. We want to find some suitable “randomness measure” such that random dense sets have high randomness measure. Now let be a positive-density set without 3-APs; we want to show that this condition is enough to guarantee that has low randomness measure.
Now for our modified-probabilistic argument. We want to show that “low-randomness” sets can’t be uniformly distributed (mod d) for every d; this guarantees the existence of an AP on which it has higher density. We pass to this AP; rinse, lather, and repeat, until our density rises above some pre-set bound ($\delta > 2/3$ will do nicely.) We also want to introduce a tool that tells us how much a set “looks like” an arithmetic progression with common difference d; this is because we need to obtain quantitative bounds for the density-increment argument, in order to ensure that the densities of our set on the sequence of APs don’t monotonically approach, say, 0.0000001.
So we’re looking for two technical tools; one that quantifies the “randomness” of a set, and one that lets us correlate our set with a candidate arithmetic progression to pass to so that our density-increment argument will actually work quantitatively. As it turns out, though, we can find one tool that will do both jobs for us; the discrete Fourier transform.
(more…)
Tags:DFT, fourier transform, GILA, math.CA, math.CO, math.NT, roth's theorem
Posted in expository, GILA | 1 Comment »
### GILA 2: The probabilistic approach
August 27, 2009
To begin, a historical curiosity. In the last post, I talked about Khinchin’s little book on number theory, which from today’s perspective is one of the earliest books to deal entirely with additive combinatorics and additive number theory. One of Khinchin’s most famous results has to do with the denominators of continued fractions; Khinchin’s original proof of the theorem was quite complicated and involved. Nowadays, the existence of Khinchin’s constant is understood as a simple consequence of the ergodicity of the Gauss-Kuzmin-Wirsing operator.
Where else do we see ergodic theory popping up to simplify complicated proofs? Why, in additive combinatorics! Although we won’t discuss it in this series, ergodic theory has become a central tool for Szemeredi-type theorems, and some results are still only provable by such methods (for instance, until recently, the density Hales-Jewett theorem was in this class.)
To top it all off, note that ergodic theory was originally motivated by problems in statistical physics. Guess who wrote one of the best-known textbooks on mathematical statistical physics? But on to the math!
We’ve seen that traditional arguments based on the pigeonhole principle and such aren’t sufficient to prove Roth’s theorem. So we’ll go to the next tool in our kit — the probabilistic method.
The probabilistic method is one of the most useful weapons in a combinatorialist’s arsenal, even having a whole textbook devoted to it. (The linked page is out of date; a third edition was published last year.) The core of the method is this — if we want to show the existence of an object with a given property, instead of constructing one directly, we instead show that a randomly chosen large enough object has the property with positive probability. Its best-known use in Ramsey theory is for proving lower bounds, where it often gives the best known results, but clever applications can give upper bounds as well, for instance in the proof of Kraft’s inequality. Of particular interest to us is the fact that it provides our first nontrivial proof in the direction of Roth’s theorem.
(more…)
Tags:math.CO, math.NT, math.PR, probabilistic method, roth's theorem
Posted in expository, GILA | 3 Comments »
### GILA 1: van der Waerden and all that
August 22, 2009
I’m beginning my GILA series announced on Thursday with a short exposition of van der Waerden’s theorem, and an attempt to extend its proof to the density version (at least in the k=3 case). Before I start, I should mention something I neglected in the last post: that this series is as much for my benefit as for that of any “interested lay audience” out there. Someone like Terry Tao could undoubtedly do a better series of this type (and, indeed, I’ll be leaning heavily on my borrowed copy of Tao and Van Vu’s Additive Combinatorics throughout) but they haven’t, so I’ll try. However, there are certain to be a number of embarrassing mistakes, unjustified statements, lines of thinking left unresolved, etc., throughout (indeed, an anonymous commenter noted two in the introductory post!), and I encourage readers to point them out in the comments.
So, on to van der Waerden’s theorem. van der Waerden’s theorem is one of the early jewels of Ramsey theory (actually predating Ramsey’s article by a couple of years) and can be viewed as a weaker version of Szemeredi’s theorem. It states simply that, for any integers $c, k > 0$, if you color the positive integers with c colors, then there exists a monochromatic k-term arithmetic progression.
Khinchin, in his wonderful book which unfortunately appears to be out of print (at least in English), gives some history of the problem:
All to whom this question [the theorem in the c = 2 case] was put regarded the problem at first sight as quite simple; its solution in the affirmative appeared to be almost self-evident. The first attempts to solve it, however, led to nought… [T]his problem, provoking in its resistance, soon became the object of general mathematical interest… I made [van der Waerden's] acquaintance, and learned the solution from him personally. It was elementary, but not simple by any means. The problem turned out to be deep; the appearance of simplicity was deceptive.
This last clause is particularly wonderful, describing as it does not only van der Waerden’s theorem but Ramsey theory and perhaps even combinatorics in general.
The usual proof (given in story-form by Zeilberger here) is not van der Waerden’s, however, but comes from Khinchin, attributed by him to a M.A. Lukomskaya in Minsk. It is a masterpiece, using nothing but a very clever inductive argument and some application of the pigeonhole principle. For the sake of self-containedness, I’ll sketch the argument below the fold.
(more…)
Tags:GILA, math.CO, math.NT, pigeonhole principle, roth's theorem, van der Waerden
Posted in expository, GILA | 1 Comment »
### GILA 0: Roth’s proof of Roth’s theorem
August 20, 2009
I’m returning from my unplanned, unannounced hiatus from blogging with a series on the standard (Fourier-theoretic) proof of Roth’s theorem on arithmetic progressions, otherwise known as the k = 3 case of Szemeredi’s theorem. I’d like to start out by noting that there’s no shortage of good expositions on the topic; a quick Google search will turn up several. This series will differ from these expositions, however, in two key ways.
First, I’ll try to highlight the key ideas of the proof, rather than get bogged down in the calculations. Although, by the end of the series, we should have at least a sketch of a complete proof of Roth’s theorem, I want to make it easy to separate the real insights from the straightfoward calculational machinery. So I’ll likely present a high-level overview of the proof in the main body of these posts, with one or several appendices fleshing out the computations needed to ensure that the overview works.
Second, and more importantly, this is to be a piece of historical fiction. Although I don’t know how Roth came upon his original proof, he was a number theorist and an expert in Diophantine approximation, and was therefore familiar with things like Fourier analysis. I want to approach the problem from the perspective of a combinatorialist roughly contemporaneous with Roth, who’d be familiar with standard combinatorial methods but not with some of the more technical tools, and attempt to re-derive the proof from that viewpoint.
The real meat of the series will start this weekend (which is why this post is numbered 0), but since this is targeted toward a Generally Interested Lay Audience, I’d like to introduce a few topics which a layman might not be familiar with below the cut.
(more…)
Tags:GILA, math.CO, math.NT, roth's theorem
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http://en.wikibooks.org/wiki/Pictures_of_Julia_and_Mandelbrot_Sets/The_Mandelbrot_set | # Pictures of Julia and Mandelbrot Sets/The Mandelbrot set
Interesting Julia set
Ugly Julia set
In appearance a Julia set can go from one extreme to the other. And if we have a family of functions containing a complex parameter c, we will observe that by far the majority of c-values the Julia set is completely without interest. In fact, the attractive Julia sets are extremely rare.
And these Julia sets are just found by considering a family of iterations and from this constructing a set in the plane that can serve as an atlas of the Julia sets, in the sense that if we find an interesting locality in this set, we can be certain that some part of the pattern at this place will be reflected in the (self-similar) structure of the Julia sets associated to the points here. Such a set is called a Mandelbrot set.
Therefore, if we have a function $f(z)$, we introduce a complex parameter c in it, usually by addition: $f(z) + c$.
## Contents
### Construction of the Mandelbrot set
The construction of the Mandelbrot set is based on the choice of two critical points $zc_1$ and $zc_2$ for the function $f(z)$: The Mandelbrot set (associated to the family $f(z) + c$ and the critical points $zc_1$ and $zc_2$) consists of the complex numbers c, such that the sequences of iteration (by $f(z) + c$) starting in $zc_1$ and $zc_2$, respectively, do not have the same terminus. This set is usually coloured black.
### Colouring the domain outside the Mandelbrot set
That a point c is lying outside the Mandelbrot set, means that the second critical point $zc_2$ is lying in the same Fatou domain (for the iteration $f(z) + c$) as the first critical point $zc_1$, and we can give c the colour of the point $zc_2$ in this Fatou domain.
In order to draw the Mandelbrot set and colour the domain outside it, we must have chosen a maximum iteration number M, a very small number $\epsilon$ (for iteration towards a finite cycle) and a very large number N (for iteration towards ∞).
If $zc_1$ = ∞ (and d ≥ 2, so that ∞ is a critical point and a (super-attracting) fixed point) we need of course not iterate $zc_1$: we iterate $zc_2$ (by $f(z) + c$) and if $|z_k|$ > N for some iteration number k < M, then c is lying outside the Mandelbrot set, and we colour c in the same way as we have coloured a z in a Fatou domain containing ∞. If we have reached the maximum iteration number M, we regard c as belonging to the Mandelbrot set.
If $zc_1$ is a finite critical point and if the iteration of $zc_1$ (by $f(z) + c$) is running to the maximum number of iterations M is reached, the terminus is most probably a finite attracting cycle that is not super-attracting (if not, there can be a fault in the colour of the pixel, but this is without significance in practice). If the last point of this iteration is z*, z* belongs to the cycle, but we must know the order and the attraction of the cycle. Therefore we continue the iteration: starting in z* and running until $|z_k - z*| < \epsilon$, then the number of iterations needed for this is the order r of the cycle, and we calculate the attraction $\alpha$ in the same way as before: 1/$\alpha$ is the product of the numbers $|f'(z_i)|$ for the r points of the cycle. We hereafter iterate $zc_2$ (by $f(z) + c$), and stop when $|z_k - z*| < \epsilon$. If this iteration runs until the maximum number of iterations M is reached, we regard c as belonging to the Mandelbrot set. If $|z_k - z*| < \epsilon$ for k < M, we colour c according to k, or rather, the corresponding real iteration number, which is found in the same way as for a Fatou domain, by dividing k by r (and taking the integral part) and from this number subtract $log(\epsilon/|z_k - z*|)/log(\alpha)$.
If the cycle contains ∞, that is, if the iteration of $zc_1$ is stopped by $|z_k|$ > N for k < M, we let ∞ be the chosen point of the cycle, and we continue the iteration until we again have $|z_k|$ > N, then the number of iterations needed to do this is the order of the cycle. We then iterate $zc_2$ (by $f(z) + c$), and stop when $|z_k|$ > N. If this iteration runs until the maximum number of iterations M is reached, we regard c as belonging to the Mandelbrot set. If $|z_k|$ > N for k < M, we colour c according to k, or rather, the corresponding real iteration number, which is found in the same way as for a Fatou domain, by dividing k by r (and taking the integral part) and from this number subtract $log(log|z_k|/log(N))/log(|d|^{r})$.
### Colouring the boundary of the Mandelbrot set
That a point c is lying outside the Mandelbrot set, means that the second critical point $zc_2$ is lying in the same Fatou domain (for the iteration $f(z) + c$) as the first critical point $zc_1$, and the estimation of the distance from $zc_2$ to the Julia set, in this Fatou domain, is an estimation of the distance from c to the boundary of the Mandelbrot set. So, the boundary of the Mandelbrot set can be coloured in the same way as a Julia set, but now the derivative of $z_k$ is not with respect to z, but with respect to c.
If we set $g(z) = f(z) + c$, we have $z_k = g(g(...g(z)))$ (the k-fold composition)(the start value z is first $zc_1$ and then $zc_2$), and we find the derivative $z'_k$ of $z_k$ with respect to c by recursion: we have $z'_{k+1} = f'(z_k)z'_k + 1$, and we find $z'_k$ successively by performing this calculation for each iteration, starting with $z'_0$ = 0, together with (and before) the calculation of the next iteration value $z_{k+1} = f(z_k) + c$, starting with z = $zc_1$ and $zc_2$, respectively.
As well as finding the point z* in the cycle by iterating $zc_1$ M times, we now also calculate the derivative z*' of z* with respect to c, and when iterating $zc_2$ towards the cycle, we now also calculate the derivative $z'_k$ of $z_k$ with respect to c. The formulas for $\delta(z)$ are for the two cases:
$\delta(z) =$limk→∞$|z_{kr} - z*|/|z'_{kr} - z*'|$ (non-super-attraction)
$\delta(z) =$limk→∞$log|z_k||z_k|/|z'_k|$ (d ≥ 2 and z* = ∞)
When the value of this number for the last iteration number is smaller than a given small number, we colour the point c dark-blue, for instance.
### Why the Mandelbrot set serves as an atlas of the Julia sets
If we choose a point c near the boundary of the Mandelbrot set, then the Julia set for $f(z) + c$ will have a (self-similar) structure that has some features in common with the Mandelbrot set at that locality. In the simple case $f(z) = z^{2} + c$ (the usual Mandelbrot set), the structure of the Julia set for c is exactly the same as the local structure of the Mandelbrot at c, but this is usually not the case for general rational functions, only that the structure of the Julia set reflects the local structure of the Mandelbrot set.
Why is this? When c is inside the Mandelbrot set, the sequence generated by $zc_2$ does not converge to the terminus of the sequence generated by $zc_1$, and this means that the two Fatou domains containing $zc_1$ and $zc_2$, respectively, are different. But when we let c pass over the boundary of the Mandelbrot set, the two sequences now have the same terminus, so that the two Fatou domains become identical. Because one of the Fatou domains has now disappeared, we can infer that the Julia set for $f(z) + c$ must change in a significant way (it becomes less connected).
It is only when c is near the boundary of the Mandelbrot set that we can predict something about the Julia set, but as there usually are several critical points, we can choose another pair and draw a new Mandelbrot set. Note that if we use two finite critical points and if we invert these, then the black is unaltered, but the colouring and the boundary can alter: the colour is determined by the value in $zc_2$ of the potential function of the Fatou domain for c containing $zc_1$. In order to get the most aesthetic colouring, we must use the value of the potential function in one and the same point (the second critical point) as c varies. When c passes the boundary of the Mandelbrot set, a Fatou domain disappears, but it is only when the second critical point leaves the Fatou domain, that we get the natural colouring and the boundary.
### The usual Mandelbrot set
For the family $f(z) = z^{2} + c$, there are two critical points, 0 and ∞, and therefore only one Mandelbrot set. This set consists of the points c such that the sequence generated by 0 (by $z^{2} + c$) remains bounded. For c outside the Mandelbrot set the sequence converges to ∞, and we can colour according to the number of iterations needed to bring the points outside a large circle with centre in origo. If we only colour according to the iteration number and if we do not draw the boundary, this circle needs only to have radius 2.
For this family, the Julia set for c has two Fatou domains when c is inside the Mandelbrot set, and one when c is outside. When c is inside the Mandelbrot set, the Julia set is connected, and when c is outside, the Julia set is disconnected (and more than that: totally disconnected - a dust cloud - because of the self-similarity). For c belonging to the boundary, the Julia set is connected, but it does not enclose an interior Fatou domain (this can be regarded as degenerated): the Julia set is just a fractal line with a "nose" and a "tail" and a "spine" connecting these two points.
The usual Mandelbrot set consists of an infinite system of cardioids and circles, all lying outside each other and some touching. When we zoom in, we find a swarm of mini-mandelbrots. Such mini-mandelbrots (possibly deformed) appear in the Mandelbrot set for every complex (differentiable) function, even for transcendental functions (see the picture in the section Julia and Mandelbrot sets for transcendental functions).
### The different types of Mandelbrot and Julia sets
$(z^{2} + z^{4}/2)/(2 - z^{2}/20) + c$
$(1 + 3z^{4})/(4z^{3}) + c$
$(1 - z^{2})/(z - 0.01z^{2} + 0.004z^{3}) + c$
Of all Mandelbrot sets the usual is the one that possesses most localities of beauty. All other Mandelbrot sets are more or less ugly in their entirety, especially when the function is not a polynomial. In return, it is in such Mandelbrot sets that we can be lucky enough to find the most interesting and original shapes.
When we draw the Mandelbrot set for different rational functions, of course some types of shape will recur, and it should be possible to classify these shapes. We cannot refer the any work in this direction, we can only state the most elementary differentiation:
1. d > 1 (m > n + 1). Then ∞ is a critical point and a super-attracting fixed point, and we usually use this as the first of the two critical points. For $f(z) = (z^{2} + z^{4}/2)/(2 - z^{2}/20) + c$ (and critical point 0), we can find this motif in the Mandelbrot set (first picture):
2. d = 1 (m = n + 1). In this case $f(z)$ is usually constructed from the Newton procedure for solving an equation $g(z) = 0$: $f(z) = z - g(z)/g'(z)$. The critical points are just the solutions to $g(z) = 0$, and we choose two having the largest distance from each other. For $g(z) = z^{4} - 1$ and thus $f(z) = (1 + 3z^{4})/(4z^{3})$, we can find this motif in the Mandelbrot set (second picture):
3. d < 1 (m < n + 1). In this case we usually use two finite critical points, and as the critical points are lying symmetrically around the x-axis (if $f(z)$ has real coefficients), we let the pair consist of conjugate numbers (of largest distance). We let the family be $f(z) = (1 - z^{2})/(z - 0.01z^{2} + 0.004z^{3}) + c$, and we zoom in at the place where the most interesting things seem to be (third picture). We choose three points on the boundary and draw their Julia sets. First a point on the thin tangent line passing through the sea horse valley. Then a point in one of the holes inside the upper black. The last point presupposes that we invert the critical points, so that we can see a part of the boundary that is not visible on this picture of the Mandelbrot set. This boundary forms a continuation downwards of the indicated vertical line in the centre.
• One Fatou domain
• Two Fatou domains
• One Fatou domain
### The parameter c
In order to get a family of iterations from the rational function f(z), we have here simply added the parameter c to f(z), so that the family is z → f(z) + c. This way is the simplest, and as we can get every Julia set in this way and as a Mandelbrot set locally is like a Julia set and as the type of the Mandelbrot sets constructed from families of the form z → f(z) + c is satisfying in every way, there is no strong reason for letting c enter in a more sophisticated way. But we can find interesting Mandelbrot sets by letting c enter in a specific way. We can transform the Mandelbrot set by replacing the family by z → $h_1(c)f(z) + h_2(c)$, for some functions $h_1(z)$ and $h_2(z)$, and we can construct Mandelbrot sets whose form in their entirety differ from those of the usual type by letting c appear in the coefficients of the rational function f(z).
Now the family is of the form z → g(z, c), and we assume that g(z, c) is rational in both z and c. The critical points can now depend on c (we denote the two chosen $zc_1(c)$ and $zc_2(c))$, and in order to draw the boundary we must (besides $g'(z, c) = \partial g(z, c)/\partial{z}$) calculate the derivative of g(z, c) with respect to c: (d/dc)g(z, c) (= $\partial g(z, c)/\partial{c}$ - see the section Terminology). And we must also calculate the derivative with respect to c of the critical points $zc_1(c)$ and $zc_2(c)$: $dzc_1(c)/dc$. The iteration is given by $z_{k+1} = g(z_k, c)$ (starting in $zc_1(c)$ and $zc_2(c)$), and the derivative $z'_k$ (with respect to c) is calculated by
$z'_{k+1} = z'_k g'(z_k, c) + (d/dc)g(z_k, c)$
starting with $z'_0 = dzc_1(c)/dc$ and $dzc_2(c)/dc$.
We let f(z), $h_1(z)$ and $h_2(z)$ be respectively $z^5$, -iz and 1, and $z^2$, i(z - 1/z)/√2 and i(z + 1/z)/√2 (first and second picture below). In the last case the four parts meet in the points ±(√2 ± i√2)/2 which correspond to the points ±i belonging to the usual Mandelbrot set.
Let us now assume that g(z, c) is the rational function given by Newton iteration for the polynomial $f(z) = (z - 1)(z^2 - c)$, that is, $g(z, c) = z - f(z)/f'(z) = (2z^3 - z^2 - c)/(3z^2 - 2z - c)$. The critical points are the solutions to the equation g'(z, c) = 0, and as $g'(z, c) = f(z)f^{(2)}(z)/f'(z)^2$, the critical points are the roots of f(z) = 0 (in our case 1 and ±√c) and the roots of $f^{(2)}(z) = 0$ (in our case 1/3). As a critical point which is a root of f(z) = 0 is a fixed point for g(z, c), the second of the critical points used in the construction of the Mandelbrot set (that is, $zc_2(c)$) must be one of the roots of $f^{(2)}(z)$. In our concrete case we construct the Mandelbrot set from only one critical point: $zc_1(c) = zc_2(c) = 1/3$, so that only the boundary of the Mandelbrot set is drawn (third picture).
We have mentioned above that if the terminus of $zc_1$ is not ∞, then it is most likely a finite attracting cycle that is not super-attracting, and if not, there can be a fault in the colour of the pixel, but this is without significance in practice. However, in our just mentioned case where g(z, c) comes from Newton iteration, all the iterations (or rather, all the Fatou domains) outside the Mandelbrot set are super-attracting, therefore we must correct the formulas for the real iteration number and for the distance function by introducing log in the formulas.
• The Mandelbrot set for $-icz^5 + 1$
• The Mandelbrot set for (i(c - 1/c)/√2)$z^2$ + (i(c + 1/c)/√2)
• The Mandelbrot set for Newton iteration for $(z - 1)(z^2 - c)$
### The drawing Mandelbrot and Julia sets in practice
A Julia set for a rational complex function is so well-defined and natural that, like with some other mathematical concepts, we are inclined to say that it belongs to nature: if they have computers in another world, they will also definitely have Julia sets. Also the definition of a Mandelbrot set is simple and obvious, and the drawing procedure must necessarily be in this way: we enter the coefficients of the two polynomials in some way, and then either some pairs of critical points are found automatically or a pair is chosen graphically by clicking in a picture where all the critical points are shown. Hereafter the Mandelbrot set appears, and we can zoom in and alter the colouring. We go to the Julia sets by pressing a key so that the point in the centre of the window can be moved by the arrows, and when we have chosen a point (usually on the boundary of the Mandelbrot set), the procedure for the Julia set is exactly the same as that for the Mandelbrot set.
When you draw a large picture, you ought to draw it at least twice as large as intended, and then reduce it so that the boundary is no longer only one colour. This will lessen the often sharp character of the boundary and it will remove dots arising from impossible calculations. | {"extraction_info": {"found_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": 122, "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": 0.9264880418777466, "perplexity_flag": "head"} |
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http://gilkalai.wordpress.com/2008/07/06/from-helly-to-cayley-iv-probability-questions/?like=1&source=post_flair&_wpnonce=8cb43e0c56 | Gil Kalai’s blog
## From Helly to Cayley IV: Probability
Posted on July 6, 2008 by
I decided to split long part III into two parts. This (truly) last part of this series deals with probabilistic problems and with combinatorial questions regarding higher Laplacians.
### 21. Higher Laplacians and their meanings
Our high dimensional extension to Cayley’s theorem reads:
$\sum |H_{d-1}(K,{\bf Z})|^2 = n^{{n-2} \choose {d}},$
The right hand side of our formula corresponds to the eigenvalues of a higher Laplacian of a complete d-dimensional complex with n vertices. In several general cases there are nice expressions for these eigenvalues - for matroidal complexes, for shifted complexes, for complete skeleta of the cubes and in other cases. There are also nice general high-dimensional matrix-tree theorems.
If $C_k(K)$ is the space of k cycles and we identify it via a certain inner product with the space of k-cocycles, then we can talk about the Laplacian defined by latex $\delta \partial+ (-1)^k \partial \delta$ where $\partial$ is the boundary operator from $C_k$ to $C_{k-1}$ and $\delta$ is the coboundary operator from $C^k$ to $C^{k+1}$.
Spectra of graphs’ Laplacians are very important, for understanding random walks on graphs, for expansion properties and they are also related to many graph parameters like the diameter. The recent series of posts on Luca Trevisan’s blog give a detailed description of these connections (See also this post on James Lee’s blog.) What about higher Laplacians? (Those do not correspond to connectivity but to higher homology groups.)
What is the analog of the random walk interpretation of the spectral gap?
What is the analog of the relation between the spectral gap and expansion properties?
What is the analog of the diameter?
### 22. Probabilistic questions
Consider the class G of all d-dimensional simplicial complexes on n labelled vertices with ${n-1} \choose {d}$ d-faces and a complete (d-1)-skeleton. Is there a substantial probability, for d>1, that a random complex in G with the property that every (d-1)-face is contained in a d-face (no isolated (d-1)-faces) is Q-acyclic? This is not the case for graphs (d=1). The probability for a graph with n vertices and n-1 edges to be a tree tends to 0 even if it has no isolated vertices. This question has a similar flavour to results regarding singularity of random matrices with 0,1 entries.
Here are other natural probabilistic questions that go back to my old paper.
Inside G consider
A) The class of collapsible complexes
B) The class of contractible complexes
C) The class of Z-acyclic complexes
D(p)) The class of Z/p-acyclic complexes
E) The class of Q-acyclic complexes
For d=1 (graphs) all classes A-E are the same. For d>2 the class C equals the class D.
We can ask if for d >1 as n tends to infinity, a random complex in E is almost surely not in D(p), and a random complex in D(p) is almost surely not in C, and a random complex in C is, for d=2, almost surely not in B, and a random complex in B is almost surely not in A.
Let me also mention that there are several recent results about random simplicial complexes by Linial and Meshulam and by Babson Hoffman and Kahle.
### 23. Russell Lyons problem: Generating random hypertrees
Find a way to generate random Q-acyclic spanning subcomplexes L of a d-dimensional simplicial complex K (with the distribution given by $|H_{d-1}^2 (L)|^2$). One way to do it is to choose a d-face based on the ratio between the weighted number of hypertrees containing and not containing this face. But we want to do something else – mimicking the beautiful ways of Broder-Aldous and of Wilson to generate a random spanning tree.
This question was raised by Russell Lyons in the context of studying determinental probability measures. (You can read more about determinental processes in this post of Terry Tao, and this survey paper by J. Ben Hough, Manjunath Krishnapur, Yuval Peres, and Bálint Virág.)
### 24. Diversion: the amazing algorithms of Aldous-Broder and of Wilson to generate random spanning trees.
The Aldous Broder algorithm goes as follows: Start a random walk on the edges of the graph and add to your tree any edge visited which does not create a cycle, the distribution on the resulting spanning trees is uniform.
The Wilson algorithmgoes as follows: Mark a vertex as a root. (at a later stage the root will be a subtree.) Choose an arbitrary vertex not in the root and make a random walk until hitting the root. Delete all cycles created in this walk and add the remaining path to the root. Returning this process also leads to a uniform random spanning tree.
The paths obtained by Wilson algorithm are called loop erased random walk; we can regard them as a certain random 1-cycle of whose boundary is a presecribed 0-cycle. Something analogous in higher dimension is quite desirable.
### Like this:
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### 6 Responses to From Helly to Cayley IV: Probability
1. Igor Carron says:
And Jame Lee’s entry would be:
http://tcsmath.wordpress.com/2008/06/15/eigenvalue-multiplicity-and-growth-of-groups/
Igor.
2. Gil says:
Thanks, Igor!
3. Dominic Dotterrer says:
Larry Guth (possibly inspired by M. Gromov and A. Naor) has convinced me that higher expansion constants can be interpreted in the following way:
For a graph, we observe the infimum (over all rectilinear maps to the line) of the maximum (over points in the line) of the number of edges intersecting the preimage of a point in the line.
This is of course, a measure of connectivity closely related to the cheeger/expansion constant.
For a complex, take again infimum (over all rectilinear maps to R^k) of the maximum (over points in the R^k) of the number of of k-faces intersecting the preimage of a point in the affine space.
For the same reason, this constant is a measure of k-connectivity. Thus it should measure “closeness” to a phase-change in the kth cohomology. For this reason, estimates of this constant should be possible using the first positive eigenvalue (i.e. closeness to a phase-change in the dimension of the kernel of the Laplace-Beltrami). But, admittedly, I haven’t yet completed this estimate, though for moral reasons, I am confident it can be done.
Actually writing out higher Laplacians for simplicial complexes shows that \$\delta d\$ acts as a random walk operator on the k-th homology of k-skeleton on the complex. This indicates that a suitable continuous analog could be accomplished by analyzing brownian motion on the loop space of a riemannian manifold (as opposed to getting estimates of the zeroth Laplacian by analyzing brownian motion on the manifold itself).
I am very glad for your post; these things seem to be on my mind quite a bit these days.
Thanks,
Dominic
4. Gil Kalai says:
Thank you, Dominic! I think that a formal theory relating high Laplacians to “hyper” expansion (related to k-connectivity) and to various random walks will be a great progress.
5. Pingback: Plans and Updates « Combinatorics and more
6. Pingback: A Beautiful Garden of Hypertrees « Combinatorics and more
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http://researchinpractice.wordpress.com/tag/galois/ | # Research in Practice
## The History of Algebra, part II: Unsophisticated vs. Sophisticated Tools
Friday, Jul 9 2010
Ben Blum-Smith 6:11 pm
Math ed bloggers love Star Wars. This post is extremely long, and involves a fair amount of math, so in the hopes of keeping you reading, I promise a Star Wars reference toward the end. Also, you can still get the point if you skip the math, though that would be sad.
The historical research project I gave myself this spring in order to prep my group theory class (which is over now – why am I still at it?) has had me working slowly through two more watershed documents in the history of math:
Disquisitiones Arithmeticae
by Carl Friedrich Gauss
(in particular, “Section VII: Equations Defining Sections of a Circle”)
and
Mémoire sur les conditions de résolubilité des équations par radicaux
by Evariste Galois
I’m not done with either, but already I’ve been struck with something I wanted to share. Mainly it’s just some cool math, but there’s a pedagogically relevant idea in here too -
Take-home lesson: The first time a problem is solved the solution uses only simple, pre-existing ideas. The arguments and solution methods are ugly and specific. Only later do new, more difficult ideas get applied, which allow the arguments and solution methods to become elegant and general.
The ugliness and specificity of the arguments and solution methods, and the desire to clean them up and generalize them, are thus a natural motivation for the new ideas.
This is just one historical object lesson in why “build the machinery, then apply it” is a pedagogically unnatural order. Professors delight in using the heavy artillery of modern math to give three-sentence proofs of theorems once considered difficult. (I’ve recently taken courses in algebra, topology, and complex analysis, with three different professors, and deep into each course, the professor gleefully showcased the power of the tools we’d developed by tossing off a quick proof of the fundamental theorem of algebra.) Now, this is a very fun thing to do. But if the goal is to make math accessible, then this is not the natural order.
The natural order is to try to answer a question first. Maybe we answer it, maybe we don’t. But the desire for and the development of the new machinery come most naturally from direct, hands-on experience with the limitations of the old machinery. And that means using it to try to answer questions.
I’m not saying anything new here. But I just want to show you a really striking example from Gauss. (Didn’t you always want to see some original Gauss? No? Okay, well…)
* * * * *
I am reading a 1966 translation of the Disquisitiones by Arthur A. Clarke which I have from the library. An original Latin copy is online here. I don’t read Latin but maybe you do.
I’m focusing on the last section in the book, but at one point Gauss makes use of a result he proved much earlier:
Article 42. If the coefficients $A, B, C, \dots, N; a, b, c, \dots, n$ of two functions of the form
$P=x^m+Ax^{m-1}+Bx^{m-2}+Cx^{m-3}+\dots+N$
$Q=x^{\mu}+ax^{\mu-1}+bx^{\mu-2}+cx^{\mu-3}+\dots+n$
are all rational and not all integers, and if the product of $P$ and $Q$ is
$x^{m+\mu}+\mathfrak{A}x^{m+\mu-1}+\mathfrak{B}x^{m+\mu-2}+etc.+\mathfrak{Z}$
then not all the coefficients $\mathfrak{A}, \mathfrak{B}, \dots, \mathfrak{Z}$ can be integers.
Note that even the statement of Gauss’ proposition here would be cleaned up by modern language. Gauss doesn’t even have the word “polynomial.” The word “monic” (i.e., leading coefficient 1) would also have been handy. In modern language he could have said, “The product of two rational monic polynomials is not an integer polynomial if any of their coefficients are not integers.”
But this is not the most dramatic difference between Gauss’ statement (and proof – just give me a sec) and the “modern version.” On page 400 of Michael Artin’s Algebra textbook (which I can’t stop talking about only because it is where I learned like everything I know), we find:
(3.3) Theorem. Gauss’s Lemma: A product of primitive polynomials in $\mathbb{Z}[x]$ is primitive.
The sense in which this lemma is Gauss’s is precisely the sense in which it is really talking about the contents of Article 42 from Disquisitiones which I quoted above.
Huh?
First of all, what’s $\mathbb{Z}[x]$? Secondly, what’s a primitive polynomial? Third and most important, what does this have to do with the above? Clearly they both have something to do with multiplying polynomials, but…
Okay. $\mathbb{Z}[x]$ is just the name for the set of polynomials with integer coefficients. (Apologies to those of you who know this already.) So a polynomial in $\mathbb{Z}[x]$ is really just a polynomial with integer coefficients. This notation was developed long after Gauss.
More substantively, a “primitive polynomial” is an integer polynomial whose coefficients have gcd equal to 1. I.e. a polynomial from which you can’t factor out a nontrivial integer factor. E.g. $4x^2+4x+1$ is primitive, but $4x^2+4x+2$ is not because you can take out a 2. This idea is from after Gauss as well.
So, “Gauss’s Lemma” is saying that if you multiply two polynomials each of whose coefficients do not have a common factor, you will not get a common factor among all the coefficients in the product.
What does this have to do with the result Gauss actually stated?
That’s an exercise for you, if you feel like it. (Me too actually. I feel confident that the result Artin states has Gauss’s actual result as a consequence; less sure of the converse. What do you think?) (Hint, if you want: take Gauss’s monic, rational polynomials and clear fractions by multiplying each by the lcm of the denominators of its coefficients. In this way replace his original polynomials with integer polynomials. Will they be primitive?)
Meanwhile, what I really wanted to show you are the two proofs. Original proof: ugly, long, specific, but containing only elementary ideas. Modern proof: cute, elegant, general, but involving more advanced ideas.
Here is a very close paraphrase of Gauss’ original proof of his original claim. Remember, $P$ and $Q$ are monic polynomials with rational coefficients, not all of which are integers, and the goal is to prove that $PQ$‘s coefficients are not all integers.
Demonstration. Put all the coefficients of $P$ and $Q$ in lowest terms. At least one coefficient is a noninteger; say without loss of generality that it is in $P$. (If not, just switch the roles of $P$ and $Q$.) This coefficient is a fraction with a denominator divisible by some prime, say $p$. Find the term in $P$ among all the terms in $P$ whose coefficient’s denominator is divisible by the highest power of $p$. If there is more than one such term, pick the one with the highest degree. Call it $Gx^g$, and let the highest power of $p$ that divides the denominator of $G$ be $p^t$. ($t \geq 1$ since $p$ was chosen to divide the denominator of some coefficient in $P$ at least once.). The key fact about the choice of $Gx^g$ is, in Gauss’s words, that its “denominator involves higher powers of $p$ than the denominators of all fractional coefficients that precede it, and no lower powers than the denominators of all succeeding fractional coefficients.”
Gauss now divides $Q$ by $p$ to guarantee that at least one term in it (at the very least, the leading term) has a fractional coefficient with a denominator divisible by $p$, so that he can play the same game and choose the term $\Gamma x^{\gamma}$ of $Q/p$ with $\Gamma$ having a denominator divisible by $p$ more times than any preceding fractional coefficient and at least as many times as each subsequent coefficient. Let the highest power of $p$ dividing the denominator of $\Gamma$ be $p^{\tau}$. (Having divided the whole of $Q$ by $p$ guarantees that $\tau \geq 1$, just like $t$.)
I’ll quote Gauss word-for-word for the next step:
“Let those terms in $P$ which precede $Gx^g$ be $'Gx^{g+1}$, $''Gx^{g+2}$, etc. and those which follow be $G'x^{g-1}$, $G''x^{g-2}$, etc.; in like manner the terms which precede $\Gamma x^{\gamma}$ will be $'\Gamma x^{\gamma+1}$, $''\Gamma x^{\gamma+2}$, etc. and the terms which follow will be $\Gamma'x^{\gamma-1}$, $\Gamma''x^{\gamma-2}$, etc. It is clear that in the product of $P$ and $Q/p$ the coefficient of the term $x^{g+\gamma}$ will
$= G\Gamma + 'G\Gamma' + ''G\Gamma'' + etc.$
$+ '\Gamma G' + ''\Gamma G'' + etc.$
“The term $G\Gamma$ will be a fraction, and if it is expressed in lowest terms, it will involve $t+\tau$ powers of $p$ in the denominator. If any of the other terms is a fraction, lower powers of p will appear in the denominators because each of them will be the product of two factors, one of them involving no more than $t$ powers of $p$, the other involving fewer than $\tau$ such powers; or one of them involving no more than $\tau$ powers of $p$, the other involving fewer than $t$ such powers. Thus $G\Gamma$ will be of the form $e/(fp^{t+\tau})$, the others of the form $e'/(f'p^{t+\tau-\delta})$ where $\delta$ is positive and $e, f, f'$ are free of the factor $p$, and the sum will
$=\frac{ef'+e'fp^{\delta}}{ff'p^{t+\tau}}$
The numerator is not divisible by $p$ and so there is no reduction that can produce powers of $p$ lower than $t+\tau$.”
(This is on pp. 25-6 of the Clarke translation.)
This argument guarantees that the coefficient of $x^{g+\gamma}$ in $PQ/p$, expressed in lowest terms, has a denominator divisible by $p^{t+\tau}$. Thus the coefficient of the same term in $PQ$ has a denominator divisible by $p^{t+\tau-1}$. Since $t$ and $\tau$ are each at least 1, this means the denominator of this term is divisible by $p$ at least once, and so a fraction. Q.E.D.
Like I said – nasty, right? But the concepts involved are just fractions and divisibility. Compare a modern proof of “Gauss’ Lemma” (the statement I quoted above from Artin – a product of primitive integer polynomials is primitive).
Proof. Let the polynomials be $P$ and $Q$. Pick any prime number $p$, and reduce everything mod $p$. $P$ and $Q$ are primitive so they each have at least one coefficient not divisible by $p$. Thus $P \not\equiv 0 \mod{p}$ and $Q \not\equiv 0 \mod{p}$. By looking at the leading terms of $P$ and $Q$ mod $p$ we see that the product $PQ$ must be nonzero mod $p$ as well. This implies that $PQ$ contains at least one coefficient not divisible by $p$. Since this argument works for any prime $p$, it follows that there is no prime dividing every coefficient in $PQ$, which means that it is primitive. Q.E.D.1
Clean and quick. If you’re familiar with the concepts involved, it’s way easier to follow than Gauss’s original. But, you have to first digest a) the idea of reducing everything mod $p$; b) the fact that this operation is compatible with all the normal polynomial operations; and c) the crucial fact that because $p$ is prime, the product of two coefficients that are not $\equiv 0 \mod{p}$ will also be nonzero mod $p$.
Now Gauss actually had access to all of these ideas. In fact it was in the Disquisitiones Arithmeticae itself that the world was introduced to the notation “$a \equiv b \mod{p}$.” So in a way it’s even more striking that he didn’t think to use them here when they would have cleaned up so much.
What bugged me out and made me excited to share this with you was the realization that these two proofs are essentially the same proof.
What?
I’m not gonna spell it out, because what’s the fun in that? But here’s a hint: that term $Gx^g$ that Gauss singled out in his polynomial $P$? Think about what would happen to that term (in comparison with all the terms before it) if you a) multiplied the whole polynomial by the lcm of the denominators to clear out all the fractions and yield a primitive integer polynomial, and then b) reduced everything mod p.
(If you are into this sort of thing, I found it to be an awesome exercise, that gave me a much deeper understanding of both proofs, to flesh out the equivalence, so I recommend that.)
* * * * *
What’s the pedagogical big picture here?
I see this as a case study in the value of approaching a problem with unsophisticated tools before learning sophisticated tools for it. To begin with, this historical anecdote seems to indicate that this is the natural flow. I mean, everybody always says Gauss was the greatest mathematician of all time, and even he didn’t think to use reduction mod $p$ on this problem, even though he was developing this tool on the surrounding pages of the very the same book.
In more detail, why is this more pedagogically natural than “build the (sophisticated) machine, then apply it”?
First of all, the machine is inscrutable before it is applied. Think about being handed all the tiny parts of a sophisticated robot, along with assembly instructions, but given no sense of how the whole thing is supposed to function once it’s put together. And then trying to follow the instructions. This is what it’s like to learn sophisticated math ideas machinery-first, application-later. I felt this way this spring in learning the idea of Brouwer degree in my topology class. Now that everything is put together, I have a strong desire to go back to the beginning and do the whole thing again knowing what the end goal is. The ideas felt so airy and insubstantial the first time through. I never felt grounded.
Secondly, the quick solution that is powered by the sophisticated tools loses something if it’s not coupled with some experience working on the same problem with less sophisticated tools. The aesthetic delight that professors take in the short and elegant solution of the erstwhile-difficult problem comes from an intimacy with this difficulty that the student skips if she just learns the power tools and then zaps it. Likewise, if the goal is to gain insight into the problem, the short, turbo-powered solution often feels very illuminating to someone (say, the professor) who knows the long, ugly solution, but like pure magic, and therefore not illuminating at all, to someone (say, a student) who doesn’t know any other way. There is something tenuous and spindly about knowing a high-powered solution only.
Here I can cite my own experience with Gauss’s Lemma, the subject of this post. I remember reading the proof in Artin a year ago and being satisfied at the time, but I also remember being unable to recall this proof (even though it’s so simple! maybe because it’s so simple!) several months later. You read it, it works, it’s like poof! done! It’s almost like a sharp thin needle that passes right through your brain without leaving any effect. (Eeew… sorry that was gross.) The process of working through Gauss’ original proof, and then working through how the proofs are so closely related, has made my understanding of Artin’s proof far deeper and my appreciation of its elegance far stronger. Before, all I saw was a cute short argument that made something true. I now see in it the mess that it is elegantly cleaning up.
I’ve had a different form of the same experience as I fight my way through Galois’ paper. (I am working through the translation found in Appendix I of Harold Edwards’ 1984 book Galois Theory. This is a great way to do it because if at any point you are totally lost about what Galois means, you can usually dig through the book and find out what Edwards thinks he means.) I previously learned a modern treatment of Galois theory (essentially the one found in Nathan Jacobson’s Basic Algebra I – what a ridiculous title from the point of view of a high school teacher!). When I learned it, I “followed” everything but I knew my understanding was not where I wanted it to be. Here the words “spindly” and “tenuous” come to mind again. The arguments were built one on top of another till I was looking at a tall machine with a lot of firepower at the very top but supported by a series of moving parts I didn’t have a lot of faith in.
An easy mark for Ewoks, and I knew it.
This version of Galois theory was all based on concepts like fields, automorphisms, vector spaces, separable and normal extensions, of which Galois himself had access to none. The process of fighting through Galois’ original development of his theory and trying to understand how it is related to what I learned before has been slowly filling out and reinforcing the lower regions of this structure for me. Coupling the sophisticated with the less sophisticated approach has given the entire edifice some solidity.
Thirdly, and this is what I feel like I hear folks (Shawn Cornally, Dan Meyer, Alison Blank, etc.) talk about a lot, but it bears repeating, is this:
If you attack a problem with the tools you have, and either you can’t solve it, or you can solve it but your solution is messy and ugly, like Gauss’s solution above (if I may), then you have a reason to want better tools. Furthermore, the way in which your tools are failing you, or in which they are being inefficient, may be a hint to you for how the better tools need to look.
Just as an example, think about how awesome reduction mod $p$ is going to seem if you are already fighting (as Gauss did) with a whole bunch of adding stuff up some of which is divisible by $p$ and some of which is not. What if you could treat everything divisible by $p$ as zero and then summarily forget about it? How convenient would that be?
I want to bring this back to the K-12 level so let me give one other illustration. A major goal of 7th-9th grade math education in NY (and elsewhere) is getting kids to be able to solve all manner of single-variable linear equations. The basic tool here is “doing the same thing to both sides.” (As in, dividing both sides of the equation by 2, or subtracting 2x from both sides…) For the kids this is a powerful and sophisticated tool, one that takes a lot of work to fully understand, because it involves the extremely deep idea that you can change an equation without changing the information it is giving you.
There is no reason to bring out this tool in order to have the kiddies solve $x+7=10$. It’s even unnatural for solving $4x-21=55$. Both of these problems are susceptible to much less abstract methods, such as “working backwards.” The “both sides” tool is not naturally motivated until the variable appears on both sides of the equation. I used to let students solve $4x-21=55$ whatever way made sense to them, but then try to impose on them the understanding that what they had “really done” was to add 21 to both sides and then divide both sides by 4, so that later when I gave them equations with variables on both sides, they’d be ready. This was weak because I was working against the natural pedagogical flow. They didn’t need the new tool yet because I hadn’t given them problems that brought them to the limitations of the old tool. Instead, I just tried to force them to reimagine what they’d already been doing in a way that felt unnatural to them. Please, if a student answers your question and can give you any mathematically sound reason, no matter how elementary, accept it! If you would like them to do something fancier, try to give them a problem that forces them to.
Basically this whole post adds up to an excuse to show you some cool historical math and a plea for due respect to be given to unsophisticated solutions. There is no rush to the big fancy general tools (except the rush imposed by our various dark overlords). They are learned better, and appreciated better, if students, teachers, mathematicians first get to try out the tools we already have on the problems the fancy tools will eventually help us answer. It worked for Gauss.
[1] This is the substance of the proof given in Artin but I actually edited it a bit to make it (hopefully) more accessible. Artin talks about the ring homomorphism $\mathbb{Z}[x] \longrightarrow \mathbb{F}_p[x]$ and the images of $P$ and $Q$ (he calls them $f$ and $g$) under this homomorphism.
ADDENDUM 8/10/11:
I recently bumped into a beautiful quote from Hermann Weyl that I had read before (in Bob and Ellen Kaplan’s Out of the Labyrinth, p. 157) and forgotten. It is entirely germane.
Beautiful general concepts do not drop out of the sky. To begin with, there are definite, concrete problems, with all their undivided complexity, and these must be conquered by individuals relying on brute force. Only then come the axiomatizers and systematizers and conclude that instead of straining to break in the door and bloodying one’s hands one should have first constructed a magic key of such and such a shape and then the door would have opened quietly, as if by itself. But they can construct the key only because the successful breakthrough enables them to study the lock front and back, from the outside and from the inside. Before we can generalize, formalize and axiomatize there must be mathematical substance. | {"extraction_info": {"found_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": 128, "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": 0.953599214553833, "perplexity_flag": "head"} |
http://math.stackexchange.com/questions/234704/approximating-a-sum-of-two-binomial-distributions | # Approximating a sum of two binomial distributions
A club basketball team will play a 60-game season. Thirty-two of these games are against class A teams, and 28 are against class B teams. The outcomes of all the games are independent. The team will win each game against a class A opponent with probability .5 and it will win each game against a class B opponent with probability .65. Let X denote its total number of victories in the season.
(a) What is the relationship between $X_A$, $X_B$, and $X$? Is $X$ a binomial random variable?
(b) Approximate the probability that the team wins 45 or more games this season.
What I did: $$X=X_A+X_B \\ X_A \text{ is a binomial random variable with } n=32 \text{ and } p=0.5 \\ X_B \text{ is a binomial random variable with } n=28 \text{ and } p=0.65 \\ p_X(k) = \sum_{a=0}^k P(X_A=a)P(X_B=k-a)$$ Can I express $p_X(k)$ as a binomial distribution somehow? Also, I think it is possible to approximate $P(X\ge45)$ using a normal distribution by the Central Limit Theorem, but I'm not sure how to apply it.
-
## 1 Answer
$X$ is not a binomial random variable. One way to see this is using the moment generating function.
You can approximate $X$ by a normal random variable having the same mean and variance.
You may want to use the continuity correction, if your instructor has mentioned that.
-
I know the mean for $X$ is $32(.5)+28(.65) = 34.2$, but how do I calculate the variance? – woaini Nov 11 '12 at 7:22
Variance of the sum of independent random variables is the sum of the variances. Do you know (or can you look up) the formula for variance of a binomial random variable? – Robert Israel Nov 11 '12 at 8:07
Okay, so I found that the variance is $(32)(.5)(.5) + (28)(.65)(.35)=14.37$. Does applying the continuity correction mean that $P(X\ge45)=P(X>44)$? – woaini Nov 11 '12 at 8:24
Yes. The continuity correction would mean using $P(X > 44.5)$ (i.e. we want to include $X=45$ but not $X=44$). – Robert Israel Nov 12 '12 at 0:41 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 15, "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": 0.9245662093162537, "perplexity_flag": "head"} |
http://physics.stackexchange.com/questions/7475/magnetic-susceptibility-in-1-ev/7486 | # 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 units??
-
In one place it is given in emu/mol, whatever that may be. – Georg Mar 23 '11 at 22:06
Excellent question :) – David Zaslavsky♦ Mar 23 '11 at 23:12
At any rate, whatever you were taught, it's clear that $\chi_\perp$, defined as a coefficient in equation 2 of that paper, has the unit of inverse energy, just like they indicate. $\int d^3 x$ cancels against $abc$, one time derivative cancels against $\partial_t$, the other $\partial_t$ combined with one $\hbar$ gives one energy which has to be cancelled for $S$ to have the right dimension of the other $\hbar$. So maybe you should have already asked a different question about the previous texts such as eqn 2. $\chi_\perp$ is apparently normalized differently than you think. – Luboš Motl May 4 '11 at 11:40
## 1 Answer
Magnetic susceptibility is a static constant relating the amount of magnetization to the applied magnetic field; it is dimensionless. Transverse susceptibility is a measure of the magnetic response of a material to an alternating field; it is not dimensionless. The transverse susceptibility is often used when talking to the ferromagnetic resonance (like I think they are doing in the paper you cite) of a material.
So the short answer is that MS is dimensionless and TS is not.
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Ok, this is something I didn't know. Anyway, as Georg already mentioned, the authors use the value given in emu/mol to evaluate $\chi _{\perp}$. How should I convert between these units? – Pie86 Mar 24 '11 at 8:37
I could not find the TS given in emu/mol. emu/mol are the units of molar susceptibility. Molar susceptibility is postulated in Curie's law for paramagnetic material. Though I must admit this is a dense paper (and not in my field) so in order to keep it from taking up my morning I've just been skimming it. – AOA Mar 24 '11 at 16:34
It's in the caption of "Table 1" (near the end). Anyway I want to thank you for your help and the precious information about TS. – Pie86 Mar 24 '11 at 20:10
'emu' should only be used as a shorthand IMHO. It is difficult to convert without context, and is an unfortunate cgs anachronism. It stands for "electromagnetic unit" and can mean simply the volume of a field (here it usually means cm^3) or it can mean magnetic moment (erg/G). I suspect that if I was more familiar with operating an NMR I would be able to give you a better crafted answer. – AOA Mar 25 '11 at 3:44 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 13, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": 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": 0.9513428807258606, "perplexity_flag": "middle"} |
http://physics.stackexchange.com/tags/string-theory/new | # Tag Info
## New answers tagged string-theory
0
### Question on Type HO/HE string theory
I just found that my question (2) is trivially easy and that indeed it is: $$\begin{array}{l} m = \sqrt {\frac{{2\pi T}}{{{c_0}}}\left( {B + {{\tilde N}_{II}} - {a_B} - {{\tilde a}_{II}}} \right)} \\ {\rm{ }} = \sqrt {\frac{{2\pi T}}{{{c_0}}}\left( {B + {{\tilde N}_{II}} - 1 - {{\tilde a}_{II}}} \right)} \end{array}$$ This is because: ...
0
### Question on the Hagedorn tower in Type I string theory
Here is my solution. For the Ramond Ramond Sector, $${m_\rm{I}} = \frac{{\left[ {\hat a_ + ^\mu ,\hat{\tilde a}_ + ^\nu } \right]}}{2}{m_{{\rm{IIB}}}}$$ For the Neveu-Schwarz Neveu-Schwarz Sector, $${m_\rm{I}} = \frac{{\left[ {\hat d_ {-1/2} ^\mu ,\hat{\tilde d}_{-1/2} ^\nu } \right]}}{2}{m_{{\rm{IIB}}}}$$ For the Ramond Neveu-Schwarz Sector or the ...
0
### Is decoherence even possible in anti de Sitter space?
Decoherence is more than anything a matter of what you define to be the "environment". The environment is supposed to be external to the system of interest and entangling interaction with it produces decoherence. If the environment in question is a part of the adS space then the subsystem can certainly decohere. If what you are asking is whether the space as ...
1
### Energy Functional
I suppose $f$ is just an arbitrary scalar function on the manifold. I'm not well-versed with the concept of Ricci flow, so I'll try to give a simple operational answer. I also don't understand what exactly you're looking for. The Ricci scalar $R$ roughly represents the amount of energy stored in spacetime (as curvature). The dilaton is a scalar field which ...
2
### Question on the Hagedorn tower in Type I string theory
Hagedorn spectrum just means that the density of states varies exponentially with the energy/mass. $m^2$ (asymptotically) given by the "level" (N) of the state (upto a sqrt). The number of states at level $N$ corresponds to the possible partitions of $N$ into different oscillator modes. That means that the number of states at level $N$ will increase ...
1
### Mathematical concept of supersymmetry
I would really recommend a study in QFT before going on to study SUSY. QFT has many quirks that make supersymmetry a very interesting expansion of the regular framework. You'd miss out on all that as you just had to believe the facts presented w/o following the thought that lead to the results in detail. On the Mathematical level you will need Grassmann ...
3
### How can string theory work without supersymmetry?
If you spend some time looking in detail at the arguments that string theory requires supersymmetry, you'll find that they are not watertight. (How could they be, since we still can't say/don't know precisely what string theory is?) Basically, some string theorists argue that that the usual classification depends too strongly on choosing nearly trivial ...
1
### How can string theory work without supersymmetry?
p-adic strings or the adelic approach created by B.Dragovich don't require SUSY at all. At least, not the usual SUSY symmetry... Non-critical string theory, the so-called Liouville theory, is based on the hypothesis of non-imposing the condition that critical strings with fermions (superstrings) impose on the space-time dimension due to internal ...
0
### Flavour diagonal SUSY breaking
Generally, in the MSSM one works with the "minimal flavor violation" paradigm that states that all flavor violation originates in the SM Yukawa sector. This paradigm is ad hoc, but explains why no huge SUSY contributions to FCNC observables are seen. There are models that go beyond minimal flavor violation and some that give an explanation for the ...
3
### Gauging discrete symmetries
I think I got the answer now. The main idea is this: When we gauge continuous symmetries we identify all the states $$A^\mu=A^\mu+\partial^\mu\chi$$ (which are continuously many) as a unique physical state. When we gauge a discrete symmetry (let's assume it's generated by $\theta$) we identify all the states $$|\Psi\rangle=\theta^n|\Psi\rangle$$ where ...
3
### How do I find constraints on the Nambu-Goto Action?
I managed to find a quasi-systematic way to do this. The idea that allowed me to do this was inspired by Noether's Theorem. Re-parameterization invariance is a symmetry of the system, a symmetry much stronger than an ordinary global symmetry. Similarly, however, a constraint is also a conserved quantity, but it is something much stronger than that. ...
0
### Fundamental equation(s) of string theory?
What I want to say here is related to user1504's comment. As Lenny Susskind explains in this and this lecture, how to describe the scattering behavior of particles is nearly the definition of string theory. So formulas for scattering amplitudes can in some way be considered as fundamental equations defining the theory. Very schematically, the equation to ...
1
### SUSY's Critical role in String Theory
"Falsifying supersymmetry" is a phrase that has to be properly qualified. Our ability to falsify with experiment is limited. We can rule out the existence of supersymmetry only at accessible energy/distance/density scales. LHC, for example, is not able to resolve physics at distance scales much smaller than \$\frac{\hbar c}{7\mbox{ TeV}} \simeq ...
0
### SUSY's Critical role in String Theory
Actually, no. The supersymmetric transformations are elegant and simple ways of extending the Bosonic String theory to fermions, but if supersymmetry is falsified somehow, then maybe all of the discovered superstring theory would have to be discarded, but a new one may emerge... It would just use different supersymmetric transformations with different ...
3
### Spectra of the Type II String theories
The NS-NS sector is the same in type IIA and IIB, but the R-NS and NS-R sectors differ. The type IIA theory is non-chiral, so the R-NS and NS-R fields transform in $\mathbf{8}_s \otimes \mathbf{8}_v$ and $\mathbf{8}_v \otimes \mathbf{8}_s'$, where $\mathbf{8}_s$ and $\mathbf{8}_s'$ are the two eight-dimensional spinor representations of $SO(8)$. Type IIB, on ...
1
### Critical dimension in quantization of p-branes
From what we understand today, p-branes are honest degrees of freedom, on equal footing with strings. Shop you have a good question. But I don't think anyone had so far managed to consistently quantize a p-brane. Loosely, a brane has much more degrees of frerdom than a string and it's difficult to get them under control. So quantizing it is a technical ...
0
### Where does quantum mechanics come from? [closed]
"Theory of everything" is not actually a theory of everything. Physical theories reach to the depths of matter, but they leave blank spots behind the front line of knowledge. That is, having studied the bricks in every detail, we could still not understand the buildings made of those bricks. There are such things as chaos and emergent phenomena possible, ...
1
### Where does quantum mechanics come from? [closed]
An "acceptable" theory of everything is quite a matter of taste. Since all your experiences are grounded in classical physics, you feel that quantum mechanics is unnatural and seek to "understand" it, probably in terms of your classical notions. For eg: Do you ever question Newton's first law... why should objects have a property called inertia? Some ...
1
### Relation between Higgs boson and force of gravitation
Glad to see enthusiasm. I'm sad to say you've been let down by the popular press. Here's the word from a real life particle physicist: I would be very much happy and grateful to know if someone can explain me a little about the relation between Higgs Boson and Force of Gravitation(Gravitons if exists). There is none. The Higgs field gives mass to some ...
3
### AdS/RCFT examples?
Yes, gravity duals have been found for such theories. You can find at least a few examples by picking a particular rational CFT X (,e.g., a minimal model, WZW, Ising) and googling 'AdS dual of X'. For example: http://arxiv.org/abs/1011.2986 http://arxiv.org/pdf/1111.1987 http://arxiv.org/abs/1011.5900
6
### Supersymmetry and non-compact $R$-symmetry group?
Noncompact internal symmetries – and R-symmetry is an internal symmetry (it doesn't transform positions in the spacetime) – are unacceptable in a physical theory because they would lead to negative-norm states. Consider the $i$-th superpartner of a bosonic particle state, $|i\rangle$, where $i=1,2,\dots,N$. The inner product $\langle i|j\rangle$ of such ...
2
### Question about charge conservation at vertices of web diagram
The symbols $\alpha,\beta$ in these toric geometry diagrams refer to the two 1-dimensional cycles of a torus, $T^2$, and the labels which are combinations of $\alpha,\beta$ correspond to the 1-cycles of the torus that degenerate at the given line of the diagram. If $c$ degenerates, so does $-c$, so all these labels are pretty much unoriented and the ...
4
### Fundamental equation(s) of string theory?
I've long been interested in this, but the impression I get is (speaking as a strict amateur with a reasonable understanding of QM and relativity) there is simply nothing like e.g. the Schrodinger equation or Einstein's field equation in string theory. String theory is developed by writing down the action (which is the area of the string world sheet), using ...
0
### Starting examples of Maldacena duality
One review I found helpful relates Type IIB superstrings on a maximally supersymmetric plane wave background to N=4 SYM: Lectures on the Plane–Wave String/Gauge Theory Duality by Jan Christoph Plefka It might be more conversational than what you're looking for, but this review by Polchinski has some great discussion and probably deserves the mention: ...
5
### What does this “Witten's Dog” Feynman diagram in Futurama episode mean? [closed]
I would guess that the professor is explaining his/the(?) theory that dark matter is neutrinos, produced via a scattering process he calls "Witten's dog". It is funny because the neutrinos are coming out of the dog's butt. In the Standard Humor Classification, this is known as a "poop joke".
2
### What does this “Witten's Dog” Feynman diagram in Futurama episode mean? [closed]
There is a Feynman diagram for particle scattering which looks like a dog... imagine each of the "tubes" in that picture shrunk to zero radius. Like @dilaton says, this looks like a scattering (world sheet) diagram for closed strings. But with random (nonsensical?) particle operator insertions (maybe they're supposed to represent fleas?). And afaik, the ...
0
### What does this “Witten's Dog” Feynman diagram in Futurama episode mean? [closed]
Agree, It seems there is no physical meaning. Here is another link: http://en.wikipedia.org/wiki/Mars_University
Top 50 recent answers are included | {"extraction_info": {"found_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": 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": 0.9292100667953491, "perplexity_flag": "middle"} |
http://mathhelpforum.com/math-puzzles/209475-logic-help-print.html | # Logic help.
Printable View
• December 9th 2012, 07:18 PM
hisajesh
Logic help.
2 men and 7 boys can do a piece of work in 14 days,
3 men and 8 boys can do it in 11 days.
8 men and 6 boys can do this 3 times of work in how many days?
Logic: I found either the first or the second sentence is redundant, but when I ignored one and tried solving, it gives me wrong answers, Kindly help how to approach?
• December 9th 2012, 09:39 PM
Prove It
Re: Logic help.
Let x represent the amount of work that can be done by a man, and let y represent the amount of work that can be done by a boy. Then $\displaystyle \begin{align*} 2x + 7y = 14 \end{align*}$ and $\displaystyle \begin{align*} 3x + 8y = 11 \end{align*}$. Solve these two equations simultaneously for x and y, then use this to answer the remainder of your question.
• December 9th 2012, 10:52 PM
Soroban
Re: Logic help.
Hello, hisajesh!
I assume that all the men work at the same constant rate
and that all the boys work at the same constant rate.
Quote:
2 men and 7 boys can do a piece of work in 14 days,
3 men and 8 boys can do it in 11 days.
8 men and 6 boys can do this 3 times of work in how many days?
Let $m$ = the number of days it takes one man to do the job..
In one day, a man can do $\tfrac{1}{m}$ of the job.
In one day, 2 men can do $\tfrac{2}{m}$ of the job.
Let $b$ = the number of days it takes one boy to do the job.
In one day, a boy can do $\tfrac{1}{b}$ of the job.
In one day, 7 boys can do $\tfrac{7}{b}$ of the job.
So, in one day, 2 men and 7 boys can do $\tfrac{2}{m} + \tfrac{7}{b}$ of the job.
. . But this equals $\tfrac{1}{14}$ of the job.
Hence: . $\tfrac{2}{m} + \tfrac{7}{b} \:=\:\tfrac{1}{14}$
In one day, 3 men can do $\tfrac{3}{m}$ of the job.
In one day, 8 boys can do $\tfrac{8}{b}$ of the job.
So in one day, 3 men and 8 boys can do $\tfrac{3}{m} + \tfrac{8}{b}$ of the job.
. . But this equals $\tfrac{1}{11}$ of the job.
Hence: . $\tfrac{3}{m} + \tfrac{8}{b} \:=\:\tfrac{1}{11}$
We have a system of equations: . $\begin{Bmatrix}\frac{2}{m} + \frac{7}{b} \:=\:\frac{1}{14} & [1] \\ \\[-4mm] \frac{3}{m} + \frac{8}{b} \:=\:\frac{1}{11} & [2] \end{Bmatrix}$
$\begin{array}{ccccccc}\text{Multiply [1] by 3:} & \frac{6}{m} + \frac{21}{b} &=& \frac{3}{14} \\ \\[-4mm] \text{Multiply [2] by -}2\!: & \text{-}\frac{6}{m} - \frac{16}{b} &=& \text{-}\frac{2}{11} \end{array}$
Add: . $\tfrac{21}{b} - \tfrac{16}{b} \:=\:\tfrac{3}{14} - \tfrac{2}{11} \quad\Rightarrow\quad \tfrac{5}{b} \:=\:\tfrac{5}{154} \quad\Rightarrow\quad b \:=\:154$
Substitute into [1]: . $\tfrac{2}{m} + \tfrac{7}{154} \:=\:\tfrac{1}{14} \quad\Rightarrow\quad \tfrac{2}{m} \:=\:\tfrac{2}{77} \quad\Rightarrow\quad m \:=\:77$
Now we know the following:
In one day, a man can do $\tfrac{1}{77}$ of the jpb.
In one day, 8 men can do $\tfrac{8}{77}$ of the job.
In $x$ days, 8 men can do $\tfrac{8x}{77}$ of the job.
In one day, a boy can do $\tfrac{1}{154}$ of the job.
In one day, 6 boys can do $\tfrac{6}{154}$ of the job.
In $x$ days, 6 boys can do $\tfrac{6x}{154}$ of the job.
In $x$ days, 8 men and 6 boys can do: $\tfrac{8x}{77} + \tfrac{6x}{154}$ of the job.
But this equals 3 jobs.
Hence: . $\frac{8x}{77} + \frac{6x}{154} \:=\:3$
Multiply by 77: . $8x + 3x \:=\:231 \quad\Rightarrow\quad 11x \:=\:231 \quad\Rightarrow\quad x \:=\:21$
Answer: 21 days.
All times are GMT -8. The time now is 10:51 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": 32, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9234158992767334, "perplexity_flag": "middle"} |
http://math.stackexchange.com/questions/97250/consecutive-coin-toss-with-static-tosses | # Consecutive Coin Toss with static tosses
I'm writing an algorithm for a coin toss problem. But I have a problem understanding the calculation given.
Here is the question:
You have an unbiased coin which you want to keep tossing until you get N consecutive heads. You've tossed the coin M times and surprisingly, all tosses resulted in heads. What is the expected number of additional tosses needed until you get N consecutive heads?
If N = 2 and M = 0, you need to keep tossing the coin until you get 2 consecutive heads. It is not hard to show that on average, 6 coin tosses are needed.
If N = 2 and M = 1, you need 2 consecutive heads and have already have 1. You need to toss once more no matter what. In that first toss, if you get heads, you are done. Otherwise, you need to start over, as the consecutive counter resets, and you need to keep tossing the coin until you get N=2 consecutive heads. The expected number of coin tosses is thus 1 + (0.5 * 0 + 0.5 * 6) = 4.0
If N = 3 and M = 3, you already have got 3 heads, so you do not need any more tosses.
Now my problem is understanding the calculation: `1 + (0.5 * 0 + 0.5 * 6) = 4.0` when N = 2 and M = 1. I understood how they got the 6 (which is basically calculating it when M = 0, formula here).
Now what if I'm going to calculate `N = 3, M = 1` or `N = 3, M = 2`?
Could someone write this calculation in a formula for me please? What is the `1`? What is `(0.5 * 0 + 0.5 * 6)`?
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So, what is the final formula? (N-M) + (2^N - 1) Is this correct? – JJ Liu Jan 7 '12 at 23:24
Im still not sure – Shawn Mclean Jan 7 '12 at 23:41
1
The general formula is $2^{N+1}-2^{M+1}$. – Byron Schmuland Jan 7 '12 at 23:41
@ByronSchmuland really? so N=3, M=1 will be 12? – JJ Liu Jan 8 '12 at 0:10
## 4 Answers
The reasoning behind the calculation $1+\frac12\cdot 0+\frac12\cdot 6$ is as follows. You definitely have to toss the coin one more time, no matter what; that’s the initial $1$ term. With probability $1/2$ you get a head, and in that case you’re done: you need $0$ more tosses. That’s the $\frac12\cdot 0$ term: with probability $1/2$ you use $0$ extra tosses beyond the first. But with probability $1/2$ you get a tail, and in that case you are in effect starting over, as if you’d not tossed the coin at all. In this case you already know that the expected number of tosses to get two consecutive heads is $6$, so you know that with probability $1/2$ you’ll need $6$ extra tosses beyond the first one; this is the $\frac12\cdot 6$ term.
Now suppose that $N=3$ and $M=1$. You’ll definitely need at least $1$ more toss. With probability $1/2$ it’s a tail, and you’ll be starting over. In that case the expected number of flips after this first one is $2^{3+1}-2=14$, giving a $\frac12\cdot 14$ term (analogous to the $\frac12\cdot 6$ term in the original problem). With probability $1/2$ you’ll get a head, and at this point you’ll be solving the $N=3,M=2$ problem. Suppose that you’ve already done it and discovered that the expected number of flips is $x$; then with probability $1/2$ it will take another $x$ flips after the first one, so you get a $\frac12x$ term, for a total of $1+\frac12\cdot14+\frac12x$ expected extra flips.
I’ll leave it to you to try the $N=3,M=2$ problem using these ideas and substitute the result for $x$.
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so `1+0.5*14 = 8` is correct for N = 3, M = 1 and `1+0.5*14+0.5*8 = 12` for N = 3 and M = 2? – Shawn Mclean Jan 7 '12 at 22:29
My last calculation seems wrong, the number should be lower. I understand it when it comes to M = 1 but when it gets higher I don't know what to do. – Shawn Mclean Jan 7 '12 at 22:40
so basically this is formula 1 + 0.5y + 0.5x where y is expected number of flips if we start over and x is the expected in the next step? – lukas Jan 9 '12 at 0:12
@lukas: Yes. After the first flip, the expected number of additional flips is $$P(\text{tail})\cdot(\text{ expected number of flips starting from }0)+P(\text{head})\cdot(\text{ exp. nr. of additional flips starting from }M+1)$$ – Brian M. Scott Jan 9 '12 at 2:14
– lukas Jan 9 '12 at 2:57
They are doing what's called conditioning.
If $\Bbb E(X)$ denotes the expected value of $X$ and if $A$ and $B$ are disjoint and exhaustive events (either $A$ or $B$ must occur, but they cannot both occur at the same time), then $$\Bbb E(X)=P(A) \Bbb E( X\text{ given }A) +P(B) \Bbb E( X\text{ given }B)$$
The 1 is there because you know the number of additional flips needed is at least one. Then, they compute the number of (even more) additional flips needed to obtain three heads in a row. Towards this end they use the conditioning formula above with $A$ and $B$ as below:
Case A: that flip was a head. This occurs with probability 1/2 and in this case, no more flips are needed.
Case B: that flip was a tail. This occurs with probability 1/2, and in this case the expected number of additional flips is 6 (since it's as if you start over).
So overall, the expected number of additional flips is
$$1+ \underbrace{ (1/2) }_{ \text{prob of }\atop\text{ case A}}\cdot \underbrace{\vphantom{(/} 0 }_{\text{exp. num. of add. }\atop\text{flips in case A}}+ \underbrace{ (1/2) }_{\text{prob of} \atop \text{case B} }\cdot \underbrace{6\vphantom{(/}}_{\text{exp. num. of add.}\atop \text{flips in case B}}.$$
I think it's easier to understand if you don't start with "1":
Let $A$ be the event that the second flip is heads and $B$ be the event that the second flip is tails. Note if $A$ occurs, then the expected number of additional flips past the first is 1. If $B$ occurs, the expected number of additional flips past the first flip is $1+6$.
The expected number of additional flips (past the first flip) is
$$\underbrace{ (1/2) }_{ \text{prob of }\atop\text{ case A}}\cdot \underbrace{\vphantom{(/} 1 }_{\text{exp. num. of add. }\atop\text{flips in case A}}+ \underbrace{ (1/2) }_{\text{prob of} \atop \text{case B} }\cdot \underbrace{7\vphantom{(/}}_{\text{exp. num. of add.}\atop \text{flips in case B}}.$$
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Thank you, could you do this example: `N = 3, M = 2` so I can see if this formula is expanded? – Shawn Mclean Jan 7 '12 at 22:24
@Lol coder For $N=3$ and $M=2$, It's almost exactly the same argument; except that you first need to find the expected number of flips to obtain three heads in a row (replace the "6" above with this number, and it goes through). Try your hand at it please. If you run into trouble, I'll be glad to help. (Certain other cases have an "expanded formula", where you condition on three or more events...) – David Mitra Jan 7 '12 at 22:39
Its supposed to be 8 based on the examples but when I calculate it I get 8.5. Here is my calculation: `1 + 0.5 * 1 + 0.5 * 14` – Shawn Mclean Jan 7 '12 at 22:44
@Lol coder $1+0.5*0+0.5*14$ (assuming that the expected number of flips to get three heads in a row is 14). – David Mitra Jan 7 '12 at 22:51
and with M = 1, `2 + 0.5 * 0 + 0.5 * 14` which is 9 right? – Shawn Mclean Jan 7 '12 at 23:06
show 3 more comments
Let the expected number of coin flips be x. The case analysis goes as follows:
a. If the first flip is a tails, then we have wasted one flip. The probability of this event is 1/2 and the total number of flips required is x+1
b. If the first flip is a heads and second flip is a tails, then we have wasted two flips. The probability of this event is 1/4 and the total number of flips required is x+2
c. If the first flip is a heads and second flip is also heads, then we are done. The probability of this event is 1/4 and the total number of flips required is 2.
Adding, the equation that we get is - x = (1/2)(x+1) + (1/4)(x+2) + (1/4)2
Solving, we get x = 6.
Thus, the expected number of coin flips for getting two consecutive heads is 6.
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I know how to do this part, the confusing part is the calculation: `1 + (0.5 * 0 + 0.5 * 6) = 4.0` due to already having 1 heads. – Shawn Mclean Jan 7 '12 at 22:17
– Abhishek Sakhaparia Jan 7 '12 at 22:19
I posted a derivation of the generalized solution to this problem here.
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http://mathforum.org/mathimages/index.php?title=Volume_of_Revolution&diff=7465&oldid=7401 | Volume of Revolution
From Math Images
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| | volume of solid= <math>{\pi\over 5} units^3</math> | | volume of solid= <math>{\pi\over 5} units^3</math> |
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| | | + | |
| | | + | ==References== |
| | | + | Bread image http://mathdemos.gcsu.edu/mathdemos/sectionmethod/sectionmethod.html<br> |
| | | + | Revolving image http://mathdemos.gcsu.edu/mathdemos/sectionmethod/sectionmethod.html |
| | | + | |
| | | | |
| | |AuthorName=Lizah Masis | | |AuthorName=Lizah Masis |
Revision as of 14:02, 7 July 2009
Solid of revolution
This image shows a solid of revolution
Solid of revolution
Field: Calculus
Created By: Lizah Masis
Basic Description
This image shows the solid formed after revolving the region bounded by $y=x^2$, $y=0$,$x=0$ and $x=1$
A More Mathematical Explanation
Note: understanding of this explanation requires: *Pre-calculus and elementary calculus
[Click to view A More Mathematical Explanation]
When finding the volume of revolution of solids, in many cases the problem is not with the calculus, [...]
[Click to hide A More Mathematical Explanation]
When finding the volume of revolution of solids, in many cases the problem is not with the calculus, but with actually visualizing the solid. To find the volume of a solid like a cylinder, usually we use the formula ${\pi} {r^2} h$. Alternatively we can imagine chopping up the cylinder into thin cylindrical plates, much like slicing up bread, computing the volume of each slice, each of which is ${\Delta x }$ units thick , then summing up the volumes of all the slices.
The disc method is much like slicing up bread and computing the volume of each slice http://mathdemos.gcsu.edu/mathdemos/sectionmethod/sectionmethod.html
This is the idea behind computing the volume of solids whose shapes are complicated. If we are given a function which decribes the shape of a solid, we can plot the function, then revolve the resultant plane area about a fixed axis to obtain the original solid, now called the the solid of revolution. The area can be bounded by many curves of almost any form. The volume of the solid can then be computed using the disc method.
Note: There are other ways of computing the volumes of complicated solids other than the disc method.
In the disc method, we imagine chopping up the solid into thin cylindrical plates, each
${\Delta x }$ units thick, calculating the volume of each plate, then summing up the volumes of all plates.
For example, let's consider a region bounded by $y=x^2$, $y=0$,$x=0$ and $x=1$
<-------Plotting the graph of this area,
If we revolve this area about the x axis ($y=0$), then we get the main image on the right hand side of the page
This image shows a plane area being revolved to create a solid http://curvebank.calstatela.edu/volrev/volrev.htm
To find the volume of the solid using the disc method:
Volume of one disc = ${\pi} y^2{\Delta x}$ where $y$- which is the function- is the radius of the circular cross-section and $\Delta x$ is the thickness of each disc
Volume of all dics:
Volume of all discs = ${\sum}{\pi}y^2{\Delta x}$, with $X$ ranging from 0 to 1
This is also the same as:
$\int_0^1 {\pi}y^2\,dx ={\pi}\int_0^1 (x^2)^2\, dx$
Evaluating this intergral,
${\pi}\int_0^1 x^4 dx$
=$[{{x^5\over 5} + C|}_0^1] {\pi}$
=$[{1\over 5} + {0\over 5}] {\pi}$
=${\pi}\over 5$
volume of solid= ${\pi\over 5} units^3$
References
Bread image http://mathdemos.gcsu.edu/mathdemos/sectionmethod/sectionmethod.html
Revolving image http://mathdemos.gcsu.edu/mathdemos/sectionmethod/sectionmethod.html
Teaching Materials
There are currently no teaching materials for this page. Add teaching materials.
Leave a message on the discussion page by clicking the 'discussion' tab at the top of this image page. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 23, "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": 0.7940955758094788, "perplexity_flag": "middle"} |
http://math.stackexchange.com/questions/194605/calculate-cosv-frac-pi6-when-cos-v-frac23 | Calculate $\cos(v+\frac{\pi}{6})$ when $\cos v = -\frac{2}{3}$
I am to calculate $\cos(v+\frac{\pi}{6})$ when $\cos v = -\frac{2}{3}$ and $0 \lt v \lt \pi$.
I know I can change $\cos(v+\frac{\pi}{6})$ into
$$\cos v \times \cos \frac{\pi}{6} - \sin v \times \sin \frac{\pi}{6}$$
but I fail to see how this helps me, or for that matter any other step I could take. Am I completely off?
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2 Answers
No, you’re doing fine. You now have $\cos\left(v+\frac{\pi}6\right)$ expressed in terms of four quantities: $\cos v$, $\cos\frac{\pi}6$, $\sin\frac{\pi}6$, and $\sin v$. You know (or should know) the numerical values of the first three, so all that’s left is to determine $\sin v$. Now $\cos v$ is negative, so $v$ is in the second or third quadrant. But you also know that $0<v<\pi$, so in fact $v$ is in the second quadrant. Now use the fact that $\sin^2 v+\cos^2v=1$ to get $\sin^2v$, and decide whether you want the positive or negative square root of this for an angle in the second quadrant.
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This helped a lot, I found the answer and it seems to be correct ($-\frac{2\sqrt{3}+\sqrt{5}}{6}$). Thank you! – Quispiam Sep 12 '12 at 10:07
Hint:
$\cos(x+y)=\cos x\cos y-\sin x\sin y$
and
$\sin x=\sqrt {1-\cos^2 x}$ (as $0<x<\pi$ and $\sin x$ is positive in this interval)
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http://calculus7.org/2012/05/08/two-subspaces/ | being boring
## Two subspaces
Posted on 2012-05-08 by
The title is borrowed from a 1969 paper by Paul Halmos. Two subspaces $M$ and $N$ of a Hilbert space $H$ are said to be in generic position if all four intersections $M\cap N$, $M^\perp\cap N$, $M\cap N^\perp$, $M^\perp\cap N^\perp$ are trivial. It may be easier to visualize the condition by writing it as $(M\cup M^\perp)\cap (N\cup N^\perp)=\{0\}$. The term “generic position” is due to Halmos, but the concept was considered before: e.g., in 1948 Dixmier called it “position p”.
Let us consider the finite-dimensional case: $H$ is either $\mathbb R^n$ or $\mathbb C^n$. The dimension count shows that there are no pairs in generic position unless
1. $n=2k$, and
2. $\mathrm{dim}\, M=\mathrm{dim}\, N=k$.
Assume 1 and 2 from now on.
In the simplest case $H=\mathbb R^2$ the situation is perfectly clear: two lines are in generic position if the angle between them is different from $0$ and $\pi/2$. Any such pair of lines is equivalent to the pair of graphs $\{y=0, y=kx\}$ up to rotation. Halmos proved that the same holds in general: there exists a decomposition $H=H_x\oplus H_y$ and a linear operator $T\colon H_x\to H_y$ such that the generic pair of subspaces if unitarily equivalent to $\{y=0, y=Tx\}$.
If we have a preferred orthonormal basis $e_1,\dots,e_n$ in $H$, it is natural to pay particular attention to coordinate subspaces, which are spanned by some subset of $\{e_1,\dots,e_n\}$. Given a subspace $M$, can we find a coordinate subspace $N$ such that $M$ and $N$ are in generic position? The answer is trivially no if $M\cup M^\perp$ contains some basis vector. When $n=2$ this is the only obstruction, as is easy to see:
Lines in generic position
In higher dimensions… later
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http://math.stackexchange.com/questions/257093/iterative-model-fitting | # Iterative model fitting
I have a sequence of points $\{(x_k,y_k,z_k)\}$ and I need to fit some $2D$ model $P(x,y)$ that approximates $z$ in some sense.
The $z_k$$'s$ are noisy samples of some $2D$ function $z_k = f(x,y) + n(x,y)$.
The noise $n$ is mainly shot noise ($std(n(x,y)) \approx \sqrt{f(x,y)}$).
The noise in $x_k,y_k$ can be neglected.
The data is being received sequentially and I can't "step back", i.e. once I moved to data point k+1 I can't get data point k again.
For $1D$ illustration look at the following figure:
I get the points one by one together with some noise component (First I get (1,3+noise), then (2,-2+noise) and so on and once I got (2,-2+noise) I can't access (1,3+noise) anymore unless I saved it somewhere).
I need a method to find\approximate the correct model parameters (In this case the polynomial coefficients: (1,-8,10) ).
Objectives:
I have two main objectives:
1. Good approximation.
2. Low memory requirements.
Other:
I have used least squares polynomial approximation and it's o.k. but if $C$ is the number of terms in my polynomial (Or, more generally, the number of free parameters in my model) then it requires $O(C^2)$ memory (to store the covariance matrix). I would like to be able to use only $O(C)$ memory.
Polynomial isn't mandatory.
I can give up optimality and settle for "good" approximation.
Even more, bounded error is preferred over mean error.
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## 1 Answer
Actually, least squares polynomial fit will do the trick. It can be done in $O(C)$ memory (And $O(C)$ operations) as long as you use orthogonal polynomials.
In this case, the covariance matrix is diagonal (By definition of orthogonality) and there is no need in storing non diagonal elements.
Generally:
Use $orthonormal$ basis so there is no need in normalization.
Carefully choose your basis according to your application needs (For example, If your problem is periodic, use sines and cosines)
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http://gmatclub.com/mwiki/index.php?title=Geometry&oldid=6020 | # Geometry
### From GMATClub
Revision as of 15:35, 25 July 2008 by Dzyubam (Talk | contribs)
(diff) ← Older revision | Current revision (diff) | Newer revision → (diff)
GMAT Study Guide - a prep wikibook
index - edit - roadmap
## Lines
"Line" is a basic concept in geometry. It refers to a straight line extending in both directions.
### Intersecting lines, vertical angles
When two lines intersect, the angles created by these lines possess special qualities. The opposite angles are equal in measure. These opposite angles are also known as vertical angles. The sum of adjacent angles equals 180 degrees.
If the two lines intersect at right angle (90 degrees), these lines are perpendicular. All 4 angles, created by these lines are right angles.
### Parallel lines
The two lines are parallel if they don't intersect. Parallel lines never intersect, no matter how far they extend. When two parallel lines are intersected by the third line, two intersections with vertical angles are created.
## Polygons
Polygon is a plane figure consisting of 3 or more line segments connected to form a closed space. These line segments are named sides; points where the sides meet are named vertices.
Convex polygon is a polygon in which each of the interior agnles is less than $180^o$.
On GMAT, the most widely used kinds of polygons are triangles and quadrilaterals.
The sum of all interior angles of a polygon is found with this formula:
$\Large (n - 2) * 180^o$
Where $n$ is the number of vertices of a polygon. Thus, the sum of angles of a triangle equals $180^o$. Quadrilateral's sum of angles equals $360^o$.
The perimeter of a polygon is the sum of lengths of its sides.
The area of a polygon is the area enclosed within the sides of a polygon.
## Triangles
### Definition
Triangle is a polygon consisting of three vertices and three line segments connecting these vertices.
### Types of triangles
Equilateral triangle is a triangle with all three sides equal in length. All three angles of an equilateral triangle are equal to $60^o$.
Isosceles triangle is a triangle with two sides equal in length. The two angles opposite to the equal sides are equal as well.
Right triangle has one angle equal to $90^o$. Hypotenuse is the side of a right triangle opposite to the right angle. Other two sides are called catheti (singular - cathetus).
Obtuse triangle has one of the angles equal to more than $90^o$ (obtuse angle).
Acute triangle has one of the angles equal to less than $90^o$ (acute angle).
### Properties
• Sum of three angles of any triangle equals $180^o$.
• Sum of lengths of two sides always exceeds the length of the third side.
Two triangles are similar if angles of one triangle are equal to the corresponding angles of the other. Lengths of the corresponding sides of similar triangles are proportional.
### Area
Area of a triangle is generally computed as follows:
$S=\frac{1}{2}ah_a$, where $a$ is a side and $h_a$ is the height or altitude of a triangle dropped from the vertex opposite to side $a$.
Here are some additional formulas:
$S=\sqrt{p(p-a)(p-b)(p-c)}$ - for all triangles ($p$ is a half-perimeter, $a$, $b$ and $c$ are lengths of the sides)
$S=\frac{a^2\sqrt{3}}{4}$ - for equilateral triangle with side $a$
$S=\frac{1}{2}ab$ - for right triangle with catheti $a$ and $b$
### Pythagorean theorem
A right triangle
Pythagorean theorem states that for any right triangle the square of length of the hypotenuse equals the sum of the squared lengths of the catheti. If $c$ is the hepotenuse and $a$ and $b$ are the catheti of the right triangle then:
$\Large a^2 + b^2 = c^2$
The backward statement is also true: if the sides of a triangle satisfy the given equation then the triangle is a right triangle.
There are some special right triangles which are worth remembering (it might save you some time on the test day):
• a 3-4-5 triangle is a right triangle with hypotenuse equal to 5 and catheti of 4 and 3 units respectively $(4^2 + 3^2 = 5^2)$. Note that you may encounter right triangles with sides which are multiples of 3, 4 and 5, respectively. The most common of them is a 6-8-10 triangle
• a 5-12-13 triangle has a hypotenuse equal to 13 and catheti of 5 and 12 respectively $(5^2 + 12^2 = 13^2)$.
In a right triangle with one of the angles equal to $30^o$ the shorter cathetus equals half the length of the hypotenuse.
Also note that in any right triangle the length of the median dropped to the hypotenuse is half the length of the hypotenuse.
### Lines of triangles
BD is a median of $\tiny\triangle$ABC. CD=AD. $\small S\tiny\triangle \small ABD=S \tiny\triangle \small CBD$
Median is a line segment that connects a vertex and the midpoint of the opposite side. Median divides a triangle into two smaller triangles of equal area.
{{#x:vspace|15em}}
## Quadrilaterals
Quadrilateral is a four-sided polygon. It has four vertices and four sides. We'll consider the most popular quadrilaterals here. These are squares, rectangles, parallelograms, trapezoids, rhombuses.
Square ABCD.
AB=BC=CD=AD.
AO=BO=CO=DO.
### Squares
Square is a quadrilateral with all sides and all angles equal. Each angle of a square measures $90^o$.
Square is also called regular quadrilateral.
#### Properties
• Opposite sides of a square are parallel
• All sides of a square are equal
• Diagonals of a square are equal
• Diagonals form a right angle and bisect each other
• Angle formed by a diagonal and an adjacent side equals $45^o$
• All squares can be inscribed into a circle (squares are cyclic quadrilaterals)
Area = $a^2$, where $a$ is the length of the side.
Perimeter = 4*a.
Length of a diagonal equals $\sqrt2*a$.
Parallelogram ABCD.
AB=CD.
AD=BC.
AE=EC, BE=ED.
### Parallelograms
Parallelogram is a quadrilateral with opposite sides parallel and equal in length.
#### Properties
• Opposite sides of a parallelogram are equal and parallel
• Diagonals of a parallelogram are bisecting each other
• Opposite angles of a parallelogram are equal
• Two angles adjacent to the same side sum up to $180^o$
Area = $a*h_a$, where $a$ is the side and $h_a$ is the height drawn down to the side $a$.
Perimeter = $2a + 2b$, where $a$ and $b$ are different sides of parallelogram.
Rectangle ABCD.
AB=CD.
AD=BC.
AC=BD.
### Rectangles
Rectangle is a parallelogram with right angles.
#### Properties
• Opposite sides of a rectangle are equal and parallel
• Diagonals of a rectangle are equal and bisecting each other
• All angles of a rectangle are right
Area = $a*b$, where $a$ and $b$ are different sides of a rectangle.
Perimeter = $2a + 2b$.
{{#x:vspace|3em}}
Rhombus ABCD.
AB=BC=CD=AD.
AK=KC, BK=KD.
### Rhombuses
Rhombus is a parallelogram with all sides equal.
#### Properties
• All sides are equal
• Diagonals of a rhombus intersect under right angle
Area = $a*h$, where $a$ is the side and $h$ is the height drawn down to the side $a$.
Area = $\frac{1}{2}*c*d$, where $c$ and $d$ are diagonals of a rhombus.
Perimeter = $4*a$.
Trapezoid ABCD.
### Trapezoids
Trapezoid is a quadrilateral with two opposite sides parallel.
#### Properties
• Two of the four sides are parallel
Area = $\frac{b+d}{2}*h$, where $b$ and $d$ are the parallel sides and $h$ is the height of the trapezoid.
Perimeter = $a + b + c + d$.
{{#x:vspace|3em}}
## Circles
Circle with center O. Radius = OB = OA = OC. DC is the chord. AB is the diameter. Central angle COB is subtended by the arc CEB. Line AF is tangent to the circle. $\angle$OAF is right.
### Definitions
A circle is a set of points on a plane equidistant from a certain point (called the center of the circle).
A radius is a segment connecting the center of the circle and the point on the circle.
A chord is a segment connecting two points on the circle.
A diameter is a chord passing through the center of the circle.
An arc is any part of a circle.
A central angle of a circle is the angle whose vertex is the center of the circle and the sides pass through the two points on the circle.
A line is tangent to the circle if it has only one point common with the circle.
### Area and circumference
The circumference is a perimeter of a circle. It is equal to $2\pi r$, where $r$ is the radius of a circle. $\pi$ is approximately 3.14.
The area of a circle is equal to $\pi r^2$.
The length of an arc is equal to $\frac{x}{360}$ of the circumference of the circle, where $x$ is the central angle subtended by the endpoints of the arc. If $\angle$COB from the picture above equals 45 degrees, then the length of the arc CEB equals $\frac{45}{360}=\frac{1}{8}$ of the circumference of the circle.
$\angle$COB is the central angle.
$\angle$CDB = $\angle$CAB = $\frac{1}{2}\angle$COB.
$\angle$CEB + $\angle$CDB = $\small 180^\circ$.
$\angle$CAB = $\angle$CEB = $\angle$CDB = $\small 90^\circ$
### Angles
Here are some important properties of the inscribed angles:
• A central angle equals twice an inscribed angle if they are subtended by the same chord (if both angles are on the same side of the chord). The image on the left illustrates it, $\angle$COB = 2$\angle$CDB = 2$\angle$CAB.
• Two inscribed angles subtended by the same chord and on the same side of the chord are equal (on the left: $\angle$CAB = $\angle$CDB).
• Two inscribed angles subtended by the same chord and on the opposite sides of the chord are supplemental (the sum of the two angles equals $\normal 180^\circ$). See the image on the left: $\angle$CAB + $\angle$CEB = $\normal 180^\circ$.
• All inscribed angles subtended by a diameter equal $\normal 90^\circ$ (see the image on the right).
## Coordinate geometry
### Equation
Equation of a line is $y=mx+b$, where m=slope and b=y intercept.
Equation of a circle is $(x - a)^2 + (y-b)^2 = r^2$, where (a,b) is the center and $r$ is the radius.
Equation of a circle is $x^2+y^2=r^2$ if (0,0) is the center.
• Points that solve the equation of a line are in the same line
• Given a point and slope, equation of the line can be found
• Given the equation, x and y intercepts can be found
Intercept
• Y intercept is the value of y when x is 0
• X intercept is the value of x when y is 0
### Distance
Distance between two points = $\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}$.
### Midpoint
$Xm=\frac{x_1+x_2}{2}$
$Ym=\frac{y_1+y_2}{2}$
### Slope
Slope $m =\frac{(y_1-y_2)}{(x_1-x_2)}$.
• A straight line with a -ve slope passes through II and IV quadrants
• A straight line with a +ve slope passes through I and III quadrants
• If the slope is 1 the angle formed by the line is 45 degrees.
• If the slope of a line is n, the slope of a line perpendicular to it is its -ve reciprocal, -1/n.
• If a line is horizontal, slope=0, equation is y=b.
• If a line is vertical, slope is not defined, equation is x=a, where a is x-intercept.
• Parallel lines have same slope. | {"extraction_info": {"found_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": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 101}, "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": 0.8979863524436951, "perplexity_flag": "middle"} |
http://mathhelpforum.com/discrete-math/178786-finding-min-max-number-edges-connected-graph.html | # Thread:
1. ## Finding min and max number of edges of a connected graph
A simple graph of 25 vertices and 6 connected components.
What is the minimum and maximum number of edges can the graph have?
Is there a formula to calculate the min and max? 25 vertices is too huge to draw and there are several possibilities for minimum number as well as maximum number of edges.
Thanks.
2. Originally Posted by xEnOn
A simple graph of 25 vertices and 6 connected components. What is the minimum and maximum number of edges can the graph have?
A single vertex is a degenerate component. Think of five degenerate components and one with twenty vertices.
If the one with twenty vertices is tree, does that give a minimum?
Does $\mathcal{K}_{20}$ give a maximum?
3. ohh...a graph has minimum edges when it is a tree. So Edges = 20-1 = 19, is the minimum number of edges.
But for the maximum, how I can calculate the maximum edges when there are 20 vertices? It can have infinite number of edges wouldn't it?
4. A simple graph (which I presume this is or the q won't make sense) allows only one edge per vertex pair, and no loops.
5. Then the maximum number of edges for a graph of 20 vertices could be $20!$ ?
Like Vertex A could join Vertex B, could also join Vertex C and form a complete graph?
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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": 2, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": 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": 0.928704023361206, "perplexity_flag": "middle"} |
http://mathhelpforum.com/algebra/138558-arithmetic-progression.html | # Thread:
1. ## Arithmetic progression
the $N^{th}$ term of an A.P is denoted by $U_{n}$ and the sum of the $1^{st}n$ terms by $S_{N}$ in a certain A.P ;; $U_{5}+U_{16}=44$ and $S_{18}=3S_{10}.$calculate the value of the first term and the common difference.
i have been trying to get two equations from the given equation using the rules for arithmetic progression but i have been unable to do so..
i really need some help coming up with these equations.
2. Originally Posted by sigma1
the $N^{th}$ term of an A.P is denoted by $U_{n}$ and the sum of the $1^{st}n$ terms by $S_{N}$ in a certain A.P ;; $U_{5}+U_{16}=44$ and $S_{18}=3S_{10}.$calculate the value of the first term and the common difference.
i have been trying to get two equations from the given equation using the rules for arithmetic progression but i have been unable to do so..
i really need some help coming up with these equations.
Using the formula or the n-th term in an A.P., for all $n\,,\,\,u_n=u_1+(n-1)d\Longrightarrow 44=u_5+u_{16}=$ $u_1+4d+u_1+15d=2u_1+19d\Longrightarrow (I)\,\,2u_1+19d=44$ , and using the formula for the sum of consecutive elements in an A.P. we get
$3S_{10}=3\cdot \frac{10}{2}\left(2u_1+9d\right)=\frac{18}{2}\left (2u_1+17d\right)=S_{18}$ $\Longrightarrow 10u_1+45d=6u_1+51d\Longrightarrow (II)\,\,2u_1-3d=0$ , and there you have two linear eq's in two unknowns: $u_1\,\,\,and\,\,\,d$
Tonio | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 17, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9622083306312561, "perplexity_flag": "head"} |
http://mathoverflow.net/revisions/40586/list | ## Return to Answer
2 added 78 characters in body
Here is a complete answer; I think it is more or less what Steve wrote in his comment, except I don't understand the appearance of $\mathbb{R}$ there. If $I$ is the infinite index set, let $L=\mathbb{Z}^{(I)}\subset P$ be the obvious free submodule. Then $\mathrm{Ext}^1(P,\mathbb{Z})=\mathrm{Ext}^1(P/L,\mathbb{Z})$.
EDIT: the last formula is wrong, see Martin's and Steve's comments below.
Now $P/L$ has a big divisible subgroup $D$, whose inverse image in $P$ consists of maps $I\to\mathbb{Z}$ converging to zero in $\widehat{\mathbb{Z}}$ (the profinite completion of $\mathbb{Z}$). (For instance, if $I=\mathbb{N}$ take the sequence $n\mapsto n!$). Since $P/L$ is torsion-free (imediate), $D$ is a nonzero $\mathbb{Q}$-vector space. Since $D$ is divisible it is a direct summand of $P/L$; hence, $P/L$ admits $\mathbb{Q}$ as a direct summand. But it is well known (and easy to see) that $\mathrm{Ext}^1(\mathbb{Q}/\mathbb{Z},\mathbb{Z})\cong\widehat{\mathbb{Z}}$, hence $\mathrm{Ext}^1(\mathbb{Q},\mathbb{Z})=\widehat{\mathbb{Z}}/{\mathbb{Z}}\neq0$.
1
Here is a complete answer; I think it is more or less what Steve wrote in his comment, except I don't understand the appearance of $\mathbb{R}$ there. If $I$ is the infinite index set, let $L=\mathbb{Z}^{(I)}\subset P$ be the obvious free submodule. Then $\mathrm{Ext}^1(P,\mathbb{Z})=\mathrm{Ext}^1(P/L,\mathbb{Z})$.
Now $P/L$ has a big divisible subgroup $D$, whose inverse image in $P$ consists of maps $I\to\mathbb{Z}$ converging to zero in $\widehat{\mathbb{Z}}$ (the profinite completion of $\mathbb{Z}$). (For instance, if $I=\mathbb{N}$ take the sequence $n\mapsto n!$). Since $P/L$ is torsion-free (imediate), $D$ is a nonzero $\mathbb{Q}$-vector space. Since $D$ is divisible it is a direct summand of $P/L$; hence, $P/L$ admits $\mathbb{Q}$ as a direct summand. But it is well known (and easy to see) that $\mathrm{Ext}^1(\mathbb{Q}/\mathbb{Z},\mathbb{Z})\cong\widehat{\mathbb{Z}}$, hence $\mathrm{Ext}^1(\mathbb{Q},\mathbb{Z})=\widehat{\mathbb{Z}}/{\mathbb{Z}}\neq0$. | {"extraction_info": {"found_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": 0.9569548964500427, "perplexity_flag": "head"} |
http://math.stackexchange.com/questions/43973/proving-that-a-function-is-bounded/43975 | # Proving that a function is bounded
This is one part of an exercise in my homework, which for some reason I can't think of any way to prove.
$\displaystyle f(x,y)=\frac{xy^2}{x^2+y^4}$, if $(x,y)\neq (0,0)$ and $0$ otherwise.
I'm trying to prove that this function is bounded. I have figured that I only need to prove it for $x\geq 0$, since $f(x,y)=-f(-x,y)$, but I can't really get around to why this is bounded near $0$.
Thanks for the help.
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Have you tried letting $z=y^2$? – Kimball Jun 8 '11 at 0:01
## 1 Answer
Hint: $(x-y^2)^2 \ge 0$ ${}{}{}{}{}{}{}{}{}$
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+1. This is a really elegant hint. You might want to replace $>$ with $\ge$ – user17762 Jun 8 '11 at 0:06
@Sivaram: I noticed it about the same time. Fixed. – Ross Millikan Jun 8 '11 at 0:08
1
– Aaron Jun 8 '11 at 0:12
@Aaron I'm well aware of that, thanks. Great hint, the answer comes practically for free. – Leonardo Fontoura Jun 8 '11 at 0:29
It took me a second to see what you were getting at, but I really appreciate this hint. Nicely constructed. – mixedmath♦ Jun 8 '11 at 1:18 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 11, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9590991735458374, "perplexity_flag": "middle"} |
http://mathoverflow.net/questions/107491?sort=oldest | Mackey(also Green and Tambara) functors and Greenlees-May
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This is somewhat related to a question that I asked on Math.SE but, sadly, received no response. I apologize ahead of time if this is not appropriate for MO. Feel free to vote to close if this is the case.
I really have two (distinct) questions. The first is regarding a paper by Greenlees and May and the second is more of a "big-picture" question with no explicit relationship to their paper.
Let $G$ be a finite group.
In their paper Some Remarks On the Structure of Mackey functors , Greenlees and May define the functor:
$R: GMod \rightarrow M[G]$ where $GMod$ is the category of finite left $G$-modules and $M[G]$ is the category of $G$ Mackey functors by:
$RV(G/H) = V^H$ where $V$ is a $G$ module and $V^H$ is the $H$ fixed point set of $V$.
In their main theorem(Thm. 12) they consider the map $\eta: M \rightarrow RM(G/H_{j,k})$.
Question 1: In Theorem 12, why are $coker(\eta)$ and $ker(\eta)$ in $\mathcal{A}$? Unfortunately I don't see how this clearly follows from the induction hypothesis at the moment.
Question 2: In general, What are some examples of added benefits (aside from additional structure) that one obtains when it is known that you have a Green or Tambara functor rather than just a Mackey functor?
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2 Answers
I thank you for the careful reading and apologize for the concision. This is a downwards induction on the size of subgroups. Using the explicit description of the $RV$ given top of page 239 and the conventions recalled at the bottom of page 240, we arrived at the description of the relevant $RV$ given just below Prop. 7. The map $\eta$ is the identity on $M(G/H_{j,k})$ and since $M(G/H_{j',k'}) = 0$ for $j'< j$ and for $j'=j$ and $k'< k$, the same is true of $RM(G/H_{j,k})$. Therefore $Ker(\eta)$ and $Coker(\eta)$ can only be non-zero on $G/H_{j',k'}$ where $j'>j$ or $j'=j$ and $k'>k$ and hence they are in $\mathcal{A}$ by the induction hypothesis.
For the second question, I see no obvious relationship between our additive description of Mackey functors and any multiplicative structure that might be present. Anything anyone can say about that would be welcome. I've not thought hard about the question.
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@PeterMay Thank you for your response! That makes much more sense now. Personally, the definition on the bottom of page 238 (definition 1) of $RV$ in terms of the $H-$fixed points seems more natural. Is there is a way to see $ker$ and $coker$ being in $\mathcal{A}$ without appealing to the explicit construction given on the top of page 239? My second question is not directly related to your paper. I was simply asking (in general) as to what one gains from having a Green or Tambara functor instead of a run-of-the-mill Mackey functor. I will edit as to make this distinction clear. Thanks! – confusedmath Sep 19 at 2:07
@PeterMay Also, I was wondering if there are analogous theorems for Green/Tambara functors. Are all Green (resp.Tambara) functors built from "simpler" Green (resp. Tambara) functors? – confusedmath Sep 19 at 2:16
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That is not what one expects from analogy with simpler structures. It would be of interest is to compute the box product'' $RV\Box RW$ in terms of the additive description of Mackey functors that Greenlees and I gave. The point is that a Green functor $M$ is specified by a product $M\Box M \to M$. The calculation should not be hard in the simplest cases (cyclic groups of prime order say), and as far as I know has not been done.
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http://math.stackexchange.com/questions/176019/about-poisson-equation | # About Poisson Equation.
I want to solve $- \Delta u = f$ in $\Omega$ with $u = \phi$ on $\partial \Omega$. But if I have the solutions of (1) and (2) below : $$- \Delta u_1 = f \; \text{in } \Omega , \; u_1 = 0 \; \text{on } \partial \Omega \tag{1}$$ $$- \Delta u_2 = 0 \; \text{in } \Omega , \; u_2 = \phi \; \text{on } \partial \Omega \tag{2}$$ Then how can I solve the problem $- \Delta u = f$ in $\Omega$ with $u = \phi$ on $\partial \Omega$ by using $u_1 , u_2$ ?
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2
Hint: The Laplacian $\Delta$ is a linear operator. – Rahul Narain Jul 27 '12 at 22:34
## 1 Answer
The solution is $v=u_1+u_2$. In fact, we have $$-\Delta v= -\Delta u_1 - \Delta u_2 = f \quad \mbox{in} \Omega$$ and $$v= 0 +\phi = \phi\quad \mbox{on} \quad \partial \Omega.$$
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http://math.stackexchange.com/questions/264771/cross-product-and-pseudovector-confusion?answertab=oldest | # Cross product and pseudovector confusion.
So called pseudovectors pop up in physics when discussing quantities defined by cross products, such as angular momentum $\mathbf L=\mathbf r\times\mathbf p$. Under the active transformation $\mathbf x \mapsto \mathbf{-x}$, we claim that such a vector gets mapped to itself because $\mathbf{-r} \times \mathbf{-p} = \mathbf r\times\mathbf p$. (Or under the equivalent passive transformation, a pseudovector turns into to its negative.) But it seems like we're just pretending that a linear transformation $T$ preserve cross products, so that $T(\mathbf a \times \mathbf b) = T(\mathbf a) \times T(\mathbf b)$, and then when things don't go as expected we label the result as a pseudovector. Is there more to the story?
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This question might be more appropriate to the physics SE. – Fabian Dec 24 '12 at 23:13
As far as I know, the difference between vectors and pseudovectors appears when one considered inversion (or mirror reflection). – Fabian Dec 24 '12 at 23:15
1
Yes, there is more to the story. If you want to fully formalize those things in a convenient mathematical framework you need to move into the realm of the differential forms. Try looking for the keyword "pseudovector" in the book "The Geometry of Physics" by Theodore Frankel. – Giuseppe Negro Dec 24 '12 at 23:30
Thanks, I will take a look at Frenkel's book, although I don't really have any experience with differential geometry yet. – yuval Dec 25 '12 at 0:15
## 2 Answers
In three dimensions, pseudovectors are a simple way to treat bivectors, oriented planar subspaces. True vectors are oriented linear subspaces with a weight (their magnitudes); bivectors are planar instead of linear. The normal vectors to these oriented subspaces are what we usually call pseudovectors, and it is for this reason that various operations (like reflections or inversions through the origin) produce "wrong" results.
Notationally, we deal directly with a bivector by forming a wedge product of vectors. That is, the bivector formed by vectors $a,b$ is $a \wedge b$. Given a linear operator $\underline T$, we define the action of the linear operator on a bivector by the following law:
$$\underline T(a \wedge b) \equiv \underline T(a) \wedge \underline T(b)$$
Let us consider the simple case of $\underline T(a) = -a$ for any $a$. Then the associated bivector transforms as $\underline T(a \wedge b) = -a \wedge -b = a \wedge b$, as you observe. Doing it this way--by defining the action of a linear operator on a bivector--makes it sensible, rather than saying simply that pseudovectors transform differently from regular vectors. Here, you build the operator according to a specific rule, and the result is deterministic.
Note that we can continue to build things with wedges that traditional formulations of vector algebra and calculus tend to gloss over. We can define the action of a linear operator on three vectors wedged together.
$$\underline T(a \wedge b \wedge c) = \underline T(a) \wedge \underline T(b) \wedge \underline T(c)$$
The quantity $a \wedge b \wedge c$ is called a trivector or pseudoscalar. In three dimensions, there is only one linearly independent unit trivector, $\hat x \wedge \hat y \wedge \hat z$. The action of $\underline T$ on this object is very interesting. It happens that
$$\underline T(\hat x \wedge \hat y \wedge \hat z) = \hat x \wedge \hat y \wedge \hat z \, \det \underline T$$
This can be taken as a definition of the determinant, defined in a wholly geometric way.
Ultimately, though, yes, linear operators should act individually on the vectors that make them up--they should preserve wedge products. Cross products are related to wedges, however, and so most of the time, applying a linear operator to preserve crosses is sensible, but there are some times (inversion and reflections being among them) that it is not.
Edit: about the relations between operators on duals and duals of operators. The Hodge star is much, much better treated in geometric algebra as multiplication by the pseudoscalar. We define $i \equiv \hat x \wedge \hat y \wedge \hat z$ and make sense of expressions like $\star a = i a$ and $\star (a \wedge b) = -i (a \wedge b)$ through the geometric product. Here are the canonical properties of the geometric product:
• $\hat u \hat u = 1$ for some unit vector $\hat u$
• $\hat u \hat v = - \hat v \hat u$ for two orthogonal unit vectors $\hat u, \hat v$
• $(ab)c = a(bc)$--that is, associativity--for three vectors $a, b, c$
You should be able to show then that $i = \hat x \hat y \hat z$ and that $\star a = i a$ captures the Hodge star operation on a vector.
Now, why bother with this stuff? Because it makes formulas that would be ugly and clumsy with the Hodge star very simple. There exists a simple formula relating the adjoint (in Euclidean space, the transpose) of an operator with the inverse. That is,
$$\overline T^{-1}(a) = [\underline T(i)]^{-1} \underline T(ia)$$
for any multivector $a$, where $\overline T$ is the adjoint operator to $\underline T$. Written with Hodge stars, we would need a term of $(-1)^k$ that would alternate based on grade, and it would all be a royal mess. This formula, however, written in geometric algebra, is entirely simple.
Now then, rotations and reflections all belong to the group of orthogonal linear operators, obeying $\overline T^{-1} = \underline T$, so for rotations and reflections we get instead,
$$\underline T(a) = \frac{1}{\det \underline T} i^{-1} \underline T(ia)$$
or, more simply,
$$(\det \underline T) i \underline T(a) = \underline T(ia)$$
In Hodge star notation, for any vector $a$,
$$(\det \underline T) \star[\underline T(a)] = \underline T(\star a)$$
For a rotation, the determinant is $+1$, and as such, the $i$ just pulls out. Rotating the vector and then finding the dual is the same as rotating the dual. For an inversion, the determinant is $-1$, and you can see how the inversion of the vector gets canceled by the determinant's factor.
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Very helpful reply, thanks! I read up a bit on wedge products and their relation to the cross product via the Hodge dual. Is there any meaningful relationship between, say, $T(\star(\mathbf a \wedge \mathbf b))$ and $\star(T(\mathbf a \wedge \mathbf b))$? – yuval Dec 25 '12 at 23:10
@yuval I've added a section on orthogonal operators and operations of duals compared to duals of operators. The core result is that $(\det \underline T) \star[\underline T(a)] = \underline T(\star a)$ for any vector $a$ when $\underline T$ is orthogonal. – Muphrid Dec 26 '12 at 2:10
@Muphrid: Nicely done. (+1) – oen Dec 26 '12 at 19:59
Very informative! – rschwieb Dec 27 '12 at 18:22
Here's a another way of looking at it. Under active transformation $$\begin{eqnarray*} {\bf r}\times{\bf p} &=& r_i p_j \epsilon_{i j k} {\bf e}_k \qquad \textrm{(component expression for cross product)} \\ &\to& (-r_i) (-p_j) \epsilon_{i j k} {\bf e}_k \qquad \textrm{(only components change under active transformation)}\\ &=& {\bf r}\times{\bf p}. \end{eqnarray*}$$ Above $\epsilon_{i j k}$ is the Levi-Civita symbol, ${\bf e}_i$ is the basis, and Einstein's summation convention is used.
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Updates on my research and expository papers, discussion of open problems, and other maths-related topics. By Terence Tao
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## 254A, Lecture 3: Minimal dynamical systems, recurrence, and the Stone-Čech compactification
13 January, 2008 in 254A - ergodic theory, math.DS, math.GN, math.LO | Tags: almost periodicity, lamplighter group, recurrence, Stone-Cech compactification, syndetic sets, ultrafilter | by Terence Tao | 53 comments
We now begin the study of recurrence in topological dynamical systems $(X, {\mathcal F}, T)$ – how often a non-empty open set U in X returns to intersect itself, or how often a point x in X returns to be close to itself. Not every set or point needs to return to itself; consider for instance what happens to the shift $x \mapsto x+1$ on the compactified integers $\{-\infty\} \cup {\Bbb Z} \cup \{+\infty\}$. Nevertheless, we can always show that at least one set (from any open cover) returns to itself:
Theorem 1. (Simple recurrence in open covers) Let $(X,{\mathcal F},T)$ be a topological dynamical system, and let $(U_\alpha)_{\alpha \in A}$ be an open cover of X. Then there exists an open set $U_\alpha$ in this cover such that $U_\alpha \cap T^n U_\alpha \neq \emptyset$ for infinitely many n.
Proof. By compactness of X, we can refine the open cover to a finite subcover. Now consider an orbit $T^{\Bbb Z} x = \{ T^n x: n \in {\Bbb Z} \}$ of some arbitrarily chosen point $x \in X$. By the infinite pigeonhole principle, one of the sets $U_\alpha$ must contain an infinite number of the points $T^n x$ counting multiplicity; in other words, the recurrence set $S := \{ n: T^n x \in U_\alpha \}$ is infinite. Letting $n_0$ be an arbitrary element of S, we thus conclude that $U_\alpha \cap T^{n_0-n} U_\alpha$ contains $T^{n_0} x$ for every $n \in S$, and the claim follows. $\Box$
Exercise 1. Conversely, use Theorem 1 to deduce the infinite pigeonhole principle (i.e. that whenever ${\Bbb Z}$ is coloured into finitely many colours, one of the colour classes is infinite). Hint: look at the orbit closure of c inside $A^{\Bbb Z}$, where A is the set of colours and $c: {\Bbb Z} \to A$ is the colouring function.) $\diamond$
Now we turn from recurrence of sets to recurrence of individual points, which is a somewhat more difficult, and highlights the role of minimal dynamical systems (as introduced in the previous lecture) in the theory. We will approach the subject from two (largely equivalent) approaches, the first one being the more traditional “epsilon and delta” approach, and the second using the Stone-Čech compactification $\beta {\Bbb Z}$ of the integers (i.e. ultrafilters).
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http://mathhelpforum.com/advanced-algebra/91495-finding-basis.html | # Thread:
1. ## Finding a basis
Can anyone verify that I am doing this correctly?
I have to find a basis for
$W=\{(a_{1}, a_{2}, a_{3}, a_{4}, a_{5} \in F^{5} | a_{1}-a_{3}-a_{4}=0\}$
So, does this mean that $a_{2}$ and $a_{5}$ can be any numbers?
I began by setting them both equal to one, giving me (0, 1, 0, 0 ,0) and (0, 0, 0, 0, 1)
Then, by letting $a_{1}=1$, this gave me $1=a_{3}+a_{4}$. Then choosing $a_{3}=0$ or $a_{3}=1$, I obtain $a_{4}=1$ or $a_{4}=0$, respectively.
So that gave me the vectors (1, 0, 0, 1, 0) and (1, 0, 1, 0, 0).
Repeating this process for a different $a$ gave me
(0, 0, 1, 1, 0).
That's a total of 5 vectors, which is how many the basis should have, correct? Can anyone verify this solution? Is this a correct method to use? I just feel weird "choosing" the a's to be 1 or 0 and stuff.
2. Originally Posted by paupsers
Can anyone verify that I am doing this correctly?
I have to find a basis for
$W=\{(a_{1}, a_{2}, a_{3}, a_{4}, a_{5} \in F^{5} | a_{1}-a_{3}-a_{4}=0\}$
So, does this mean that $a_{2}$ and $a_{5}$ can be any numbers? yes!
I began by setting them both equal to one, giving me (0, 1, 0, 0 ,0) and (0, 0, 0, 0, 1)
Then, by letting $a_{1}=1$, this gave me $1=a_{3}+a_{4}$. Then choosing $a_{3}=0$ or $a_{3}=1$, I obtain $a_{4}=1$ or $a_{4}=0$, respectively.
So that gave me the vectors (1, 0, 0, 1, 0) and (1, 0, 1, 0, 0).
Repeating this process for a different $a$ gave me
(0, 0, 1, 1, 0).
That's a total of 5 vectors, which is how many the basis should have, correct? Can anyone verify this solution? Is this a correct method to use? I just feel weird "choosing" the a's to be 1 or 0 and stuff.
no, it's not quite correct. from the relation $a_1-a_3-a_4=0$ you get $a_1=a_3+a_4.$ so an element of $W$ is in the form:
$(a_3+a_4,a_2,a_3,a_4,a_5)=a_2e_1+a_3e_2+a_4e_3+a_5 e_4,$ where $e_1=(0,1,0,0,0), \ e_2=(1,0,1,0,0), \ e_3=(1,0,0,1,0),$ and $e_4=(0,0,0,0,1).$
the set $\{e_1,e_2,e_3,e_4 \}$ is a basis for $W.$
3. Originally Posted by paupsers
Can anyone verify that I am doing this correctly?
I have to find a basis for
$W=\{(a_{1}, a_{2}, a_{3}, a_{4}, a_{5} \in F^{5} | a_{1}-a_{3}-a_{4}=0\}$
So, does this mean that $a_{2}$ and $a_{5}$ can be any numbers?
I began by setting them both equal to one, giving me (0, 1, 0, 0 ,0) and (0, 0, 0, 0, 1)
Then, by letting $a_{1}=1$, this gave me $1=a_{3}+a_{4}$. Then choosing $a_{3}=0$ or $a_{3}=1$, I obtain $a_{4}=1$ or $a_{4}=0$, respectively.
So that gave me the vectors (1, 0, 0, 1, 0) and (1, 0, 1, 0, 0).
Repeating this process for a different $a$ gave me
(0, 0, 1, 1, 0).
That's a total of 5 vectors, which is how many the basis should have, correct? Can anyone verify this solution? Is this a correct method to use? I just feel weird "choosing" the a's to be 1 or 0 and stuff.
No. A basis for all of $R^5$ must have 5 vectors but this is a subspace, not all of $R^5$. Typically each equation restricting values reduces the dimension by 1. Since this space is restricted by one equation, we would expect the dimension to be 4 and expect a basis to contain four vectors.
Yes, $a_2$ and $a_5$ have no restrictions so (0, 1, 0, 0, 0) and (0, 0, 0, 0, 1) are basis vectors for it. NonCommAlg solved the single equation for $a_1$. You could, of course, solve for either $a_3$ or $a_4$ as functions of the other two.
For example, solving for $a_3= a_1+ a_4$, we can see that if $a_1= 1$ and $a_4= 0$, then $a_3= 1+ 0= 1$ so (1, 0, 1, 0, 0) is another basis vector. If $a_1= 0$ and $a_4= 0$, then $a_3= 0+ 1= 1$ so (0, 0, 1, 1, 0) is a fourth basis vector. Those four vectors, {(0, 1, 0, 0, 0), (0, 0, 0, 0, 1), (1, 0, 1, 0, 0), (0, 0, 1, 1, 0)} form a basis for this subspace slightly different from the one NonCommAlgebra gave. | {"extraction_info": {"found_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": 52, "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": 0.9377951622009277, "perplexity_flag": "head"} |
http://en.wikipedia.org/wiki/Metric_spaces | # Metric space
(Redirected from Metric spaces)
In mathematics, a metric space is a set where a notion of distance (called a metric) between elements of the set is defined.
The metric space which most closely corresponds to our intuitive understanding of space is the 3-dimensional Euclidean space. In fact, the notion of "metric" is a generalization of the Euclidean metric arising from the four long-known properties of the Euclidean distance. The Euclidean metric defines the distance between two points as the length of the straight line segment connecting them. Other metric spaces occur for example in elliptic geometry and hyperbolic geometry, where distance on a sphere measured by angle is a metric, and the hyperboloid model of hyperbolic geometry is used by special relativity as a metric space of velocities.
A metric space also induces topological properties like open and closed sets which leads to the study of even more abstract topological spaces.
## History
This section requires expansion with: Reasons for generalizing the Euclidean metric, first non-Euclidean metrics studied, consequences for mathematics. (August 2011)
Maurice Fréchet introduced metric spaces in his work Sur quelques points du calcul fonctionnel, Rendic. Circ. Mat. Palermo 22 (1906) 1–74.
## Definition
A metric space is an ordered pair $(M,d)$ where $M$ is a set and $d$ is a metric on $M$, i.e., a function
$d \colon M \times M \rightarrow \mathbb{R}$
such that for any $x, y, z \in M$, the following holds:
1. $d(x,y) \ge 0$ (non-negative),
2. $d(x,y) = 0\,$ iff $x = y\,$ (identity of indiscernibles),
3. $d(x,y) = d(y,x)\,$ (symmetry) and
4. $d(x,z) \le d(x,y) + d(y,z)$ (triangle inequality) .
The first condition follows from the other three, since:
$2d(x,y) = d(x,y) + d(y,x) \ge d(x,x) = 0.$
The function $d$ is also called distance function or simply distance. Often, $d$ is omitted and one just writes $M$ for a metric space if it is clear from the context what metric is used.
## Examples of metric spaces
• Ignoring mathematical details, for any system of roads and terrains the distance between two locations can be defined as the length of the shortest route connecting those locations. To be a metric there shouldn't be any one-way roads. The triangle inequality expresses the fact that detours aren't shortcuts. Many of the examples below can be seen as concrete versions of this general idea.
• The real numbers with the distance function $d(x,y) = \vert y - x \vert$ given by the absolute difference, and more generally Euclidean $n$-space with the Euclidean distance, are complete metric spaces. The rational numbers with the same distance also form a metric space, but are not complete.
• The positive real numbers with distance function $d(x,y) =\vert \log(y/x) \vert$ is a complete metric space.
• Any normed vector space is a metric space by defining $d(x,y) = \lVert y - x \rVert$, see also metrics on vector spaces. (If such a space is complete, we call it a Banach space.) Examples:
• The Manhattan norm gives rise to the Manhattan distance, where the distance between any two points, or vectors, is the sum of the differences between corresponding coordinates.
• The maximum norm gives rise to the Chebyshev distance or chessboard distance, the minimal number of moves a chess king would take to travel from $x$ to $y$.
• The British Rail metric (also called the Post Office metric or the SNCF metric) on a normed vector space is given by $d(x,y) = \lVert x \rVert + \lVert y \rVert$ for distinct points $x$ and $y$, and $d(x,x) = 0$. More generally $\lVert . \rVert$ can be replaced with a function $f$ taking an arbitrary set $S$ to non-negative reals and taking the value $0$ at most once: then the metric is defined on $S$ by $d(x,y) = f(x) + f(y)$ for distinct points $x$ and $y$, and $d(x,x) = 0$. The name alludes to the tendency of railway journeys (or letters) to proceed via London (or Paris) irrespective of their final destination.
• If $(M,d)$ is a metric space and $X$ is a subset of $M$, then $(X,d)$ becomes a metric space by restricting the domain of $d$ to $X \times X$.
• The discrete metric, where $d(x,y) = 0$ if $x=y$ and $d(x,y) = 1$ otherwise, is a simple but important example, and can be applied to all non-empty sets. This, in particular, shows that for any non-empty set, there is always a metric space associated to it. Using this metric, any point is an open ball, and therefore every subset is open and the space has the discrete topology.
• A finite metric space is a metric space having a finite number of points. Not every finite metric space can be isometrically embedded in a Euclidean space.[1][2]
• The hyperbolic plane is a metric space. More generally:
• If $M$ is any connected Riemannian manifold, then we can turn $M$ into a metric space by defining the distance of two points as the infimum of the lengths of the paths (continuously differentiable curves) connecting them.
• If $X$ is some set and $M$ is a metric space, then, the set of all bounded functions $f \colon X \rightarrow M$ (i.e. those functions whose image is a bounded subset of $M$) can be turned into a metric space by defining $d(f,g) = \sup_{x \in X} d(f(x),g(x))$ for any two bounded functions $f$ and $g$ (where $\sup$ is supremum.[3] This metric is called the uniform metric or supremum metric, and If $M$ is complete, then this function space is complete as well. If X is also a topological space, then the set of all bounded continuous functions from $X$ to $M$ (endowed with the uniform metric), will also be a complete metric if M is.
• If $G$ is an undirected connected graph, then the set $V$ of vertices of $G$ can be turned into a metric space by defining $d(x,y)$ to be the length of the shortest path connecting the vertices $x$ and $y$. In geometric group theory this is applied to the Cayley graph of a group, yielding the word metric.
• The Levenshtein distance is a measure of the dissimilarity between two strings $u$ and $v$, defined as the minimal number of character deletions, insertions, or substitutions required to transform $u$ into $v$. This can be thought of as a special case of the shortest path metric in a graph and is one example of an edit distance.
• Given a metric space $(X,d)$ and an increasing concave function $f \colon [0,\infty) \rightarrow [0,\infty)$ such that $f(x) = 0$ if and only if $x=0$, then $f \circ d$ is also a metric on $X$.
• Given an injective function $f$ from any set $A$ to a metric space $(X,d)$, $d(f(x), f(y))$ defines a metric on $A$.
• Using T-theory, the tight span of a metric space is also a metric space. The tight span is useful in several types of analysis.
• The set of all $m$ by $n$ matrices over some field is a metric space with respect to the rank distance $d(X,Y) = \mathrm{rank}(Y - X)$.
• The Helly metric is used in game theory.
## Open and closed sets, topology and convergence
Every metric space is a topological space in a natural manner, and therefore all definitions and theorems about general topological spaces also apply to all metric spaces.
About any point $x$ in a metric space $M$ we define the open ball of radius $r > 0$ about $x$ as the set
$B(x;r) = \{y \in M : d(x,y) < r\}.$
These open balls form the base for a topology on M, making it a topological space.
Explicitly, a subset $U$ of $M$ is called open if for every $x$ in $U$ there exists an $r > 0$ such that $B(x;r)$ is contained in $U$. The complement of an open set is called closed. A neighborhood of the point $x$ is any subset of $M$ that contains an open ball about $x$ as a subset.
A topological space which can arise in this way from a metric space is called a metrizable space; see the article on metrization theorems for further details.
A sequence ($x_n$) in a metric space $M$ is said to converge to the limit $x \in M$ iff for every $\epsilon>0$, there exists a natural number N such that $d(x_n,x) < \epsilon$ for all $n > N$. Equivalently, one can use the general definition of convergence available in all topological spaces.
A subset $A$ of the metric space $M$ is closed iff every sequence in $A$ that converges to a limit in $M$ has its limit in $A$.
## Types of metric spaces
### Complete spaces
A metric space $M$ is said to be complete if every Cauchy sequence converges in $M$. That is to say: if $d(x_n, x_m) \to 0$ as both $n$ and $m$ independently go to infinity, then there is some $y\in M$ with $d(x_n, y) \to 0$.
Every Euclidean space is complete, as is every closed subset of a complete space. The rational numbers, using the absolute value metric $d(x,y) = \vert x - y \vert$, are not complete.
Every metric space has a unique (up to isometry) completion, which is a complete space that contains the given space as a dense subset. For example, the real numbers are the completion of the rationals.
If $X$ is a complete subset of the metric space $M$, then $X$ is closed in $M$. Indeed, a space is complete iff it is closed in any containing metric space.
Every complete metric space is a Baire space.
### Bounded and totally bounded spaces
Diameter of a set.
A metric space M is called bounded if there exists some number r, such that d(x,y) ≤ r for all x and y in M. The smallest possible such r is called the diameter of M. The space M is called precompact or totally bounded if for every r > 0 there exist finitely many open balls of radius r whose union covers M. Since the set of the centres of these balls is finite, it has finite diameter, from which it follows (using the triangle inequality) that every totally bounded space is bounded. The converse does not hold, since any infinite set can be given the discrete metric (one of the examples above) under which it is bounded and yet not totally bounded.
Note that in the context of intervals in the space of real numbers and occasionally regions in a Euclidean space Rn a bounded set is referred to as "a finite interval" or "finite region". However boundedness should not in general be confused with "finite", which refers to the number of elements, not to how far the set extends; finiteness implies boundedness, but not conversely.
### Compact spaces
A metric space M is compact if every sequence in M has a subsequence that converges to a point in M. This is known as sequential compactness and, in metric spaces (but not in general topological spaces), is equivalent to the topological notions of countable compactness and compactness defined via open covers.
Examples of compact metric spaces include the closed interval [0,1] with the absolute value metric, all metric spaces with finitely many points, and the Cantor set. Every closed subset of a compact space is itself compact.
A metric space is compact iff it is complete and totally bounded. This is known as the Heine–Borel theorem. Note that compactness depends only on the topology, while boundedness depends on the metric.
Lebesgue's number lemma states that for every open cover of a compact metric space M, there exists a "Lebesgue number" δ such that every subset of M of diameter < δ is contained in some member of the cover.
Every compact metric space is second countable,[4] and is a continuous image of the Cantor set. (The latter result is due to Pavel Alexandrov and Urysohn.)
### Locally compact and proper spaces
A metric space is said to be locally compact if every point has a compact neighborhood. Euclidean spaces are locally compact, but infinite-dimensional Banach spaces are not.
A space is proper if every closed ball {y : d(x,y) ≤ r} is compact. Proper spaces are locally compact, but the converse is not true in general.
### Connectedness
A metric space $M$ is connected if the only subsets that are both open and closed are the empty set and $M$ itself.
A metric space $M$ is path connected if for any two points $x, y \in M$ there exists a continuous map $f\colon [0,1] \to M$ with $f(0)=x$ and $f(1)=y$. Every path connected space is connected, but the converse is not true in general.
There are also local versions of these definitions: locally connected spaces and locally path connected spaces.
Simply connected spaces are those that, in a certain sense, do not have "holes".
### Separable spaces
A metric space is separable space if it has a countable dense subset. Typical examples are the real numbers or any Euclidean space. For metric spaces (but not for general topological spaces) separability is equivalent to second countability and also to the Lindelöf property.
## Types of maps between metric spaces
Suppose (M1,d1) and (M2,d2) are two metric spaces.
### Continuous maps
Main article: Continuous function (topology)
The map f:M1→M2 is continuous if it has one (and therefore all) of the following equivalent properties:
General topological continuity
for every open set U in M2, the preimage f -1(U) is open in M1
This is the general definition of continuity in topology.
Sequential continuity
if (xn) is a sequence in M1 that converges to x in M1, then the sequence (f(xn)) converges to f(x) in M2.
This is sequential continuity, due to Eduard Heine.
ε-δ definition
for every x in M1 and every ε>0 there exists δ>0 such that for all y in M1 we have
$d_1(x,y)<\delta \Rightarrow d_2(f(x),f(y))< \varepsilon.$
This uses the (ε, δ)-definition of limit, and is due to Augustin Louis Cauchy.
Moreover, f is continuous if and only if it is continuous on every compact subset of M1.
The image of every compact set under a continuous function is compact, and the image of every connected set under a continuous function is connected.
### Uniformly continuous maps
The map ƒ : M1 → M2 is uniformly continuous if for every ε > 0 there exists δ > 0 such that
$d_1(x,y)<\delta \Rightarrow d_2(f(x),f(y))< \varepsilon \quad\mbox{for all}\quad x,y\in M_1.$
Every uniformly continuous map ƒ : M1 → M2 is continuous. The converse is true if M1 is compact (Heine–Cantor theorem).
Uniformly continuous maps turn Cauchy sequences in M1 into Cauchy sequences in M2. For continuous maps this is generally wrong; for example, a continuous map from the open interval (0,1) onto the real line turns some Cauchy sequences into unbounded sequences.
### Lipschitz-continuous maps and contractions
Given a number K > 0, the map ƒ : M1 → M2 is K-Lipschitz continuous if
$d_2(f(x),f(y))\leq K d_1(x,y)\quad\mbox{for all}\quad x,y\in M_1.$
Every Lipschitz-continuous map is uniformly continuous, but the converse is not true in general.
If K < 1, then ƒ is called a contraction. Suppose M2 = M1 and M1 is complete. If ƒ is a contraction, then ƒ admits a unique fixed point (Banach fixed point theorem). If M1 is compact, the condition can be weakened a bit: ƒ admits a unique fixed point if
$d(f(x), f(y)) < d(x, y) \quad \mbox{for all} \quad x \ne y \in M_1$.
### Isometries
The map f:M1→M2 is an isometry if
$d_2(f(x),f(y))=d_1(x,y)\quad\mbox{for all}\quad x,y\in M_1$
Isometries are always injective; the image of a compact or complete set under an isometry is compact or complete, respectively. However, if the isometry is not surjective, then the image of a closed (or open) set need not be closed (or open).
### Quasi-isometries
The map f : M1 → M2 is a quasi-isometry if there exist constants A ≥ 1 and B ≥ 0 such that
$\frac{1}{A} d_2(f(x),f(y))-B\leq d_1(x,y)\leq A d_2(f(x),f(y))+B \text{ for all } x,y\in M_1$
and a constant C ≥ 0 such that every point in M2 has a distance at most C from some point in the image f(M1).
Note that a quasi-isometry is not required to be continuous. Quasi-isometries compare the "large-scale structure" of metric spaces; they find use in geometric group theory in relation to the word metric.
## Notions of metric space equivalence
Given two metric spaces (M1, d1) and (M2, d2):
• They are called homeomorphic (topologically isomorphic) if there exists a homeomorphism between them (i.e., a bijection continuous in both directions).
• They are called uniformic (uniformly isomorphic) if there exists a uniform isomorphism between them (i.e., a bijection uniformly continuous in both directions).
• They are called isometric if there exists a bijective isometry between them. In this case, the two metric spaces are essentially identical.
• They are called quasi-isometric if there exists a quasi-isometry between them.
## Topological properties
Metric spaces are paracompact[5] Hausdorff spaces[6] and hence normal (indeed they are perfectly normal). An important consequence is that every metric space admits partitions of unity and that every continuous real-valued function defined on a closed subset of a metric space can be extended to a continuous map on the whole space (Tietze extension theorem). It is also true that every real-valued Lipschitz-continuous map defined on a subset of a metric space can be extended to a Lipschitz-continuous map on the whole space.
Metric spaces are first countable since one can use balls with rational radius as a neighborhood base.
The metric topology on a metric space M is the coarsest topology on M relative to which the metric d is a continuous map from the product of M with itself to the non-negative real numbers.
## Distance between points and sets; Hausdorff distance and Gromov metric
A simple way to construct a function separating a point from a closed set (as required for a completely regular space) is to consider the distance between the point and the set. If (M,d) is a metric space, S is a subset of M and x is a point of M, we define the distance from x to S as
$d(x,S) = \inf\{d(x,s) : s \in S \}$ where $\inf$ represents the infimum.
Then d(x, S) = 0 if and only if x belongs to the closure of S. Furthermore, we have the following generalization of the triangle inequality:
$d(x,S) \leq d(x,y) + d(y,S),$
which in particular shows that the map $x\mapsto d(x,S)$ is continuous.
Given two subsets S and T of M, we define their Hausdorff distance to be
$d_H(S,T) = \max \{ \sup\{d(s,T) : s \in S \} , \sup\{ d(t,S) : t \in T \} \}$ where $\sup$ represents the supremum.
In general, the Hausdorff distance dH(S,T) can be infinite. Two sets are close to each other in the Hausdorff distance if every element of either set is close to some element of the other set.
The Hausdorff distance dH turns the set K(M) of all non-empty compact subsets of M into a metric space. One can show that K(M) is complete if M is complete. (A different notion of convergence of compact subsets is given by the Kuratowski convergence.)
One can then define the Gromov–Hausdorff distance between any two metric spaces by considering the minimal Hausdorff distance of isometrically embedded versions of the two spaces. Using this distance, the set of all (isometry classes of) compact metric spaces becomes a metric space in its own right.
## Product metric spaces
If $(M_1,d_1),\ldots,(M_n,d_n)$ are metric spaces, and N is the Euclidean norm on Rn, then $\Big(M_1\times \ldots \times M_n, N(d_1,\ldots,d_n)\Big)$ is a metric space, where the product metric is defined by
$N(d_1,...,d_n)\Big((x_1,\ldots,x_n),(y_1,\ldots,y_n)\Big) = N\Big(d_1(x_1,y_1),\ldots,d_n(x_n,y_n)\Big),$
and the induced topology agrees with the product topology. By the equivalence of norms in finite dimensions, an equivalent metric is obtained if N is the taxicab norm, a p-norm, the max norm, or any other norm which is non-decreasing as the coordinates of a positive n-tuple increase (yielding the triangle inequality).
Similarly, a countable product of metric spaces can be obtained using the following metric
$d(x,y)=\sum_{i=1}^\infty \frac1{2^i}\frac{d_i(x_i,y_i)}{1+d_i(x_i,y_i)}.$
An uncountable product of metric spaces need not be metrizable. For example, $\mathbf{R}^\mathbf{R}$ is not first-countable and thus isn't metrizable.
### Continuity of distance
It is worth noting that in the case of a single space $(M,d)$, the distance map $d\colon M\times M \rightarrow R^+$ (from the definition) is uniformly continuous with respect to any of the above product metrics $N(d,d)$, and in particular is continuous with respect to the product topology of $M\times M$.
## Quotient metric spaces
If M is a metric space with metric d, and ~ is an equivalence relation on M, then we can endow the quotient set M/~ with the following (pseudo)metric. Given two equivalence classes [x] and [y], we define
$d'([x],[y]) = \inf\{d(p_1,q_1)+d(p_2,q_2)+\dotsb+d(p_{n},q_{n})\}$
where the infimum is taken over all finite sequences $(p_1, p_2, \dots, p_n)$ and $(q_1, q_2, \dots, q_n)$ with $[p_1]=[x]$, $[q_n]=[y]$, $[q_i]=[p_{i+1}], i=1,2,\dots, n-1$. In general this will only define a pseudometric, i.e. $d'([x],[y])=0$ does not necessarily imply that $[x]=[y]$. However for nice equivalence relations (e.g., those given by gluing together polyhedra along faces), it is a metric. Moreover if M is a compact space, then the induced topology on M/~ is the quotient topology.
The quotient metric d is characterized by the following universal property. If $f:(M,d)\longrightarrow(X,\delta)$ is a metric map between metric spaces (that is, $\delta(f(x),f(y))\le d(x,y)$ for all x, y) satisfying f(x)=f(y) whenever $x\sim y,$ then the induced function $\overline{f}\colon M/\sim\longrightarrow X$, given by $\overline{f}([x])=f(x)$, is a metric map $\overline{f}\colon (M/\sim,d')\longrightarrow (X,\delta).$
A topological space is sequential if and only if it is a quotient of a metric space.[7]
## Generalizations of metric spaces
• Every metric space is a uniform space in a natural manner, and every uniform space is naturally a topological space. Uniform and topological spaces can therefore be regarded as generalizations of metric spaces.
• If we consider the first definition of a metric space given above and relax the second requirement, we arrive at the concepts of a pseudometric space or a dislocated metric space.[8] If we remove the third or forth, we arrive at a quasimetric space, or a semimetric space.
• If the distance function takes values in the extended real number line R∪{+∞}, but otherwise satisfies all four conditions, then it is called an extended metric and the corresponding space is called an $\infty$-metric space. If the distance function takes values in some (suitable) ordered set (and the triangle inequality is adjusted accordingly), then we arrive at the notion of generalized ultrametric.[9]
• Approach spaces are a generalization of metric spaces, based on point-to-set distances, instead of point-to-point distances.
• A continuity space is a generalization of metric spaces and posets, that can be used to unify the notions of metric spaces and domains.
### Metric spaces as enriched categories
The ordered set $(\mathbb{R},\geq)$ can be seen as a category by requesting exactly one morphism $a\to b$ if $a\geq b$ and none otherwise. By using $+$ as the tensor product and $0$ as the identity, it becomes a monoidal category $R*$. Every metric space $(M,d)$ can now be viewed as a category $M*$ enriched over $R *$:
• Set $\operatorname{Ob}(M*):=M$
• For each $X,Y\in M$ set $\operatorname{Hom}(X,Y):=d(X,Y)\in \operatorname{Ob}(R^*)$
• The composition morphism $\operatorname{Hom}(Y,Z)\otimes \operatorname{Hom}(X,Y)\to \operatorname{Hom}(X,Z)$ will be the unique morphism in $R*$ given from the triangle inequality $d(y,z)+d(x,y)\geq d(x,z)$
• The identity morphism $0\to \operatorname{Hom}(X,X)$ will be the unique morphism given from the fact that $0\geq d(X,X)$.
• Since $R*$ is a strict monoidal category, all diagrams that are required for an enriched category commute automatically.
See the paper by F.W. Lawvere listed below.
## Notes
1. Pascal Hitzler and Anthony Seda, Mathematical Aspects of Logic Programming Semantics. Chapman and Hall/CRC, 2010.
2. Pascal Hitzler and Anthony Seda, Mathematical Aspects of Logic Programming Semantics. Chapman and Hall/CRC, 2010.
## References
• Victor Bryant, Metric Spaces: Iteration and Application, Cambridge University Press, 1985, ISBN 0-521-31897-1.
• Dmitri Burago, Yu D Burago, Sergei Ivanov, A Course in Metric Geometry, American Mathematical Society, 2001, ISBN 0-8218-2129-6.
• Athanase Papadopoulos, Metric Spaces, Convexity and Nonpositive Curvature, European Mathematical Society, 2004, SBN 978-3-03719-010-4.
• Mícheál Ó Searcóid, Metric Spaces, Springer Undergraduate Mathematics Series, 2006, ISBN 1-84628-369-8.
• Lawvere, F. William, "Metric spaces, generalized logic, and closed categories", [Rend. Sem. Mat. Fis. Milano 43 (1973), 135—166 (1974); (Italian summary)
This is reprinted (with author commentary) at Reprints in Theory and Applications of Categories Also (with an author commentary) in Enriched categories in the logic of geometry and analysis. Repr. Theory Appl. Categ. No. 1 (2002), 1–37. | {"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": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 179, "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": 0.9017434120178223, "perplexity_flag": "head"} |
http://math.stackexchange.com/questions/103208/relations-between-various-definitions-of-a-radon-measure | # Relations between various definitions of a Radon measure
The following various definitions of a Radon measure seem to be given for the Borel sigma algebra of different types of topological spaces: general, Hausdorff, locally compact, or locally compact Hausdorff.
I was wondering if the definitions are related in some way?
Can these definitions or most of them be unified?
References are appreciated! Thanks and regards!
1. From Measure Theory, Volumes 1-2 by Vladimir I. Bogachev
Let $X$ be a topological space. A Borel measure $\mu$ on $X$ is called a Radon measure if for every $B$ in $B(X)$ and $ε>0$, there exists a compact set $K_ε ⊂ B$ such that $|\mu|(B - K_ε) <ε$.
2. From Wikipedia:
On the Borel $σ$-algebra of a Hausdorff topological space $X$, a measure is called a Radon measure if it is
• locally finite, and
• inner regular.
3. From ncatlab
If $X$ is a locally compact Hausdorff topological space, a Radon measure on $X$ is a Borel measure on $X$ that is
• finite on all compact sets,
• outer regular on all Borel sets, and
• inner regular on open sets.
4. From planetmath
Let $X$ be a Hausdorff space. A Borel measure $\mu$ on $X$ is said to be a Radon measure if it is:
• finite on compact sets,
• inner regular (tight).
5. From Wikipedia's Radon measures on locally compact spaces
When the underlying measure space is a locally compact topological space, the definition of a Radon measure can be expressed in terms of continuous linear functionals on the space of continuous functions with compact support.
-
I was wondering whether tag [geometric-measure-theory] applies here? Why? – Tim Jan 29 '12 at 16:37
## 2 Answers
Schwartz (Radon measures on arbitrary topological spaces and cylindrical measures, 1973) defines Radon measures as comprising two measures. The first is the measure given in version 3 above and the second is the essential measure defined as locally finite, tight measure He then shows that each can generate the other. On LCH spaces, version 3 equivalent to version 5. Prinz (Regularity of Riesz measures, 1986, Proc Amer Math Soc) calls version 3 a "Riesz" measure and the locally finite, tight version a "Radon" measure and refers to Schwartz to give their duality.
-
I will concentrate on comparing (3) and (4). The definition (1) is meant for finite signed measures, whereas all the other definitions are meant for arbitrary positive measures; (1) is equivalent to (4) in the case of finite positive measures. (2) appears to be equivalent to (4) ["locally finite" can mean "finite on compact sets", although it is sometimes taken to mean "finite on the elements of some topological basis"; these are equivalent in the LCH (locally compact Hausdorff) case]. Finally, (5) does not appear to be a definition at all, but rather a description of a definition.
Now then,
i) In the case of a second countable LCH space, every locally finite measure satisfies both (3) and (4) (Theorem 7.8 of [1]). This is the most commonly considered scenario in applications, which is why almost no one bothers to carefully sort out the differences between the different definitions.
ii) In the case of a sigma-compact LCH space, (3) and (4) are equivalent. The forward direction is Corollary 7.6 of [1]; the backwards direction follows from the forward direction together with (iv) below (but I'm sure there is an easier proof).
iii) (3) and (4) are not equivalent in general, even for LCH metrizable spaces (Exercise 7.12 of [1]).
iv) In an LCH space, there is a bijection between
A) measures satisfying (3),
B) measures satisfying (4), and
C) positive linear functionals on the space of continuous functions with compact support.
(The Riesz representation theorem gives either (A)<->(C) or (B)<->(C), depending on where you look; (A)<->(B) is in the Schwarz book mentioned by Joe Lucke; see also Exercise 7.14 of [1])
[1] G. B. Folland, Real Analysis: Modern Techniques and Their Applications
Note: In [1], "Radon" refers to measures satisfying (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": 16, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 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": 0.9057478904724121, "perplexity_flag": "head"} |
http://mathoverflow.net/revisions/25229/list | Return to Question
2 added 11 characters in body; deleted 16 characters in body
For example, I take differentiability, analyticity, and algebraicity(of a function). All(more or less) imply continuity. So when we define a differentiable function on $\mathbb R^n$ or an analytic function on $\mathbb C^n$, or a regular map on an affine space, we do not explicitly require that the functions are continuous. It follows automatically from the stronger condition.
But, when I look at the definitions in books of a global structure using sheaf theory, it is almost always for a global picturedefinition of a morphism, ie define on a differentiable manifold or an analytic space, or an abstract algebraic variety, the definition of a morphism requires a priori that the map be continuous, and then one requires that there is additionally a morphism of sheaves of algebras(of the suitable type of structure sheaves, depending on the local model used).
Why is this so? Is it something done for fancy, or is there a real need for the extra continuity assumption? I mean could things go wrong if this assumption is dropped?
1
Why is continuity required for sheaf-theoretic definitions of a structure on a space
For example, I take differentiability, analyticity, and algebraicity(of a function). All(more or less) imply continuity. So when we define a differentiable function on $\mathbb R^n$ or an analytic function on $\mathbb C^n$, or a regular map on an affine space, we do not explicitly require that the functions are continuous. It follows automatically from the stronger condition.
But, when I look at the definitions in books of a global structure using sheaf theory, it is almost always a global picture, ie define a differentiable manifold or an analytic space, or an abstract algebraic variety, the definition of a morphism requires a priori that the map be continuous, and then one requires that there is additionally a morphism of sheaves of algebras(of the suitable type of structure sheaves, depending on the local model used).
Why is this so? Is it something done for fancy, or is there a real need for the extra continuity assumption? I mean could things go wrong if this assumption is dropped? | {"extraction_info": {"found_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": 0.8787779211997986, "perplexity_flag": "head"} |
http://mathhelpforum.com/advanced-algebra/51718-prove-group.html | # Thread:
1. ## Prove that in a group
(xy)^(-1) = b^(-1)a^(-1),
keeping in mind the group may not be commutative. Thanks.
2. Originally Posted by universalsandbox
(xy)^(-1) = b^(-1)a^(-1),
keeping in mind the group may not be commutative. Thanks.
To show $(ab)^{-1} = b^{-1}a^{-1}$ you need to show $(ab)(b^{-1}a^{-1}) = e$.
Can you do that now?
3. Originally Posted by ThePerfectHacker
To show $(ab)^{-1} = b^{-1}a^{-1}$ you need to show $(ab)(b^{-1}a^{-1}) = e$.
Can you do that now?
Could you add a little more process to that please? This stuff is very new to me. Thanks.
4. Originally Posted by universalsandbox
Could you add a little more process to that please? This stuff is very new to me. Thanks.
Say that $x^{-1} = y$. What does that mean? It means that $xy = e$ and $yx=e$. That is the definition of what it means to be a multiplicative inverse. In your problem you have $x=ab$ and $y=b^{-1}a^{-1}$. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 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": 0.3, "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": 0.9320101141929626, "perplexity_flag": "head"} |
http://mathforum.org/mathimages/index.php?title=Compass_%26_Straightedge_Construction_and_the_Impossible_Constructions&diff=15823&oldid=13187 | # Compass & Straightedge Construction and the Impossible Constructions
### From Math Images
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| | |ImageName=Creating a regular hexagon with a ruler and compass | | |ImageName=Creating a regular hexagon with a ruler and compass |
| | |Image=HexagonConstructionAni.gif | | |Image=HexagonConstructionAni.gif |
| | |ImageIntro=This image shows the step by step construction of a hexagon inscribed in the circle using a compass and a unmarked straightedge. | | |ImageIntro=This image shows the step by step construction of a hexagon inscribed in the circle using a compass and a unmarked straightedge. |
| - | |ImageDescElem=Many have learned about compass and straightedge construction in a very simplified manner. Nonetheless, it is one of the most interesting aspect of geometry a very hands-on experience. It makes us to seriously appreciate geometry in its purest and simplest form. With a compass and a unmarked straightedge, the possible constructions are almost unlimited. However, it is always the case that many geometry textbooks only touche on the subject in an introductory and perfunctory manner, not reaching the deeper issues surrounding the subject. On the other hand, information on the subject from the internet has great breadth but little depth. Hence, this page is in intended to introduce compass and straightedge construction to those to whom the subject is new and also give a fair amount of in-depth analysis on the subject. Hopefully as a result, this page can serve as a gateway that leads the abler and more interested to places where they can find more information on the subject. | + | |ImageDescElem= |
| | | + | Let's assume we only have a compass and a unmarked straightedge. What can we construct and how can we construct them? That were the problems that Euclid pondered not only because those were probably the only instruments that he had at his time but also he wanted to build his theorems with as few assumptions, or {{EasyBalloon|Link=axioms|Balloon=In traditional logic, an axiom or postulate is a proposition that is not proved or demonstrated but considered to be either self-evident, or subject to necessary decision. Therefore, its truth is taken for granted, and serves as a starting point for deducing and inferring other (theory dependent) truths.}}, as possible.<ref>Peterson, 2003</ref> With these two simple tools, he managed to build myriad of theorems in both plane and solid geometry .<ref>Hartshorne, 2000</ref> These theorem are as good as they were some 2000 years ago and it is this enduring quality of Euclid's work that inspired this page. In the main image of this page, we want to divide the circle into six equal arcs and then connect consecutive points to form the hexagon. It seems to be a fairly simple construction but you should be prompted to ask two questions: Are other polygons constructible and is every polygon constructible, that is able to be constructed using only compass and straightedge? To extend the question, what are constructible and what are not? That is the problem that is resolved in this page. |
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| - | Let's assume we only have a compass and a unmarked straightedge. How could we construct certain geometric shapes and prove certain theorems? That was the problems that Euclid pondered not only because those were probably the only instrument that he had at his time but also he wanted to construct his theorems with as few assumptions, or axioms, as possible. In this picture, we want to divide the circle into six equal arcs and then connect consecutive points to form the hexagon. It seems to be a fairly simple construction but it should prompt you to ask a question: is every polygon constructible, that is able to be constructed using only compass and straightedge? To extend the question, what are constructible and what is not? The issue will be addressed in a later subsection. | + | |ImageDesc===What is Compass & Straightedge Constructions== |
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| - | |ImageDesc===The Compass and straightedge== | + | |
| - | [[Image:Compass.jpg|border|right|150px]] | + | |
| - | A Compass is a tool which can be used for drawing circles. It has two legs, one end of which is fixed on the plane of construction and the other end is of desirable distance away and maintains the distance throughout the construction. A normal form of compass is shown on the right but there are many other variants, crude or precise. For example, a crude form will be pinning a thread of string on the plane and fixing a pencil/pen at a certain distance away from the pin. We usually see this kind of crude compass in military field operations. Look out for such a compass in HBO's TV series '''''Band of Brothers''''' second episode when the paratroopers were trying to figure out where they would be dropped on D-Day based on the distances of their preparatory flights. | + | |
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| - | A straightedge is a tool which can be used for drawing straight lines, or segments thereof. It should be noted that the difference between a straightedge and a ruler is that the former has no graduations on it (and does not allow any markings as well) while the latter has divisions on them according to certain unit system, be it the metric or the British customary system. | + | |
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| - | | + | |
| - | ==What is Compass&Straightedge Constructions== | + | |
| | | | |
| | ==='''Introduction'''=== | | ==='''Introduction'''=== |
| | | + | We start by familiarizing ourselves with Euclid's three Postulates in his books '''''Elements'''''. |
| | | | |
| - | Compass & Straightedge Construction is the construction of points, lengths, angles, and other geometric figures using only an idealized straightedge and compass. The straightedge is infinite in length, has no markings on it and only one edge. The compass collapses when lifted from the page, so may not be directly used to transfer distances. However, it turns out that this restriction makes no difference due to the Compass Equivalence Theorem which was stated as Proposition II of Book I of Euclid's Elements. It stated that from a given point, it was possible to construct a straight line equal to a given straight line using collapsible compass. Euclid's proof for the Compass Equivalence Theorem will be presented after the section of Basic Construction. | + | <blockquote> |
| | | + | {{{!}} |
| | | + | {{!}}'''''Let it be granted |
| | | + | {{!}}- |
| | | + | {{!}}'''''1. that a straight line may be drawn from any one point to any other point; |
| | | + | {{!}}- |
| | | + | {{!}}'''''2. that a line segment may be extended into a straight line; |
| | | + | {{!}}- |
| | | + | {{!}}'''''3. that given any straight line segment, a circle may be described having the segment as radius and one endpoint as center.<ref>Taylor, 1895, p. 14</ref> |
| | | + | {{!}}} |
| | | + | </blockquote> |
| | | | |
| - | We start by familiarizing ourselves with Euclid' three Postulates in his book Elements | + | {{{!}} |
| | | + | {{!}} |
| | | + | It should be carefully noted that Euclid started with two given points and produced a line segment, from where he could extend into a straight line if he pleased. Then ONLY from the original two points, he could use one point as center and ONLY spread the legs of compass the distance of the line segment to produce a circle. Even though many translations say that a circle can be described at any center with any radius, we must take it with a pinch of salt. He could not specify any points and lengths other than what was already given, that is to say he could not claim "I wanted to spread the legs of the compass <math>\pi</math> centimeters apart (or any specified denominations) with the center of the circle half way between the two given points". In all, Compass and Straightedge Constructions only allow us to start with points (and hence lengths) we have been given(or constructed from given points), and create the ones we don't. |
| | | + | {{!}}- |
| | | + | {{!}} |
| | | + | Thus, we define Compass & Straightedge Construction as the construction of points, lengths, angles, and circles using only ideal straightedge and compass. A straightedge is infinite in length, has no markings on it and only one edge. A compass has two legs, one end of which is fixed on the plane of construction and the other end is of given distance away and maintains the distance throughout the construction. It collapses when lifted from the page, so may not be '''directly''' used to transfer distances. However, it turns out that this restriction makes no difference due to the Compass Equivalence Theorem which was stated as Proposition II of Book I of Euclid's Elements. It stated that from a given point, it was possible to construct a line segment equal to a given line segment using collapsible compass in any desirable direction. Euclid's proof for the Compass Equivalence Theorem will be presented after the section of Basic Construction. Since Euclid has proven this using only the three postulates, then he did not have to use a collapsible compass any more.<ref>Peterson, 2003</ref> |
| | | + | {{!}}} |
| | | | |
| - | <blockquote> "Let it be granted | + | ===Some Basic Constructions=== |
| - | # that a straight line may be drawn from any one point to any other point; | + | The constructions below are some basic ones from where many more constructions are possible and they are by no means exhaustive. In the figures below, what we are given are in blue; intermediate steps are in dotted black; the resulting products are in red. The proofs for these constructions are relatively simple and only require the knowledge of [[congruent triangles]]. Euclid derived the theories on congruency and congruent triangles directly from his Postulates. Try proving the theorems yourself! |
| - | # that a line segment may be extended into a straight line; | + | |
| - | # that given any straight line segment, a circle may be described having the segment as radius and one endpoint as center. " </blockquote> | + | |
| | | | |
| | | + | ===='''Line Segment Bisection'''==== |
| | | + | [[Image:CS1.png|border|550px|center]] |
| | | + | {{SwitchPreview|ShowMessage=Click here to show construction|HideMessage=Click here to hide construction|PreviewText=Given points <math>{\color{Gray}A}</math> and <math>{\color{Gray}B}</math> and the straight line passing through it. Construct a line that bisects line segment <math>AB</math>.|FullText=Given points <math>A</math> and <math>B</math> and the straight line passing through it. Construct a line that bisects line segment <math>AB</math>. |
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| - | It should be carefully noted that Euclid started with two given points and produced a line segment, from where he could extend into a straight line if he pleases. Then from the original two points, he could use one point as center and spread the legs of compass the distance of the line segment to produce a circle. He could not (so couldn't we)however, choose any two points as he please. The three postulates was stated in ancient Greek and both its old and modern translations have not been consistent with each other. What is stated above is an assortment of different translations that give the least confusions about what could and could not be done with a compass and a straightedge. Throughout his book, Euclid used only these three operations to do all his plane geometry: the drawing of straight line through two given points, and circle with a given center to pass through a given point. The terms ''Euclidean construction'', ''construction'', ''construct'' and ''constructible'' all refer to Euclid's three operations repeated any finite number of times. | + | # Draw a circle centered at point <math>A</math> with radius equals <math>AB</math>. |
| - | | + | |
| - | | + | |
| - | ===Some Basic Construction=== | + | |
| - | The constructions below are some basic ones from where many more constructions and operations are possible and they are by no means exhaustive. The resulting products are in red. The proofs for these constructions are relatively simple and only require the knowledge of congruent triangles. Try proving the theorems yourself! | + | |
| - | | + | |
| - | ===='''Line Bisection'''==== | + | |
| - | [[Image:CS1.png|border|700px|center]] | + | |
| - | | + | |
| - | Given points <math>A</math> and <math>B</math> and the straight line passing through it. Construct a line that bisects line segment <math>AB</math>. | + | |
| - | | + | |
| - | # Draw a circle centered at point <math>A</math>. | + | |
| | # Next, draw a circle centered at point <math>B</math> with the same radius. | | # Next, draw a circle centered at point <math>B</math> with the same radius. |
| - | # The two circles intersect at points <math>C</math> and <math>D</math>(Of course, you have to make sure they intersect in the first place. One way to do this is to let the radius be equal to <math>AB</math>). | + | # Where the two circles intersect, call those points <math>C</math> and <math>D</math>. |
| | # Draw a line through points <math>C</math> and <math>D</math>. | | # Draw a line through points <math>C</math> and <math>D</math>. |
| - | # <math>CD</math> intersects <math>AB</math> at the midpoint. | | |
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| - | It should be noted that <math>CD \perp AB</math> as well. Line <math>CD</math> is the perpendicular bisector of line segment <math>AB</math>. | + | Line <math>CD</math> intersects line segment <math>AB</math> at the midpoint. It should be noted that <math>CD \perp AB</math> as well. Line <math>CD</math> is the perpendicular bisector of line segment <math>AB</math>.}} |
| | | + | |
| | | | |
| | ===='''Angle Bisection'''==== | | ===='''Angle Bisection'''==== |
| - | [[Image:CS2.png|border|700px|center]] | + | [[Image:CS2.png|border|550px|center]] |
| - | Given angle <math>\angle AOB</math>, construct a line that bisects the angle. | + | {{SwitchPreview|ShowMessage=Click here to show construction|HideMessage=Click here to hide construction|PreviewText=Given angle <math>{\color{Gray}\angle AOB}</math>, construct a line that bisects the angle.|FullText=Given angle <math>\angle AOB</math>, construct a line that bisects the angle. |
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| - | # Construct a circle centered at point <math>O</math> with radius <math>OA</math>. This circle should intersect at points <math>A</math> and <math>B</math>. | + | # Construct a circle centered at point <math>O</math> with radius <math>OA</math>. This circle intersects <math>OA</math> and <math>OB</math> at point <math>A</math> and <math>B</math>. |
| - | # Keeping the same radius, draw a circle at points <math>A</math> and <math>B</math>. They must intersect each other at <math>C</math>. | + | # Keeping the same radius, draw a circle at point <math>A</math> and <math>B</math> respectively. Where they intersect each other, call it point <math>C</math>. |
| - | # Draw a line through points <math>C</math> and <math>O</math>. This line bisects <math>\angle AOB</math>. | + | # Draw a line through point <math>C</math> and <math>O</math>. This line bisects <math>\angle AOB</math>. |
| | | + | }} |
| | | | |
| | | + | ===='''Perpendicular Through a Point'''==== |
| | | + | [[Image:Perp.png|border|450px|center]] |
| | | + | {{SwitchPreview|ShowMessage=Click here to show construction|HideMessage=Click here to hide construction|PreviewText=Given a point <math>{\color{Gray}M}</math> on a line <math>{\color{Gray}AB}</math>, construct a line that is perpendicular to the given line through <math>M</math>.|FullText=Given a point <math>M</math> on a line <math>AB</math>, construct a line that is perpendicular to the given line through <math>M</math>. |
| | | | |
| - | ===='''Perpendicular through a point'''==== | + | # Draw a circle centered at <math>M</math> with radius <math>MB</math> (or <math>MA</math>. It is your choice). |
| - | [[Image:CS3.png|border|600px|center]] | + | # Where the circle intersects the original line, construct a perpendicular bisector (see the line segment bisector construction above).}} |
| - | Given a point on a line, construct a line that is perpendicular to the given line through the given point. | + | |
| - | | + | |
| - | # Draw a circle centered at <math>M</math>. | + | |
| - | # Where the circle intersects the original line, construct a perpendicular bisector. | + | |
| | | | |
| | | + | ===='''Parallel Line'''==== |
| | | + | [[Image:CS4.png|border|550px|center]] |
| | | + | {{SwitchPreview|ShowMessage=Click here to show construction|HideMessage=Click here to hide construction|PreviewText=Given two points, <math>{\color{Gray}A}</math> and <math>{\color{Gray}B}</math> and the straight line passing through them, construct a line that is parallel to the given line through another given point <math>C</math>.|FullText=Given two points, <math>A</math> and <math>B</math> and the straight line passing through them, construct a line that is parallel to the given line through another given point <math>C</math>. |
| | | | |
| - | ===='''Parallel'''==== | + | # Draw a circle at <math>A</math>, crossing <math>C</math>. Where the circle <math>A</math> intersects <math>AB</math>, call the point <math>D</math>. |
| - | [[Image:CS4.png|border|750px|center]] | + | # Centered at points <math>C</math> and <math>D</math>, draw circles crossing point <math>A</math>. Where these two circles intersect each other, call it <math>E</math>. |
| - | Given two points and the straight line passing through, construct a line that is parallel to the given line through another given point. | + | # Draw a line through point <math>C</math> and <math>E</math>. |
| - | | + | |
| - | # Draw a circle at <math>A</math>, crossing <math>C</math>. Circle <math>A</math> intersects <math>AB</math> at <math>D</math>. | + | |
| - | # Centered at points <math>C</math> and <math>D</math>, draw circles crossing point <math>A</math>. These two circles intersect each other at <math>E</math>. | + | |
| - | # Connect points <math>C</math> and <math>E</math> with a line. <math>CE</math> is parallel to <math>AB</math> | + | |
| | | | |
| | | + | <math>CE</math> is parallel to <math>AB</math>}} |
| | | | |
| | ===='''Tangent Line to a Circle'''==== | | ===='''Tangent Line to a Circle'''==== |
| - | [[Image:CS5.png|border|750px|center]] | + | [[Image:CS5.png|border|550px|center]] |
| - | Given a circle and another given point, construct a line that is tangent to the circle. | + | {{SwitchPreview|ShowMessage=Click here to show construction|HideMessage=Click here to hide construction|PreviewText=Given a circle centered at <math>{\color{Gray}O}</math> and another given point, <math>{\color{Gray}A}</math>, construct a line that is tangent to the circle.|FullText=Given a circle centered at <math>O</math> and another given point, <math>A</math>, construct a line that is tangent to the circle. |
| | | | |
| - | # Connect the point <math>A</math>, with the center of the circle <math>O</math> | + | # Draw a line through point <math>A</math> and the center of the circle <math>O</math> |
| | # Let <math>M</math> be the midpoint of <math>OA</math>(construction omitted since we know how to construction mid point). Draw the circle centered at <math>M</math> going through <math>A</math> and <math>O</math>. | | # Let <math>M</math> be the midpoint of <math>OA</math>(construction omitted since we know how to construction mid point). Draw the circle centered at <math>M</math> going through <math>A</math> and <math>O</math>. |
| | # Let the point where the two circles meet be <math>C</math>. Draw line segment <math>AC</math>. | | # Let the point where the two circles meet be <math>C</math>. Draw line segment <math>AC</math>. |
| - | # To see that <math>AC</math> is tangent to the circle, connect <math>OC</math> | | |
| - | # <math>\angle ACO</math> is an inscribed angle in the circle about <math>M</math>, so the inscribed <math>\angle ACO</math> is <math>90^\circ</math>. So <math>OC</math> is perpendicular to <math>AC</math>. Since the tangent is perpendicular to the radius to the point of tangency, by the uniqueness of the line through <math>C</math> perpendicular to <math>OC</math>, <math>AC</math> is tangent to the circle. | | |
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| | | + | |
| | | + | To see that <math>AC</math> is tangent to the circle, connect <math>OC</math>. <math>\angle ACO</math> is an inscribed angle in the circle about <math>M</math>, so the inscribed <math>\angle ACO</math> is <math>90^\circ</math>. So <math>OC</math> is perpendicular to <math>AC</math>. Since the tangent is perpendicular to the radius to the point of tangency, by the uniqueness of the line through <math>C</math> perpendicular to <math>OC</math>, <math>AC</math> is tangent to the circle.}} |
| | | | |
| | ===='''Euclid's Proof of Compass Equivalence Theorem'''==== | | ===='''Euclid's Proof of Compass Equivalence Theorem'''==== |
| | | | |
| - | This part reconnects with the previous section about the issue of compass being collapsible. Euclid's proof is presented in its originality. My comments are contained in the parenthesis. | + | This part refers back to the previous section about the issue of compass being collapsible. Euclid's original proof is presented. Additional comments are contained in the parenthesis. |
| - | | + | |
| - | [[Image:CETpic.png|border|center|600px]] | + | |
| - | | + | |
| - | From a given point to draw a line segment equal to a given line segment. | + | |
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| - | <u>Let <math>A</math> be the given point, and <math>BC</math> the given straight line : it is required to draw from the point <math>A</math> a straight line equal to <math>BC</math>.</u> | + | [[Image:CETpic.png|border|center|450px]] |
| | | + | {{SwitchPreview|ShowMessage=Click here to show construction|HideMessage=Click here to hide construction|PreviewText=From a given point to draw a line segment equal to a given line segment.|FullText=From a given point to draw a line segment equal to a given line segment. <u>Let <math>A</math> be the given point, and <math>BC</math> the given straight line : it is required to draw from the point <math>A</math> a straight line equal to <math>BC</math>.</u> |
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| | # From the point <math>A</math> to <math>B</math> draw the straight line <math>AB</math> and on it describe the equilateral triangle <math>DAB</math>(this is done the same way as bisecting a line segment), and produce the straight lines <math>DA</math>, <math>DB</math> to <math>E</math> and <math>F</math>. | | # From the point <math>A</math> to <math>B</math> draw the straight line <math>AB</math> and on it describe the equilateral triangle <math>DAB</math>(this is done the same way as bisecting a line segment), and produce the straight lines <math>DA</math>, <math>DB</math> to <math>E</math> and <math>F</math>. |
| - | # From the center <math>B</math>, at the distance <math>BC</math>, describe the circle <math>CGH</math>, meeting <math>DF</math> at <math>G</math>. | + | # From the center <math>B</math>, at the distance <math>BC</math>(with the radius equals to <math>BC</math>), describe the circle <math>CGH</math>, meeting <math>DF</math> at <math>G</math>. |
| - | # From the center <math>D</math>, at the distance <math>DG</math>, describe the circle <math>GKL</math>, meeting <math>DE</math> at <math>L</math>. | + | # From the center <math>D</math>, at the distance <math>DG</math>(with the radius equals to <math>DG</math>), describe the circle <math>GKL</math>, meeting <math>DE</math> at <math>L</math>. |
| | # <math>AL</math> shall be equal to <math>BC</math>. | | # <math>AL</math> shall be equal to <math>BC</math>. |
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| Line 109: | | Line 104: | |
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| - | Wherefore from the given point <math>A</math> a straight line <math>AL</math> has been drawn equal to the given straight line <math>BC</math>.∎ | + | Wherefore from the given point <math>A</math> a straight line <math>AL</math> has been drawn equal to the given straight line <math>BC</math>.∎<ref>Taylor, 1895, p. 18</ref> |
| - | | + | |
| - | It should be noted that from <math>AL</math>, we could "duplicate" <math>AL</math> in all directions by construct a circle centered at <math>A</math> with radius <math>AL</math>. | + | |
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| | | + | It should be noted that from <math>AL</math>, we could "duplicate" <math>AL</math> in all directions by construct a circle centered at <math>A</math> with radius <math>AL</math>.}} |
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| | ==Algebraicization of Compass & Straightedge Constructions== | | ==Algebraicization of Compass & Straightedge Constructions== |
| | | | |
| - | ==='''A simple derivation'''=== | + | ==='''What is Algebraicization?'''=== |
| - | | + | |
| - | From the few basic constructions, you would have probably realized that the different possibilities seems infinite. However, by intuition, we know that the possibility could not be infinite. Hence, mathematician are curious to find out what are constructible and what aren't and for this purpose, the language of pure geometry seems to have "limited vocabulary". Back in ancient times, mathematicians had limited algebraic knowledge and were more familiar with geometry. But in modern times, the reverse is true. Hence, today's mathematicians go back to their familiar realm of Algebra and try to find the link between geometry and algebra. | + | |
| - | | + | |
| - | | + | |
| - | ''Algebraicization''? This is certainly a strange word to you. You probably cannot even find it in a dictionary or encyclopedia entry. However show this to any mathematician or math majors, they will immediately tell you what that is all about. ''Algebraicization'' is the translation of any problem statements into algebraic problems. In the case of Compass & Straightedge construction, we algebraicize each step of a straightedge and compass construction, and consequently obtaining general results about the nature of constructibility. Hilda P. Hudson put it aptly in his lecture ''Ruler & Compasses'', | + | |
| - | | + | |
| - | <blockquote>"each step of a ruler and compass construction is equivalent to a certain analytical process; it is found that the power to use a ruler corresponds exactly to the power to solve linear equations, and the power to use compasses to the power to solve quadratics. For this reason, problems that can be solved with ruler only are called linear problems, and those that can be solved with ruler and compasses are called quadratic problems. Since each step of a ruler and compass construction is equivalent to the solution of an equation of the first or second degree, we consider that these algebraic processes can lead to , when combined in every possible way, and that enables us to answer the question before us and say that those problems and those problems alone can be solved by ruler only, which can be made to depend on a linear equation, whose root can be calculated by carrying out rational operations only; and that those problems and those problems alone can be solved by ruler and compasses, which can be made to depend on an algebraic equation, whose degree must be a power of 2, and whose roots can be calculated by carrying out rational operations together with the extraction of square roots only."</blockquote> | + | |
| - | | + | |
| - | In order to algebraicize straightedge and compass construction, we begin by choosing a specific length to be considered one unit. Then, every time we construct a figure, we think of it instead as constructing a set of ''numbers'' representing the lengths of the constructed line segments.Compass & straightedge construction only allow us to construct points that are at intersections of lines and/or circles. This means that every time we do a construction, we can use equations of circles and lines from coordinate geometry to figure out what numbers are being constructed. In this way, a geometric process is translated into an algebraic process. | + | |
| - | | + | |
| | | | |
| - | Firstly, we define "1" on a straight line by specifying the length between any two point. But some will say, hey, in Euclid's postulates, he did not say you can choose any points; a point is the intersection of two lines, two circles or one line and one circle. That is a good observation. To have a line, we have to have given points. Then we draw the straight line. Choose one of the two given points, and use that as the center for our compass. Now, spread the legs of the compass by a distance between the two given points. The intersection is a new point. With the new point, we now have one arbitrarily chosen points. | + | From the few basic constructions, you would have probably realized that the different possibilities seems infinite. Hence, mathematician are curious to find out what are constructible and what aren't and for this purpose, the language of pure geometry seems to have "limited vocabulary"<ref>Hudson, 1916, p. 3</ref>. Back in ancient times, mathematicians had limited algebraic knowledge and were more familiar with geometry. But in modern times, the reverse is true. Hence, today's mathematicians go back to their familiar realm of Algebra and try to find the link between geometry and algebra. |
| | | | |
| | | | |
| - | Note, the "1" does not necessarily have to measure meter nor foot since we have chosen two random points. It could measure a new length of your choosing. Then, once you have chosen that length to be "1", you have to stick to this specification throughout your construction. | + | ''Algebraicization'' is the translation of any problem statements into algebraic problems. In the case of Compass & Straightedge construction, we algebraicize each step of a straightedge and compass construction, and consequently obtaining general results about the nature of constructibility. Hilda P. Hudson put it aptly in his lecture '''''Ruler & Compasses''''', |
| | | | |
| | | + | <blockquote>''"each step of a ruler and compass construction is equivalent to a certain analytical process; it is found that the power to use a ruler corresponds exactly to the power to solve linear equations, and the power to use compasses to the power to solve quadratics...... Since each step of a ruler and compass construction is equivalent to the solution of an equation of the first or second degree, we consider that these algebraic processes can lead to , when combined in every possible way, and that enables us to answer the question before us......"''<ref>Hudson, 1916, p. 3</ref></blockquote> |
| | | | |
| - | Next, it is very obvious that we could construct all the integers, that is <math>\cdots -3,-2,-1,0,1,2,3,\cdots</math> (or <math>x</math> = <math>\{x|- \infty < x < \infty,x \in \mathbb{Z}\}</math>). How so? Well, once we have the "1", all we have to do is to use the Compass Equivalence Theorem finite number of times to duplicate the length "1" that we previously defined. Now, that means that we could have any two random integers, <math>a</math> and <math>b</math>, and for the sake of this discussion and clarity, we are talking about positive integers here. Next, I will show that from <math>a</math> and <math>b</math>, we could construct <math>a \pm b</math>, <math>a \times b</math> and <math>\frac {a}{b}</math>. | + | Hudson lectured on this in the early 20th century and certain phrases of his could potentially cause confusion. The take-away from this paragraph is that in order to algebraicize straightedge and compass construction, we begin by designating a given point as the origin and the coordinates of another given point (we are given two points at least) as <math>(1,0)</math> or <math>(0,1)</math>. Thus we have established the Cartesian Coordinates. Then, every time we construct a straight line or a circle, we think of it instead as adding a new equation into a system of equations. These equations represent the coordinates of all the points on the line or circle, but that is easy since we all know the expression for a line and a circle as <math>y = ax + b</math> and <math>(x-m)^2 + (y-n)^2 = r^2</math>. However, the only times we can pinpoint a point (and find its coordinates as a result) is when a line intersects with a line, or a circle, or a circle intersects with another circle in which case we can pinpoint 2 points. We then conclude that only those coordinates of the points of intersections are constructible. In this way, a geometric process is translated into an algebraic process. |
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| - | To construct <math>a \pm b</math>, we will use <math>a</math> as the center and use <math>b</math> as radius. The two points of intersection with the line will be <math>a+b</math> and <math>a-b</math>. | + | ==='''A Simple Derivation'''=== |
| - | [[Image:A+-b.png|center|border|700px]] | + | |
| | | | |
| | | + | Firstly, we define "1" on a straight line as stated previously. Then, once you have chosen that point to be <math>(1,0)</math>, you have to stick to this specification throughout your construction. Next, it is very obvious that we could construct all the integers, that is <math>\cdots -3,-2,-1,0,1,2,3,\cdots</math> (or <math>x</math> = <math>\{x|- \infty < x < \infty,x \in \mathbb{Z}\}</math>). How so? Well, once we have the "1", all we have to do is to use the Compass Equivalence Theorem finite number of times to duplicate the length "1" that we previously defined. Now, that means that we could have any two random integers, <math>a</math> and <math>b</math>, and for the sake of this discussion and clarity, we are talking about positive integers here. Next, it is shown that from <math>a</math> and <math>b</math>, we could construct <math>a \pm b</math>, <math>a \times b</math>, <math>\frac {a}{b}</math> and <math>\sqrt {a}</math>. |
| | | | |
| - | To construct <math>a \times b</math>, we have <math>0</math>, <math>1</math>, <math>a</math> and <math>b</math> on the straight line. | + | {{{!}}border="1" |
| | | + | {{!}}align="center"{{!}}<math>a \pm b</math>{{!}}{{!}}align="center"{{!}}<math>a \times b</math> |
| | | + | {{!}}- |
| | | + | {{!}}align="center"{{!}}[[Image:A+-b2.png|center|border|300px]]{{!}}{{!}}align="center"{{!}}[[Image:Achub.png|center|border|400px]] |
| | | + | {{!}}- |
| | | + | {{!}}To construct <math>a \pm b</math>, we will use <math>a</math> as the center and use <math>b</math> as radius. The two points of intersection with the line will be <math>a+b</math> and <math>a-b</math>.{{!}}{{!}}To construct <math>a \times b</math>, we have <math>0</math>, <math>1</math>, <math>a</math> and <math>b</math> on the straight line <math>l_0</math>. |
| | | | |
| - | # Draw any straight line through <math>0</math>, call it <math>l_1</math>. | + | # Draw a straight line through <math>0</math>, call it <math>l_1</math>. <math>l_1</math> could be constructed in many ways. For example, it could be the tangent line to circle centered at <math>a</math> with radius equals to the line segment connecting <math>b</math> to <math>a</math>. |
| | # Construct circle centered at <math>0</math> with radius <math>b</math>, intersecting <math>l_1</math> at <math>B</math>. | | # Construct circle centered at <math>0</math> with radius <math>b</math>, intersecting <math>l_1</math> at <math>B</math>. |
| | # Connect <math>B</math> and <math>1</math>, call it <math>l_2</math>. | | # Connect <math>B</math> and <math>1</math>, call it <math>l_2</math>. |
| Line 149: | | Line 139: | |
| | # Construct circle centered at <math>0</math> with radius <math>0A</math>, intersecting <math>l_0</math> at a point. | | # Construct circle centered at <math>0</math> with radius <math>0A</math>, intersecting <math>l_0</math> at a point. |
| | | | |
| - | The distance between <math>0</math> and that point is <math>ab</math>. | + | The distance between <math>0</math> and that point is <math>a \times b</math>. |
| | | + | {{!}}- |
| | | + | {{!}}align="center"{{!}}<math>\frac {a}{b}</math>{{!}}{{!}}align="center"{{!}}<math>\sqrt {a}</math> |
| | | + | {{!}}- |
| | | + | {{!}}align="center"{{!}}[[Image:Axb.png|center|border|400px]]{{!}}{{!}}align="center"{{!}}[[Image:Sqrta2.png|center|border|350px]] |
| | | + | {{!}}- |
| | | + | {{!}}To construct <math>\frac {a}{b}</math>, we have <math>0</math>, <math>1</math>, <math>a</math> and <math>b</math> on the straight line. |
| | | | |
| - | [[Image:Atimesb3.png|center|border|600px]] | + | # Draw straight line through <math>0</math>, call it <math>l_1</math>. Construction of <math>l_1</math> is the same as above. |
| - | | + | |
| - | | + | |
| - | Similarly, we could construct <math>\frac {a}{b}</math>. | + | |
| - | | + | |
| - | # Draw any straight line through <math>0</math>, call it <math>l_1</math>. | + | |
| | # Construct circle centered at <math>0</math> with radius <math>b</math>, intersecting <math>l_1</math> at <math>B</math>. | | # Construct circle centered at <math>0</math> with radius <math>b</math>, intersecting <math>l_1</math> at <math>B</math>. |
| | # Connect <math>B</math> and <math>1</math>, call it <math>l_2</math>. | | # Connect <math>B</math> and <math>1</math>, call it <math>l_2</math>. |
| Line 163: | | Line 154: | |
| | | | |
| | The distance between <math>0</math> and that point is <math>\frac {a}{b}</math>. | | The distance between <math>0</math> and that point is <math>\frac {a}{b}</math>. |
| | | + | {{!}}{{!}} |
| | | + | Therefore, it has been proven that we could construction all the rational numbers since <math>a</math> and <math>b</math> are any arbitrary integers. The natural question to ask right now is that what else is possible to construct? It is not hard to think of numbers that are not rational. For example, <math>\sqrt{2}</math> is constructible. Construct a unit square and the diagonal is of length <math>\sqrt{2}</math>. So is it possible to construct <math>\sqrt{a}</math> given any constructible number <math>a</math>? It turns out that we could. See below for method. |
| | | | |
| - | [[Image:Ab.png|center|border|750px]] | + | To construct <math>\sqrt {a}</math>, we start with <math>0</math>, <math>1</math>, <math>a</math>. |
| | | | |
| - | I will leave the proofs to you since they are very simple using similar triangles. | + | # Construct <math>a+1</math>. |
| | | + | # Construct <math>\frac {1}{2}(a+1)</math>. |
| | | + | # Construct circle at <math>\frac {1}{2}(a+1)</math> with radius <math>\frac {1}{2}(a+1)</math>. |
| | | + | # Draw perpendicular through <math>a</math>, intersecting the circle at point <math>A</math>. |
| | | | |
| | | + | <math>aA=\sqrt {a}</math>.Again, I will leave the proof to you as well using similar triangles. |
| | | + | {{!}}} |
| | | | |
| - | Therefore, it has been proven that we could construction all the rational numbers since <math>a</math> and <math>b</math> are any arbitrary integers. | | |
| - | | | |
| - | | | |
| - | The natural question to ask right now is that what else is possible to construct? It is not hard to think of numbers that are not rational. For example, <math>\sqrt{2}</math> is constructible. Construct a unit square and the diagonal is of length <math>\sqrt{2}</math>. So is it possible to construct <math>\sqrt{a}</math> given any constructible number <math>a</math>? It turns out that we could. See below for method. | | |
| - | | | |
| - | [[Image:Sqrta.png|center|border|650px]] | | |
| - | | | |
| - | # Construct <math>a+1</math> | | |
| - | # Construct <math>\frac {1}{2}(a+1)</math> | | |
| - | # Construct circle at <math>\frac {1}{2}(a+1)</math> with radius <math>\frac {1}{2}(a+1)</math> | | |
| - | # Draw perpendicular through <math>a</math>, intersecting the circle at point <math>A</math> | | |
| - | # <math>aA=\sqrt {a}</math> | | |
| - | | | |
| - | Again, I will leave the proof to you as well using similar triangles. | | |
| | | | |
| | | + | I will leave the proofs to you since they are very simple using similar triangles. |
| | | | |
| | Next, we moved to the general solution of the problem. | | Next, we moved to the general solution of the problem. |
| Line 198: | | Line 183: | |
| | Since <math>x_1</math>, <math>x_2</math>, <math>y_1</math> and <math>y_2</math> are constant we can express this as <math>ax+by=p</math> which is the general expression of a straight line. | | Since <math>x_1</math>, <math>x_2</math>, <math>y_1</math> and <math>y_2</math> are constant we can express this as <math>ax+by=p</math> which is the general expression of a straight line. |
| | | | |
| - | [[Image:Arbitrarypoint.png|center|border|650px]] | + | [[Image:Arbitrarypoint.png|center|border|500px]] |
| | | | |
| | | | |
| Line 217: | | Line 202: | |
| | \end{cases}</math> | | \end{cases}</math> |
| | | | |
| - | To solve for the points of intersection, we only need the operations of addition, subtraction, multiplication and division along with the <u>extraction of square roots</u>. Therefore, from this analysis, we have turned geometric problem into algebraic problem and come to the conclusion that '''a number is constructible if and only if it may be obtained from the integers by repeated use of addition, subtraction, multiplication, division and the extraction of square roots'''. | + | Similarly, if you there are two circles intersecting, then the two points of intersections have to satisfy the two equations of the circles. To solve for the points of intersection, we only need the operations of addition, subtraction, multiplication and division along with the <u>extraction of square roots</u>. Therefore, from this analysis, we have turned geometric problem into algebraic problem and come to the conclusion that '''a number is constructible if and only if it may be obtained from the integers by repeated use of addition, subtraction, multiplication, division and the extraction of square roots'''.<ref>Bryant, & Sangwin, 2008, p. 77</ref> |
| | | | |
| | | | |
| Line 225: | | Line 210: | |
| | What I have presented above is a simplified version of the derivation towards the theorem. To see a rigorous proof of this theorem at a college level, refer to the text below which is mainly taken from I. N. Herstein's ''Topics in Algebra, Second Edition''. You need some knowledge in Linear Algebra and/or Abstract Algebra. Also see [http://en.wikipedia.org/wiki/Constructible_number Constructible Numbers]. You should not be discouraged should you find it hard to understand. Instead, you should be marveled by the simplicity and elegance of the algebraic proof. | | What I have presented above is a simplified version of the derivation towards the theorem. To see a rigorous proof of this theorem at a college level, refer to the text below which is mainly taken from I. N. Herstein's ''Topics in Algebra, Second Edition''. You need some knowledge in Linear Algebra and/or Abstract Algebra. Also see [http://en.wikipedia.org/wiki/Constructible_number Constructible Numbers]. You should not be discouraged should you find it hard to understand. Instead, you should be marveled by the simplicity and elegance of the algebraic proof. |
| | | | |
| - | {{HideShow|1=We have proven that if <math>a</math> and <math>b</math> are constructible numbers, then so are <math>a \pm b</math>, <math>ab</math>, and when <math>b \ne 0</math>, <math>\frac {a}{b}</math>. Therefore, the set of constructible numbers form a subfield, <math>W</math>, of the [http://en.wikipedia.org/wiki/Field_(mathematics)#Constructible_numbers field] of real numbers. | + | {{SwitchPreview|ShowMessage=Click here to show proof|HideMessage=Click here to hide proof|PreviewText=We have proven that if <math>{\color{Gray}a}</math> and <math>{\color{Gray}b}</math> are constructible numbers,|FullText=<blockquote>We have proven that if <math>a</math> and <math>b</math> are constructible numbers, then so are <math>a \pm b</math>, <math>ab</math>, and when <math>b \ne 0</math>, <math>\frac {a}{b}</math>. Therefore, the set of constructible numbers form a subfield, <math>W</math>, of the [http://en.wikipedia.org/wiki/Field_(mathematics)#Constructible_numbers field] of real numbers. |
| | | | |
| | | | |
| Line 251: | | Line 236: | |
| | | | |
| | # '''If <math>a</math> is constructible then <math>a</math> lies in some extension of the rationals of degree a power of 2.''' | | # '''If <math>a</math> is constructible then <math>a</math> lies in some extension of the rationals of degree a power of 2.''' |
| - | # '''If the real number <math>a</math> satisfies an irreducible polynomial over the field of rational numbers of degree <math>k</math>, and if <math>k</math> is not a power of 2, then <math>a</math> is not constructible.''' }} | + | # '''If the real number <math>a</math> satisfies an irreducible polynomial over the field of rational numbers of degree <math>k</math>, and if <math>k</math> is not a power of 2, then <math>a</math> is not constructible.'''<ref>Herstein, 1975, p. 229</ref></blockquote>}} |
| | | | |
| | | + | |other=A little Geometry and Some Abstract Algebra |
| | | + | |AuthorName=Wikipedia |
| | | + | |AuthorDesc=Wikipedia, Powerpoint and Flash |
| | | + | |SiteName=Compass and Straightedge Constructions |
| | | + | |SiteURL=http://en.wikipedia.org/wiki/Compass_and_straightedge_constructions |
| | | + | |Field=Geometry |
| | | + | |WhyInteresting= <br> |
| | | + | ==What is Impossible to Construct (of course, using compass and straightedge alone)?== |
| | | | |
| | | + | Below is the brief introduction of a few of the impossible constructions. Remember that a number is constructible if and only if it may be obtained from the integers by repeated use of addition, subtraction, multiplication, division and the extraction of square roots. |
| | | | |
| - | ==='''What is Impossible to Construct (of course, using compass and straightedge alone)?'''=== | + | <ol> |
| - | | + | <li><math>\pi</math> cannot be obtained from the integers by repeated use of addition, subtraction, multiplication, division and the extraction of square roots. In fact, <math>\pi</math> belongs to a special class of numbers called the transcendental number that does not satisfy any rational polynomials. In other words, <math>\pi</math> is not a solution of any polynomials with rational coefficients. Too see complete proof that <math>\pi</math> is {{EasyBalloon|Link=transcendental|Balloon=a number (possibly a complex number) which is not algebraic—that is, it is not a root of a non-constant polynomial equation with rational coefficients.<ref>Wikipedia (Transcendental number)</ref>}} |
| - | Now, having derived what is possible to construct, I will briefly introduce a few of the impossible constructions. | + | , see [http://en.wikipedia.org/wiki/Transcendental_number Transcendental number] and [http://sprott.physics.wisc.edu/pickover/trans.html The 15 Most Famous Transcendental Numbers].</li> |
| - | | + | <li>From the above impossible construction, it follows that it is impossible to "square the circle (that is to construct a square that has the same area as a given circle)" because given a circle with radius 1, which is constructible, the area of the circle will be <math>\pi</math> and we have to construct square with sides equal to <math>\sqrt \pi</math> which is not constructible. Due to this exception, there is no general method to square the circle.</li> |
| - | # <math>\pi</math> is transcendental since it does not satisfy any rational polynomials. Too see complete proof that <math>\pi</math> is transcendental, see [http://en.wikipedia.org/wiki/Transcendental_number Transcendental number] and [http://sprott.physics.wisc.edu/pickover/trans.html The 15 Most Famous Transcendental Numbers]. | + | <li>We could not double the volume of a given cube. Say we start with cube of volume 1, which is constructible. Then we have to construct cube of volume 2, which means we have to construct sides of <math>\sqrt [3]{2}</math> which is impossible to construct. So we can't double the cube.</li> |
| - | # From the above impossible construction, it follows that it is impossible to "square the circle(that is to construct a square that has the same area as a given circle" because given a circle with radius <math>r</math> and hence area <math>\pi r^2</math>, we can not construct <math>\sqrt {\pi} r</math>. | + | <li>We generally can not trisect any given angle because the process involves taking cube root. For example, it is impossible to trisect <math>60^\circ</math>. See below for proof. For more, refer to [http://www.jimloy.com/geometry/trisect.htm#curves Trisection of an Angle]for explanation in great detail. Proof of <math>60^\circ</math> is impossible to trisect. {{HideShowThis|ShowMessage=Click here to show proof|HideMessage=Click here to hide proof|HiddenText=If we could trisect <math>60^\circ</math> by compass and straightedge, then the length <math>a = \cos 20^\circ</math> would be constructible. Since <math>\cos 3\theta = 4\cos^3\theta - 3 \cos \theta</math>. Substituting <math>\theta = 20^\circ</math> and <math>\cos60^\circ=\frac {1}{2}</math>, we obtain <math>4a^3-3a=\frac {1}{2}</math>. Thus <math>a</math> is a root of a cubic polynomial over the rational field. Since this polynomial is irreducible over the rational field and its degree is 3, <math>a</math> is not constructible. Thus <math>60^\circ</math> cannot be trisected.<ref>Herstein, 1975, p. 230</ref>}} |
| - | # We could not double the volume of a given cube because we could not construct <math>\sqrt[3]{2}</math>. | + | </li> |
| - | # We generally can not trisect any given angle because the process involves taking cube root. For example, it is impossible to trisect <math>60^\circ</math>. See below for proof. For more, refer to [http://www.jimloy.com/geometry/trisect.htm#curves Trisection of an Angle]for explanation in great detail. | + | <li>There are certain polygons that are impossible to construct. See [http://en.wikipedia.org/wiki/Constructible_polygon Constructible polygon] for more detail. </li> |
| - | # There are certain polygons that are impossible to construct. See [http://en.wikipedia.org/wiki/Constructible_polygon Constructible polygon] for more detail. | + | </ol> |
| | | | |
| | Number 2, 3 and 4 are the so-called [http://mathworld.wolfram.com/GeometricProblemsofAntiquity.html Geometric Problems of Antiquity]. Though they have been proven impossible to construct with straightedge and compass, it does not deter amateur mathematicians to come up with false proofs even today. | | Number 2, 3 and 4 are the so-called [http://mathworld.wolfram.com/GeometricProblemsofAntiquity.html Geometric Problems of Antiquity]. Though they have been proven impossible to construct with straightedge and compass, it does not deter amateur mathematicians to come up with false proofs even today. |
| | | | |
| - | Proof that <math>60^\circ</math> is impossible to trisect. | | |
| | | | |
| - | If we could trisect <math>60^\circ</math> by compass and straightedge, then the length <math>a = \cos 20^\circ</math> would be constructible. Since <math>\cos 3\theta = 4\cos^3\theta - 3 \cos \theta</math>. Substituting <math>\theta = 20^\circ</math> and <math>\cos60^\circ=\frac {1}{2}</math>, we obtain <math>4a^3-3a=\frac {1}{2}</math>. Thus <math>a</math> is a root of a cubic polynomial over the rational field. Since this polynomial is irreducible over the rational field and its degree is 3, <math>a</math> is not constructible. Thus <math>60^\circ</math> cannot be trisected. | | |
| | | | |
| - | |other=A little Geometry and Some Abstract Algebra | | |
| - | |AuthorName=Wikipedia | | |
| - | |AuthorDesc=Wikipedia, Powerpoint and Flash | | |
| - | |SiteName=Compass and Straightedge Constructions | | |
| - | |SiteURL=http://en.wikipedia.org/wiki/Compass_and_straightedge_constructions | | |
| - | |Field=Geometry Algebra | | |
| - | |WhyInteresting= | | |
| | |ImageRelates= | | |ImageRelates= |
| | *):http://hptgn.tripod.com/ | | *):http://hptgn.tripod.com/ |
| | *):http://en.wikipedia.org/wiki/Compass_and_straightedge_constructions | | *):http://en.wikipedia.org/wiki/Compass_and_straightedge_constructions |
| | | + | *):http://www.mathopenref.com/tocs/constructionstoc.html |
| | | + | *):http://mathforum.org/library/drmath/view/66052.html |
| | | + | *):http://mathforum.org/library/drmath/view/52601.html |
| | | + | *):http://www.ams.org/notices/200004/fea-hartshorne.pdf |
| | | + | =Notes= |
| | | + | <references/> |
| | |References= | | |References= |
| - | |ToDo=zzz | + | #Peterson, D. (2003, November 21). Collapsible Compass. Retrieved from The Math Forum: http://mathforum.org/library/drmath/view/66052.html |
| - | |InProgress=Yes | + | #Hartshorne, Robin . (2000). Teaching geometry according to euclid. NOTICES OF THE AMS, 47(4), 460-465. |
| | | + | #Taylor, H. M. (1895). Euclid's elements of geometry. Cambridge: Cabridge University Press. |
| | | + | #Hudson, H. P. (1916). Ruler & compass. London: Longmans Green & Company, Inc.. |
| | | + | #Bryant, John, & Sangwin, Christopher. (2008). How Round is your circle?. Princeton & Oxford: Princeton Univ Pr. |
| | | + | #Herstein, I. N. (1975). Topics in algebra. John Wiley & Sons Inc. |
| | | + | #Wikipedia (Transcendental number). (n.d.). Transcendental number. Retrieved from Wikipedia: http://en.wikipedia.org/wiki/Transcendental_number |
| | | + | |InProgress=No |
| | }} | | }} |
## Current revision
Creating a regular hexagon with a ruler and compass
Field: Geometry
Image Created By: Wikipedia
Website: Compass and Straightedge Constructions
Creating a regular hexagon with a ruler and compass
This image shows the step by step construction of a hexagon inscribed in the circle using a compass and a unmarked straightedge.
# Basic Description
Let's assume we only have a compass and a unmarked straightedge. What can we construct and how can we construct them? That were the problems that Euclid pondered not only because those were probably the only instruments that he had at his time but also he wanted to build his theorems with as few assumptions, or axiomsIn traditional logic, an axiom or postulate is a proposition that is not proved or demonstrated but considered to be either self-evident, or subject to necessary decision. Therefore, its truth is taken for granted, and serves as a starting point for deducing and inferring other (theory dependent) truths., as possible.[1] With these two simple tools, he managed to build myriad of theorems in both plane and solid geometry .[2] These theorem are as good as they were some 2000 years ago and it is this enduring quality of Euclid's work that inspired this page. In the main image of this page, we want to divide the circle into six equal arcs and then connect consecutive points to form the hexagon. It seems to be a fairly simple construction but you should be prompted to ask two questions: Are other polygons constructible and is every polygon constructible, that is able to be constructed using only compass and straightedge? To extend the question, what are constructible and what are not? That is the problem that is resolved in this page.
# A More Mathematical Explanation
Note: understanding of this explanation requires: *A little Geometry and Some Abstract Algebra
[Click to view A More Mathematical Explanation]
## What is Compass & Straightedge Constructions
### Introduction
We start by familiarizing [...]
[Click to hide A More Mathematical Explanation]
## What is Compass & Straightedge Constructions
### Introduction
We start by familiarizing ourselves with Euclid's three Postulates in his books Elements.
Let it be granted
1. that a straight line may be drawn from any one point to any other point;
2. that a line segment may be extended into a straight line;
3. that given any straight line segment, a circle may be described having the segment as radius and one endpoint as center.[3]
It should be carefully noted that Euclid started with two given points and produced a line segment, from where he could extend into a straight line if he pleased. Then ONLY from the original two points, he could use one point as center and ONLY spread the legs of compass the distance of the line segment to produce a circle. Even though many translations say that a circle can be described at any center with any radius, we must take it with a pinch of salt. He could not specify any points and lengths other than what was already given, that is to say he could not claim "I wanted to spread the legs of the compass $\pi$ centimeters apart (or any specified denominations) with the center of the circle half way between the two given points". In all, Compass and Straightedge Constructions only allow us to start with points (and hence lengths) we have been given(or constructed from given points), and create the ones we don't.
Thus, we define Compass & Straightedge Construction as the construction of points, lengths, angles, and circles using only ideal straightedge and compass. A straightedge is infinite in length, has no markings on it and only one edge. A compass has two legs, one end of which is fixed on the plane of construction and the other end is of given distance away and maintains the distance throughout the construction. It collapses when lifted from the page, so may not be directly used to transfer distances. However, it turns out that this restriction makes no difference due to the Compass Equivalence Theorem which was stated as Proposition II of Book I of Euclid's Elements. It stated that from a given point, it was possible to construct a line segment equal to a given line segment using collapsible compass in any desirable direction. Euclid's proof for the Compass Equivalence Theorem will be presented after the section of Basic Construction. Since Euclid has proven this using only the three postulates, then he did not have to use a collapsible compass any more.[4]
### Some Basic Constructions
The constructions below are some basic ones from where many more constructions are possible and they are by no means exhaustive. In the figures below, what we are given are in blue; intermediate steps are in dotted black; the resulting products are in red. The proofs for these constructions are relatively simple and only require the knowledge of congruent triangles. Euclid derived the theories on congruency and congruent triangles directly from his Postulates. Try proving the theorems yourself!
#### Line Segment Bisection
Given points ${\color{Gray}A}$ and ${\color{Gray}B}$ and the straight line passing through it. Construct [...]
Given points $A$ and $B$ and the straight line passing through it. Construct a line that bisects line segment $AB$.
1. Draw a circle centered at point $A$ with radius equals $AB$.
2. Next, draw a circle centered at point $B$ with the same radius.
3. Where the two circles intersect, call those points $C$ and $D$.
4. Draw a line through points $C$ and $D$.
Line $CD$ intersects line segment $AB$ at the midpoint. It should be noted that $CD \perp AB$ as well. Line $CD$ is the perpendicular bisector of line segment $AB$.
#### Angle Bisection
Given angle ${\color{Gray}\angle AOB}$, construct a line that bisects the angle.
Given angle $\angle AOB$, construct a line that bisects the angle.
1. Construct a circle centered at point $O$ with radius $OA$. This circle intersects $OA$ and $OB$ at point $A$ and $B$.
2. Keeping the same radius, draw a circle at point $A$ and $B$ respectively. Where they intersect each other, call it point $C$.
3. Draw a line through point $C$ and $O$. This line bisects $\angle AOB$.
#### Perpendicular Through a Point
Given a point ${\color{Gray}M}$ on a line ${\color{Gray}AB}$, construct a line that is perpendicular to th [...]
Given a point $M$ on a line $AB$, construct a line that is perpendicular to the given line through $M$.
1. Draw a circle centered at $M$ with radius $MB$ (or $MA$. It is your choice).
2. Where the circle intersects the original line, construct a perpendicular bisector (see the line segment bisector construction above).
#### Parallel Line
Given two points, ${\color{Gray}A}$ and ${\color{Gray}B}$ and the straight line passing through them, con [...]
Given two points, $A$ and $B$ and the straight line passing through them, construct a line that is parallel to the given line through another given point $C$.
1. Draw a circle at $A$, crossing $C$. Where the circle $A$ intersects $AB$, call the point $D$.
2. Centered at points $C$ and $D$, draw circles crossing point $A$. Where these two circles intersect each other, call it $E$.
3. Draw a line through point $C$ and $E$.
$CE$ is parallel to $AB$
#### Tangent Line to a Circle
Given a circle centered at ${\color{Gray}O}$ and another given point, ${\color{Gray}A}$, construct a line [...]
Given a circle centered at $O$ and another given point, $A$, construct a line that is tangent to the circle.
1. Draw a line through point $A$ and the center of the circle $O$
2. Let $M$ be the midpoint of $OA$(construction omitted since we know how to construction mid point). Draw the circle centered at $M$ going through $A$ and $O$.
3. Let the point where the two circles meet be $C$. Draw line segment $AC$.
To see that $AC$ is tangent to the circle, connect $OC$. $\angle ACO$ is an inscribed angle in the circle about $M$, so the inscribed $\angle ACO$ is $90^\circ$. So $OC$ is perpendicular to $AC$. Since the tangent is perpendicular to the radius to the point of tangency, by the uniqueness of the line through $C$ perpendicular to $OC$, $AC$ is tangent to the circle.
#### Euclid's Proof of Compass Equivalence Theorem
This part refers back to the previous section about the issue of compass being collapsible. Euclid's original proof is presented. Additional comments are contained in the parenthesis.
From a given point to draw a line segment equal to a given line segment.
From a given point to draw a line segment equal to a given line segment. Let $A$ be the given point, and $BC$ the given straight line : it is required to draw from the point $A$ a straight line equal to $BC$.
1. From the point $A$ to $B$ draw the straight line $AB$ and on it describe the equilateral triangle $DAB$(this is done the same way as bisecting a line segment), and produce the straight lines $DA$, $DB$ to $E$ and $F$.
2. From the center $B$, at the distance $BC$(with the radius equals to $BC$), describe the circle $CGH$, meeting $DF$ at $G$.
3. From the center $D$, at the distance $DG$(with the radius equals to $DG$), describe the circle $GKL$, meeting $DE$ at $L$.
4. $AL$ shall be equal to $BC$.
Because the point $B$ is the center of the circle $CGH$, $BC$ is equal to $BG$. And because the point $D$ is the center of the circle $GKL$, $DL$ is equal to $DG$ and $DA$, $DB$ parts of them are equal therefore the remainder $AL$ is equal to the remainder $BG$.
But it has been shewn that $BC$ is equal to $BG$ ; therefore $AL$ and $BC$ are each of them equal to $BG$. But things which are equal to the same thing are equal to one another. Therefore $AL$ is equal to $BC$.
Wherefore from the given point $A$ a straight line $AL$ has been drawn equal to the given straight line $BC$.∎[5]
It should be noted that from $AL$, we could "duplicate" $AL$ in all directions by construct a circle centered at $A$ with radius $AL$.
## Algebraicization of Compass & Straightedge Constructions
### What is Algebraicization?
From the few basic constructions, you would have probably realized that the different possibilities seems infinite. Hence, mathematician are curious to find out what are constructible and what aren't and for this purpose, the language of pure geometry seems to have "limited vocabulary"[6]. Back in ancient times, mathematicians had limited algebraic knowledge and were more familiar with geometry. But in modern times, the reverse is true. Hence, today's mathematicians go back to their familiar realm of Algebra and try to find the link between geometry and algebra.
Algebraicization is the translation of any problem statements into algebraic problems. In the case of Compass & Straightedge construction, we algebraicize each step of a straightedge and compass construction, and consequently obtaining general results about the nature of constructibility. Hilda P. Hudson put it aptly in his lecture Ruler & Compasses,
"each step of a ruler and compass construction is equivalent to a certain analytical process; it is found that the power to use a ruler corresponds exactly to the power to solve linear equations, and the power to use compasses to the power to solve quadratics...... Since each step of a ruler and compass construction is equivalent to the solution of an equation of the first or second degree, we consider that these algebraic processes can lead to , when combined in every possible way, and that enables us to answer the question before us......"[7]
Hudson lectured on this in the early 20th century and certain phrases of his could potentially cause confusion. The take-away from this paragraph is that in order to algebraicize straightedge and compass construction, we begin by designating a given point as the origin and the coordinates of another given point (we are given two points at least) as $(1,0)$ or $(0,1)$. Thus we have established the Cartesian Coordinates. Then, every time we construct a straight line or a circle, we think of it instead as adding a new equation into a system of equations. These equations represent the coordinates of all the points on the line or circle, but that is easy since we all know the expression for a line and a circle as $y = ax + b$ and $(x-m)^2 + (y-n)^2 = r^2$. However, the only times we can pinpoint a point (and find its coordinates as a result) is when a line intersects with a line, or a circle, or a circle intersects with another circle in which case we can pinpoint 2 points. We then conclude that only those coordinates of the points of intersections are constructible. In this way, a geometric process is translated into an algebraic process.
### A Simple Derivation
Firstly, we define "1" on a straight line as stated previously. Then, once you have chosen that point to be $(1,0)$, you have to stick to this specification throughout your construction. Next, it is very obvious that we could construct all the integers, that is $\cdots -3,-2,-1,0,1,2,3,\cdots$ (or $x$ = $\{x|- \infty < x < \infty,x \in \mathbb{Z}\}$). How so? Well, once we have the "1", all we have to do is to use the Compass Equivalence Theorem finite number of times to duplicate the length "1" that we previously defined. Now, that means that we could have any two random integers, $a$ and $b$, and for the sake of this discussion and clarity, we are talking about positive integers here. Next, it is shown that from $a$ and $b$, we could construct $a \pm b$, $a \times b$, $\frac {a}{b}$ and $\sqrt {a}$.
$a \pm b$ $a \times b$
To construct $a \pm b$, we will use $a$ as the center and use $b$ as radius. The two points of intersection with the line will be $a+b$ and $a-b$. To construct $a \times b$, we have $0$, $1$, $a$ and $b$ on the straight line $l_0$. Draw a straight line through $0$, call it $l_1$. $l_1$ could be constructed in many ways. For example, it could be the tangent line to circle centered at $a$ with radius equals to the line segment connecting $b$ to $a$. Construct circle centered at $0$ with radius $b$, intersecting $l_1$ at $B$. Connect $B$ and $1$, call it $l_2$. Construct $l_3$ that is parallel to $l_2$, intersecting $l_1$ at $A$. Construct circle centered at $0$ with radius $0A$, intersecting $l_0$ at a point. The distance between $0$ and that point is $a \times b$.
$\frac {a}{b}$ $\sqrt {a}$
To construct $\frac {a}{b}$, we have $0$, $1$, $a$ and $b$ on the straight line. Draw straight line through $0$, call it $l_1$. Construction of $l_1$ is the same as above. Construct circle centered at $0$ with radius $b$, intersecting $l_1$ at $B$. Connect $B$ and $1$, call it $l_2$. Construct circle centered at $0$ with radius $a$, intersecting $l_1$ at $A$. Construct $l_3$ that is parallel to $l_2$, intersecting $l_0$ at a point. The distance between $0$ and that point is $\frac {a}{b}$. Therefore, it has been proven that we could construction all the rational numbers since $a$ and $b$ are any arbitrary integers. The natural question to ask right now is that what else is possible to construct? It is not hard to think of numbers that are not rational. For example, $\sqrt{2}$ is constructible. Construct a unit square and the diagonal is of length $\sqrt{2}$. So is it possible to construct $\sqrt{a}$ given any constructible number $a$? It turns out that we could. See below for method. To construct $\sqrt {a}$, we start with $0$, $1$, $a$. Construct $a+1$. Construct $\frac {1}{2}(a+1)$. Construct circle at $\frac {1}{2}(a+1)$ with radius $\frac {1}{2}(a+1)$. Draw perpendicular through $a$, intersecting the circle at point $A$. $aA=\sqrt {a}$.Again, I will leave the proof to you as well using similar triangles.
I will leave the proofs to you since they are very simple using similar triangles.
Next, we moved to the general solution of the problem.
Assume we have two points $P_1$ and $P_2$ with coordinates $(x_1, y_1)$ and $(x_2, y_2)$. Take an arbitrary point $X (x,y)$ on the line.
By similar triangle, $\frac {x_1-x}{x-x_2} = \frac {y-y_1}{y_2-y}$.
Rearranging the above we have
$x(y_1-y_2)+y(x_2-x_1)=x_2y_1-x_1y_2$.
Since $x_1$, $x_2$, $y_1$ and $y_2$ are constant we can express this as $ax+by=p$ which is the general expression of a straight line.
Now, if we have two lines specified by four given points, $P_1$,....,$P_4$ with coordinates $(x_i,y_i)$. The intersection of the two lines, $P (x,y)$ will satisfy two equations
$\begin{cases} x(y_1-y_2)+y(x_2-x_1)=x_2y_1-x_1y_2\\ x(y_3-y_4)+y(x_4-x_3)=x_4y_3-x_3y_4\\ \end{cases}$
You may say that the there might not be a solution. True the two lines do not have to intersect. But if they do, we only need the operations of addition, subtraction, multiplication and division to find the point.
Now, we move onto circle. Say we have circle centered at some point $P_5$ with coordinates $(x_5, y_5)$ and radius $r$. We know that the explicit expression for a circle is $(x-x_5)^2+(y-y_5)^2=r^2$. Hence, if that circle intersects with one of the straight lines, then the points of intersection will satisfy
$\begin{cases} x(y_1-y_2)+y(x_2-x_1)=x_2y_1-x_1y_2\\ (x-x_5)^2+(y-y_5)^2=r^2\\ \end{cases}$
Similarly, if you there are two circles intersecting, then the two points of intersections have to satisfy the two equations of the circles. To solve for the points of intersection, we only need the operations of addition, subtraction, multiplication and division along with the extraction of square roots. Therefore, from this analysis, we have turned geometric problem into algebraic problem and come to the conclusion that a number is constructible if and only if it may be obtained from the integers by repeated use of addition, subtraction, multiplication, division and the extraction of square roots.[8]
### A Rigorous Proof
What I have presented above is a simplified version of the derivation towards the theorem. To see a rigorous proof of this theorem at a college level, refer to the text below which is mainly taken from I. N. Herstein's Topics in Algebra, Second Edition. You need some knowledge in Linear Algebra and/or Abstract Algebra. Also see Constructible Numbers. You should not be discouraged should you find it hard to understand. Instead, you should be marveled by the simplicity and elegance of the algebraic proof.
We have proven that if ${\color{Gray}a}$ and ${\color{Gray}b}$ are constructible numbers,
We have proven that if $a$ and $b$ are constructible numbers, then so are $a \pm b$, $ab$, and when $b \ne 0$, $\frac {a}{b}$. Therefore, the set of constructible numbers form a subfield, $W$, of the field of real numbers. In particular, since $1 \in W$, $W$ must contain $F_0$, the field of rational numbers. If $w \in W$, we can reach $w$ from the rational field by a finite number of constructions. Let $F$ be any subfield of the field of the field of real numbers. Consider all the points $(x,y)$ in the real Euclidean plane both of whose coordinates $x$ and $y$ are in $F$; we call the set of these points the plane of $F$. Any straight line joining two points in the plane of $F$ has an equation of the form $ax+by+c=0$ where $a$,$b$,$c$ are all in $F$. Moreover,any circle having as center a point in the plane of $F$ and having as radius an element of $F$ has equation of the form $x^2+y^2+ax+by+c=0$, where all of $a$, $b$, $c$ are in $F$. Given two lines in $F$ which intersect in the real plane, then their intersection point is a point in the plane of $F$. On the other hand, the intersection of a line in $F$ and a circle in $F$ need not yield a point in the plane of $F$. But, using the fact that the equation of a line in $F$ is of the form $ax+by+c=0$ and that of a circle in F is of the form $x^2+y^2+dx+ey+f=0$, where $a$, $b$, $c$, $d$, $e$, $f$ are all in $F$, we can show that when a line and circle of $F$ intersect in the real plane, they intersect either in a point in the plane of $F$ or in the plane of $F(\sqrt {r})$ for some positive $r$ in $F$. Finally, the intersection of two circles in $F$ can be realized as that of a line in $F$ and a circle in $F$, for if these two circles are $x^2+y^2+a_1x+b_1y+c_1=0$ and $x^2+y^2+a_2x+b_2y+c_2=0$, then their intersection is the intersection of either of these with the line $(a_1-a_2)x+(b_1-b_2)y+(c_1-c_2)=0$, so also yields a point either in the plane of $F$ or of $F(\sqrt{r})$ for some positive $r$ in $F$. Thus lines and circles of $F$ lead us to points either in $F$ or in quadratic extensions of $F$. If we now are in $F(\sqrt{r_1})$ intersect in in points in the plane of $F(\sqrt{r_1},\sqrt{r_2})$ where $r_2$ is a positive number in $F(\sqrt{r_1})$. A point is constructible from $F$ is we can find real numbers $r_1,...,r_n$, such that $r_1^2 \in F$, $r_2^2 \in F(r_1)$, $r_3^2 \in F(r_1,r_2),...,$$r_n^2 \in F(r_1,...,r_{n-1})$, such that the point is in the plane of $F(r_1,...,r_n)$. Conversely, if $r \in F$ is such that $\sqrt{r}$ is real then we can realize $r$ as an intersection of lines and circles in $F$. Thus a point is constructible from $F$ is and only if we can find a finite number of real numbers $r_1,...,r_n$, such that
1. $[F(r_1:F)] = 1$ or $2$;
2. $[F(r_1,...,r_i):F(r_1,...,r_{i-1}]=1$ or $2$ for $i=1,2,...,n$;
and such that our point lies in the plane of $F(r_1,...,r_n)$. We have defined a real number $a$ to be constructible if by use of straightedge and compass we can construct a line segment of length $a$. But this translates, in terms of the discussion above, into: $a$ is constructible if starting from the plane of the rational numbers, $F_0$, we can imbed a in a field obtained from $F_0$ by a finite number of quadratic extensions. And therefore,
1. If $a$ is constructible then $a$ lies in some extension of the rationals of degree a power of 2.
2. If the real number $a$ satisfies an irreducible polynomial over the field of rational numbers of degree $k$, and if $k$ is not a power of 2, then $a$ is not constructible.[9]
# Why It's Interesting
## What is Impossible to Construct (of course, using compass and straightedge alone)?
Below is the brief introduction of a few of the impossible constructions. Remember that a number is constructible if and only if it may be obtained from the integers by repeated use of addition, subtraction, multiplication, division and the extraction of square roots.
1. $\pi$ cannot be obtained from the integers by repeated use of addition, subtraction, multiplication, division and the extraction of square roots. In fact, $\pi$ belongs to a special class of numbers called the transcendental number that does not satisfy any rational polynomials. In other words, $\pi$ is not a solution of any polynomials with rational coefficients. Too see complete proof that $\pi$ is transcendentala number (possibly a complex number) which is not algebraic—that is, it is not a root of a non-constant polynomial equation with rational coefficients. , see Transcendental number and The 15 Most Famous Transcendental Numbers.
2. From the above impossible construction, it follows that it is impossible to "square the circle (that is to construct a square that has the same area as a given circle)" because given a circle with radius 1, which is constructible, the area of the circle will be $\pi$ and we have to construct square with sides equal to $\sqrt \pi$ which is not constructible. Due to this exception, there is no general method to square the circle.
3. We could not double the volume of a given cube. Say we start with cube of volume 1, which is constructible. Then we have to construct cube of volume 2, which means we have to construct sides of $\sqrt [3]{2}$ which is impossible to construct. So we can't double the cube.
4. We generally can not trisect any given angle because the process involves taking cube root. For example, it is impossible to trisect $60^\circ$. See below for proof. For more, refer to Trisection of an Anglefor explanation in great detail. Proof of $60^\circ$ is impossible to trisect.
If we could trisect $60^\circ$ by compass and straightedge, then the length $a = \cos 20^\circ$ wou [...]
If we could trisect $60^\circ$ by compass and straightedge, then the length $a = \cos 20^\circ$ would be constructible. Since $\cos 3\theta = 4\cos^3\theta - 3 \cos \theta$. Substituting $\theta = 20^\circ$ and $\cos60^\circ=\frac {1}{2}$, we obtain $4a^3-3a=\frac {1}{2}$. Thus $a$ is a root of a cubic polynomial over the rational field. Since this polynomial is irreducible over the rational field and its degree is 3, $a$ is not constructible. Thus $60^\circ$ cannot be trisected.[11]
5. There are certain polygons that are impossible to construct. See Constructible polygon for more detail.
Number 2, 3 and 4 are the so-called Geometric Problems of Antiquity. Though they have been proven impossible to construct with straightedge and compass, it does not deter amateur mathematicians to come up with false proofs even today.
# Teaching Materials
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# About the Creator of this Image
Wikipedia, Powerpoint and Flash
# Notes
1. ↑ Peterson, 2003
2. ↑ Hartshorne, 2000
3. ↑ Taylor, 1895, p. 14
4. ↑ Peterson, 2003
5. ↑ Taylor, 1895, p. 18
6. ↑ Hudson, 1916, p. 3
7. ↑ Hudson, 1916, p. 3
8. ↑ Bryant, & Sangwin, 2008, p. 77
9. ↑ Herstein, 1975, p. 229
10. ↑ Wikipedia (Transcendental number)
11. ↑ Herstein, 1975, p. 230
# References
1. Peterson, D. (2003, November 21). Collapsible Compass. Retrieved from The Math Forum: http://mathforum.org/library/drmath/view/66052.html
2. Hartshorne, Robin . (2000). Teaching geometry according to euclid. NOTICES OF THE AMS, 47(4), 460-465.
3. Taylor, H. M. (1895). Euclid's elements of geometry. Cambridge: Cabridge University Press.
4. Hudson, H. P. (1916). Ruler & compass. London: Longmans Green & Company, Inc..
5. Bryant, John, & Sangwin, Christopher. (2008). How Round is your circle?. Princeton & Oxford: Princeton Univ Pr.
6. Herstein, I. N. (1975). Topics in algebra. John Wiley & Sons Inc.
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http://mathoverflow.net/questions/52874/statistics-of-extended-gcd/53637 | ## Statistics of Extended GCD
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
Consider a coprime pair of integers $a, b.$ As we all know ("Bezout's theorem") there is a pair of integers $c, d$ such that $ac + bd=1.$ Consider the smallest (in the sense of Euclidean norm) such pair $c_0, d_0$, and consider the ratio $\frac{\|(c_0, d_0)\|}{\|(a, b)\|}.$ The question is: what is the statistics of this ratio as $(a, b)$ ranges over all visible pairs in, for example, the square $1\leq a \leq N, 1 \leq b \leq N?$
Experiment shows the following amazing histogram:
EDIT by popular demand: the histogram is for an experiment for $N=1000.$ The $x$ axis is the ratio, the $y$ axis is the number of points in the bin. The total number of points is $1000000/\zeta(2),$ so there are $100$ bins each with around $6000$ points.
But no immediate proof leaps to mind.
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Can you label your axes? I take it $x$-axis is the ratio, $y$ axis is number of pairs, $N$ is some moderately large number? – Gerry Myerson Jan 22 2011 at 22:11
@Gerry: yes, you guessed right. $N$ is 1000 in this histogram. – Igor Rivin Jan 22 2011 at 22:31
It looks as if you've got a bin-width of about 0.003, so about 166 bins and somewhat over 6000 cases per bin. It would be less distracting from the point of your question if you said all this explicitly. – Michael Hardy Jan 22 2011 at 23:31
Perhaps rewrite it as $|ac-bd|=1$ with everything positive and examine for each $c,d$ which pairs $a,b$ it "works" for. $d/c$ should be a next to the last convergent to $a/b$ so... – Aaron Meyerowitz Jan 22 2011 at 23:53
2
Concerning the importance of labeling axes, see xkcd.com/833 – Gerry Myerson Jan 23 2011 at 5:44
show 4 more comments
## 5 Answers
I did a little experiment. Fix $a=29$, let $b=1,2,\dots,28$. So, you get 28 data points. Well, these points are already extremely regularly distributed. Taking just the first half, $1\le b\le14$, and rearranging the ratios in increasing order, they are (to three decimals) $$.034,.069,.103,.138,.172,.207,.242,.275,.310,.345,.379,.414,.448,.483$$ To three decimals, and modulo round-off errors, these are the numbers $1/29,2/29,\dots,14/29$, which is to say they are about as regularly distributed as possible. The ratios for $15\le b\le28$ are essentially the same numbers - in fact, the ratio for $(a,b)$ seems to be pretty nearly the ratio for $(b-a,b)$.
If what's happening for 29 happens in general, I think it would explain the original histogram.
EDIT: So I think I see what's going on. We're looking at the numbers $$\sqrt{c^2+d^2\over a^2+b^2}$$ But $b$ is very close to $-ac/d$ (since $ac+bd=1$), so these numbers are very close to $$\sqrt{c^2+d^2\over a^2+(ac/d)^2}$$ which simplifies to $|d|/a$. For fixed $a$, as $b$ runs through the units modulo $a$, so does $d$, since $bd\equiv1\pmod a$. So our ratios are as uniformly distributed as the fractions $|d|/a$, which is very.
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The bijection (reversing continued fractions) in my remark explains this. – David Feldman Jan 23 2011 at 5:35
"Explains" in what sense? Proves that for fixed $a$ you get pretty nearly exactly the numbers $1/a$, $2/a$, etc.? I don't see how reversing continued fractions gets information this specific. – Gerry Myerson Jan 23 2011 at 5:51
Because denominators stay invariant on reversing the continued fraction (if you do it right). Look at your own example in that notation. – David Feldman Jan 23 2011 at 5:57
@Gerry: this explains many things! A question: why are $|d|/a$ "very uniformly distributed"? – Igor Rivin Jan 23 2011 at 15:20
@Gerry: presumably, one can write down an error term (for deviation from uniformity), which probably has much to do with @Bill's mixing rate for horocycle flow... – Igor Rivin Jan 23 2011 at 15:22
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.
Here's a more geometric formulation of your question:
On the torus $\mathbb R^2/\mathbb Z^2$, consider a long simple closed geodesic $\overline {(0,0)(a,b)}$. It cuts the torus into a thin cylinder; the cylinder is joined to itself by a twist by some angle to form the torus. What is the distribution of the angle of the twist?
From this perspective, perhaps your intuition tells you that the angles of twist should tend toward the uniform distribution, as the homotopy classes of geodesics are chosen uniformly with longer and longer lengths.
To get a rigorous argument, we can think about the problem from the opposite direction. Begin by starting from a long thin annulus of area 1, and ask what are the ways to glue it together to form a torus? You can glue it by any angle; however, the torus you get is not usually isometric to the square torus; but it is isometric to $\mathbb E^2$ modulo some lattice of area 1.
The space of lattices up to similarity in $\mathbb E^2$ together with a choice of positively oriented generators is the Teichmüller space for the torus, and can be identified with the hyperbolic plane, in the upper halfspace model: make the first vector go from $0$ to $1$ along the $x$-axis, and the second vector will be a point $z$ in upper half space. Change of generators acts by fractional linear transformations, and preserves the hyperbolic metric; the quotient is the space of isometry classes of Euclidean tori of area 1, the moduli space of the torus.
Twisting an annulus is a well-studied operation, the horocycle flow, on the moduli space. The action preserves a probably familiar tiling of the hyperbolic plane by ideal triangles whose vertices are at rational points on the bounding line, completed to make $\mathbb {RP}^1$, and whose edges connect pairs of slopes corresponding to your equation. The points in Teichmüller space that give square toruses are the midpoints of the edges. (Actually, the edges are infinitely long, so they don't have an obvious definition of midpoint, but the triangles have altitudes whose feet we can call the midpoints: these are the square toruses).
In any case: if you look at all lattice vectors with length between say $N$ and $2N$, and ask for the distribution of angles among these, this is equivalent to taking all points representing square lattices in a band in Teichmüller space between horocycles that appear in upper half space as a rectangle bounded by horizontal lines at height $1/N$ and $1/2N$ and vertical lines $x = \pm 1/2$; the question is the distribution of $x$-coordinates of points representing square toruses within that band. That this tends to the uniform distribution follows from ergodicity of the horocycle flow, a well-known fact whose history probably predates the sources I'm familiar with, so I won't try to give the attribution.
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@Bill: This is certainly much of the answer, but a couple of questions: ergodicity implies that any reasonable function will have a limiting distribution, but the one I had picked is uniform. This implies that the ratio I defined is the $x$-coordinate of square toruses. This is not immediately obvious to me (but I have not completely absorbed what you said) A second point is that $1000$ is not a very large number, and yet the distribution is essentially perfectly uniform. So, the mixing must be very rapid, which is not an obvious fact. ($N=100$ gives a very similar histogram, by the way...) – Igor Rivin Jan 23 2011 at 2:00
@Bill: By the way, a brief googling seems to show that the ergodicity of horocycle flow is a lot more recent than one might have thought ('60s-70s). – Igor Rivin Jan 23 2011 at 2:01
This is a very nice answer. I like it a lot. – Eric Naslund Jan 23 2011 at 2:50
1
I think ergodicity of the Horocycle flow goes back to Hedlund in the 1930s. Unique ergodicity (which might be needed here) was shown by Furstenberg in the 1970s (en.wikipedia.org/wiki/…) – Anthony Quas Jan 23 2011 at 3:11
@Igor: You can consider averages of the delta measure of square toruses smeared out into a small ball. Boundary issues work out since averages converge over horocycles, not just over the area, it works out. The horocycle flow mixes particularly quickly, as indicated by unique ergodicity. That's why the bins are so even. It's closely related to self-encoding sequences. @Quas: Thanks. I was thinking Hedlund was likely involved, but I suspect there are earlier versions not phrased as ergodicity, but perhaps in terms of elliptic functions or Poincare series or modular forms or whatnot. – Bill Thurston Jan 23 2011 at 3:33
Roughly:
Suppose you have a fraction $a/b$ and you expand it as a simple continued fraction $1/c_1+1/c_2\cdots+1/c_{n-1}+1/c_n$. Now truncate the last convergent and collapse $1/c_1+1/c_2\cdots+1/c_{n-1}$ to get, say, $a'/b'$. Now consider the simple continued fraction expansion of $b'/b$. As I recall, this will (usually? perhaps I need a hypothesis to avoid degenerate cases?) equal the reverse of the continued fraction of $a/b$.
For complicated fractions the beginning and end of a continued fraction should be almost uncorrelated. Also for a random fraction the convergents have a known distribution (Gauss-Kuzmin) and reversing the continued fraction doesn't change the distribution of the convergents, so you get the uniform distribution back.
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This is certainly very interesting. Any reference for the first paragraph? I don't think I have seen this... – Igor Rivin Jan 23 2011 at 4:37
Igor, I discovered that as an undergraduate, always assumed that I'd rediscovered the wheel and never bothered to look it up so I have no reference. But I believe it follows now that I'm thinking about it afresh just from the looking at continued fractions from the matrix point of view (as products of matrices $\begin{array}{c} c_i & 1 \\ 1 & 0 \end{array}$) and then just taking the adjoint (which reverses products and switches present numerator and historical denominator). – David Feldman Jan 23 2011 at 4:48
Just to be clear, there are always two ways to end a simple continued fraction, either ...$c$ or ...$c-1, 1$ with $c>1$. These have distinct reversals, one in the interval $[0,.5]$ and the other in the interval $[.5,1]$. So to get a "bijection" you have to pick an interval on the one hand and a normal form on the other. – David Feldman Jan 23 2011 at 6:26
1
I remember now that my original interest in continued fraction reversal had to do with music theory! The ratio of two frequencies comes to one's ears as a real number, but its "meaning" in Western music requires interpreting that number as representing a nearby rational. Small denominators make the interpretation unambiguous, but more complex fractions ("intervals") require context in order to acquire a unique interpretation ("function"). Add a bass ($\alpha$ for frequences $a\alpha$ and $b\alpha$ does it, but $\alpha$ may sound so low as not to work in a practical way. – David Feldman Jan 23 2011 at 22:53
... But juxtaposing two competing interpretations (a "modulation" via a "pivot chord" one has two $\alpha$'s and even in the wrong register, the ratio between the $\alpha$'s has the reverse continued fraction of the ratio between $a$ and $b$ and thus pins down the least audible information that ratio meant to carry. – David Feldman Jan 23 2011 at 22:55
Here is an attempt to give a somewhat finer grained view of the distribution. The set of ratios `$\sqrt{\frac{s^2+t^2}{a^2+b^2}} \subset(0,\frac{1}{2})$` are essentially the values in the first half of the Farey sequence `$\lbrace \frac{p}{q} | \gcd(p,q)=1,\ 2p \le q<N\ \rbrace$`. This has already been pointed out but I'll give a simple (if less nuanced) justification. Then I'll mention how that sequence is and is not smoothly distributed.
Instead of looking at all the relatively prime pairs $(a,b)$ with $1 \le a,b\le N$ I'll just consider those with `$a<b$` since order is irrelevant for the question asked and $(a,b)=(1,1)$ is an extreme outlier. There are non-negative integers $s,t$ with $|as-bt|=1$ and just one such pair with $\sqrt{\frac{s^2+t^2}{a^2+b^2}}<\frac{1}{2}$. This ratio turns out to be very close to $\frac{t}{s}$. Then $(0,0),(t,s),(a,b)$ and $(t+a,s+b)$ are corners of a long thin parallelogram with area 1 and (thus) no integer points on its boundary or interior. Because the sides are very nearly parallel, the ratio `$\sqrt{\frac{s^2+t^2}{a^2+b^2}}$` of their lengths is quite close to $\frac{t}{a}$ and even closer to $\frac{s}{b}$ (in fact they are convergents to the continued fraction for that irrational number). So that set of ratios is quite close to the lower half of the set of fractions
LATER For fixed $b$ the discrepency between $\sqrt{\frac{s^2+t^2}{a^2+b^2}}$ and $\frac{s}{b}$ is almost exactly $\frac{1}{b(2b^2+1)}$ for $\frac{1}{b}$ and increases to $\frac{2}{b(4b-1)}$ for $\frac{b-1}{b}$
Let $\mathcal{H}_N=\lbrace \frac {p}{q} |\frac{p}{q}\le \frac{1}{2} ,\gcd(p,q)=1,q \le N \rbrace$ The letter $\mathcal{H}$ is because this is half a Farey sequence. It is known that $P(N)=|\mathcal{H}_N|=\frac{3N^2}{2\pi^2}+O(N\log N)$. How evenly spaced are these? There are $P \approxeq \frac{0.15}{N^2}$ points in an interval of width $1/2$ so perfectly even spacing would put the kth point at $\frac{k}{2P}\approx\frac{3.3k}{N^2}$. However a fraction $\frac{p}{q}$ with $q$ small will be about $\frac{1}{qN}$ from the next nearest points. Hence the largest point other than $\frac{1}{2}$ is $\frac{1}{2}-\frac{1}{N}$ (replace N by N-1 in the even case) and the smallest point is $\frac{1}{N}$ which seems far from $\frac{3.3}{N^2}$ These empty zones force other points closer together, the first few points are only separated by about $\frac{1}{N^2}$. I can't resist an attempt to put in a picture of Ford Circles. A disk of radius $\frac{1}{q^2}$ is centered at $(\frac{p}{q},\frac{1}{q^2}).$ Disks are either disjoint or tangent. One can see the enforced distance around fractions with small denominators. On the other hand, each disk is put into the largest gap present (albeit not exactly at the center). This is a subtle topic. I'll just mention that a conjecture about the distance (in the $\ell_1$ or $\ell_2$ norm) between the sorted vector of Farey fractions and the evenly spaced vector $[0,1/2P,1/4P,\cdots$ is equivalent to the Riemann Hypothesis.
With the appropriate bin size and placement things might come out fairly even. The chart by the OP uses 100 bins for roughly 608,382 points. As I said, the results should be essentially the same as for $\mathcal{H}_N$. The very first bin is smashed against the y axis but it is below average by 213 and the next two bins are over by about 148 and 30 respectively. It is easier to see that the bin containing $\frac{1}{3}$ ($0.330<1/3<0.335$) is deficient from the average by about 95 points (by my calculations) the bin before it is about average but the one after is up by about 68 points. The last bin is under by 33 and the one before it over by 24. My other answer discussed an example made with rounding rather than truncation and a number of bins ($2520=8 \cdot 9 \cdot 5 \cdot 7$) that put simple fractions in the center of a bin. This allowed more choppy behavior.
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This is more a few comments than an answer (since the question seems well answered). I assume that the pair $(a,b)=(1,1)$ was discarded, it would give a value $\frac{\sqrt{2}}{2}$ outside the range of the rest.
Taking instead the $10^6$ points in a quarter disk of radius 1128 gives almost the same bin sizes (maybe not surprising since there is a good overlap). This includes the features that the bin 0-0.005 is smaller than the rest and that 0.330-0.335 is rather deficient and then 0.335-0.34 is higher then average.
This is an indication of a slight repulsion from simple fractions. I repeated the experiment using 2520 bins and using rounding. I also used $0 \le a,b \le 5000$ giving about the same expected number of points per bin: 6030 or in my case 3015 since I only used `$a<b$` (the situation being symmetric.)
The least filled bins were $\tiny{[0,0],[1/3,2187],[1/4,2479],[1/6,2761],[2/5,2773],[1/5,2774],[275/1008,2865],[229/1008,2875],[323/840,2891],[229/1260,2893],[1/8,2895]}$ $\tiny{[3/8,2897],[1/7,2897],[3/7,2900],[2/7,2902],[97/840,2906],[155/1008,2910],[1/10,2913],[3/10,2915],[37/630,2925],[59/560,2927],[139/315,2928]}$ $\tiny{[221/560,2936],[127/720,2939],[1/9,2939],[4/9,2940],[199/630,2940],[2/9,2940],[877/5040,2947],[611/1680,2948],[1/315,2948],[157/315,2951]}$ $\tiny{[31/360,2952],[229/2520,2956],[97/1260,2956],[5/14,2959],[1643/5040,2960],[3/14,2960],[1/14,2962]}$
Here 275/1008 is nearly 3/11 and 229/1008 nearly 5/22.
The most filled bins were
$\tiny{[277/720, 3123], [83/720, 3124], [229/840, 3139], [1259/2520, 3146], [191/840, 3151], [1/1680, 3212], [1259/5040, 3227], [1261/5040, 3229]}$ ${\tiny [1/5040, 3281], [1681/5040, 3316], [1679/5040, 3316], [1/2520, 3656], [2519/5040, 4141]}$
The final 5 are the bins adjacent to 0,1/2 and 1/3 (a bin for 1/2 would also be empty)
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That's very interesting. Nothing was discarded, but I don't think you see outliers on the histogram (notice that the pair (29, 1) would give something very close to one). I will look at the reference Gerry Myerson suggested to see what their results implied (to a peasant like myself, it is not a priori obvious why the units mod $n$ should be very evenly distributed mod a very smooth (highly composite) $n,$ but numerical experiments indicate that the distribution is very even indepdendently of the nature of $n.$). – Igor Rivin Jan 24 2011 at 15:34
Is the case a,b=1,1 on the chart? Wouldn't it be out at 0.7? You can see that your bin at 0 is quite short. Your bin containing 1/3 is kind of short and the next one kind of high. You might expect that 97 bins would be even smoother. Maybe not though since bins with 1/3 1/4 1/5 2/5 etc near their centers might be low. I bet 90 bins with rounding would be choppier. – Aaron Meyerowitz Jan 24 2011 at 18:38
@Igor actually for 1,29 one would use 1*1+29*0=1 for a ratio of 1/sqrt(842) which is quite close to 1/(29+1/58) and close enough to 1/29. – Aaron Meyerowitz Jan 24 2011 at 21:22
@Aaron: yes, of course... – Igor Rivin Jan 29 2011 at 3:06 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 135, "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": 0.9523546099662781, "perplexity_flag": "middle"} |
http://mathoverflow.net/questions/87424?sort=oldest | ## Critical points in Hilbert space [closed]
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Let $f$ be a $C^1$ functional on a Hilbert space $X$, and $Y$ a closed subspace of $X$. Suppose the restriction of $f$ on $Y$ has a critical point $x_0 \in Y$.
Q: Is $x_0$ a critical point of $f$?
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Of course not. Take e.g. $f$ a non-zero linear functional and Y=ker(f). – Pietro Majer Feb 3 2012 at 12:55
(or also, whatever X and f, and Y=(0)). – Pietro Majer Feb 3 2012 at 13:20
## 1 Answer
Especially in Calculus of Variations and Mechanics, a submanifold $Y$ of a manifold $X$ is usually called "a natural constraint" for a functional $f$ on $X$, if the special circumstance that you are considering does happen: constrained critical points of $f$ on $Y$ are free critical points: $\operatorname{crit} (f_{|Y})\subset \operatorname{crit}(f)$.
An important situation when this is true arises when $Y$ is defined as fixed point set of a group of symmetries of $f$. The corresponding principle, claiming that $Y$ is a natural constraint, is known since old times as a popular proverb. In the late 70' Richard Palais has made this into a deep and beautiful theorem, stating the right hypotheses for its validity and also providing counter-examples to the naive claim of the principle (all popular proverbs are 95% true but never 100% true!). [ R. Palais, The Principle of Symmetric Criticality, Comm. Math. Phys., Vol. 69, Number 1 (1979), 19-30 ]
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http://unapologetic.wordpress.com/2012/08/28/the-jordan-chevalley-decomposition/?like=1&_wpnonce=122935e828 | # The Unapologetic Mathematician
## The Jordan-Chevalley Decomposition
We recall that any linear endomorphism of a finite-dimensional vector space over an algebraically closed field can be put into Jordan normal form: we can find a basis such that its matrix is the sum of blocks that look like
$\displaystyle\begin{pmatrix}\lambda&1&&&{0}\\&\lambda&1&&\\&&\ddots&\ddots&\\&&&\lambda&1\\{0}&&&&\lambda\end{pmatrix}$
where $\lambda$ is some eigenvalue of the transformation. We want a slightly more abstract version of this, and it hinges on the idea that matrices in Jordan normal form have an obvious diagonal part, and a bunch of entries just above the diagonal. This off-diagonal part is all in the upper-triangle, so it is nilpotent; the diagonalizable part we call “semisimple”. And what makes this particular decomposition special is that the two parts commute. Indeed, the block-diagonal form means we can carry out the multiplication block-by-block, and in each block one factor is a constant multiple of the identity, which clearly commutes with everything.
More generally, we will have the Jordan-Chevalley decomposition of an endomorphism: any $x\in\mathrm{End}(V)$ can be written uniquely as the sum $x=x_s+x_n$, where $x_s$ is semisimple — diagonalizable — and $x_n$ is nilpotent, and where $x_s$ and $x_n$ commute with each other.
Further, we will find that there are polynomials $p(T)$ and $q(T)$ — each of which with no constant term — such that $p(x)=x_s$ and $q(x)=x_n$. And thus we will find that any endomorphism that commutes with $x$ with also commute with both $x_s$ and $x_n$.
Finally, if $A\subseteq B\subseteq V$ is any pair of subspaces such that $x:B\to A$ then the same is true of both $x_s$ and $x_n$.
We will prove these next time, but let’s see that this is actually true of the Jordan normal form. The first part we’ve covered.
For the second, set aside the assertion about $p$ and $q$; any endomorphism commuting with $x$ either multiplies each block by a constant or shuffles similar blocks, and both of these operations commute with both $x_n$ and $x_n$.
For the last part, we may as well assume that $B=V$, since otherwise we can just restrict to $x\vert_B\in\mathrm{End}(B)$. If $\mathrm{Im}(x)\subseteq A$ then the Jordan normal form shows us that any complementary subspace to $A$ must be spanned by blocks with eigenvalue $0$. In particular, it can only touch the last row of any such block. But none of these rows are in the range of either the diagonal or off-diagonal portions of the matrix.
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Posted by John Armstrong | Algebra, Linear Algebra
## 3 Comments »
1. [...] now give the proof of the Jordan-Chevalley decomposition. We let have distinct eigenvalues with multiplicities , so the characteristic polynomial of [...]
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2. [...] Now that we’ve given the proof, we want to mention a few uses of the Jordan-Chevalley decomposition. [...]
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3. [...] first thing we do is take the Jordan-Chevalley decomposition of — — and fix a basis that diagonalizes with eigenvalues . We define to be the [...]
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### Why is Near Field Communication (NFC) range limited to about 20cm?
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### Is it true that any system of accelerating charges will radiate?
I was recently told by a physics teacher that "any system of charges in which at least some of the charges are executing some sort of accelerated motion, will radiate and lose energy". This refers 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": 4, "mathjax_display_tex": 1, "mathjax_asciimath": 2, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9441329836845398, "perplexity_flag": "middle"} |
http://math.stackexchange.com/questions/278631/eigenvalues-and-equalizers/278668 | # Eigenvalues and equalizers
I am asked to relate eigen{value,vectors} with equalizers in the category of matrices. However, in the category of matrices $g\circ f$ means the (matrix) product $fg$. And $\lambda$ is an eigenvalue of $A$ if there exists $v$ such that $Av=\lambda v$.
To relate this with equalizers, I was expecting the definition of eigenvalue to be such that $vA=v\lambda$.
Am thinking right?
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## 1 Answer
Yes. The eigenspace associated with the eigenvalue $\lambda$ is usually defined as the kernel of $f - \lambda \operatorname{id}$, but is equally well defined as the equaliser of $f$ and $\lambda \operatorname{id}$. Accordingly, you can define eigenvalue as a scalar $\lambda$ such that the equaliser of $f$ and $\lambda \operatorname{id}$ is non-zero. (Beware: a scalar $\lambda$ such that $f - \lambda \operatorname{id}$ is non-invertible is not necessarily an eigenvalue if your vector spaces are not finite-dimensional!)
- | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 16, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "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": 0.9253086447715759, "perplexity_flag": "head"} |
http://mathoverflow.net/questions/54929/multiplication-tables-for-hg-p/54973 | ## Multiplication tables for H*(G/P)?
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
Hi everyone. My recent work has me developing software to compute in $H^\ast(G/P)$, where $G$ is a complex connected semisimple algebraic group and $P$ is a standard parabolic subgroup (usually, $B$ or a maximal $P$). While my programs are built on sound theory, one can never be too sure. It's always good practice to check your work.
I'm looking for references of multiplication tables of these cohomology rings.
In particular, I'm interested in cases where $G$ is NOT simply-laced (that is, Lie type $B_n$, $C_n$, $F_4$, $G_2$), though simply-laced tables would be nice too. Any tables would depend on a choice of additive basis for $H^\ast(G/P)$. I typically use cohomology classes either Hom-dual or Poincare dual to the usual Schubert varieties living in $G/P$, and I like to parameterize my Schubert varieties with $W^P$, the minimal length coset representatives of $W/W_P$ where $W$ is the Weyl group and $W_P$ is the Weyl group of the Levi associated to $P$. Tables using this convention would be great. Of course, tables in any basis would be fine. :D
Thanks so much.
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## 2 Answers
You may consult with Duan Haibao's papers on arxiv.
Also, I have a Maple realization of my own, available at
http://www.staff.uni-mainz.de/semenov/software.html
Unfortunately, there is no documentation, and the coding style is ugly, but examples may help.
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Actually, I was using Haibao's result's for my program (which I've written in Mathematica). – brandyn Feb 11 2011 at 15:47
### 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.
Some people have written code to do this, but not necessarily the most efficient possible way. Alex Yong wrote one that works for general type, and it's linked in the references for the paper by H. Thomas and A. Yong, "A combinatorial rule for (co)minuscule Schubert calculus" (in Adv. Math., or on the arXiv). As for published tables, for small rank some are given in Griffeth-Ram, "Affine Hecke algebras and the Schubert calculus". (I also wrote down the table for $G_2$ in my thesis.) There are probably many other sources, these are just the ones that come to 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": 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": 0.9235631823539734, "perplexity_flag": "middle"} |
http://physics.stackexchange.com/questions/3495/does-decoherence-need-non-determinism | # Does decoherence need non-determinism?
Consider standard quantum mechanics, but forget about the collapse of the wavefunction. Instead, use decoherence through interaction with the environment to bring the evolving quantum state into an eigenstate (rspt arbitrarily close by). Question: Can this theory be fundamentally deterministic?
If one takes into account that the variables of the environment are not known, then the evolution is of course 'undetermined' in a probabilistic sense, but that isn't the question. Question is if fundamentally quantum mechanics with environmentally induced decoherence can be deterministic. Note that I'm not saying it has to be. I might be mistaken, but it seems to me decoherence could be followed by an actual non-deterministic process still, so the decoherence alone doesn't settle the question of determinism or non-determinism. Question is if one still needs a non-deterministic ingredient?
Update: Please note that I asked whether the evolution can fundamentally be deterministic, or whether it has to be non-deterministic. It is clear to me that for all practical purposes it will appear non-deterministic. Note also that my question does not refer to the prepared state after tracing out the environmental degrees of freedom, but to the full evolution of system and environment. Does one need a non-deterministic ingredient to reproduce quantum mechanics, or can it with the help of decoherence be only apparently non-deterministic yet fundamentally deterministic?
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What definition of determinism are you using? – Holowitz Jan 21 '11 at 14:45
If you know the state at time t_1, you can in principle calculate everything that's going to happen, or did happen at time t_2. – WIMP Jan 23 '11 at 10:32
You need to rephrase your question if both myself and Matt have misunderstood your intention. Are you now asking if you can deterministically collapse to a particular eigenstate? The answer to that is no, because it violates linearity. – Joe Fitzsimons Jan 23 '11 at 10:56
@ Joe: I just updated the question, hope it's clearer now? You don't need to collapse exactly to a particular eigenstate, just arbitrarily close by (as I already stated in my original question). – WIMP Jan 23 '11 at 11:15
I've posted an updated answer to answer the question as I now understand it. – Joe Fitzsimons Jan 23 '11 at 12:22
## 5 Answers
Short answer to "Can this theory be fundamentally deterministic?": No. Decoherence is the diagonalization of the density matrix in a preferred basis, with the off-diagonals vanishing at late times. Since you can get the same final diagonal matrix from several possible initial pure states of the system under consideration, there's a necessary loss of information and irreversibility. (I'm guessing this is what you meant by non-deterministic)
A bit more detail: Decoherence proceeds by the rapid establishment of entanglement-induced correlations between the system and the infinite degrees of freedom of the decohering environment. The second law prevents this process from being reversible (since S has to always increase, and S is zero for the pure state, while it is greater than zero for the decohered mixed state). If you take the second law to be fundamental, then the non-determinism here is fundamental too.
UPDATE: The updated question now refers to the full evolution of system + environment, in other words, the entire universe. Since there's nothing else for the universe to entangle with, it will remain in a pure state and evolve deterministically for ever if it always was in a pure state. I however don't know if the universe is in a pure state or a mixed state. Anyone?
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I would answer the same thing, so plus one point. ;-) – Luboš Motl Jan 21 '11 at 19:29
Another way of saying this is the process is not unitary. – Lawrence B. Crowell Jan 22 '11 at 3:34
2
This isn't technically correct. Decoherence can be the result of an entirely deterministic (and unitary process), since you only care about the reduced density matrix for the system in question. Open quantum systems -do- decohere, even though the entire process is unitary. The larger wavefunction of the entire system is still pure, but the state of the local system becomes mixed. – Joe Fitzsimons Jan 22 '11 at 6:26
My understanding is that the evolution turns non-unitary once you've traced out the environmental degrees of freedom. That's not what I'm asking for. Also, time evolution doesn't need to be unitary to be deterministic. – WIMP Jan 23 '11 at 10:34
1
I concur with the criticisms of this answer: you are making the mistake of only considering the subsystem in question, not the total system. The subsystem loses information, yes, but it's lost to the total system, which may well be reversible as a whole. – Greg Graviton Jan 23 '11 at 14:19
show 2 more comments
I don't disagree with the other answers, but I want to try to use different words:
Evolution of a quantum state is deterministic in the sense that it is given by the Hamiltonian. Quantum mechanics doesn't need anything beyond unitary evolution. So in that sense, the answer is deterministic.
However, decoherence means that eventually a quantum state may evolve into a superposition of very nearly orthogonal states, which for large enough systems will resemble to arbitrarily high precision the answer you might get from assuming that there is a nondeterministic, nonunitary process of "wavefunction collapse."
For all practical purposes, since we are large classical observers, we can observe only one such nearly-orthogonal combination, and the interference with other possible outcomes will be unmeasurably small. So, in this sense the outcome is "nondeterministic."
If this seems counterintuitive to you, consider something analogous about classical statistical mechanics and thermodynamics. If I start with a collection of gas molecules all bunched up in one corner of the room, it is a very atypical (low-entropy) state under any natural coarse-graining of phase space. Now, by entirely reversible interactions, it can become a typical (high-entropy) state with molecules scattered all over the room. This process appears to have lost information, in the sense that I would have to do many, many difficult measurements to ascertain that a short time before this was a very special state indeed. But really, the underlying physics is deterministic, so in principle the final state remembers where it came from, although for all practical purposes if I tried to evolve it backwards I would never discover the right answer. (To be clear, I'm not claiming a very sharp analogy here. But I'm saying that the notion that microscopically deterministic evolution can be consistent with apparent or "for all practical purposes" loss of determinism in real observations is something that might be more intuitive in this context.)
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Good answer! – Joe Fitzsimons Jan 22 '11 at 6:28
It doesn't seem counterintuitive to me, but that wasn't my question. I'm not asking "for all practical purposes." I know that for all practical purposes it will appear non-deterministic in the sense that we won't be able to predict what's going to happen -- too many variables in the environment. I'm asking if the time evolution of the system is fundamentally 'in principle' deterministic. That of course can only be before you've averaged over the variables of the environment. – WIMP Jan 23 '11 at 10:41
1
Then I would say yes, it is "in principle" deterministic; the Hamiltonian specifies how everything evolves. Most of my answer was just trying to explain how to reconcile this with the observation that quantum effects look nondeterministic. – Matt Reece Jan 23 '11 at 15:41
Thanks. Yes, that was also my understanding. – WIMP Jan 24 '11 at 7:40
+1 for thermodynamics mention, I think some day decoherence and 2nd law will shake hands – HDE Jan 19 '12 at 0:16
UPDATE: It seems that we have not been answering the question WIMP intended. Here is an updated answer to deal with what I now understand to be the question: Given any unknown quantum state $|\psi\rangle$, can there be any deterministic process which will make it collapse onto a particular state $|\phi \rangle$, if $|\phi \rangle\langle \psi| \neq 0$?
The answer to this question is no, because it violates the linearity of quantum mechanics, allowing us to distinguish between non-orthogonal states. This is trivial, because states orthogonal to $|\phi \rangle$ will have zero probability of collapsing onto it. This may not seem like a big deal, but it turns out that linearity is fundemantal to quantum mechanics on many levels. If we remove this constraint, then entanglement can be used to signal, and hence create problems with causality. No signalling seems one of the most fundamental features of physics, showing up in many independent theories (electromag, quantum mechanics, relativity, etc.).
To see how this can be done, consider an entangled state $\frac{1}{\sqrt{2}}(|01\rangle - |10\rangle)$. This is the anti-symmetric state: for any basis $\sigma$ a measurement resulting in outcome $m$ will leave the other qubit in the opposite eigenstate of $\sigma$. Thus, if you could deterministically collapse onto the state $|0\rangle$ then you can be sure the your half of the EPR pair was not left in state $|1\rangle$ after the measurement on the other half. So, for Alice to communicate with Bob, she need only choose to measure in the $X$ or $Z$ basis. Measuring in $X$ will mean Bob receives the output $|0\rangle$ with probability 1, where as measuring in $Z$ will return result $|1\rangle$ with probability $\frac{1}{2}$. Although this is probabilistic, you can repeat the process arbitrarily many times to get exponentially close to perfect communication. This instantaneous communication breaks causality.
If you allow all states to collapse to the target state, then the only solution is a channel which swaps the state with another ancilla system. Systems which can perform such deterministic collapse can always be used to signal, as well as allowing all sorts of additional weirdness like efficient solutions to PSPACE-complete problems in computation and time travel. As a result, this is totally impossible within the current framework of physical theories, and there are very substantial reasons to believe that it is a feature of any physical theory that is valid in our world.
The answer is no, if by deterministic you mean possessing a local hidden variable interpretation. This follows directly from the observed violations of Bell's inequality, which ever interpretation of quantum mechanics you choose (what you are referring to is known as the Everett interpretation of quantum mechanics).
Bell's inequality works as follows: Given to possible local measurement operators ($A_i$ and $B_i$) at each of two localions $i \in \{1,2\}$, what is the maximum value of the expectation value of $\langle A_1 B_1 + A_1 B_2 + A_2 B_1 - A_2 B_2\rangle$. What Bell showed was that this can take on a value of at most 2 for any local hidden variables theory. However quantum mechanics allows it to take on values up to $2\sqrt{2}$, and many experiments have recorded violations of this inequality, showing values in the range $2 < v \leq 2\sqrt{2}$. This essentially rules out a local hidden variable model.
If, however, you mean can the unitary interaction of two particles give rise to decoherence, then the answer is yes, as follows: Imagine two particles initially in the state $1/\sqrt{2}(|0\rangle + |1\rangle)$. Now imagine they interact via an Ising interaction. After a certain time, they will be in the joint state $1/2(|00\rangle - |10\rangle - |01\rangle + |11\rangle)$. This is still a pure state, and so no decoherence has occurred. However, imagine one of these particles moves off far away (into the environment). If we only have access to one of these particles, then its reduced density matrix will be $1/2(|0\rangle \langle 0|+|1\rangle \langle 1|)$, which is simply a classical random distribution over the two orthogonal states, the same as would occur do to a collapse of the wavefunction.
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Thanks for the reply. Even though you've correctly interpreted the meaning of deterministic, you haven't answered my question. You're saying the theory can't be deterministic because that would be in conflict with experimental tests on Bell's inequality. That's not correct. The theory could also be non-local instead, you just need to violate one of the assumptions to prove the theorem. – WIMP Jan 23 '11 at 10:38
1
@WIMP: The first line of my answer reads: "if by deterministic you mean possessing a local hidden variable interpretation". Certainly global hidden variables can be made to work, but then they always can. – Joe Fitzsimons Jan 23 '11 at 10:42
Yes, I know. But that non-local hidden variable theories are not excluded by Bell's theorem isn't an answer to my question. If you want to pursue that line of thought, the question would then be whether decoherence, rspt the entanglement with the environment is a sort of non-locality that spoils the assumptions to Bell's inequality and thus, isn't excluded by experiment. Also, I was wondering about the possibility of deterministic evolution from a theoretical rather than an experimental point of view. – WIMP Jan 23 '11 at 10:53
@WIMP: Can you please revise the question to make it clear exactly what you are asking? It's currently not clear either from the question or subsequent comments. – Joe Fitzsimons Jan 23 '11 at 10:59
@ Joe: Thanks for the update. Two things: First, you don't need to collapse exactly into |0>, you just need to get close enough so it would 'for all practical purposes' appear to be an eigenstate, see question & update. Second, that the evolution is fundamentally deterministic doesn't mean that you could in practice deterministically collapse. (Besides this, I didn't ask if it leads to instantaneous messaging as you seem to think. Also, instantaneous messaging doesn't necessarily cause problems with causality, but that's a different point.) – WIMP Jan 24 '11 at 9:13
show 1 more comment
Not sure if you've ever heard of the De Broglie-Bohm pilot wave interpretation of QM, but it is a fundamentally deterministic interpretation.
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Yes, I've heard of it. But my question wasn't if there is any deterministic interpretation of QM, but if the standard interpretation with decoherence instead of collapse can be deterministic. – WIMP Jan 24 '11 at 7:45
My answer is NO. There is a problem in many QM considerations that we have an isolated system obeying (deterministic!) Schrödinger equation that is a subject of mystical "measurements" that introduce non-deterministic behavior. But one can't make a measurement and leave system isolated (indeed saying that system is isolated is always an approximation in QM, and much heavier than in CM). In fact measurement is an act of introducing interactions with measuring apparatus, measurer, coffee measurer drinks, ... -- so the measurement result can be in theory calculated, but would involve inaccessible and enormous amounts of information. This makes it practically non-deterministic, but fundamentally it is no better than classical chaos.
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Actually, you are arguing that the answer is yes, rather than no. I haven't asked whether it is practically non-deterministic, but whether it can be fundamentally deterministic though appear non-deterministic (much like chaos indeed). – WIMP Jan 24 '11 at 7:48
My answer to "Does decoherence need non-determinism?" is No (-; – mbq♦ Jan 24 '11 at 10:19 | {"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": 26, "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": 0.9388725161552429, "perplexity_flag": "middle"} |
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http://www.physicsforums.com/showthread.php?p=3565509 | Physics Forums
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## How does GR handle metric transition for a spherical mass shell?
Quote by PeterDonis Actually, since I posted the assessment in #8, I owe you a mea culpa, Q-reeus; after thinking this over and thinking about DaleSpam's comment, I think my assessment in #8 was incorrect, because I was comparing apples and oranges. The Earth is not a hollow sphere made of steel. The case you just gave in the above quote is a much better test case, and as we'll see, it does *not* support the claim that pressure is negligible. I did say in #8 that the key factor is not whether pressure is small compared to energy density, but whether the ratio of pressure to energy density is small compared to the "correction factor" in the metric coefficients. So let's estimate those numbers for a "toy globe". Let's suppose our "toy globe" is a spherical shell made of steel; its total radius is 1 meter, and the shell is 0.1 meter thick (with a hollow interior). We'll assume that the stress components inside the shell are of the same order as atmospheric pressure, or about 10^5 pascals. (They may actually be somewhat larger, but we'll use the lower bound since that will make the pressure/energy density ratio as small as possible.) The mass density of steel is about 8000 kg/m^3. Energy density of the shell = 8000 kg/m^3 x c^2 = 7 x 10^19 J/m^3 Ratio of pressure to energy density = about 10^-15 Volume of the shell = 4/3 pi x (1^3 - 0.9^3) = about 1 m^3 (how convenient!). Mass of the shell = 8000 kg Correction to metric coefficient at shell surface = 2GM / c^2 r = (2 x 6.67 x 10^-11 x 8000) / (9 x 10^16 x 1) = about 10^-23 So I was wrong in post #8; for a typical object made of ordinary materials, the pressure is *not* negligible; the pressure/energy density ratio is many orders of magnitude *larger* than the correction to the metric coefficients for the object. So there is plenty of "room" for the spatial parts of the stress-energy tensor inside the object to "correct" the metric to flat from the outside to the inside of the shell. I should note, though, that even the above computation is not really "correct", in that I haven't actually tried to compute any components of the field equation; I've just done a quick order of magnitude estimate of the same quantities I estimated in post #8, to show that the relationship I talked about there doesn't hold for an ordinary object.
Still struggling to reply effectively to your two very helpful earlier posts, but will deal with this one now. I can't see any mea culpa. First, had in mind the relevant stresses in the globe are those owing only to gravitational self-interaction (i.e. - it's floating out there in space). Vastly smaller than and nothing to do with any atmosperic pressure. But even so, let's take the 'huge' atmospheric pressure relevant figure of ~ 10-15. This is the ratio of relevant contributions to that very tiny metric distortion figure of ~ 10-23. So rho contributes a fraction (1-10-15), while p's contribute a fraction 10-15. Isn't that still the only important consideration on this? How can something 10-15 times smaller than the other be overwhelmingly dominant?! I have a feeling there is some thought of mixing up assumed elastic, material distortions with the underlying, vastly smaller metric ones. The relative contribution of p to the latter is always insignificant for 'steel globes', no? Am I missing something here? More later.
Quote by PeterDonis There is a definite meaning to "radial distance", yes: in principle, we could line up our tiny measuring objects from some given sphere with area A, all the way to the center of the whole scenario, and count them. The radial distance measured this way, if we start from a sphere in one of the regions where K > 1 (EV or NV), will be larger than the area A divided by 4 pi. However, the exact relationship between the two will be complicated, because it has to take into account how K varies from the sphere with area A all the way to the center of the scenario. It's simpler to state things in terms of the differential area and volume as I did because those quantities are "local", at least in terms of radial movement (and since we're assuming spherical symmetry and time independence, everything can vary only as a function of radial movement), so K can be assumed constant for any given pair of spheres with areas A and A + dA. But the radial distance, as I've defined it above, is *not* necessarily the same as the radial coordinate r. You can define r to always be the same as the radial distance, but that may not be the easiest definition to work with. More on that in a future post when I talk about coordinates.
See now I wrongly claimed to have 'perfectly understood' your previous #13 posting. Being unfamiliar with K and J as standard objects in GR, had thought them merely convenient and arbitrary symbols used on an ad hoc basis here. Alright I now finally get it that 'r' as SC's radial coordinate is not really equivalent to physical r, but a derived quantity based on K operating on A. Very confusing but I suppose something has to go in the interests of 'local invariance'. Does this not create a circular definition paradox though: r defined in terms of A, but A expressed in terms of r? Area and volume have to be based on linear measure somehow, so is it not still down to picking a linear length measure as 'the' proper yardstick? How else to climb out of this hole? My hunch is tangent spatial measure is implicitly that yardstick. Which gets to the next part.
Originally Posted by Q-reeus: "Gets back I suppose to that other thread where I asked for what a distant observer will see through a telescope - a distorted or undistorted test sphere. A sensibly and physically real Sr implies oblate spheroid will be observed." This is a more complicated question as well because it requires you to evaluate the path of the light rays from the object to the distant observer, and the "scaling factor" and time dilation factor will change through the intervening spacetime. It may well be that you are correct that the anisotropy I described (more volume between spheres A and A + dA than Euclidean geometry would lead you to expect) will be seen by a distant observer as a test sphere appearing distorted, but it's not as straightforward a question as you seem to think it is.
An important one in my mind in untangling SC 'r' from 'actual' r, and so for tangent length. Gravitationally induced optical distortion can in principle be fully accounted for, either directly with corrective optics, or by computer processing of image data, just as e.g atmospheric distortion is compensated for in modern astronomy. Redshift is abberation free in GR so not a problem. You are now quite aware the test sphere was to be considered locally stress free. So only effect distant observer determines is underlying metric distortion. Given all that, and to the extent SM is correct, am I not entitled to expect as distant observer to see a distorted test sphere as oblate spheroid as per the Sr = J (or close to it), St = 1? This is an important part of my reality check list.
But my whole point is that K and J are the "primitives". K and J are coordinate-independent; they can be directly measured in terms of local observations (K is how much the volume between two spheres with area A and A + dA exceeds the Euclidean value, and J is the observed gravitational redshift/blueshift factor at a given sphere with area A). If by "V" you mean what you have been calling the "potential", the coordinate-independent definition of that would be made in terms of J, the "gravitational redshift" factor, via the usual definition: $$J = 1 + 2 \phi$$ where $\phi$ is the potential in units where c = 1, and with the usual convention that the potential is zero at "infinity" and negative in a bound system such as this one. But this potential is not what is directly observed; that's J.
Too late now to re-edit #1, but from now on will be sticking with more standard usage. Using V as label for 'the Einstein potential' -g00 rather than the redshift factor J you have used, created possible confusion with it's more familiar use; as the Newtonian potential -GM/r. Apologies for any confusion caused.
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Quote by Q-reeus Still struggling to reply effectively to your two very helpful earlier posts, but will deal with this one now. I can't see any mea culpa. First, had in mind the relevant stresses in the globe are those owing only to gravitational self-interaction (i.e. - it's floating out there in space). Vastly smaller than and nothing to do with any atmosperic pressure.
This is true, and I didn't try to calculate what the actual stresses inside the globe due only to its own gravity would be.
Quote by Q-reeus But even so, let's take the 'huge' atmospheric pressure relevant figure of ~ 10-15. This is the ratio of relevant contributions to that very tiny metric distortion figure of ~ 10-23. So rho contributes a fraction (1-10-15), while p's contribute a fraction 10-15.
No, this is not correct. The various components of the stress-energy tensor match up with different components of the curvature, which in turn match up with different components of the metric. So the ratio of pressure to energy density may not be relevant to the effect on the metric. That's why I said in my last post that the calculation I gave there wasn't really "correct"; the correct thing to do is to look at the Einstein Field Equation and figure out how each component of the stress-energy tensor affects the curvature, and through that the metric. That's also why DaleSpam has been saying that the pressure may be relevant even though it's much smaller than the energy density. The paper he referenced gives an example where the effects of pressure are certainly not negligible.
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Quote by Q-reeus Does this not create a circular definition paradox though: r defined in terms of A, but A expressed in terms of r?
No, because I didn't express A in terms of r; I used A to *define* r. A is the "primitive", the actual observable quantity, and I defined it in terms of covering the 2-sphere with little identical objects and counting them.
Quote by Q-reeus Area and volume have to be based on linear measure somehow, so is it not still down to picking a linear length measure as 'the' proper yardstick? How else to climb out of this hole? My hunch is tangent spatial measure is implicitly that yardstick.
In the sense that to cover the 2-sphere with little identical objects and count them, you have to place the objects tangentially, yes, I suppose it is. But as I said before, that's because we are assuming spherical symmetry, so the area of the 2-sphere seems like a good benchmark, or "primitive", on which to base other things.
Quote by Q-reeus Gravitationally induced optical distortion can in principle be fully accounted for, either directly with corrective optics, or by computer processing of image data, just as e.g atmospheric distortion is compensated for in modern astronomy.
In order to account for gravitationally induced optical distortion, you first have to have a model that says what it is, so it can be compensated for. That's the part that I don't think is very simple: you have to evaluate what a null geodesic will look like as it passes through spacetime with varying curvature.
Also, since a test sphere in this spacetime, to an observer next to it, will *not* appear distorted--it will be spherical--if it does appear distorted to an observer much further away from the black hole, that distortion would have to be "gravitationally induced optical distortion", would it not? If so, you wouldn't want to "compensate" for that, since it would eliminate the effect you are interested in.
Quote by PeterDonis ...No, this is not correct. The various components of the stress-energy tensor match up with different components of the curvature, which in turn match up with different components of the metric. So the ratio of pressure to energy density may not be relevant to the effect on the metric. That's why I said in my last post that the calculation I gave there wasn't really "correct"; the correct thing to do is to look at the Einstein Field Equation and figure out how each component of the stress-energy tensor affects the curvature, and through that the metric. That's also why DaleSpam has been saying that the pressure may be relevant even though it's much smaller than the energy density. The paper he referenced gives an example where the effects of pressure are certainly not negligible.
Well in that case I have very little idea of how it goes. Got the impression from sundry sources (pop sci maybe) that pressure acts simply as an additive term - apart from obvious complications of spatial pressure/density gradients. As I surmised in #1, there I asume no 'directionality' to the contribution from stress of an element of matter any different to it's mass? Very different when motion/flow is involved of course, but that does'nt concern our situation.
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Quote by Q-reeus But even so, let's take the 'huge' atmospheric pressure relevant figure of ~ 10-15. This is the ratio of relevant contributions to that very tiny metric distortion figure of ~ 10-23. So rho contributes a fraction (1-10-15), while p's contribute a fraction 10-15. Isn't that still the only important consideration on this? How can something 10-15 times smaller than the other be overwhelmingly dominant?!
Because they are "pointing" in different directions (i.e. they affect different components of the tensor). If you were only interested in the time dilation then you would be correct that the pressure is negligible in "ordinary" situations compared to the energy density. However you are specifically interested in the spatial components of the curvature tensor, so the energy density is irrelevant. It is big, but it is in the wrong "direction". Since there is not any momentum flow the only components in the spatial directions are the various normal stress components. You cannot neglect the only source of something you are interested in regardless of how it compares to other things.
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Quote by Q-reeus Well in that case I have very little idea of how it goes. Got the impression from sundry sources (pop sci maybe) that pressure acts simply as an additive term - apart from obvious complications of spatial pressure/density gradients.
For determining the inward "pull of gravity", you are correct; the relevant quantity is $\rho + 3 p$, and the pressure p would be negligible in the case we're discussing. However, the "pull of gravity" comes under the heading of "time components" of the curvature. We're talking about "space components", where the energy density does not come into play.
Quote by DaleSpam Because they are "pointing" in different directions (i.e. they affect different components of the tensor). If you were only interested in the time dilation then you would be correct that the pressure is negligible in "ordinary" situations compared to the energy density. However you are specifically interested in the spatial components of the curvature tensor, so the energy density is irrelevant. It is big, but it is in the wrong "direction". Since there is not any momentum flow the only components in the spatial directions are the various normal stress components. You cannot neglect the only source of something you are interested in regardless of how it compares to other things.
Something is drastically not adding up here imo. Do you agree that the overwhelmingly larger matter contribution is what will determine/generate the exterior SM? And that spatial anisotropy is inherently present there - owing to that matter contribution? And that there *must* be significant change to at least one spatial metric component upon transition to the flat MM interior? Well how does all this square up with what you have been saying?
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Quote by Q-reeus Do you agree that the overwhelmingly larger matter contribution is what will determine/generate the exterior SM? And that spatial anisotropy is inherently present there - owing to that matter contribution? And that there *must* be significant change to at least one spatial metric component upon transition to the flat MM interior? Well how does all this square up with what you have been saying?
I think you're confusing the metric and curvature in the exterior vacuum region with the metric and curvature in the non-vacuum "shell" region. Everything DaleSpam and I have been saying applies only to the metric and curvature in the non-vacuum region. In the exterior vacuum region the stress-energy tensor is zero; there is no pressure or energy density, so there is no "matter contribution" at all. The curvature of spacetime in the exterior region is determined by the overall mass of the shell, via the form of the Schwarzschild metric, but that metric is a solution of the *vacuum* Einstein Field Equation (stress-energy tensor equal to zero). The exterior metric does not depend on any details of the shell's internal structure; two shells with identical overall mass, but drastically different energy density (say one is much thicker than the other, and correspondingly much less dense), would lead to the same metric in the exterior vacuum region. But within the shell, the metric would be quite different for those two cases: the spatial metric would have to change to flat over a much shorter distance for the thin shell, so the curvature components would have to change must faster as you descended through the shell.
Quote by PeterDonis No, because I didn't express A in terms of r; I used A to *define* r. A is the "primitive", the actual observable quantity, and I defined it in terms of covering the 2-sphere with little identical objects and counting them... ...In the sense that to cover the 2-sphere with little identical objects and count them, you have to place the objects tangentially, yes, I suppose it is. But as I said before, that's because we are assuming spherical symmetry, so the area of the 2-sphere seems like a good benchmark, or "primitive", on which to base other things.
What I meant was not just that r is defined as A = 4*pi*r2, but that from the SC's the θ and phi tangent components are themselves functions of r. So it still seems circular. One must somehow have a good handle on r in order to have that 'primitive' of A determined, surely.
Also, since a test sphere in this spacetime, to an observer next to it, will *not* appear distorted--it will be spherical--if it does appear distorted to an observer much further away from the black hole, that distortion would have to be "gravitationally induced optical distortion", would it not? If so, you wouldn't want to "compensate" for that, since it would eliminate the effect you are interested in.
Is there not a clear sense in which one is optical - light bending, while the other is inherent - the metric distortion as physical 'object'?
Quote by PeterDonis I think you're confusing the metric and curvature in the exterior vacuum region with the metric and curvature in the non-vacuum "shell" region. Everything DaleSpam and I have been saying applies only to the metric and curvature in the non-vacuum region. In the exterior vacuum region the stress-energy tensor is zero; there is no pressure or energy density, so there is no "matter contribution" at all. The curvature of spacetime in the exterior region is determined by the overall mass of the shell, via the form of the Schwarzschild metric, but that metric is a solution of the *vacuum* Einstein Field Equation (stress-energy tensor equal to zero). The exterior metric does not depend on any details of the shell's internal structure; two shells with identical overall mass, but drastically different energy density (say one is much thicker than the other, and correspondingly much less dense), would lead to the same metric in the exterior vacuum region. But within the shell, the metric would be quite different for those two cases: the spatial metric would have to change to flat over a much shorter distance for the thin shell, so the curvature components would have to change must faster as you descended through the shell.
Here's where the issue seems to be hitting the fan. I entirely meant shell matter's contribution to the exterior, SM region, and said so explicitly in #25. So no confusion on my part there. Agree with essentially all the rest above, but not the probable implication that stress in the shell can account for anything remotely significant re transition through shell wall. There is a severe logical chasm imo. Exterior, *presumably* anisotropic SM generated by essentially shell matter exclusively. Transition to interior flat region demanding 'steep' gradients in at least one spatial metric component. Relative pressure arbitrarily small in self-gravitating case for 'small' shell. Are we trying to pull a rabbit out of a hat, folks?
Now this would all go away if in fact hte SM is *actually* isotropic spatially, so are SC's fooling us? This is why I want that test sphere result so bad! Bed time.
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Quote by Q-reeus Now this would all go away if in fact hte SM is *actually* isotropic spatially, so are SC's fooling us? This is why I want that test sphere result so bad!
But why would the appearance of a test sphere to someone far away make any difference, when we've already established that, to an observer right next to the test sphere, it would appear spherical, *not* distorted? To me, that local observation is a much better gauge of whether space is isotropic than the observation of someone far away, particularly when the light going to the faraway observer could be distorted by the variation in gravity in between.
Perhaps I introduced confusion when I used the word "anisotropy" to refer to the fact that there is more distance between two neighboring 2-spheres in the Schwarzschild exterior region than Euclidean geometry would lead one to expect based on the areas of the spheres. The only thing that that shows to be non-isotropic is the "Euclidean-ness" of the space. I've already described in detail how that "anisotropy" goes away as you descend through the non-vacuum "shell" region; if you want a "physical" explanation of how that works, I would say it's because the non-Euclideanness of the space is due to the mass of the shell being below you, as you descend through the shell, less and less of its mass is below you. "Below" here just means "at a smaller radius", or, if I were to be careful about describing everything in terms of direct observables, "below" means "lying on a 2-sphere with a smaller area than the one you are on".
It's also worth noting, I think, that even the "non-Euclideanness" of the space, as I described it, is observer-dependent; it assumes that the non-Euclideanness is being judged using 2-spheres which are at rest relative to the shell. Observers who are freely falling through the exterior vacuum region will *not* see this anisotropy in the space that is "at rest" relative to them; in other words, if a freely falling observer were to measure the areas of two adjacent 2-spheres that were falling with him, and measure the distance between the two 2-spheres, he would find the relationship between those measurements to be exactly Euclidean.
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Quote by Q-reeus Do you agree that the overwhelmingly larger matter contribution is what will determine/generate the exterior SM? And that spatial anisotropy is inherently present there - owing to that matter contribution?
The exterior SM is a vacuum solution, so there isn't any matter contribution in the exterior SM.
Quote by Q-reeus And that there *must* be significant change to at least one spatial metric component upon transition to the flat MM interior? Well how does all this square up with what you have been saying?
That transition requires non-zero spatial components of the stress-energy tensor, i.e. stress and pressure. The large contribution of the energy density is in the wrong place to matter for the spatial components of the curvature.
Recognitions: Gold Member Science Advisor The diagram below of a "Flamm's paraboloid" may aid understanding the curvature of space (but not spacetime) outside the event horizon of a black hole. The surface represents a 2D cross-section through 4D spacetime involving the $r$ and $\phi$ coordinates only. In this diagram $r$ is the radius in the horizontal direction. "Ruler" distances in space (e.g. PeterDonis's method of counting small identical objects packed together) are represented by distances along the curved surface. (The vertical direction has no physical meaning at all.) Allen McCloud, Wikimedia Commons, CC-BY-SA-3.0The diagram goes all the way to the event horizon, where the surface becomes vertical at the bottom of the "trumpet". So, for the example being discussed in this thread, you need to slice off the bottom part of the surface. The interior of the shell could then be represented by a horizontal flat circular disk almost capping the bottom of the remaining trumpet, but there is then a gap to be bridged between the disk and the trumpet to represent the shell. I've no idea whether is possible to construct a curved surface to bridge that gap which would correctly represent the geometry within the shell. (I suspect it might not.) The mathematics of Flamm's paraboloid is discussed on Wikipedia at Schwarzschild metric: Flamm's paraboloid. P.S. Flamm's paraboloid does not represent the gravitational potential. The potential has a somewhat similar shape but it's a completely different formula.
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Quote by DrGreg The interior of the shell could then be represented by a horizontal flat circular disk almost capping the bottom of the remaining trumpet, but there is then a gap to be bridged between the disk and the trumpet to represent the shell. I've no idea whether is possible to construct a curved surface to bridge that gap which would correctly represent the geometry within the shell. (I suspect it might not.)
I was thinking of the "bridge" surface as being basically a section of a torus, with the "bottom" part merging into the horizontal circular disk representing the interior, and the "top" part having just the right slope to merge into the exterior paraboloid at the appropriate circular "cut" from it. If "torus" is interpreted generally enough (basically to allow an elliptical "cross section" as well as a circular one), would that be a plausible candidate?
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Quote by PeterDonis I was thinking of the "bridge" surface as being basically a section of a torus, with the "bottom" part merging into the horizontal circular disk representing the interior, and the "top" part having just the right slope to merge into the exterior paraboloid at the appropriate circular "cut" from it. If "torus" is interpreted generally enough (basically to allow an elliptical "cross section" as well as a circular one), would that be a plausible candidate?
It certainly sounds plausible -- that's pretty much what I had in mind myself. But on the topic of embedding an arbitrary curved manifold into a higher-dimensional Euclidean space, I am a bit out of my depth. I think, in general, there's no guarantee it can be done with just one extra dimension, i.e. you usually need more. It may be just good luck that Flamm's paraboloid works in just 3 dimensions for the exterior Schwarzschild solution.
And even if you can produce a 2D curved surface in 3D space to represent the entire "shell" spacetime, I'm not sure if the "junctions" (where there's a discontinuity in the energy-momentum-stress tensor from zero to non-zero) would need to be smooth -- maybe there would a sharp "crease" in the surface at these points?
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Quote by DrGreg It certainly sounds plausible -- that's pretty much what I had in mind myself. But on the topic of embedding an arbitrary curved manifold into a higher-dimensional Euclidean space, I am a bit out of my depth. I think, in general, there's no guarantee it can be done with just one extra dimension, i.e. you usually need more. It may be just good luck that Flamm's paraboloid works in just 3 dimensions for the exterior Schwarzschild solution. And even if you can produce a 2D curved surface in 3D space to represent the entire "shell" spacetime, I'm not sure if the "junctions" (where there's a discontinuity in the energy-momentum-stress tensor from zero to non-zero) would need to be smooth -- maybe there would a sharp "crease" in the surface at these points?
This is leading to a thought I've been having for a while, but didn't want to distract this thread. Simply, why bother with a shell at all? In this case, you have a perfectly continuous but non-smooth manifold joining the SC geometry to Minkowski geometry inside a chosen r value. The metric can be made continuous, a single global chart can be used, you just have metric derivatives undefined on one surface. This is perfectly analogous to taking a ball and cutting it in half, and treating the resulting boundary as a 2-manifold. No one would consider that a weird or inadmissable surface from a geometric point of view. As for physics, the shell-less geometry would lead to failure of EFE on the junction surface (only). It may simplify things if Q-reeus could state his fundamental issue for this simpler case - I still don't get his confusion at all. This shell-less manifold wouldn't be strictly physical, but all physics off the shell, and almost all physics going through the shell (e.g. geodesics) would be perfectly well defined.
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http://mathoverflow.net/questions/92631/unique-beltrami-differential-of-the-form-k-frac-barqq | ## Unique Beltrami Differential of the form $k\frac{\bar{q}}{q}$?
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I'm having a brain freeze.
Let $B$ be the space of complex valued measurable functions on the unit disk in the complex plane with essential supremum less than 1. Then, the universal Teichmuller space $T$ can be though of as a quotient space of $B$.
I vaguely recall reading somewhere that for every point $p \in T$ has a unique representative of the form
$\mu(z) = k \frac{\bar{q}}{q}$, where $q$ is a holomorphic function on the unit disk.
First of all, am I crazy and am remembering something that's completely off?
Secondly, if there is a statement like that, where should I be looking?
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## 2 Answers
1. Strebel discovered (in 1962) that Teichmüller's existence theorem fails for quasiconformal maps of the unit disk. Uniqueness fails too for points in $T(D^2)$ which do not have Teichmuller representatives. See Strebel's examples in the book by Gardiner and Lakic, "Quasiconformal Teichmüller Theory," p. 177-178.
2. On the other hand, the uniqueness theorem in certain sense does hold (Theorem 5, Chapter 4 of the same book). More precisely, if a quasiconformal map $f: D^2\to D^2$ is a Teichmüller map, i.e., it has Beltrami differential of the Teichmüller form $t \bar{\phi}/|\phi|$ (where $\phi$ is holomorphic), then $f$ is the unique extremal quasiconformal map in its equivalence class $[f]\in T(D^2)$. In particular, $[f]$ contains no other Teichmüller maps.
3. On third hand, Strebel proved in 1976 ("On the existence of extremal Teichmueller mappings") that for $[f]\in T(D^2)$ given by, say, smooth boundary values, the Teichmüller existence theorem does hold.
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Thank you for the reference. I'm actually looking at a beltrami differential of a very particular form: mu(z) = k \bar{q}/q, where q is the Koebe function precomposed with the map exp(2pi i z). (So it is a beltrami differential on a punctured disk.) Theorem 5 doesn't really apply in my case, but Chapter 9 of the reference you gave me (which contains the pages you gave) seems to be the right place to start. Again, thank you very much for your answer. – BrainDead Apr 3 2012 at 2:32
@BrainDead: You are welcome. Judging by your description, your $q$ is holomorphic on the upper half-plane (composition of two holomorphic functions), so Theorem 5 would apply. Maybe the problem is that your quadratic differential is unbounded: I forgot if Theorem 5 has boundedness requirement on $q$. – Misha Apr 3 2012 at 3:44
Right, by punctured disk, I meant the strip [0,1] x [0,infinity] in the upperhalf plane. And the problem is that $\phi = q^2$ does not have a finite $L^1$ norm. Theorem 3 in Chapter 9.3 (pg.183) seems to be exactly what I need. – BrainDead Apr 3 2012 at 4:59
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I think you probably referred to this fact: you can find for your point $p \in T$ a representative in the form of a Beltrami differential $\mu$ expressed as $\mu=k\frac{\bar{\phi}}{\vert \phi \vert}$, with $\phi$ a quadratic differential. But I doubt about the uniqueness, as you can always multiply $\phi$ by some positive constant...
Hubbard's book on Teichm\üller theory contains certainly all of that. Also, I think that Lehto's book has an entire chapter on Universal Teichm\üller space.
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@Sylvain: The question is about uniqueness of $\mu$ rather than uniqueness of $\phi$. Once you know $\mu$, the holomorphic quadratic differential $\phi$ is determined uniquely up to a constant positive multiple. This is the usual ambiguity in Teichmüller theory which is usually remedied by assuming that $\phi$ has unit norm (for some choice of the norm on the space of quadratic differentials). – Misha Mar 31 2012 at 1:22
Thanks for the comment! together with your answers, the whole thing is much clearer now... – Sylvain Bonnot Apr 1 2012 at 16:05 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 29, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": 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": 0.9249418377876282, "perplexity_flag": "head"} |
http://mathoverflow.net/questions/95160/solid-rings-and-tor | ## Solid Rings and Tor
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A solid ring is a ring $R$ such that the multiplication $R\otimes_{\mathbb{Z}} R \to R$ is an isomorphism.
These were classified by Bousfield and Kan; they are
1. subrings of $R\subseteq\mathbb{Q}$,
2. $\mathbb{Z}/n$,
3. products $R\times \mathbb{Z}/n$ with $R\subseteq \mathbb{Q}$ and every divisor of $n$ invertible in $R$
4. colimits of these.
I wonder how small the list gets if I put the additional constraint that $\mathrm{Tor}_{\mathbb{Z}}(R,R) = 0$.
REFERENCE: Bousfield, A. K.; Kan, D. M. The core of a ring. J. Pure Appl. Algebra 2 (1972), 73–81.
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Did you mean Tor_i = 0 for i > 0? – Jason Polak Apr 25 2012 at 14:22
Is R supposed to be $\mathbb{Q}$ on the second line? – Sean Tilson Apr 25 2012 at 14:31
1
It seems R must be {\mathbb Q}. Also, I think you must mean colimits, not limits. – Steven Landsburg Apr 25 2012 at 14:44
1
Your summary of Bousfield and Kan's results is inaccurate in a number of ways. You should probably start by reviewing their paper. I think it works out that the only solid rings with $\text{Tor}_{\mathbb{Z}}(R,R)=1$ are the localisations $\mathbb{Z}[J^{-1}]$ (for any set of primes $J$). – Neil Strickland Apr 25 2012 at 14:55
I apologize for the mangling of the classification of solid rings; fixed now, I think. – Jeff Strom Apr 26 2012 at 14:06
## 1 Answer
Let $R^t$ be the torsion submodule and consider the exact sequence
$$0\rightarrow R^t\rightarrow R \rightarrow R/R^t\rightarrow 0$$
Bousfield and Kan show that the ring on the right is a localization of ${\mathbb Z}$, hence flat over ${\mathbb Z}$, so its $Tor$ with $R$ vanishes. Thus if we $Tor$ the above with $R$, we get $Tor(R^t,R)=Tor(R,R)$.
Now tensor the exact sequence with $R^t$ instead of $R$. This gives $Tor(R^t,R^t)=Tor(R^t,R)$.
Thus $Tor(R,R)=Tor(R^t,R^t)$. But if $R^t$ is nonzero then (see Bousfield and Kan) it contains some ${\mathbb Z}/p{\mathbb Z}$ as a direct summand and hence $Tor(R^t,R^t)$ does not vanish. Thus $Tor(R,R)=0$ implies $R^t=0$. It follows (B/K 3.7) that $R$ is a localization of ${\mathbb Z}$.
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This is perfect! – Jeff Strom Apr 26 2012 at 14:06 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 32, "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": 0.8939594030380249, "perplexity_flag": "middle"} |
http://www.physicsforums.com/showpost.php?p=3783210&postcount=3 | View Single Post
Recognitions:
Science Advisor
I'm not an expert in you topic, but I'd find it interesting to chat about it until one comes along.
Quote by Hare Hi. I have a few questions about sensor specifications and its implementation in a Kalman Filter and simulation of gyroscope/accelerometer output. Abbreviation used: d - discrete c - continuous Q1: From book: Aided Navigation - Farrell (you don't need the book to understand the question).
Maybe an navigation engineer doesn't need a the book! - but this isn't the engineering section of the forum.
Section 7.2 Methodology: Detailed Example: Here $\sigma_1$, $\sigma_{bu}$ and $\sigma_{by}$ are root PSD of continuous white noise. $\sigma_2$ is std. deviation of discrete white noise. $Qc$ = diag([$\sigma_{bu}^2$ $\sigma_{by}^2$ $\sigma_1^2$ ]) $Qd$ = function of $Qc$ and sampling time $Rd$ = $\sigma_2^2$ This I understand, but not section 7.4.
I don't understand from that description, what equations relate those quantities to each other.
Section 7.4 An Alternative Approach: Same sigma values as above, plus $sigma_3$ is root PSD of continuous white noise. Qc = diag([$\sigma_{bu}^2$ $\sigma_{by}^2$ $\sigma_3^2$]) Qd = function of Qc and sampling time Rd =$\sigma_1^2$ and later Rd = diag([$\sigma_2^2$ $\sigma_1^2$]) a) Can you just use $\sigma$ in Rd no matter if it's std. deviation for discrete white noise ($\sigma_2$) or root PSD of continuous white noise($\sigma_1$)? b) Shouldn't $\sigma_1$ be "discretizised" or something?
Again, I don't know the equations that your are using.
Q2: If you get noise specification for the sensor noise from PSD method, Allan Variance method or data sheet, it is for continuous white noise. Is this correct?
For us non-engineeers, give a link to a specific data sheet where sensor noise is given. Perhaps we can learn what the specification means from an example.
Q3: How do you simulate the output of a sensor in e.g. Matlab/simulink if you have noise specification in continuous time?
Unfortunately, I don't own Matlab or simulink. I can run the free software Octave which is similar.
Q4: When we discretizise a stochastic linear system, we get: Qd = f(Qc, A, T) Rd = Rc (http://en.wikipedia.org/wiki/Discretization) I don't understand why Rd = Rc?
Are you saying that something in that article asserts Rd = Rc or are asking about material from Farell's book?
Q5: In the book: Optimal State Estimation - Kalman, Hinf and Nonlinear Approaches - 2006 - Dan Simon, section 8.1 DISCRETE-TIME AND CONTINUOUS-TIME WHITE NOISE, it says: "... discrete-time white noise with covariance Qd in a system with a sample period of T, is equivalent to continuous-time white noise with covariance Qc*$\delta$ (t) (*$\delta$ (t): dirac delta function), where Qc = Qd / T." "... Rd = Rc / T ... This establishes the equivalence between white measurement noise in discrete time and continuous time. The effects of white measurement noise in discrete time and continuous time are the same if v(k) ~ (0,Rd) v(t) ~ (0,Rc) " How does this relate to my other questions? It seems to suggest that Rd $\neq$ Rc as Q4 suggests.
I don't know if you think about white noise the way electrical engineers often do, as a type of "power spectrum". I haven't learned to think of it that way.
My intuitive understanding of white noise goes like this. Suppose you wanted to simulate a white noise at 1 second time intervals by drawing a random number from some distribution. You do this an produce a graph. Then someone (say your boss) wants you to simulate "the same" noise but at a time step of 1/10 of a second. If you draw numbers from the same distribution at each 1/10 of a second, you get another jumpy graph. Maybe you think it looks OK. But suppose your boss is using the white noise in some other function that adds it up, like the simple sum of all the white noise jumps you provide. He probably won't think that you did a good job because in a time interval of a certain size , say 60 seconds, the graph of his function will look a more jumpy with your 1/10 th second noise that it did with the 1 second noise.
Suppose you try to fix this by scaling the numbers you draw. For 1/10 a second, you draw random numbers from the same distribution that you were using and then you divide the numbers by 10. This won't please the boss either. The graph of his function won't look jumpy enough. (Intuitively, this is because the jumps he gets at 1 second intervals is now the average of 10 draws at 1/10 second intervals which has a smaller variation that one draw per second.
The way to please your boss is to scale the numbers so their variance is 1/10 of the standard deviation of the original distribution. Then, according the law for finding the variance of a sum of independent random variables, the variation he gets a 1 second intervals is the sum of 10 terms, each of which has variance that is 1/10 of the variance he had at 1 second intervals. So he sees the same variability over a given time interval as he did with your 1 second noise.
If I want to talk about the variance of white noise, in the sense of a process that takes places in continuous time rather than at discrete steps, I think of the units of measure as being something-squared per unit time. Only when both units are established (e.g. feet and seconds) is the variance of the noise well defined.
I can't tell what the texts you presented are saying about Rc=Rd. An emprical measurment of noise at an arbitrary time step has the proper units for noise at a unit time step, e.g. 2.4 ft^2/ 2 secs = 1.2 ft^2/ 1 sec. But I don't know if that is what is being asserted. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 26, "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": 0.9458849430084229, "perplexity_flag": "middle"} |
http://physics.stackexchange.com/questions/11978/impact-of-covering-glass-on-lens-performance?answertab=oldest | # Impact of covering glass on lens performance
I've seen microscope lenses optimized for 0.17mm covering glass. I don't see what needs to be optimized here? As glass does not touch the lens (as in case of oil/water immersion) - it should just affect focal distance without introducing any aberrations.
Is that correct, or covering glass will cause aberrations imbalance, and will require recalculating the lens? (I can probably only think of very slight chromatic aberration, but it's not important in my case)
Same for water immersion: If we cover sample with 1mm of water, but water does not touch the lens, will it cause any additional aberrations and require recalculating the lens?
Same for diffraction-limited laser focusing optics: I've seen some of them are optimized for laser output windows - what is the nature of this optimization?
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## 1 Answer
Any time you know there will be something in your optical system, it is generally good practice to include it in your model during the design process. Of course, in many applications these things will not make a difference. The image quality of your point-and-shoot camera will not become unacceptable if somebody chooses to take a picture through their bedroom window, for example. In the cases you cited, however, the extra aberrations produced by the cover glass or laser window can indeed be significant. I'll explain why, starting with some background.
## Aberration Theory
Wavefront aberrations are usually expressed as a polynomial function 1:
$$W = \sum_{i,j,k} W_{i j k} h^i \rho^j \phi^k$$
where $h$ is the object height (or angle, in the case of an object at infinity), and $(\rho, \phi)$ are polar coordinates in the pupil, normalized such that $h=1$ is the maximum object height and $\rho = 1$ is the edge of the pupil.
For example, $W_{040}$ is the coefficient of spherical aberration. It is often called "third-order" spherical, because lens designers are concerned with the ray angle errors, which are given by the first derivative of the wavefront error; this also distinguishes it from higher order aberrations, like 5th order spherical, $W_{060}$.
There are a number of ways that aberration coefficients can be computed. Continuing with spherical aberration as our example, the contribution from a single surface can be expressed:
$$W_{040}= -\frac{1}{8} A^2 Y_a \Delta\left(\frac{u_a}{n}\right) \rho^4$$
where $A \equiv n i_a$, $n$ is the refractive index, $i_a$ is the angle of incidence of the marginal ray (the "a-ray") on the surface, $Y_a$ is the marginal ray height on the surface, $u_a$ is the (paraxial) marginal ray angle, and $\Delta(x) \equiv x' - x$, where primed variables represent quantities after refraction at the surface, and un-primed variables are quantities immediately before refraction.2
There are a number of techniques for computing aberration coefficients for complex optical systems, such as "Seidel Sums" or the more modern and efficient technique of "G-sums,"3 but the important point here with respect to your question is this: the aberrations due to a surface depend just as much on the characteristics of the rays at the surface as they do on the shape of the surface.
So, you can see that the aberrations introduced by a flat plate can be very significant indeed. In the formalism I have presented here, a high NA microscope system would have very large values for $u_a$, and the design must absolutely include these in balancing aberrations if very high performance is to be achieved.
This should also help make clear why liquid immersion is helpful in getting to very high NA without aberrations becoming a problem. However, in your question you also ask about the case of a water droplet over the object, without being in contact with the microscope objective as in a true liquid immersion objective. In this case you must keep in mind that the curved surface of the droplet would be a refractive surface and must be included in the optical calculations if the resulting lens is to function at all!
One last note, on the brief discussion in comments about the reason for cover glass. Cover glass serves a variety of purposes, but isn't really intended as a beneficial component of the microscope imaging system. It is primarily used to hold the sample still; protect it from air; and most importantly, to hold it flat, because a high NA microscope system will have a very shallow depth of field. It would be quite inconvenient for the microscope user to have micron-scale surface variations in his sample produce focus variations between adjacent features in the image. Cover glass can be accommodated by the optical design, but if it weren't for these practical considerations the lens designer would likely not call for it.
Footnotes:
1: These are Seidel aberrations; more modern approaches use an orthogonal set of polynomials as their basis functions, such as Zernike polynomials, which are more appropriate for numerical techniques. Seidel aberrations are left over from the time when aberrations had to be calculated by hand, and are still useful and well understood by modern lens designers.
2: The marginal ray is the ray from an object point on the optical axis, which goes through the edge of the pupil/aperture stop. Also important is the chief ray, which is a ray from the edge of the field of view through the center of the pupil/aperture stop. These rays are used heavily in the geometrical analysis of optical systems. For more on this, see "Modern Optical Engineering" by W.J. Smith.
3: Again, these are covered by Smith.
- | {"extraction_info": {"found_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": 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": 0.9535908699035645, "perplexity_flag": "middle"} |
http://www.all-science-fair-projects.com/science_fair_projects_encyclopedia/Mechanical_work | All Science Fair Projects
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Mechanical work
Work (abbreviated W) is the energy transferred by a force (F) to an object as the object moves to a position (s). Work is a scalar quantity, but it can be positive or negative.
Contents
Definition
Note: Readers not familiar with multivariate calculus or vectors, please see "Simpler formulae" below)
Work is defined as the following line integral
$W = \int_{C} \vec F \cdot \vec{ds} \,\!$
where
C is the curve traversed by the object;
$\vec F$ is the force vector;
$\vec s$ is the position vector.
Units
The SI derived unit of work is the Joule, which is defined as the work done by a force of one Newton acting over a distance of one metre. The unit N m, however, is often used with the quantity of work instead of Joule.
Non-SI units of work include the erg, the foot-pound, and the foot-poundal .
Simpler formulae
In the simplest case, that of a body moving in a steady direction, and acted on by a force parallel to that direction, the work is given by the formula
$W = \mathbf{F} s \,\!$
where
F is the force and
s is the distance traveled by the object.
The work is taken to be negative when the force opposes the motion. More generally, the force and distance are taken to be vector quantities, and combined using the dot product:
$W = \vec F \cdot \vec s = |\mathbf{F}| |s| cos\phi \,\!$
where
φ is defined as the angle between the force and the displacement vector.
This formula holds true even when the force acts at an angle to the direction of travel. To further generalize the formula to situations in which the force and the object's direction of motion changes over time, it is necessary to use differentials, d, to express the infinitesimal work done by the force over an infinitesimal time, thus:
$dW = \vec F \cdot \vec{ds} \,\!$
The integration of both sides of this equation yields the most general formula, as given above.
Types of work
Forms of work that are not evidently mechanical, such as electrical work, can be considered as special cases of this principle; for instance, in the case of electricity, work is done on charged particles moving through a medium.
Heat conduction from a warmer body to a colder one is not normally considered to be a form of mechanical work, because at the macroscopic level, there is no measurable force. At the atomic level, there are forces as the atoms collide, but they average to nearly zero in bulk.
Not all forces do work. For instance, a centripetal force in uniform circular motion does not transfer energy; the speed of the object undergoing the motion remains constant. This fact is confirmed by the formula: if the vectors of force and displacement are perpendicular, their dot product is zero.
Mechanical energy
In physics, mechanical energy is one of several forms of energy. It is distinguished by the property that it can be transferred from one system to another by forces known to Newtonian mechanics. This category includes kinetic energy, Ek, and gravitational potential energy, Ep.
Conservation of mechanical energy
The conservation of mechanical energy is a principle which states that the total mechanical energy of a system in a gravitational field where the gravitation is the only force acting upon it, is constant. It is also the sum of the kinetic and potential energy. If an object with constant mass is in free fall, the total energy of position 1 will be equal position 2.
$(E_k + E_p)_1 = (E_k + E_p)_2 \,\!$
where
Ek is the kinetic energy, and
Ep is the potential energy.
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http://math.stackexchange.com/questions/237679/help-with-proof-of-chain-rule | Help with proof of chain rule
I'm following along on this proof of the chain rule. All is clear to me except the step where they say:
Differentiablility implies continuity; therefore $\Delta_u \to 0$ as $\Delta_x \to 0$ in $f\big(g(x)\big)$.
Could someone explain to a calculus newb why that's the case, in simple terms?
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2 Answers
A piece of advice: try another site as the one you linked uses what I consider pretty poor notation.
If we have $\,f(g(x))\,$ then I'd go
$$\lim_{x\to x_0}\frac{f(g(x))-f(g(x_0))}{x-x_0}=\lim_{x\to x_o}\frac{f(g(x))-f(g(x_0))}{g(x)-g(x_0)}\frac{g(x)-g(x_0)}{x-x_0}$$
And what that site says with its cumbersome (for me) notation is that $\,x\to x_0\Longrightarrow g(x)\to g(x_0)\,$ , since differentiability implies continuity.
Some care must be put to avoid the possibility of dividing by zero in the above, of course.
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(before answering your question, let me tell you that I don't really recommend that proof; the main problem is that derivatives are calculated at a point, and the notation used in the proof obscures that fact completely)
You are calculating your derivatives at some $x=x_0$. Then $\Delta x=x-x_0$; and $\Delta u=u(x)-u(x_0)$. As $u$ is differentiable, it is continuous. Continuity at $x_0$ for $u$ means exactly that $u(x)-u(x_0)\to0$ when $x-x_0\to0$, i.e. $\Delta u\to 0$ when $\Delta x\to 0$.
- | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 15, "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": 0.9697665572166443, "perplexity_flag": "head"} |
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http://mathhelpforum.com/differential-equations/201878-eigenvalues-eigenvectors.html | # Thread:
1. ## Eigenvalues and eigenvectors?
i been given this 3x3 matrix and can't find the eigenvaules and eigenvectors.
-3 -2 -2
-1 -4 -1
2 4 1
can someone show me how to do it?
2. ## Re: Eigenvalues and eigenvectors?
Have you found the characteristic equation? It has three small integer solutions. Once you know the three values for the eigenvalue, $\lambda$, find the three independent eigenvectors by solving $Av= \lambda v$. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 2, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.8561285734176636, "perplexity_flag": "middle"} |
http://mathoverflow.net/questions/59222/importance-of-poincare-recurrence-theorem-any-example/59225 | ## Importance of Poincaré recurrence theorem? Any example?
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Recently I am learning ergodic theory and reading several books about it.
Usually Poincaré recurrence theorem is stated and proved before ergodicity and ergodic theorems. But ergodic theorem does not rely on the result of Poincaré recurrence theorem. So I am wondering why the authors always mention Poincaré recurrence theorem just prior to ergodic theorems.
I want to see some examples which illustrate the importance of Poincaré recurrence theorem. Any good example can be suggested to me?
Books I am reading: Silva, Invitation to ergodic theory. Walters, Introduction to ergodic theory. Parry, Topics in ergodic theory.
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The ergodic theorems can be seen as improvements of the Poincaré recurrence theorem, because they relate the time average with the space average. However, the Poincaré theorem is treated first because it is not hard to prove and it is very nice. – subshift Mar 22 2011 at 19:33
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This isn't very related to your question, but I warmly recommend Einsiedler's & Ward's book too ("Ergodic Theory with a view towards Number Theory"). – Mark Schwarzmann Mar 22 2011 at 19:56
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The Poincare recurrence theorem has counterintuitive (or paradoxical) consequences in the development of statistical mechanics. See the discussion of the theorem and why it is famous in Petersen's Ergodic Theory (p. 34 ff). Imagine a wall divides an empty chamber in two and a gas is pumped into one side. Remove the wall and the gas diffuses. Would you expect the gas at a future point to all recollect in one half of the chamber (infinitely often)? What does the Poincare recurrence theorem say? – KConrad Mar 22 2011 at 21:09
The proof of Poincare recurrence is intuitive and quantitative. The first property for the Birkhoff ergodic theorem is debatable, the second property wrong. Also the assumption of being measure preserving is more natural than ergodicity! Hence, the Poincare recurrence theorem is the better theorem. – Helge Mar 23 2011 at 4:25
## 11 Answers
Part of the importance of the Poincare recurrence theorem is in the follow-up questions it legitimizes. Knowing that for any set $A$ of positive measure we can find a positive $n$ such that $\mu(A \cap T^{-n} A) > 0$, one could ask whether
• we can choose $n$ from some "nice" set;
• it is possible to observe "multiple" recurrence;
• there are "many" $n$ for which the result holds.
One example of "nice" is "a square": one can always find a positive $n$ such that $\mu(A \cap T^{-n^2}A) > 0$. See for example theorem 2.1 in part 6 of these notes.
An example of "multiple" is that one can always find positive integers $m$ and $n$ such that $\mu(A \cap T^{-n}A \cap T^{-m}A \cap T^{-(m+n)}A) > 0$. To prove this, iterate the Poincare recurrence theorem. A more involved example of "multiple" is given by requiring that $m = n$ in the previous expression.
Lastly, an example of "many" is given by "syndetic": it follows from Khintchine's recurrence theorem (theorem 3.3 in Petersen's "Ergodic Theory") that the set of such $n$ has bounded gaps.
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That the Poincare recurrence theorem is just the tip of the iceberg for more general multiple recurrence theorems is a good example. One role for multiple recurrence theorems can be found in the book review for Furstenberg's Recurrence in Ergodic Theory and Combinatorial Number Theory at journals.cambridge.org/article_S0143385700001887. (But note the typo in the heading of the review: Princeton is not in the state of New Yersey.) – KConrad Mar 23 2011 at 2:01
The start of the notes at math.stanford.edu/~ksound/Notes.pdf by Soundararajan, which I learned about from an answer by Frank Thorne, traces a path from the Poincare recurrence theorem to multiple recurrence theorems to applications in additive combinatorics. – KConrad Mar 25 2011 at 8:17
Does the celebrated Green-Tao Theorem dependences on the result of multiple recurrence (Furstenberg's Theorem)? – Choi Apr 4 2011 at 10:54
The Green-Tao theorem relies on Szemeredi's theorem, which is logically equivalent to Furstenberg's proof of the aforementioned multiple recurrence theorem in ergodic theory. – Mark Schwarzmann Apr 16 2011 at 23:07
### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you.
If you are number-theoretically inclined, you might want to have a look at Harry Furstenberg, Poincare recurrence and number theory, Bulletin of the American Mathematical Society 5 (1981) 211-234, which seems to be freely available on the web. Among other things, Furstenberg uses Poincare recurrence to prove van der Waerden's Theorem on arithmetic progressions: given any partition of the integers into finitely many subsets, at least one of the subsets contains arbitrarily long arithmetical progressions.
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3
There are some excellent examples of the applications of Poincare recurrence in that article, but if you'll forgive my pickiness I don't think that the proof of van der Waerden's theorem is one of them. Furstenberg's proof of van der Waerden uses a generalisation of the Birkhoff recurrence theorem, not the Poincare recurrence theorem. – Ian Morris Mar 24 2011 at 9:48
1
Guess I skimmed it too lightly. Glad to see confirmation that the article's relevant to the question, even if I got the details wrong. Thanks. – Gerry Myerson Mar 24 2011 at 11:13
This may not be a very dynamical application but still very interesting. Poincaré recurrence theorem was used in this paper
http://arxiv.org/pdf/math/0606232
to show that every amenable left-orderable group is locally indicable.
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First of all Poincare's Recurrence Theorem is important historically because it poses a serious obstacle to modeling an ideal gas in a box as a mechanical system consisting of hard spheres bouncing around. See The wikipedia article on the second law of thermodynamics but the basic idea is that such a mechanical system preserves a natural volume on its phase space (Liouville's theorem) but this is in contradiction (by Poincare's Theorem) with the fact that entropy decreases along all trajectories (assuming that entropy is a continuous function in phase space).
Also, I like this proof that any harmonic function of a compact Riemannian manifold is constant:
Take the gradient flow of your harmonic function and notice that it's volume preserving (more or less by the definition of harmonicity since divergence of the gradient is zero). Now apply Poincare's Recurrence Theorem so that you now have a dense set of recurrent orbits. However since this is a gradient flow, the value of the function decreases strictly along any non-singular orbit, this means all the recurrent orbits are singular so that the gradient has a dense set of zeros. Conclusion: the function is constant.
Last but not least Poincare's Recurrence Theorem is more or less just the Pigeonhole Principle (finite space + lots of stuff = lots of overlap) which is certainly a fundamental mathematical fact even if not so many theorems follow directly from it. See Terrence Tao's course notes on Ergodic Theory where he emphasized this point very beautifully.
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1
+1 for harmonic functions! – Margaret Friedland Feb 16 2012 at 21:50
The Poincaré recurrence theorem is sometimes useful because of the way it translates into recurrence in metric spaces. For example, a corollary of the Poincaré theorem is that for a measure-preserving transformation of a separable metric space - which need not be continuous - almost every point is recurrent in the topological sense. To see this, choose a sequence which is dense in the metric space, and consider the cover of the metric space by balls of radius $1/n$ around points in this sequence. By the Poincaré theorem almost every point belonging to one of these balls returns to that ball infinitely often, and hence returns to within distance $2/n$ of itself. Now take the intersection over $n$ to see that almost every point is recurrent.
A more general corollary which is also sometimes useful is the following: if $T \colon X \to X$ is measure-preserving, and $f$ is a measurable function from $X$ to a separable metric space, then $\liminf_{n \to \infty} d(f(T^nx),f(x))=0$ almost everywhere. This result naturally can be very useful in circumstances where one knows that a function is measurable, but no more. An example which springs to mind is the proof of Theorem 15 in "A formula with some applications to the theory of Lyapunov exponents" by Avila and Bochi, where the Poincaré theorem is applied to prove the recurrence of the measurable splittings in the multiplicative ergodic theorem.
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Instead of comparing the Poincare recurrence theorem with ergodic theorems one should rather look at the underlying notions of conservativity and ergodicity in the general context of a measure class preserving action of a countable group. If the Poincare theorem says precisely that actions with a finite invariant measure are conservative, it is somewhat misleading to identify ergodicity with presence of an ergodic theorem (as this is the case for a very limited class of actions only).
In terms of the ergodic decomposition of an action, ergodicity (of course) means that there is only one ergodic component which coincides with the whole action. On the other hand, conservativity means that there are no discrete ergodic components (i.e., ones which coincide just with action orbits) - indeed, any wandering set obviously gives rise to discrete ergodic components, and it is not hard to see that the converse is also true. From this point of view ergodicity is a strengthening of conservativity.
However, there are several classes of dynamical systems (actions), for which conservativity and ergodicity are equivalent, i.e., any system from this class is either ergodic or completely dissipative (all ergodic components are discrete $\equiv$ the whole state space is the union of translates of a certain fundamental domain"). This phenomenon is called "Hopf dichotomy", the most famous example of which (precisely the one originally studied by Hopf) is the case of geodesic flows on negatively curved manifolds (and of the associated boundary actions).
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You can also use it to prove Borel-Harish Chandra theorem that an irreducible lattice in a semi-simple Lie group (without compact factors) is Zariski dense. Here is an elementary case: Take $G=SL_n({\mathbb R})$ and $\Gamma=SL_n({\mathbb Z})$. Let $H$ be the Zariski closure of $H$ and let $f: G \to SL_m({\mathbb R})$ be a representation such that the stabilizer of $v \in {\mathbb R}^m$ is exactly $H$. The existence of $f$ follows from a general fact known as Chevalley's theorem. This allows you to consider the projective space $P({\mathbb R}^m)$ as a $G$-space. Consider the map $G/ \Gamma \to P({\mathbb R}^m)$ defined by $g \Gamma \mapsto gv$ and let $\mu$ be the push-forward of the finite $G$-invariant measure on $G/ \Gamma$, which will be $G$-invariant. Now,using the Jordan canonical form, you can see that for any unipotent element $u \in \Gamma$ $u^n \cdot w$ converges to a point in projective space. One the other hand, by Poincare recurrence, the only way you can have this is when $\mu$ charges only one point. Now since $\Gamma$ is generated by one-parameter (discrete) unipotent subgroups, you will see that the orbit has to be only one point, hence $H=G$.
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The Poincare recurrence lemma is used in the theory of measure foliations (or measured geodesic laminations) on surfaces, to construct combinatorial approximations of the foliation, as explained in "Thurston's works on surfaces" by Fathi, Laudenbach, and Poenaru. In terminology that came a little after that book, one is constructing train track approximations. This is related to the comment of Matheus regarding Rauzy induction, and the comment of Stephane Laurent about Rokhlin towers.
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There is a whole field of research focused on the consequences of Poincare recurrence theorem.
The link below provides one of the first articles, critical theory at the beginning of this
Two professionals who work extensively with these issues are: Suassol Benoit and Luis Barreira, below is a link to the page the first author:
http://www.math.univ-brest.fr/perso/benoit.saussol/articles.html
Containing several articles on this subject, with interesting results.
The following link contains a collection of articles accordingly:
http://www.im.ufrj.br/~arbieto/ensino/20091/20091.html
I hope this information is useful.
Best,
Eduardo.
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An interesting application of Poincare recurrence theorem is provided by Masur-Veech theorem on the unique ergodicity of almost every interval exchange maps. Indeed, very roughly speaking, a proof of this result goes like this. There is a natural renormalization dynamics of interval exchange maps known as Rauzy-Veech induction. By construction of an appropriate invariant finite mass measure (sometimes called Masur-Veech measure), it follows from Poincare recurrence theorem that almost every interval exchange map is recurrent with respect to the Rauzy-Veech induction and this information can be shown to imply unique ergodicity.
For more details on this, see this survey by J.-C. Yoccoz.
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Poincaré's lemma also allows to define the induced transformation on a Borel set. And then to derive results on perodic approximations based on Kakutani-Rokhlin towers. (I wanted to post my remark as a comment but the "add coment" box does not appear here...)
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http://unapologetic.wordpress.com/2009/09/24/differentials/?like=1&source=post_flair&_wpnonce=1c947633de | # The Unapologetic Mathematician
## Differentials
Okay, partial derivatives don’t work as an extension of derivation to higher-dimensional spaces. Even generalizing them to directional derivatives doesn’t give us what we want. What we need is not just the separate existence of a bunch of directional derivatives, but a single object which gives us all directional derivatives at once. To find it, let’s look back at the derivative in one dimension.
If we know the derivative of a function $f$ at a point $x$, we can use it to build a close linear approximation to the function near that point. This is what we mean when we say that the derivative is the slope of the tangent line. It says that if we move away from $x$ by an amount $t$, we can approximate the change in the function’s value
$\displaystyle f(x+t)-f(x)\approx f'(x)t$
I’m going to write the part on the right-hand side as one function: $df(x;t)=f'(x)t$. We use a semicolon here to distinguish the very different roles that $x$ and $t$ play. Before the semicolon we pick a point at which to approximate $f$. After the semicolon we pick a displacement from our starting point. The “differential” $df$ approximates how much the function will change from its value at $x$ when we move away by a displacement $t$. Importantly, for a fixed value of $x$ this displacement is a linear function of $t$. This is obvious here, since once we pick $x$, the value of $df(x;t)$ is determined by multiplying by some real number, and multiplication of real numbers is linear.
What’s less obvious is also more important: the differential $df(x;t)$ approximates the difference $f(x+t)-f(x)$. By this, we can’t just mean that the distance between the two goes to zero as $t$ does. This is obvious, since both of them must themselves go to zero by linearity and continuity, respectively. No, we want them to agree more closely than that. We want something better than just continuity.
I say that if $f$ has a finite derivative at $x$ (so the differential exists), then for every $\epsilon>0$ there is a $\delta>0$ so that if $\delta>\lvert t\rvert>0$ we have
$\displaystyle\lvert\left[f(x+t)-f(x)\right]-df(x;t)\rvert<\epsilon\lvert t\rvert$
That is, not only does the difference get small (as the limit property would say), but it gets small even faster than $\lvert t\rvert$ does. And indeed this is the case. We can divide both sides by $\lvert t\rvert$, which (since $t$ is small) magnifies the difference on the left side.
$\displaystyle\left\lvert\frac{\left[f(x+t)-f(x)\right]-df(x;t)}{t}\right\rvert=\left\lvert\frac{f(x+t)-f(x)}{t}-f'(x)\right\rvert<\epsilon$
But if we can always find a neighborhood where this inequality holds, we have exactly the statement of the limit
$\displaystyle f'(x)=\lim\limits_{t\to0}\frac{f(x+t)-f(x)}{t}$
which is exactly what it means for $f'(x)$ to be the derivative of $f$ at $x$.
So for a single-variable function, having a derivative — the limit of a difference quotient — is equivalent to being differentiable — having a differential. And it’s differentials that generalize nicely.
Now let $f$ be a real-valued function defined on some open region $S$ in $\mathbb{R}^n$. The differential of $f$ at a point $x\in S$, if it exists, is a function $df$ satisfying the properties
• The function $df$ takes two variables in $\mathbb{R}^n$. The values $df(x;t)$ are defined for every value of $t\in\mathbb{R}^n$, and for some region of $x$ values containing the point under consideration. Typically, we’ll be looking for it to be defined in the same region $S$.
• The differential is linear in the second variable. That is, given two vectors $s$ and $t$ in $\mathbb{R}^n$, and real scalars $a$ and $b$, we must have
$\displaystyle df(x;as+bt)=adf(x;s)+bdf(x;t)$
• The differential closely approximates the change in the value of $f$ as we move away from the point $x$, in the sense that for every $\epsilon>0$ there is a $\delta>0$ so that if $\delta>\lVert t\rVert>0$ we have
$\displaystyle\lvert\left[f(x+t)-f(x)\right]-df(x;t)\rvert<\epsilon\lVert t\rVert$
I’m not making any sort of assertion about whether or not such a function exists, or under what conditions it exists. More subtly, I’m not yet making any assertion that if such a function exists it is unique. All I’m saying for the moment is that having this sort of linear approximation to the function near $x$ is the right generalization of the one-variable notion of differentiability.
### Like this:
Posted by John Armstrong | Analysis, Calculus
## 19 Comments »
1. I’m still confused about the distinction between a “derivative” and a (the?) “differential”.
Is it just a matter of how they’re specified? If both exist, do they always denote the same function? Will the distinction become clearer once we move into higher-dimensional spaces?
Comment by Joe English | September 26, 2009 | Reply
2. In one dimension, the distinction is mostly semantic, but semantics are important. The differential of a function (at a point) is a linear function which, given a change in the input, gives an estimate of the resulting change in the output. It’s the best such linear estimate, as we’ll see when we establish uniqueness.
So, for a single-valued function of a single real variable what does this mean? It’s a linear transformation from $\mathbb{R}^1$ to $\mathbb{R}^1$, since changes in both the input and output are one-dimensional. And thus we can represent this transformation with a $1\times1$ matrix with a single real entry. What is that entry? The derivative of the function at that point.
That is, the derivative is a number, and the differential is the linear transformation of multiplication by that number. In one dimension it doesn’t really look like there’s any difference, but in higher dimensions it’s the second interpretation that generalizes.
Comment by | September 26, 2009 | Reply
3. [...] In light of our discussion of differentials, I want to make a point here that is usually glossed over in most treatments of multivariable [...]
Pingback by | September 28, 2009 | Reply
4. [...] if we look at a particular point we can put it into our differential and leave the second (vector) slot blank: . We will also write this simply as , and apply it to a [...]
Pingback by | September 29, 2009 | Reply
5. [...] first we take in the definition of the differential, to [...]
Pingback by | September 30, 2009 | Reply
6. [...] for the Differential To this point we’ve seen what happens when a function does have a differential at a given point , but we haven’t yet seen any conditions that tell us that any such function [...]
Pingback by | October 1, 2009 | Reply
7. [...] by defining partial derivatives and directional derivatives, as we did. But instead of defining the differential, it simply collects the partial derivatives together as the components of a vector called the [...]
Pingback by | October 5, 2009 | Reply
8. [...] how to modify the notion of the derivative of a function to deal with vector inputs by defining the differential. But what about functions that have vectors as [...]
Pingback by | October 6, 2009 | Reply
9. [...] we don’t just have a single value for the instantaneous rate of change, we have a differential . But we can use it to find directional derivatives. Specifically, we’ll consider the [...]
Pingback by | October 13, 2009 | Reply
10. [...] Just like we assembled partial derivatives into the differential of a function, so we can assemble higher partial derivatives into higher-order differentials. The [...]
Pingback by | October 16, 2009 | Reply
11. [...] themselves continuous throughout . We’ve seen that this will imply that such a function has a differential at each point of . This gives us a subalgebra of which we write as . That is, these functions have [...]
Pingback by | October 21, 2009 | Reply
12. [...] function of real variables with components . The differential, then, is itself a vector-valued function whose components are the differentials of the component [...]
Pingback by | November 11, 2009 | Reply
13. [...] spaces, we will assume that transforms small enough changes in almost linearly — is differentiable — and that the Jacobian determinant is everywhere nonzero, so we can invert the [...]
Pingback by | January 5, 2010 | Reply
14. [...] like a term of art. Really what we know is what it means for a function from to itself to be differentiable, and for that differential to be continuous, and for there to be a second differential, and so on. [...]
Pingback by | February 25, 2011 | Reply
15. [...] vectors already: a gadget that takes a tangent vector at and gives back a number. It’s the differential, which when given a vector returns the directional derivative in that direction. And we can [...]
Pingback by | April 13, 2011 | Reply
16. [...] this is obvious. The differential is a linear transformation. Since it goes between finite-dimensional vector spaces it’s [...]
Pingback by | May 4, 2011 | Reply
17. [...] the -form that takes a vector field and evaluates it on the function . And this is just like the differential of a multivariable function: a new function that takes a point and a vector at that point and gives [...]
Pingback by | July 15, 2011 | Reply
18. So, let me see if I get the generalization to higher dimensions: in a (differentiable) map from R^n to R^m, the differential will be the mxn-Jacobian matrix (evaluated at a vector dx of “infinitesimal displacements” dx_1,dx_2,…,dx_n , and the derivative is the higher-dimensional verson of the tangent-plane approximation? Also: where do these differentials “live”; are they elements of the dual of differentiable functions?
Comment by Esteban | May 5, 2013 | Reply
19. Pretty much, yeah. The differential is the Jacobian of a function from $\mathbb{R}^n$ to $\mathbb{R}$. As for where it lives, the differential $df$ at a point $p$ is in the dual of the tangent space $T_p\mathbb{R}^n$, so $df$ itself is a section of the cotangent bundle $T^*\mathbb{R}^n$ (which I don’t think I really got around to discussing properly before other demands on my time drew me further and further away from this blog).
Comment by | May 6, 2013 | Reply
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## 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": 68, "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": 0.9363921284675598, "perplexity_flag": "head"} |
http://physics.stackexchange.com/questions/35700/fields-of-steady-currents-using-electrostatics?answertab=oldest | # Fields of Steady Currents Using Electrostatics
Suppose you have a uniform ring charge rotating at constant angular velocity so that you also have a uniform ring of steady current, and thus you can use the Biot-Savart Law to compute the magnetic field. But I also remember from college that to good approximation you can use electrostatics to compute the electric field due to the charge if the current is steady, in spite of the fact that charges are moving. I never understood how this works. Can someone offer insight?
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## 1 Answer
In the stationary situation all partial derivatives with respect to time vanish (this is in a sense the definition of a stationary state). Looking at the Maxwell equation relevant to determine the electric field, you note that they are given by $$\nabla \cdot \mathbf E = \rho, \qquad \text{and} \qquad \nabla\times\mathbf{E} = -\frac1c\partial_t\mathbf B=0,$$ i.e., the same equations as in the steady state, so you can introduce a potential and reduce the set of equation to Poisson's equation.
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That's crystal clear. I was anticipating some complex cancellation of first order contributions. It would also seem I was somewhat mistaken in that given ideal sources, the reliability of electrostatics is in fact exact, not just an approximation. That's good to know. – David H Sep 5 '12 at 18:41 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 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": 0.9586700201034546, "perplexity_flag": "head"} |
http://math.stackexchange.com/questions/286333/proof-on-catalan-numbers?answertab=votes | # Proof on Catalan numbers
In the book I'm using on Catalan numbers, the author gives a scenario in order to develop the formula for Catalan numbers.
The scenario is that a boy has an empty jar. Every day he either puts in a dollar coin or takes on out for $2n$ days. At the end of the $2n$ days, the jar is empty. In how many ways can this happen?
I understand how he gets to the Catalan numbers. But there is one part where he says this:
note that if $n\gt 0$ then there had to be a first day other than the starting day when the jar was empty.
I'm having trouble understanding why this is the case. When I think of taking or putting a coin in a jar as the equality $y=x$, you could just add a coin in for the first $n$ days, then take one out for the remaining $2n - n = n$ days.
Why does there have to be a day other than the starting day in which the jar was $0$?
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## 2 Answers
The "first day other than the starting day when the jar was empty" can also be the last day, as it is in your example. That must be the case if $n=1$, for example: since there are only two days, he has to put a coin in the first day and take it out the second.
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1
Ah I see, that makes sense then. I think I was just thinking too hard. – MITjanitor Jan 25 at 2:28
The jar has always $\ge 0$ dollars in it. It starts out empty, and ends up empty. It could be empty in before the end (say you put money in for 3 straigt days, take out money for 3 straight days, and then continue your merry way for the rest of the month). It doesn't have to be empty in between. Methinks you got those two cases mixed up?
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http://mathoverflow.net/questions/65900?sort=oldest | ## A stupid question about Automorphic forms
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Okay, so an automorphic form $f$ on a reductive group $G/ \mathbb{Q}$ and arithmetic subgroup $\Gamma$ is a smooth function satisfying the following conditions:
(a) invariance with respect to left $\Gamma -$ translations.
(b) Right $K -$ finiteness.
(c) Annihilated with respect to a finite co-dimension ideal of the center $Z(\mathcal{U}(\mathfrak{g}_{\mathbb{C}}))$ ( of the complexified universal enveloping algebra).
(d) Growth conditions.
Now as I understand, it is clear why condition (a) is relevant. Condition (b) comes from the idea that representation of $G$ associated to $f$ is admissible.
My understanding is that condition (d) ensures that we can "extend" the automorphic form to the cusps of the symmetric space.
I would like to know: what exactly is the intuition behind condition (c)? (I hope it is not just to provide a framework which includes classical modular forms over upper half-plane and the Maass forms.)
Also any comments about "my understanding" of conditions (a), (b) and (d) are appreciated!
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4
Someone will probably come along with a proper answer later on, but here's something one could think about until then: imagine the case of $GL(1)$ over the rationals. If $K$ is all of $\hat{\mathbf{Z}}$ and you demand invariance, rather than finiteness, then (a) and (b) together just give you an arbitrary ($C^\infty$) function on the positive reals. The growth condition is just "doesn't grow too fast" but if you don't put any more conditions then you have a space which is perhaps far too big to be of arithmetic interest. – Kevin Buzzard May 24 2011 at 22:22
## 2 Answers
These conditions have technical/subtle interactions. It probably suffices to think about automorphic forms on a Lie group, and not think of the interaction with finite places.
For example, on each K-isotype, the Casimir element is an elliptic operator, but it is not elliptic without specifying the behavior under K. From elliptic regularity, such eigenfunctions are real-analytic.
With K-finite and z-finite, moderate growth implies that all derivatives are of moderate growth, etc. (Borel's little book talks about this for SL(2,R), and his Park City notes recapitulate this. The general argument is similar to SL(2,R), anyway.)
K-finite, z-finite, moderate-growth cuspforms are (up to adjustment of central character) provably of rapid decay, so certainly L^2. This and the previous assertion are part of the "theory of the constant term".
Dropping the moderate growth condition is useful occasionally, as in the "weak Maass form" business.
In a more elementary vein, a choice to require annihilation by a finite-codimension ideal in z is akin to looking at functions on the real line annihilated by a product (D^2-lambda_1)...(D^2-lambda_n), namely, polynomial multiples of exponentials. Certainly not every function on the line is annihilated by such, but the spectral theory (Fourier transform) decomposes a general function into a superposition of certain of these special functions. (Here the analogue of K is {1}.)
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Welcome to MO, Paul! – Victor Protsak Jun 1 2011 at 5:05
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I am no expert in the general theory, but let me share some thoughts. In some sense (c) accounts for the fact that $K$ does not have enough open subgroups (unlike at nonarchimedan places), so that the usual convolution definition of Hecke operators is too crude. Instead, one needs to work with the convolution algebra of all distributions on $G$ with support in $K$, i.e. with $U(\mathfrak{g})\otimes_{U(\mathfrak{k})}A_K$, where $\mathfrak{k}$ is the Lie algebra of $K$ and $A_K$ is the convolution algebra of finite measures on $K$. For compatibility with (b) one needs to restrict to elements of $U(\mathfrak{g})$ which commute with elements of $\mathfrak{k}$, and a simple way to achieve this is to restrict to $Z(U(\mathfrak{g}))$. In this way one can obtain a unified treatment at all places for which the adelic framework is most suitable. For example, a Hecke cusp form on $\mathrm{GL}_n$ can be characterized by $n$ parameters at each place: these parameters define the local $L$-functions whose product is the global $L$-function.
For a special case see Section 2.3 in Goldfeld: Automorphic forms and $L$-functions for the group $GL(n,\mathbb{R})$. For a brief discussion of the general theory see Borel-Jacquet: Automorphic forms and automorphic representations, Proc. Symp. Pure Math. 33 (1979), 189-202.
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Thanks, I looked up the Borel-Jacquet reference, they seem to imply that condition (c) above is necessary to ensure that the action of Hecke algebra (and not just the group) on the space of "automorphic forms" is nice ( admissible?) They refer to the work of Harish-Chandra for some general results, but I haven't looked through it. – isildur May 25 2011 at 17:47
Yes, but note that the action of the Hecke algebra itself comes from the group action (convolution by measures on the maximal compact and derivations in various directions). As usual in number theory, at archimedean places the definitions are slightly different than at nonarchimedan ones, yet a unified treatment can be achieved at all places which is beneficial and nice. – GH May 25 2011 at 18:57 | {"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": 29, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9263930320739746, "perplexity_flag": "head"} |
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# Lie algebra
In mathematics, a Lie algebra (named after Sophus Lie, pronounced "lee") is an algebraic structure whose main use lies in studying geometric objects such as Lie groups and differentiable manifolds. Lie algebras were introduced to study the concept of infinitesimal transformations.
Contents
## Definition
A Lie algebra is a type of an algebra over a field; it is a vector space g over some field F together with a binary operation [·, ·] : g × g → g, called the Lie bracket, which satisfies the following properties:
$[a x + b y, z] = a [x, z] + b [y, z], \quad [z, a x + b y] = a[z, x] + b [z, y]$
for all a, b ∈ F and all x, y, z ∈ g.
• For all x in g
$[x,x]=0 \quad$
• The Jacobi identity:
$[x,[y,z]] + [y,[z,x]] + [z,[x,y]] = 0 \quad$
for all x, y, z in g.
Note that the first and second properties together imply
[x,y] = - [y,x]
for all x, y in g ("anti-symmetry"). Conversely, the antisymmetry property implies property 2 above as long as F is not of characteristic 2.
Also note that the multiplication represented by the Lie bracket is not in general associative, that is, [[x,y],z] need not equal [x,[y,z]]. Therefore, Lie algebras are not rings or associative algebras in the usual sense, although much of the same language is used to describe them.
## Examples
1. Every vector space becomes an abelian Lie algebra trivially if we define the Lie bracket to be identically zero.
2. Euclidean space R3 becomes a Lie algebra with the Lie bracket given by the cross product of vectors.
3. If an associative algebra A with multiplication * is given, it can be turned into a Lie algebra by defining [x, y] = x * y − y * x. This expression is called the commutator of x and y. Conversely, it can be shown that every Lie algebra can be embedded into one that arises from an associative algebra in this fashion. See universal enveloping algebra.
4. Another important example of a Lie algebra comes from differential topology: the smooth vector fields on a differentiable manifold form an infinite dimensional Lie algebra when equiped with the Lie derivative as the Lie bracket. The Lie derivative identifies a vector field X with a partial differential operator acting on any smooth scalar field f by letting X(f) be the directional derivative of f in the direction of X. Then in the expression (YX)(f), the juxtaposition YX represents composition of partial differential operators. Then the Lie bracket [X, Y] is defined by
[X, Y] f = (XY − YX) f
for every smooth function f on the manifold.
This is the Lie algebra of the infinite-dimensional Lie group of diffeomorphisms of the manifold.
5. The vector space of left-invariant vector fields on a Lie group is closed under this operation and is therefore a finite dimensional Lie algebra. One may alternatively think of the underlying vector space of the Lie algebra belonging to a Lie group as the tangent space at the group's identity element. The multiplication is the differential of the group commutator, (a,b) |-> aba−1b−1, at the identity element.
6. As a concrete example, consider the Lie group SL(n,R) of all n-by-n matrices with real entries and determinant 1. The tangent space at the identity matrix may be identified with the space of all real n-by-n matrices with trace 0, and the Lie algebra structure coming from the Lie group coincides with the one arising from commutators of matrix multiplication.
For more examples of Lie groups and their associated Lie algebras, see the Lie group article.
## Homomorphisms, subalgebras, and ideals
A homomorphism φ : g → h between Lie algebras g and h over the same base field F is an F-linear map such that [φ(x), φ(y)] = φ([x, y]) for all x and y in g. The composition of such homomorphisms is again a homomorphism, and the Lie algebras over the field F, together with these morphisms, form a category. If such a homomorphism is bijective, it is called an isomorphism, and the two Lie algebras g and h are called isomorphic. For all practical purposes, isomorphic Lie algebras are identical.
A subalgebra of the Lie algebra g is a linear subspace h of g such that [x, y] ∈ h for all x, y ∈ h. The subalgebra is then itself a Lie algebra.
An ideal of the Lie algebra g is a subspace h of g such that [a, y] ∈ h for all a ∈ g and y ∈ h. All ideals are subalgebras. If h is an ideal of g, then the quotient space g/h becomes a Lie algebra by defining [x + h, y + h] = [x, y] + h for all x, y ∈ g. The ideals are precisely the kernels of homomorphisms, and the fundamental theorem on homomorphisms is valid for Lie algebras.
## Classification of Lie algebras
Real and complex Lie algebras can be classified to some extent, and this classification is an important step toward the classification of Lie groups. Every finite-dimensional real or complex Lie algebra arises as the Lie algebra of unique real or complex simply connected Lie group (Ado's theorem), but there may be more than one group, even more than one connected group, giving rise to the same algebra. For instance, the groups SO(3) (3×3 orthogonal matrices of determinant 1) and SU(2) (2×2 unitary matrices of determinant 1) both give rise to the same Lie algebra, namely R3 with cross-product.
A Lie algebra is abelian if the Lie bracket vanishes, i.e. [x, y] = 0 for all x and y. More generally, a Lie algebra g is nilpotent if the lower central series
g > [g, g] > [[g, g], g] > [[[g, g], g], g] > ...
becomes zero eventually. By Engel's theorem, a Lie algebra is nilpotent if and only if for every u in g the map
ad(u): g → g
defined by
ad(u)(v) = [u,v]
is nilpotent. More generally still, a Lie algebra g is said to be solvable if the derived series
g > [g, g] > [[g, g], [g,g]] > [[[g, g], [g,g]],[[g, g], [g,g]]] > ...
becomes zero eventually. A maximal solvable subalgebra is called a Borel subalgebra.
A Lie algebra g is called semi-simple if the only solvable ideal of g is trivial. Equivalently, g is semi-simple if and only if the Killing form K(u,v) = tr(ad(u)ad(v)) is non-degenerate; here tr denotes the trace operator. When the field F is of characteristic zero, g is semi-simple if and only if every representation is completely reducible, that is for every invariant subspace of the representation there is an invariant complement (Weyl's theorem ).
A Lie algebra is simple if it has no non-trivial ideals and is not abelian. In particular, a simple Lie algebra is semi-simple, and more generally, the semi-simple Lie algebras are the direct sums of the simple ones.
Semi-simple complex Lie algebras are classified through their root systems.
## Category theoretic definition
Using the language of category theory, a Lie algebra can be defined as an object A in the category of vector spaces together with a morphism $[\cdot,\cdot]:A\otimes A\rightarrow A$ such that
• $[\cdot,\cdot]\circ (id+\tau_{A,A})=0$
• $[\cdot,\cdot]\circ ([\cdot,\cdot]\otimes id)\circ(id+\sigma+\sigma^2)=0$
where σ is the cyclic permutation braiding $(id\otimes \tau_{A,A})\circ(\tau_{A,A}\otimes id)$. In diagrammatic form:
## References
• Humphreys, James E. Introduction to Lie Algebras and Representation Theory, Second printing, revised. Graduate Texts in Mathematics, 9. Springer-Verlag, New York, 1978. ISBN 0-387-90053-5
• Jacobson, Nathan, Lie algebras, Republication of the 1962 original. Dover Publications, Inc., New York, 1979. ISBN 0-486-63832-4
03-10-2013 05:06:04 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 7, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.8502087593078613, "perplexity_flag": "head"} |
http://mathhelpforum.com/calculus/40206-weird-integral.html | # Thread:
1. ## Weird Integral
A question that I have been given asks me to use a certain substitution to find the answer exactly. The only thing is not only do I think I can manage without it the substitution is the most random thing one I've seen.
Upper Boundary = 2, Lower Boundary = 0
Integrate: (4-x)^(1/2).dx
Use the substitution x = 2sin u to exactly evaluate this definite integral.
Should I just do it without using that substitution?
2. You just need to solve $\int_{0}^{2}{\sqrt{4-x}\,dx}.$
Make the substitution $z^2=4-x,$ and don't forget to get the new integration limits.
3. Originally Posted by Evales
A question that I have been given asks me to use a certain substitution to find the answer exactly. The only thing is not only do I think I can manage without it the substitution is the most random thing one I've seen.
Upper Boundary = 2, Lower Boundary = 0
Integrate: (4-x)^(1/2).dx
Use the substitution x = 2sin u to exactly evaluate this definite integral.
Should I just do it without using that substitution?
the x is not squared. what you wrote does not require a trig sub. a substitution of u = 4 - x will suffice
4. Originally Posted by Jhevon
the x is not squared. what you wrote does not require a trig sub. a substitution of u = 4 - x will suffice
Omg thanks for that. God your sharp yes the x is actually squared.
5. Originally Posted by Evales
A question that I have been given asks me to use a certain substitution to find the answer exactly. The only thing is not only do I think I can manage without it the substitution is the most random thing one I've seen.
Upper Boundary = 2, Lower Boundary = 0
Integrate: (4-x)^(1/2).dx
Use the substitution x = 2sin u to exactly evaluate this definite integral.
Should I just do it without using that substitution?
Originally Posted by Evales
Omg thanks for that. God your sharp yes the x is actually squared.
$\int_0^2 \sqrt{4-x^2}\,dx$
They tell you to use the substitution $x=2\sin u$. This implies that $dx=2\cos u \ \,du$
Change the limits of integration:
$2=2\sin u \implies u=\arcsin(1)=\frac{\pi}{2}$
$0=2\sin u \implies u=\arcsin(0)=0$
Thus, we have:
$\int_{0}^{\frac{\pi}{2}}2\cos u \sqrt{4-4\sin^2u}\,du$
Simplify the integrand:
$2\cos u\sqrt{4-4\sin^2u}=4\cos u \sqrt{1-\sin^2u}=4\cos u\sqrt{\cos^2u}=4\cos^2u$
We now have:
$\int_{0}^{\frac{\pi}{2}}4\cos^2u\,du$
I'll let you take it from here, noting that $\cos^2u=\frac{1+\cos (2u)}{2}$.
Hope that this helped! | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 11, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": 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": 0.9162811040878296, "perplexity_flag": "middle"} |
http://mathhelpforum.com/calculus/59471-analysis-function.html | # Thread:
1. ## Analysis Function
Is there a function such that, f is continuous on (1,5), reaches a maximum on (1,5), but does not reach a minimum on (1,5). Does such a function exist or is it impossible?
2. Originally Posted by thahachaina
Is there a function such that, f is continuous on (1,5), reaches a maximum on (1,5), but does not reach a minimum on (1,5). Does such a function exist or is it impossible?
f both reaches a maximum on (1,5) and a minimum on (1,5), never heard of that => i guess it reaches a max on (1,5), but not a minimum on (1,5) is possible.
3. Originally Posted by thahachaina
Is there a function such that, f is continuous on (1,5), reaches a maximum on (1,5), but does not reach a minimum on (1,5). Does such a function exist or is it impossible?
Hint, basically think of a function that when differentiated gives a function such that $\exists{x_1}\in(1,5)\backepsilon{f(x_1)=0}$
This sounds hard, but it isn't. Think of trigonmetric function with an altered period for one.
4. Thanks man. I kinda thought about it, but it looks to me like that function is not possible since it does not reach a maximum at (1,5). Is that right?
5. Originally Posted by thahachaina
Is there a function such that, f is continuous on (1,5), reaches a maximum on (1,5), but does not reach a minimum on (1,5). Does such a function exist or is it impossible?
$f(x)=-(x-1)(x-5)$
has a maximum at $x=3$ but has no minimum on the open interval $(1,5)$ .
(If it were a closed interval it would be a different matter)
CB | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 4, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": 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": 0.9462406039237976, "perplexity_flag": "middle"} |
http://unapologetic.wordpress.com/2009/10/16/higher-order-differentials/?like=1&_wpnonce=34114a6dfa | # The Unapologetic Mathematician
## Higher-Order Differentials
Just like we assembled partial derivatives into the differential of a function, so we can assemble higher partial derivatives into higher-order differentials. The differential measures how the function itself changes as we move around, and the higher differentials will measure how lower differentials change.
First let’s look at the second-order differential of a real-valued function $f$ of $n$ variables $x^i$. We’ll use the $dx^i$ as a basis for the space of differentials, which allows us to write out the components of the differential:
$\displaystyle df(x)=\frac{\partial f}{\partial x^i}dx^i$
So, just as we did for vector-valued functions, we’ll just take the differentials of each of these components separately, and then cobble them together.
$\displaystyle d\left[df\right](x)=\left(d\frac{\partial f}{\partial x^i}\right)dx^i=\left(\frac{\partial^2 f}{\partial x^j\partial x^i}dx^j\right)dx^i$
Now this second displacement may have nothing to do with the first, but it should be the same for all components. That is, we could write out the second differential as a function of not only the point $x$ but of two displacements $t_1$ and $t_2$ from the point:
$\displaystyle d^2f(x;t_1,t_2)=\frac{\partial^2 f}{\partial x^j\partial x^i}t_1^it_2^j$
Commonly we’ll collapse this into a function of a point and a single displacement. We just put the same vector $t$ in for both $t_1$ and $t_2$
$\displaystyle d^2f(x;t)=\frac{\partial^2 f}{\partial x^j\partial x^i}t^it^j$
Unfortunately, these higher differentials are more complicated than our first-order derivatives. In particular, they don’t obey anything like Cauchy’s invariant rule, meaning they don’t transform well when we compose functions. As an example, let’s go back and look at the polar coordinate transform again:
$\displaystyle\begin{aligned}x&=r\cos(\theta)\\y&=r\sin(\theta)\end{aligned}$
We’ve seen how we can use Cauchy’s invariant rule to rewrite differentials:
$\displaystyle\begin{aligned}dx&=\cos(\theta)dr-r\sin(\theta)d\theta\\dy&=\sin(\theta)dr+r\cos(\theta)d\theta\end{aligned}$
We can also invert the transformation and rewrite differential operators:
$\displaystyle\begin{aligned}\frac{\partial f}{\partial x}&=\cos(\theta)\frac{\partial f}{\partial r}-\frac{\sin(\theta)}{r}\frac{\partial f}{\partial\theta}\\\frac{\partial f}{\partial y}&=\sin(\theta)\frac{\partial f}{\partial r}+\frac{\cos(\theta)}{r}\frac{\partial f}{\partial\theta}\end{aligned}$
So let’s take our second-order differential
$\displaystyle d^2f=\left(\frac{\partial}{\partial x}\frac{\partial}{\partial x}f\right)(dx)^2+\left(\frac{\partial}{\partial x}\frac{\partial}{\partial y}f\right)(dx)(dy)+\left(\frac{\partial}{\partial y}\frac{\partial}{\partial x}f\right)(dy)(dx)+\left(\frac{\partial}{\partial y}\frac{\partial}{\partial y}f\right)(dy)^2$
and try to rewrite it. The nasty bit is working out all these second-order partial derivatives in terms of $r$ and $\theta$.
$\displaystyle\begin{aligned}\frac{\partial}{\partial x}\frac{\partial}{\partial x}f=&\left[\cos(\theta)\frac{\partial}{\partial r}-\frac{\sin(\theta)}{r}\frac{\partial}{\partial\theta}\right]\left(\cos(\theta)\frac{\partial f}{\partial r}-\frac{\sin(\theta)}{r}\frac{\partial f}{\partial\theta}\right)\\=&\cos(\theta)\frac{\partial}{\partial r}\left(\cos(\theta)\frac{\partial f}{\partial r}-\frac{\sin(\theta)}{r}\frac{\partial f}{\partial\theta}\right)-\frac{\sin(\theta)}{r}\frac{\partial}{\partial\theta}\left(\cos(\theta)\frac{\partial f}{\partial r}-\frac{\sin(\theta)}{r}\frac{\partial f}{\partial\theta}\right)\\=&\left(\cos(\theta)^2\frac{\partial^2f}{\partial r^2}+\frac{\cos(\theta)\sin(\theta)}{r^2}\frac{\partial f}{\partial\theta}-\frac{\cos(\theta)\sin(\theta)}{r}\frac{\partial^2f}{\partial r\partial\theta}\right)\\&+\left(\frac{\sin(\theta)^2}{r}\frac{\partial f}{\partial r}-\frac{\sin(\theta)\cos(\theta)}{r}\frac{\partial^2f}{\partial\theta\partial r}+\frac{\sin(\theta)\cos(\theta)}{r^2}\frac{\partial f}{\partial\theta}+\frac{\sin(\theta)^2}{r^2}\frac{\partial^2f}{\partial\theta^2}\right)\\=&\cos(\theta)^2\frac{\partial^2f}{\partial r^2}-2\frac{\cos(\theta)\sin(\theta)}{r}\frac{\partial^2f}{\partial r\partial\theta}+\frac{\sin(\theta)^2}{r^2}\frac{\partial^2f}{\partial\theta^2}\\&+\frac{\sin(\theta)^2}{r}\frac{\partial f}{\partial r}+2\frac{\cos(\theta)\sin(\theta)}{r^2}\frac{\partial f}{\partial\theta}\end{aligned}$
$\displaystyle\begin{aligned}\frac{\partial}{\partial x}\frac{\partial}{\partial y}f=\frac{\partial}{\partial y}\frac{\partial}{\partial x}f=&\cos(\theta)\sin(\theta)\frac{\partial^2f}{\partial r^2}+\frac{\cos(\theta)^2-\sin(\theta)^2}{r}\frac{\partial^2f}{\partial r\partial\theta}-\frac{\sin(\theta)\cos(\theta)}{r^2}\frac{\partial^2f}{\partial\theta^2}\\&-\frac{\sin(\theta)\cos(\theta)}{r}\frac{\partial f}{\partial r}+\frac{\sin(\theta)^2-\cos(\theta)^2}{r^2}\frac{\partial f}{\partial\theta}\end{aligned}$
$\displaystyle\begin{aligned}\frac{\partial}{\partial y}\frac{\partial}{\partial y}f=&\sin(\theta)^2\frac{\partial^2f}{\partial r^2}+2\frac{\sin(\theta)\cos(\theta)}{r}\frac{\partial^2f}{\partial r\partial\theta}+\frac{\cos(\theta)^2}{r^2}\frac{\partial^2f}{\partial\theta^2}\\&+\frac{\cos(\theta)^2}{r}\frac{\partial f}{\partial r}-2\frac{\cos(\theta)\sin(\theta)}{r^2}\frac{\partial f}{\partial\theta}\end{aligned}$
After that it’s no trouble at all to transform the differential terms
$\displaystyle\begin{aligned}(dx)^2&=\cos(\theta)^2(dr)^2-2r\sin(\theta)\cos(\theta)(dr)(d\theta)+r^2\sin(\theta)^2(d\theta)^2\\(dx)(dy)=(dy)(dx)&=\cos(\theta)\sin(\theta)(dr)^2+r(\cos(\theta)^2-\sin(\theta)^2)(dr)(d\theta)-r^2\sin(\theta)\cos(\theta)(d\theta)^2\\(dy)^2&=\sin(\theta)^2(dr)^2+2r\sin(\theta)\cos(\theta)(dr)(d\theta)+r^2\cos(\theta)^2(d\theta)^2\end{aligned}$
Let’s just work out the component that goes with $(d\theta)^2$ when we put these all together
$\displaystyle\begin{aligned}\left(r^2\cos(\theta)^2\sin(\theta)^2-2r^2\cos(\theta)^2\sin(\theta)^2+r^2\cos(\theta)^2\sin(\theta)^2\right)&\frac{\partial^2f}{\partial r^2}\\+\left(-2r\cos(\theta)\sin(\theta)^3+2r(\cos(\theta)\sin(\theta)^3-\cos(\theta)^3\sin(\theta))+2r\cos(\theta)^3\sin(\theta)\right)&\frac{\partial^2f}{\partial r\partial\theta}\\+\left(\sin(\theta)^4+2\cos(\theta)^2\sin(\theta)^2+\cos(\theta)^4\right)&\frac{\partial^2f}{\partial\theta^2}\\+\left(r\sin(\theta)^4+2r\sin(\theta)^2\cos(\theta)^2+r\cos(\theta)^4\right)&\frac{\partial f}{\partial r}\\+\left(2\cos(\theta)\sin(\theta)^3+2(\cos(\theta)^3\sin(\theta)-\cos(\theta)\sin(\theta)^3)-2\cos(\theta)^3\sin(\theta)\right)&\frac{\partial f}{\partial\theta}\\=&\frac{\partial^2f}{\partial\theta^2}+r\frac{\partial f}{\partial r}\end{aligned}$
Which has an extraneous term! If an invariance rule held, we should just get $\frac{\partial^2f}{\partial\theta^2}$.
The difference comes from the way that the differential operators themselves change as we move our point around. Increasing $\theta$ by a little bit means something different at the point $(x,y)=(1,0)$ than it does at the point $(x,y)=(0,1)$. This doesn’t really matter when we’re talking about first-order differentials because we’re never putting two differential operators together, and so we never get any measurement of how an operator changes from point to point. We will eventually learn how to compensate for this effect, but that will wait until we have a significantly more general approach.
### Like this:
Posted by John Armstrong | Analysis, Calculus
## 4 Comments »
1. [...] Differentials and Composite Functions Last time we saw an example of what can go wrong when we try to translate higher differentials the way we did [...]
Pingback by | October 19, 2009 | Reply
2. [...] the mixed second partials are equal, since they’re both continuous, and we can define the second differential. These functions form a subalgebra of (and thus a further subalgebra of ) which we write as . [...]
Pingback by | October 21, 2009 | Reply
3. [...] analogue of the second derivative for multivariable functions is the second differential . This function assigns to every point a bilinear function of two displacement vectors and , and [...]
Pingback by | November 24, 2009 | Reply
4. [...] to itself to be differentiable, and for that differential to be continuous, and for there to be a second differential, and so on. We even introduced classes of functions to describe this whole tower, where consists [...]
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## About this weblog
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http://mathoverflow.net/questions/20107?sort=newest | ## Surprising Analogue of Q
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
I was describing Manish Kumar's work a few weeks ago to a fellow graduate student, and she stumped me with a big-picture question I couldn't answer.
Manish Kumar proved that the commutator subgroup of $\pi_1(\mathbb{A}^1_K)$, where $K$ is a characteristic $p$, algebraically closed field, is pro-finite free. He proved this, in fact, for any smooth affine curve over $K$.
(He proved this for algebraically closed fields of char $p$ which are uncountable in his thesis: http://www.math.msu.edu/~mkumar/Publication/thesis.pdf; and without the cardinality restriction in: http://arxiv.org/PS_cache/arxiv/pdf/0903/0903.4472v2.pdf)
As I explained to my colleague, this is a geometric analogue of Shafarevich's conjecture, that $Gal(\mathbb{Q}^{ab})$ is pro-finite free. Indeed $Gal(\mathbb{Q}^{ab})=\pi_1^c(Spec(\mathbb{Q}))$ ($c$ stands for the commutator subgroup). But why is $\mathbb{A}^1_K$, for $K$ an algebraically closed characteristic $p$ field (or indeed, any smooth affine curve over $K$), an analogue of $Spec(\mathbb{Q})$? Usually $\mathbb{A}^1_K$ (for $K$ an algebraically closed field) is an analogue of $Spec(\mathbb{Z})$. I came up with some partial explanations, but no full heuristic. Can you think of one?
-
## 3 Answers
One can see that the commutator subgroup of the topological fundamental group of a complex curve is free for elementary reasons, but this is a pretty weak analogy. In fact I don't have a good heuristic of why it should be true, other than the fact that is. I was Manish's adviser, and I was pretty surprised by the result when he proved it.
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### 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.
Alright, I think I should write my 2 cents here:
Obviously $Spec(\mathbb{Q})$ and $\mathbb{A}_K$ are not directly analogous, but they do appear to be in relation to this problem. It seems that they are related through the intermediary $Spec(F)$ where $F$ is a function field over an algebraically closed field. Shafarevich's conjecture holds for $Spec(F)$ (this is an earlier result).
If we take any smooth affine curve, $C$, over an algebraically closed field of positive characteristic; and use the result that $\pi_1^c(C)$ is pro-finite free. Every abelian covering of $C$ will give an abelian extension of $\kappa(C)$ ($C$'s function field). However, there's no reason to think we get all abelian extensions of $\kappa(C)$ that way. However $\kappa(C)$ is also $\kappa(D)$ for different smooth affine curves, so may use their abelian unramified covers.
To make some order of this, start with an abelian extension of $\kappa(C)$, $L$. We may take $C$'s normalization in $L$. This may be branched at some points in $C$, but we may discard those. So any abelian extension of $\kappa(C)$ comes from an abelian unramified cover of some possibly different smooth affine curve whose function field is $\kappa(C)$.
It seems, however, extraordinary to expect that since $\pi_1^c(Spec(\kappa(C)))$ is pro-finite free, $\pi_1^c(C)$ should be; for any affine curve $C$. Is there some secret motivation for thinking this that I'm missing?
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1
I would not necessarily say that one is free "since" the other is free. – JSE Apr 9 2010 at 1:24
I guess I'm not sure I agree they're analogous. First of all, extensions of Q can be ramified anywhere, while covers of A^1_K are unramified away from the infinite place. Q is much more like Spec K(T), the generic point of A^1_K -- but even here, to get a good analogy, you perhaps want K to be finite instead of algebraically closed.
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http://mathhelpforum.com/calculus/94467-finding-relative-maximum-minimum.html | Thread:
1. Finding Relative Maximum and Minimum
Q1. Test the Function $g(x)=6x^3-(108)x^2+(630)x-2$ for relative maximum and minimum. Find the critical values. Use the 2nd derivative test if possible.
My solution:
$g'(x)=18x^2-216x+630$ if follows that the critical values of g are x1= -7 and x2=-5.
*
To find the critical numbers, I set g'(x) =0. This is where I think I went wrong.
First, I factored out a "6"
$6(3x^2-36x+105)$
Then, I factored out a 3 and that's how I arrive on my answer of -7 and -5.
Q.2 (Related to Q1)
Since $g"(x)=36x-216$ <-- My guess, we have that g"(x1)= ? greater or less than 0 and g"(x2) = ? greater or less than 0. (I need to find out values for x1 and x2)
Therefore, by the 2nd derivative test, there is a either a relative max or min when x=x1 and there is a relative max or min when x=x2. (Choose either or)
TIA
2. Originally Posted by mvho
Q1. Test the Function $g(x)=6x^3-(108)x^2+(630)x-2$ for relative maximum and minimum. Find the critical values. Use the 2nd derivative test if possible.
My solution:
$g'(x)=18x^2-216x+630$ if follows that the critical values of g are x1= -7 and x2=-5.
* To find the critical numbers, I set g'(x) =0. This is where I think I went wrong.
First, I factored out a "6"
$6(3x^2-36x+105)$
Then, I factored out a 3 and that's how I arrive on my answer of -7 and -5.
Q.2 (Related to Q1)
Since $g"(x)=36x-216$ <-- My guess, we have that g"(x1)= ? greater or less than 0 and g"(x2) = ? greater or less than 0. (I need to find out values for x1 and x2)
Therefore, by the 2nd derivative test, there is a either a relative max or min when x=x1 and there is a relative max or min when x=x2. (Choose either or)
TIA
What exactly did you do after you factored out the three? You were on the right track.
1. take the derivative.
2. Set it to zero
3. the solution set are your critical numbers.
3. I am such an Idiot!! The critical values are 5 and 7. I originally had it factored down to $(x-7)(x-5)$
x-7=0
x=7
DUH!
I am a bit confused on the 2nd part though. I know my 2nd derivative is correct:
$g"(x)=36x-216$ but am unsure about this whole " we have that g"(x1)=_____ greater OR less than 0 and g"(x2)=________ greater or less than 0.
Therefore, by the second derivative test there is a relative max or min when x=x1 and there is a relative max or min when x=x2.
Can someone explain that in Simpler terms?
4. Once you've found your critical numbers (by setting the first derivative to zero and solving for x), you can apply the second derivative test by putting those critical values into the second derivative. If the value is positive then the function at the given point is a relative minimum, and vise versa for a negative value returned from the second derivative test. Hope that helps.
5. Originally Posted by mvho
I am a bit confused on the 2nd part though. I know my 2nd derivative is correct:
$g"(x)=36x-216$ but am unsure about this whole " we have that g"(x1)=_____ greater OR less than 0 and g"(x2)=________ greater or less than 0.
Therefore, by the second derivative test there is a relative max or min when x=x1 and there is a relative max or min when x=x2.
Can someone explain that in Simpler terms?
I had the same question recently. Check this out.
http://www.mathhelpforum.com/math-he...tive-test.html
6. Once you've found your critical numbers (by setting the first derivative to zero and solving for x), you can apply the second derivative test by putting those critical values into the second derivative. If the value is positive then the function at the given point is a relative minimum, and vise versa for a negative value returned from the second derivative test. Hope that helps.
Ok, that is much easier to understand.
So, since my critical numbers were 5 and 7 in this case, I simply plug them into my g"(x)=36x-216.
When I plug in "5", the value is negative so I have a maximum
When I plug in "7", the value is positive so I have a minimum.
7. Exactly 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": 11, "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": 0.9189534187316895, "perplexity_flag": "middle"} |
http://cnr.lwlss.net/CircleObjectivesR/ | # Testing optimisation algorithms: Part II - Implementing objective function in R
22:16 Wednesday 7th September 2011
In Part I of this series, I presented the reasons I'm interested in optimisation, and gave a mathematical statement of the example constrained optimisation problem I will solve. In this post, I am going to present two practical implementation algorithms, coded in R. I'll also present an R function for visualising the resultant circles.
Code for optimisation and visualisation can be found at the end of this post. It's fairly well commented, so just some general comments first:
Most optimisation functions take a single vector of parameter values as an argument. This is a bit unwieldy for our problem, since we have N triplets of x,y and $r$. Here I arbitrarily define a parameter vector z, of length 3N and split it up (consistently) into vectors for x, y and r.
R has a convenient combinatorics facility, so I use that to generate a unique list of all possible circle pairs to optimise the calculations required for constraint $g_5$.
I'm using function closures here to avoid defining global variables. In R, a function closure is a function which returns another function, together with its own environment for storing variables etc., which can be initialised within the outer function before returning the inner function. For example, I have created a function createObj which initialises the number of circles N, and the rectangle dimensions (W and H), generates a unique list of circle pairs (cpairs) and creates a function newObj which only depends on z, but nevertheless has access to private copies of N, W, H and clist.
Here, constraints are embedded into the objective function. A numerical value quantifying constraint violation is subtracted from the objective function we are trying to maximise. There are two alternative versions of the objective function closures in the code below: createObj is "soft", the constraint violation values are the sum of the square of the magnitude of violations, createHardObj is "hard", any single violated constraint, no matter how small, returns in an area of -Infinity. In the first case, a derivative based algorithm will always be able to figure out a path towards satisfying constraints if it finds itself in a region where they are violated, but small violations may be tolerated in the final solution. In the latter case, all constraints must be completely obeyed and so any solution will be perfect, but on the other hand, there are no hints about how to improve from an invalid set of parameters, and so finding a solution is much more difficult.
The drawCircles function can plot circles to screen, or in vector graphics format (to a .pdf for example). Here are some outputs generated with this function (initial guess on left, optimised result on right):
Finally, the optimisation routine presented here just uses the L-BFGS-B method which is a pretty good, bounded, derivative-based optimiser, which is made available within the optim function which is standard in R installations (it's part of the core). It's fairly slow for more than 100 circles, and the results are not perfect, as circles are often forced to the edge of the box with negligble radii. Nevertheless, it's a good start. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.8952609896659851, "perplexity_flag": "middle"} |
http://mathhelpforum.com/pre-calculus/50605-graphing-composite-functions.html | # Thread:
1. ## Graphing Composite Functions
Hello, I am given two graphics, F(x), G(x) and I am supposed to graph g(f(x))
I am a bit rusty with composite functions, so I was hoping someone could give me some general information on how to do this? From what I recall, you need to take the range of f(x) and input it as the domain of g(x)? I know both functions domains must be satisfied. It looks like f(x) D = [-3,3] R = [-1,3] and g(x) D = [-3,3] R = [-2, 2]
I hope this made sense, thank you
2. You're right for a composition $g(f(x))$ f's range is the domain of g....so for example, let:
$<br /> f:[-3,3]\rightarrow\mathbb{R}, f(x)=x+1<br />$
$<br /> g:\mathbb{R}\rightarrow\mathbb{R}, g(x)=x^2<br />$
Then the range of f is $[f(-3),f(3)] = [-2,4]$ then since g is composed with f, (also written $g\circ f(x)$) it's domain is the range of f, [-2,4]. Then the range of g is $[g(-2),g(4)] = [(-2)^2,4^2] = [4,16]$
So we have $f:[-3,3]\rightarrow[-2,4]$ and $g:[-2,4]\rightarrow[4,16]$ and for a compostion $f_1\circ f_2\circ ...\circ f_n(x)$ you need only know the domain of $f_n$ to know all other domains/ranges.
However, the best appraoch to graphing would be to write out the composition as a single expression, e.g: $g\circ f(x) = (x+1)^2$ then choose a set of points from your domain e.g: $[-3,3]\in\mathbb{Z}$, calculate a table of values, plot the points and interpolate them into a smooth curve...just as you would with the graph of a normal function.
EDIT: Hmmm.. can someone please tell me the LaTeX for the composition symbol, the little circle? I thoubht it was: \circ
... I also tried \o and \O, but they didn't work either.
3. Originally Posted by Greengoblin
You're right for a composition $g(f(x))$ f's range is the domain of g....so for example, let:
$<br /> f:[-3,3]\rightarrow\mathbb{R}, f(x)=x+1<br />$
$<br /> g:\mathbb{R}\rightarrow\mathbb{R}, g(x)=x^2<br />$
Then the range of f is $[f(-3),f(3)] = [-2,4]$ then since g is composed with f, (also written $g\circ f(x)$) it's domain is the range of f, [-2,4]. Then the range of g is $[g(-2),g(4)] = [(-2)^2,4^2] = [4,16]$
So we have $f:[-3,3]\rightarrow[-2,4]$ and $g:[-2,4]\rightarrow[4,16]$ and for a compostion $f_1\circ f_2\circ...\circ f_n(x)$ you need only know the domain of $f_1$ to know all other domains/ranges.
However, the best appraoch to graphing would be to write out the composition as a single expression, e.g: $g\circ f(x) = (x+1)^2$ then choose a set of points from your domain e.g: $[-3,3]\in\mathbb{Z}$, calculate a table of values, plot the points and interpolate them into a smooth curve...just as you would with the graph of a normal function.
EDIT: Hmmm.. can someone please tell me the LaTeX for the composition symbol, the little circle? I thoubht it was: \circ
... I also tried \o and \O, but they didn't work either.
Hi, yes it is \circ
but if you don't put any space between \circ and f, it will be understood as \circf, which is an unknown command
4. Thanks, its working now. hoorah! | {"extraction_info": {"found_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": 24, "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": 0.9587552547454834, "perplexity_flag": "middle"} |
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http://physics.stackexchange.com/questions/20480/should-i-heat-my-room-when-im-not-here-energy-efficiently-speaking?answertab=active | # Should I heat my room when I'm not here, energy-efficiently speaking?
I was wondering as it's getting cold : is it better for my electricity bill to shut down completely my (electric) heater during day, and to turn it on again when I come home (then it will have to heat the room from something like 5°C to around 20°C), or should I keep the temperature around, say, 15°C while I'm away ?
I would naively guess it's better off, but I'm also wondering about inertia of the walls for instance...
-
## 4 Answers
This problem is very simple, but it's easy to overcomplicate by looking at too small a scale. At every second – no matter what the heater does – you waste money by heating the outside of your house. The rate of heating – and thus the rate at which you waste money – is given by Newton's Law of Cooling. So
$$\text{Wasted money} \propto \int (T_\text{in}-T_\text{out})\;dt$$
The lower your house's temperature, the less money you waste – no matter what. So set the thermostat to the lowest practical temperature when you're away.
-
5
But there is also a requirement that it is habitable when you are in it. If it takes 6hours to reach 'normal' temperature when you get home in the evening that's not a solution – Martin Beckett Feb 3 '12 at 18:40
Hence the "practical" bit I added. Basically, you'd want to lower the temperature a bit at first, then keep lowering it until you find a good trade-off between livability and savings. If you're comfortable shutting the heater off entirely, that's obviously the ideal – you don't need to measure your house's heat capacity in a series of silly experiments. – rdhs Feb 3 '12 at 18:54
1
If the temperature drops below practical temperatures in a normal work day, you should really worry about the insulation of your house. – Bernhard Feb 3 '12 at 19:24
3
There is another concern--besides how long it takes to come back up to habitable--in setting "practical": condensation is very bad. Make sure that doesn't happen. – dmckee♦ Feb 4 '12 at 3:11
Also in temperate climates during the day T_out may be high enough not to need much heating. So it would be a different comfort question if you are a night worker or come back at night. I know people who turn on the heat by phone two hours before going to their country cottage. – anna v Feb 4 '12 at 9:23
This is a very easy question to answer, though as some have pointed out, it's also easy to get confused by making it too hard. If you think about it correctly, the answer is immediately apparent.
Let me ask you an exactly equivalent question:
Let us suppose I have a leaky bucket, one with a hole in the bottom. I'm going to leave for the day, and I'm trying to decide whether I waste less water by keeping the bucket full all the time I'm gone, or whether it makes more sense for me just to let the bucket run dry and then refill it when I get home.
The answer is obvious: I waste far more water trying to keep the bucket filled. I should turn the water to the bucket off until I get home, then refill it. Note that this is true for any amount of time I'm gone; I always come out ahead when I don't keep the bucket full, because the pressure differential from having a full bucket leads to the water exiting the bucket faster than if the bucket were only partially full.
Similarly, you should turn off the heat to your home while you're away (assuming there is no other reason to keep your house warm, e.g. a pet or cold-sensitive items), then reheat it when you get back. You will always save money this way, since as the heat differential between the inside of the house and the outside lessens, your heat loss will in turn lessen.
This is true in all but the most contrived cases, such as, "I have a well-insulated heat source in my house that will lose its insulative properties and dump all its heat if the temperature drops below X degrees". Ignoring such silly "gotcha" scenarios, the answer is good for all realistic cases.
-
IMO it depends on the latitude. In northern regions and also up in mountains the pipes may freeze in the house and the expense of repairs will we worse than keeping the thermostat at a higher than freezing level. – anna v Feb 4 '12 at 9:19
1
@annav I believe that your comment is already covered by the proviso: assuming there is no other reason to keep your house warm, e.g. a pet or cold-sensitive items. – Paul Wagland Feb 7 '12 at 7:34
Let keep the problem as simple as possible. To start with some assumptions.
• The outside, or ambient temperature $T_a$ is constant.
• The temperature of your house $T_h$, is uniform over for your whole house, but a function of time $T_h(t)$. (Also known as an ideally stirred tank)
• The flow of heat out of your house is proportional can be cast into one heat transfer coefficient $h$ ($h$ will be smaller when your house is insulated better)
No heating The budget of heat for your house, contains two terms, the change in amount of heat in your house, and the flow out of your house.
$$c\frac{dT_h(t)}{dt}=-h(T_h(t)-T_a)$$ Where c is a constant related to the amount of heat that can be stored in your house (including the amount of heat stored in the walls, the size of your house, etc). This is easy to solve $T_h(t)=T_a+(T_h(0)-T_a)e^{-\frac{h}{c} t}$
Now, if you come back at your home, you want to heat your house from $T_h(t)$ back to $T_h(0)$. Your electrical heater should gives this amount of heat, which is just $c(T_h(t)-T_h(0))$, which equals to
$$Q_a = c \Big(T_a-T_h(0)\Big)\Big(1-e^{-\frac{h}{c}t_{away}}\Big)$$ where $Q$ is the amount of heat delivered by the heater and $t_{away}$ the time you've been absent.
Heater on When you keep your heater on the whole time. The amount of heat delivered by your heater, is constant and equal to $$Q_b=-h t_{away} (T_h(0)-T_a)$$
Now, if you are away for a short period of time, you can linearize the exponent, and you see that the amount of heat that you need to put in your house by the heater is the same to first order. However, as your away-time increases, the amount of electricity you need in the first case will reach a constant value, while if you canstantly heat your house the electricity you need will increase linearly.
The heat capacity of the walls you mention, will in principle affect the constants $c$ (related to total heat capacity) and $h$ (related to heat lose), will only change the timescale for which both approaches are more or less equivalent, but it will not change the physics or equations.
So, to conclude: you should determine the values of $c$ and $h$ of your house. You can measure h by trying the second method once. And then c by trying out the first method. Good luck!
-
The answer to your question depends on a number of variables that are not so easy to estimate.
If you have relatively weak isolation in the walls and windows it is better to lower the temperature during the day quite a bit, how much depends also on how long you are gone.
If the heat capacity of the room and the things in the room is high it will take a long time in the evening to heat up again, so you don't want to cool it down too much.
If you have tight sealing windows and doors lowering the temperature significantly during the day will put your room temperature below the condensation point of the water vapor, so you will have wet surface on which mold will grow on quickly.
There are quite a few engineers working on these problems but there is no answer from the pure physics side what is better. It all depends on what you want to achieve.
To estimate an answer you can use an electricity meter connected to your heater and measure the room temperature over time when you switch the heater off, the power you need to keep the temperature constant and the heating rate and maximum power when you warm up your room again. With these values you can work out the heat capacity as $$C \propto {\tau}{\kappa}$$ with the relaxation time constant $\tau$ and the thermal conductivity $\kappa$ (details for example in this Calorimeter description).
-
That's an interesting point you make about the dew point of vapor water, I didn't consider this side of the house-heating problem ! – Zonko Feb 3 '12 at 14:10
– Vincent Apr 9 '12 at 12:41 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 17, "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": 0.3, "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": 0.9592647552490234, "perplexity_flag": "middle"} |
http://unapologetic.wordpress.com/2008/05/22/ | # The Unapologetic Mathematician
## Matrices II
With the summation convention firmly in hand, we continue our discussion of matrices.
We’ve said before that the category of vector space is enriched over itself. That is, if we have vector spaces $U$ and $V$ over the field $\mathbb{F}$, the set of linear transformations $\hom(U,V)$ is itself a vector space over $\mathbb{F}$. In fact, it inherits this structure from the one on $V$. We define the sum and the scalar product
$\left[S+T\right](u)=S(u)+T(u)$
$\left[cT\right](u)=cT(u)$
for linear transformations $S$ and $T$ from $U$ to $V$, and for a constant $c\in\mathbb{F}$. Verifying that these are also linear transformations is straightforward.
So what do these structures look like in the language of matrices? If $U$ and $V$ are finite-dimensional, let’s pick bases $\left\{e_i\right\}$ of $U$ and $\left\{f_j\right\}$ of $V$. Now we get matrix coefficients $s_i^j$ and $t_i^j$, where $i$ indexes the basis of $U$ and $j$ indexes the basis of $V$. Now we can calculate the matrices of the sum and scalar product above.
We do this, as usual, by calculating the value the transformations take at each basis element. First, the sum:
$\left[S+T\right](e_i)=S(e_i)+T(e_i)=s_i^jf_j+t_i^jf_j=(s_i^j+t_i^j)f_j$
and now the scalar product:
$\left[cT\right](e_i)=cT(e_i)=(ct_i^j)f_j$
so we calculate the matrix coefficients of the sum of two linear transformations by adding the corresponding matrix coefficients of each transformation, and the matrix coefficients of the scalar product by multiplying each coefficient by the same scalar.
Posted by John Armstrong | Algebra, Linear Algebra | 1 Comment
## 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": 27, "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": 0.8970556259155273, "perplexity_flag": "head"} |
http://mathhelpforum.com/algebra/103090-find-values.html | # Thread:
1. ## find values
hi, i've got the question:
the polynomials f(x) and g(x) are defined by $f(x) = x^3 + px^2 - x + 5$, $g(x) = x^3 - x^2 + px + 1$ where p is a constant. when f(x) and g(x) are divided by x - 2 the remainder is R in each case. find the values of p and R
i got it down to
R = f(2) = 8 + 4p - 2 + 5 = 11 + 4p and
R = g(2) = 8 - 4 + 2p + 1 = 5 + 2p but i'm not sure if either of those are right.
can someone show me the way to do it please? thanks
2. Originally Posted by mark
hi, i've got the question:
the polynomials f(x) and g(x) are defined by $f(x) = x^3 + px^2 - x + 5$, $g(x) = x^3 - x^2 + px + 1$ where p is a constant. when f(x) and g(x) are divided by x - 2 the remainder is R in each case. find the values of p and R
i got it down to
R = f(2) = 8 + 4p - 2 + 5 = 11 + 4p and
R = g(2) = 8 - 4 + 2p + 1 = 5 + 2p but i'm not sure if either of those are right.
can someone show me the way to do it please? thanks
I would start by dividing each by x-2, then set the remainders equal.
3. $\frac{x^{3}+px^{2}-x+5}{x-2}=\underbrace{\frac{4p+11}{x-2}}_{\text{remainder}}$
$\frac{x^{3}-x^{2}+px+1}{x-2}=\underbrace{\frac{2p+5}{x-2}}_{\text{remainder}}$
$4p+11=2p+5\Rightarrow p=-3$
$R=-1$
Therefore, the remainder in each case would be $\frac{-1}{x-2}$
$\frac{x^{3}-x^{2}-3x+1}{x-2}=\boxed{\frac{-1}{x-2}}+x^{2}+x-1$
$\frac{x^{3}-3x^{2}-x+5}{x-2}=\boxed{\frac{-1}{x-2}}+x^{2}-x-3$
4. Galactus, your latexing method is as good as that of soroban, excellent!
5. ok i understand it now up to $\frac {-1}{x - 2}$, when you get to that, wouldn't x have to be 3 for the final answer to be -1? how do you know the value of x?
6. Yes, p=-3. Sub that in and we get R=-1 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 12, "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": 0.9418404698371887, "perplexity_flag": "head"} |
http://mathhelpforum.com/trigonometry/76143-finding-angle-x-radians.html | # Thread:
1. ## finding and angle x in radians?
Find an angle x in radians such that 8pi<x<9pi and cosx=1/2
What is this question asking for? I came up with pi/3, but I don't think that's in between 8 and 9 pi. Can anyone explain? Thanks
2. Originally Posted by captaintoast87
Find an angle x in radians such that 8pi<x<9pi and cosx=1/2
What is this question asking for? I came up with pi/3, but I don't think that's in between 8 and 9 pi. Can anyone explain? Thanks
Remember that $\cos{x}$ is periodic with a a period of $2\pi$ which means you get your original answer of $\frac{\pi}{3}$ and add 4 lots of $2\pi$ (this is 4 periods) to give you an answer of $8\pi + \frac{\pi}{3} = \frac{25\pi}{3}$ | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 5, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9791679978370667, "perplexity_flag": "middle"} |
http://gilkalai.wordpress.com/2009/01/16/telling-a-simple-polytope-from-its-graph/?like=1&source=post_flair&_wpnonce=c1ce40f6e1 | Gil Kalai’s blog
## Telling a Simple Polytope From its Graph
Posted on January 16, 2009 by
Peter Mani (a photograph by Emo Welzl)
## Simple polytopes, puzzles
Micha A. Perles conjectured in the ’70s that the graph of a simple $d$-polytope determines the entire combinatorial structure of the polytope. This conjecture was proved in 1987 by Blind and Mani and shortly afterwards I gave a simple proof which I like very much to present.
Let $P$ and $Q$ be two simple d-dimensional polytopes. Let $\psi$ be a bijection from the vertices of $P$ to the vertices of $Q$ which maps edges to edges. In other words, $\psi$ is an isomorphism between the graph $G(P)$ of $P$ to the graph $G(Q)$ of $Q$. Does $\psi$ extend uniquely to a combinatorial isomorphism between $P$ and $Q$? The answer is ‘yes’ but the argument is not so easy. There are also several related open problems.
Eric Friedman
A quick reminder about simple polytopes. A $d$-polytope $P$ is simple if every vertex $v$ of $P$ is contained in $d$ edges. (Equivalently, if the dual polytope $P^*$ is simplicial.) Thus the graph $G(P)$ of a simple $d$-polytope $P$ is a $d$-regular graph. The crucial property of simple polytopes that we use is that for every vertex $v$ of $P$ and every collection of $k$ edges of $P$ containing $v$ there is a unique $k$-face of P which contains $v$ and these $k$ edges.
Among the famous simple polytopes are the cubes and simplices (in any dimension), polygons in the plane, and the dodecahedron in dimension 3 (but not the icosahedron or octahedron).
More background on polytopes can be found at the end of the post. (It is taken from this post presenting five open problems regarding polytopes.).
## Good orderings, unique sink acyclic orientations, and shelling.
Consider a total ordering $<$ of the vertices of a simple $d$-polytope $P$ . A vertex $v$ is a local maximum if for every neighbor $w$ of $v$, $w<v$. Of course, the maximum vertex in $P$ with respect to the ordering is a local maximum. An ordering of the vertices of $P$ is called good, if for every face $F$ of $P$ every local maximum in $F$ is the global maximum in $F$.
( Just to make it clear: a vertex $v$ in $F$ is a local maximum in $F$, if for every neighbor $w$ of $v$ in $F$, $w<v$. The global maximum in $F$ is the maximum vertex of $F$ with respect to $<$.)
There are several variations of this notion. Instead of talking about the entire ordering we can talk just about the orientation it induces on edges of the graph. We orient an edge $\{v,u\}$ from $v$ to $u$ if $v < u$. This is an acyclic orientation of the graph and being good or bad depends only on the orientation. Good orientations are called “unique-sink acyclic orientations”. Also, as we will mention later, good orderings of the vertices of a simple polytope are closely connected to shelling order on the facets of the dual polytope.
## The proof: part I
Let $G$ be the graph of a simple $d$-polytope $P$.
### The crucial claim
Given an ordering $<$ of the vertices of the polytope we define the degree of a vertex $v$, denoted by $deg_<(v)$ as the number of neighbors of $v$ which are smaller than $v$ with respect to the ordering. Let $h_k^<$ be the number of vertices of degree $k$. Consider the following expression:
$f^< =$ $h_0^< +2 h_1^< + \dots + 2^kh_k^< + \cdots +2^nh_n^<$.
Let $f$ be the number of nonempty faces of $P$.
Claim: 1) $f^< \ge f$
2) Equality holds if and only if $<$ is a good ordering.
Proof: Let $<$ be an ordering of the vertices of $G$. We will count pairs $(F,v)$ where $F$ is a face of $P$ and $v$ is a vertex in $F$ which is a local maximum with respect to $<$.
The first thing to note is that if $v$ is a vertex of degree $k$ then it is a local maximum in precisely $2^k$ faces. Therefore the number of pairs is precisely $f^< =$ $h_0^< +2 h_1^<+ \dots + 2^kh_k^< + \cdots +2^nh_n^<$.
On the other hand, every face $F$ of the polytope has a vertex which is a local maximum with respect to $<$. This is the vertex of $F$ which is the global maximum! If $<$ is a good ordering then every face $F$ contributes a unique vertex $v$ which is a local maximum and therefore $f^< = f$. If $<$ is not a good ordering then every face $F$ contributes at least one local maximum and at least one face $F$ contributes more than one! Therefore $f^< > f$. Sababa!
### Telling the number of faces and the good orderings
Now we can tell the total number of faces of $P$ by looking at the graph $G$. We compute the value $f^<$ for every ordering of the vertices of $G$. The minimum value over all orderings $<$ is the value of $f$ – the number of non-empty faces of $P$.
We can tell also the good orderings. We compute the quantity $f^<$ for every ordering $<$ of the vertices of the polytope and identify the minimum value $f$. Now, $f$ must be the number of faces of the polytope $P$. By the claim an ordering $<$ is a good ordering if and only if $f^<=f$.
## The proof: part II
### Telling the faces
Let $F$ be a $k$-dimensional face of $P$ and let $H$ be its graph. $F$ is itself a $k$-dimensional polytope. What do we know about $H$? It is a k-regular graph. In fact it is an induced subgraph of $G(P)$. (Recall that $H$ is an induced subgraph of a graph $G$ if $H$ consists of a subset of the vertices of $G$ and all the edges of $G$ on these vertices.)
Claim: An induced connected k-regular subgraph H of G(P) is the graph of a face if and only if its vertices come first in some good ordering of $P$.
Proof: There are two directions to prove. We start by showing that the graph of a $k$-face $F$ is an induced connected $k$-regular subgraph of $G$ and there is a good ordering that starts with the vertices of $F$. Suppose that $F$ is a $k$-face of $P$. The graph of a face is always an induced subgraph of the graph of the polytope. Next, there is a linear functional $\phi$ on $R^d$ whose minimum on the polytope $P$ is attained precisely at $F$. (This is essentially the definition of a face: a face is the intersection of the polytope with a supporting hyperplane, and we can take $\phi$ to be a linear functional that vanishes on $H$ and hence also on $F$.) Now we can perturb $\phi$ a little so that it will attain different values on the vertices of $F$, and such that all these values are still smaller than the values on all other vertices.
Now suppose that $H$ is an induced connected $k$-regular graph and that $<$ is a good ordering such that the vertices of $H$ are smaller with respect to $<$ than all other vertices. Let $w$ be the largest vertex of $H$ w.r.t. $<$. Now $deg(w) = k$ and therefore $w$ is a local maximum in the $k$-face $F$ which is spanned by the $k$ edges of $H$ containing $w$. Since $<$ is a good ordering, $w$ is a the global maximum vertex of $F$. It follows that all the vertices of $F$ are vertices of $H$. (Because the vertices of $H$ are precisely all the vertices that are smaller than $w$ with respect to the ordering.) Therefore $G(F)$, the graph of $F$, is a $k$-regular subgraph of $H$. But note that the only $k$-regular subgraph of a connected $k$-regular graph is the graph itself! Therefore, the vertices of $H$ are the vertices of the face $F$. Sababa!
## Further results
There are various other results about reconstructions of the entire combinatorial structure of a $d$-polytope from partial information. My chapter on graphs and skeleta of polytopes in the “Handbook of discrete and computational geometry” contains several further such results. See also the new edition of Grunbaum’s classic book “Convex Polytopes“.
Let me mention two results: Hassler Whitney proved that the graph of a every 3-dimensional polytope determines its faces. The faces are simply the induced non-separating cycles. A theorem of Micha A. Perles asserts that the [$d/2$]-dimensional skeleton of a simplicial $d$-polytope determines the entire combinatorial structure.
## Friedman’s result
The algorithm to tell the polytope from a graph is simple but terrible. To start you have to consider all the orderings of the vertices of the polytope. All other theorems about reconstruction of the entire combinatorial structure of a polytope from partial information give simple polynomial algorithms.
Is there a polynomial time algorithm for telling the polytope from its graph? Eric Friedman proved a startling theorem asserting that there is a polynomial algorithm! (It deserves a whole separate post.)
## A dual form: puzzles.
Let $K$ be a pure $d$-dimensional simplicial complex. The puzzle of $K$, denoted by $puzz(K)$ is the graph whose vertices are the facets of $K$ (a facet =a maximal face) and two facets are adjacent if their intersection is $(d-1)$-dimensional. (The puzzle is precisely the same graph we associated to families of sets in the longish series of posts about Hirsch’s conjecture.)
Blind-Mani’s theorem stated in a dual form asserts that simplicial polytopes are determined by their puzzles and this is the way Blind and Mani stated it. We can talk also about puzzles where the pieces are themselves complicated polytopes which come with instructions how to glue two pieces locally, and ask if there is a unique global solution. (If the pieces are simplices, then all ways to glue them along facets are the same.) This was studied by Michael Joswig.
The first part of the proof described above asserts that for shellable simplicial complexes, the shelling orders, the total number of faces, and (by a slight extension of the proof) the f-vector are determined by the puzzle. In fact, the proof gives the following claim.
Claim: Let $K$ and $L$ be two pure $d$-dimensional simplicial complexes. Suppose that $puzz(K) =puzz(L)$ and that $K$ is shellable. Then $f_i(K) \ge f_i(L)$ for every $i$.
It is not always true that for shellable complexes the entire complex is detemined by the puzzle. Here are two 2-dimensional shellable complexes with the same puzzle. (A reminder of the notion of shellability is given below.)
## Problems:
### Spherical puzzles
Conjecture 1: A spherical puzzle always has a unique solution!
(By a spherical puzzle we mean a puzzle of a triangulation of a $d$-dimensional sphere.)
Conjecture 2: The puzzle of a $d$-dimensional Cohen-Macaulay simplicial complex $K$ determines the $f$-vector of $K$.
Conjecture 3:Let $K$ and $L$ be two pure $d$-dimensional simplicial complexes. Suppose that $puzz(K)=puzz(L)$ and that $K$ is Cohen-Macaulay. Then $f_i(K) \ge f_i(L)$ for every $i$.
The Cohen-Macaulay property can be regarded as an algebraic generalization of “shellability” (just like vanishing homology is an algebraic generalization of “collapsibility”). Triangulations of spheres are Cohen-Macaulay simplicial complexes.
### Characterizing the facets.
Conjecture 4: Let $G$ be the graph of a simple 4-dimensional polytope. Then a 3-regular induced subgraph $H$ of $G$ is the graph of a facet if and only if it is planar and non-separating.
An even stronger conjecture proposed by Perles (where $H$ is not assumed to be planar) was refuted by Christian Haase and Gunter M. Ziegler. Here is a modelby Nikolaus Witte of their example.
Conjecture 5: Let $G$ be the graph of a 4-dimensional spherical puzzle. Then a 3-regular induced subgraph of $G$ corresponds to the link of a vertex if and only if it is planar and non-separating.
We can also ask similar questions for puzzles of simplicial manifolds. Puzzles of general triangulated closed surfaces do not determine the triangulations; note that the puzzle of the 6-vertex triangulation of the real projective plane (which is the Peterson graph) has more automorphisms than the entire triangulation has!
And here is a conjecture (that we will not explain here) which is a little analogous to the conjecture that the f-vectors of Cohen-Macaulay complexes are determined by the puzzles.
### Kazhdan-Lusztig polynomials.
Conjecture (Lusztig): For intervals of the Bruhat order of Coxeter groups the Kazhdan-Lusztig polynomial can be read from the Bruhat interval. (In other words, two isomorphic intervals have the same Kazhdan-Lusztig polynomials.)
## Reminder: shellability
We say that a $d$-dimensional pure polyhedral complex is shellable if its maximal faces (facets) can be ordered by a sequence $F_1,F_t,\dots ,F_t$ such that for every $k$, $1 < k \le t$, the intersection of $F_k$ with the union of $F_1,\dots,F_{k-1}$ is homeomorphic to a $(d-1)$-dimensional ball or sphere. This condition implies that the entire complex is homeomorphic to a sphere or a ball. (We mentioned shellability here and here.)
For simplicial complexes, shellability is especially simple and becomes a purely combinatorial condition. The intersection of $F_k$ with the previous facets should be a (nonempty) union of some (or all) of its own facets. In a shelling sequence of facets, adding $F_k$ to the simplicial complex $K_k$ generated by $\{F_1,F_2,\dots,F_{k-1} \}$ has a very simple form. We add to $K_k$ a subset $S_k \subset F_k$ and all sets that contain $S_k$ which are contained in $F_k$. If $|S_k| =m$ we say that adding $F_k$ is a shelling step of type $m$.
If $P$ is a simplicial polytope then a shelling order on the facets of $P$ amounts to good ordering on the vertices of the dual polytope. The type of a facet in the shelling is the degree of the corresponding vertex in the good ordering.
The fact that the boundary complex every $d$-polytope is shellable was assumed in 19th century proofs of Euler’s theorem in high dimension but it was first proved by Bruggesser and Mani in 1970. Peter Mani once told me that he found a crucial ingredient of the proof in a dream.
## Reminder: background
A convex polytope is the convex hull of a finite set of points in an Euclidean space. A proper faceof a polytope P is the intersection of P with a supporting hyperplane. The empty face and P itself are regarded as trivial faces. The convex hull of a set of points which affinely span a d-dimensional space is a d-dimensional polytope or briefly a d-polytope. Faces of polytopes are themselves polytopes, a k-face is a short way to say k-dimensional face. 0-faces are called vertices, 1-faces are called edges and (d-1)-faces of a d-polytope are called facets. The set of faces of a polytope is a POSET (=partially ordered set) which is a “lattice”, ”atomic”, and “graded”. For a polytope P $f_k(P)$ denotes the number of k-faces of P. The f-vector of P is the vector $(f_{-1}(P),f_0(P),f_2(P), \cdots , f_d(P))$. Two d-polytopes P and Q are combinatorially equivalent if there is a bijection between the faces of P and the faces of Q which preserves the order relation.
The simplest d-polytope is the simplex: the convex hull of d+1 affinely independent points. The face lattice (=POSET of faces) of a simplex is just the Boolean lattice. A d-polytope is simplicial if all its proper faces are simplicies. A polytope is simple if its dual is simplicial or, equivalently, if every vertex is included in precisely d edges. A d-polytope is k-simplicial if all its k-faces are simplices. (Here d>k.) The d-cube is the convex hull of all vextors of length d with entries +1/-1.
The five open questions in this post belong to the combinatorial theory of polytopes that deal with the set of faces of polytopes. (There are also interesting metrical and arithmetic questions about polytopes.)
If P is a convex polytope which contains the origin in its interior, the polar dual of P is the set of all points y so that $<x,y> \le 1$ for every $x \in P$. There is a 1-1 order reversing bijection between the faces of P and the faces of its polar.
A polytope P is centrally symmetric if whenever x belongs to P so is -x.
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### 4 Responses to Telling a Simple Polytope From its Graph
1. Gil Kalai says:
Let me also mention that Friedman’s polynomial algorithm uses an earlier result by Joswig, Kaibel, and Korner who provided a “polynomial certificate” for the problem. Joswig, Kaibel, and Korner introduced a pair of combinatorial optimization problems which is used in Friedman’s proof.
2. Daniel Gertsch says:
The p-polytope proof can be considered different: A reduction of A to A’ is simply the reduction of G(p)=p :=: G(Q)=Q to G(x1,x2,x3,…)=x1,x2,x3,… and therefore n-1 in complexity and n-1=x for n=x+1 which means that the polytope is not determined if it determines not the figure with vertices 1 in addition. But this is a contradiction for P(X)=X and P(X)=X+1 for complexity n of P(X)=X or X+1 and invaluable for proof.
3. Pingback: Is Backgammon in P? | Combinatorics and more
4. Pingback: Remote Blogging: Efficiency of the Simplex Method: Quo vadis Hirsch conjecture? | Combinatorics and more
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http://soffer801.wordpress.com/2012/05/16/a-proof-from-erdos/ | #### Andy Soffer
UCLA Mathematics Ph.D Student
# A proof from Erdős
May 16, 2012
I have a friend from HCSSiM who reminded me of this proof, and a few others, that he put together for MIT Splash. Thanks Ben!
Let $\{p_1,\dots,p_n\}$ be the set of primes less than some number $N$. Pick a number $x$ which is less than $N$. We can break up $x$ into two parts:
• Let $a^2$ be the largest square dividing $x$
• Let $b$ be $x/a^2$.
Though I didn’t seem to mention factoring $x$, we are using unique factorization here in a fundamental way. We know a few things. We know $x$ is only divisible by the primes $p_1,\dots,p_n$, since all others are bigger than $N$. We know $b$ is not divisible by any perfect squares, lest $a^2$ not be maximal.
How many posibilities are there for $a$ and $b$? Since $a^2\le x\le N$, there are certainly no more than $\sqrt N$. For $b$, there can be no more than $2^n$, as there are $2^n$ choices for the set of primes which divides $b$.
Each number less than $N$ works like this, and so we get the inequality
$N\le 2^n\sqrt N$.
Solving for $n$, we get
$n\ge\frac12\log_2N$.
We now have a lower bound on the number of primes less than $N$. This lower bound grows towards infinity as $N$ gets big, and so therefore, the number of primes must do so as well. | {"extraction_info": {"found_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": 29, "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": 0.9547980427742004, "perplexity_flag": "head"} |
http://math.stackexchange.com/questions/175804/another-question-in-graph-theory | Another question in Graph Theory
Question: Let $H$ be a graph of order 10 such that $3\le d(v)\le5$ for each vertex $v$ in $H$ [where $d(v)$ is the degree of $v$]. Not every vertex is of even degree. No two odd-degree vertices are of the same degree. What is the size [number of edges] of $H$?
Answer: Size of $H$ is 20
let $V(H)=\{{a,b,c,d,e,f,g,h,i,j\}}$, then $$d(a)=3, d(b)=4, d(c)=4, d(d)=4, d(e)=4, d(f)=4, d(g)=4, d(h)=4, d(i)=4,$$ $$d(j)=5$$
Sum is 40
hence size is 20
Am I correct?
-
Yes, you only have the possible degrees 3, 4 and 5. There's at least one odd degree vertex (so there must be at least two), but no two odd-degree vertices can have the same degree, so there must be exactly one of degree 3 and one of 5. Then the rest must be degree 4. – Luke Mathieson Jul 27 '12 at 10:48
1
There's a step missing from the argument in Luke's argument. It's also necessary to show that such a graph does indeed exist; presumably by drawing it (or describing it fully). – user22805 Jul 27 '12 at 11:37
a little bit more practice will clear my concepts... thank you!!! – Intellectual_ Jul 27 '12 at 11:38
1 Answer
You've done all the hard work. Now all that remains is to show that such a graph exists.
Solvitur ambulando:
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http://physics.stackexchange.com/questions/56056/work-done-to-tighten-a-screw | # Work done to tighten a screw [closed]
We use a wrench to turn nuts on bolts because they require less force. Consider a hexagonal nut 1 cm in diameter. We can tighten this nut with one of two wrenches, wrench A with lever arm 10 cm and wrench B with lever arm 20 cm. Both wrenches have a very small mass, so you may neglect their masses in this problem. What is the ratio of the total work it would take to tighten the nut one full turn with wrench A to the total work it would take with wrench B?
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Could you show what you have tried already to solve the question?! – michielm Mar 6 at 11:21
Please see our homework policy. We expect homework problems to have some effort put into them, and deal with conceptual issues. If you edit your question to explain (1) What you have tried, (2) the concept you have trouble with, and (3) your level of understanding, I'll be happy to reopen this. (Flag this message for ♦ attention with a custom message, or reply to me in the comments with `@Manishearth` to notify me) – Manishearth♦ Mar 6 at 16:03
## closed as too localized by Mark Eichenlaub, Ϛѓăʑɏ βµԂԃϔ, akhmeteli, David Zaslavsky♦Mar 6 at 19:22
This question is unlikely to help any future visitors; it is only relevant to a small geographic area, a specific moment in time, or an extraordinarily narrow situation that is not generally applicable to the worldwide audience of the internet. For help making this question more broadly applicable, see the FAQ.
## 1 Answer
Assuming it requires the same torque to tighten the nut then the work done will be the same. Work done in a rotational situation is $$Wd = \int \tau \rm{d}\alpha.$$
Both wrenches will turn the same $\alpha$ in one rotation and so the work done is the same.
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http://www.physicsforums.com/showthread.php?s=60133cbd6442f915027a0433f5918350&p=4192777 | Physics Forums
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## What will the gravitational field inside the Earth be?
There is a question: If we dig a channel from the North pole of the Earth to the South pole of the Earth, then we release a ball into this channel from the North pole, what will be the motion of this be like?
I think this question is related to the inner gravitational field of the Earth
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
Recognitions: Science Advisor Ignoring friction and air resistance, etc., it would oscillate from pole to pole.
With a pole to pole length of time of about 42 minutes!
## What will the gravitational field inside the Earth be?
The force will change linearly with the distance from the centre.
This is connected with the shell theorem, i.e. there is no net force acting on a body inside an uniformly dense sphere.
As the ball gets deeper under the surface, the layers above it stop exerting gravitational force, and all that matters is the mass underneath.
That mass gets smaller with the third power of distance(volume of a sphere) as the ball goes down, but at the same time it is getting closer to the centre of mass attracting it, which force is inversely proportional to the distance squared(Newton's law of gravity).
$$F=\frac{GMm}{r^2}$$
$$M=ρV$$
$$V=\frac{4}{3}πr^3$$
$$F=\frac{\frac{4}{3}πr^3 ρGm}{r^2}$$
$$F=\frac{4}{3}πr ρGm$$
$$a=r \frac{4}{3} πρG$$
G is constant and we can assume the density of Earth ρ to be constant as well, so we have a linear relationship between acceleration and distance.
So the ball starts falling by being accelerated by g=9,81 m/s2, then the acceleration falls to 0 in the centre of the Earth just as the velocity reaches maximum.
Then it gets slowed down more and more the farther away from the centre it gets. As it reaches the surface on the other side, the velocity is again 0 and the acceleration is again g.
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Quote by Bandersnatch The force will change linearly with the distance from the centre.
That does assume uniform density, whereas the density is sure to be higher towards the centre. But on that assumption (and ignoring the spin of the Earth), the linear relationship would make it simple harmonic motion.
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Quote by haruspex That does assume uniform density, whereas the density is sure to be higher towards the centre.
It most certainly is. Which results in gravitational pull of Earth actually increasing almost half the way to the center, and only then beginning to drop.
Wikipedia has this graph that illustrates the effect.
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http://mathoverflow.net/revisions/34640/list | ## Return to Question
2 added 52 characters in body
Can someone please explain the difference between local rigidity and infinitesimal rigidity? Does either version of rigidity imply the other?
In particular, I'm thinking about Weil's rigidity theorem for hyperbolic metrics on manifolds of dimension $\geq 3$. I've seen it referred to as both local and infinitesimal, which further adds to my confusion about the distinction.
1
# Local vs. infinitesimal rigidity
Can someone please explain the difference between local rigidity and infinitesimal rigidity? In particular, I'm thinking about Weil's rigidity theorem for hyperbolic metrics on manifolds of dimension $\geq 3$. I've seen it referred to as both local and infinitesimal, which further adds to my confusion about the distinction. | {"extraction_info": {"found_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": 0.9238908290863037, "perplexity_flag": "middle"} |
http://mathhelpforum.com/algebra/203613-need-easy-way-solve-print.html | # need a easy way to solve it
Printable View
• September 17th 2012, 09:23 AM
Petrus
need a easy way to solve it
For which integers are x^5+4x^4+3x+1 divisible by 5?
i have solved this prob and there is none but i did it on a "hard way"
i did make it to x(x^3(x+4)+3)+1 and try put like x to 1,2,3 etc.. is there any easy way to solve it?
(sorry for bad english:/)
• September 17th 2012, 05:28 PM
Wilmer
Re: need a easy way to solve it
Wolfram gives this as solution:
solve x^5+4*x^4+3*x+1-5*k=0 for x - Wolfram|Alpha
What d'heck does it mean...Plato?
• September 17th 2012, 07:15 PM
Kiwi_Dave
Re: need a easy way to solve it
Let's plase all of the integers into one of 5 different classes:
x = 5a + 0, or
x = 5a + 1, or
x = 5a + 2, or
x = 5a + 3, or
x = 5a + 4
for each type of number I would try substituting for x in your original equation and determining if it is possible for the result to be divisible by 5. For example:
Let X = 5a + 0. Yor equation becomes:
$(5a)^5+4(5a)^4+3(5a)+1 = 5(5^4a^5+4.5^3.a^4+3a)+1$ and this cannot be divisible by 5 so the enture class of numbers x=5a+0 cannot give a result that is divisible by 5.
Let X = 5a + 1. Yor equation becomes:
$(5a+1)^5+4(5a+1)^4+3(5a+1)+1$
When you multiply out the first bracket only the last term is not divisible by 5, the last term is 1
When you multiply out the second bracket only the last term is not divisible by 5, the last term is 4
When you multiply out the third bracket only the last term is not divisible by 5, the last term is 3
so we can say that for some k
$(5a+1)^5+4(5a+1)^4+3(5a+1)+1 = 5k + 1+4+3+1=5k'+4$ and this cannot be divisible by 5 so the enture class of numbers x=5a+1 cannot give a result that is divisible by 5.
Now you can examine the remaining three classes of integers in the same way.
• September 17th 2012, 07:53 PM
MaxJasper
Re: need a easy way to solve it
$f({x})\text{=}x^5+4x^4+3x+1$
for k=1,2,3,....n
$\text{Mod}(f(k), 5) = \{4,3,2,1,1, 4,3,2,1,1, 4,3,2,1,1, ....\}$
• September 17th 2012, 08:18 PM
richard1234
Re: need a easy way to solve it
Just check x = 0,1,2,3,4. If x = k works, then any x that is congruent to k (mod 5) also works.
All times are GMT -8. The time now is 10:31 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": 5, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9261651635169983, "perplexity_flag": "middle"} |
http://math.stackexchange.com/questions/1901/why-is-fundamental-group-insufficient-to-classify-manifolds/1907 | # Why is fundamental group insufficient to classify manifolds?
I have heard that some issues in group theory prevent classifying all manifolds upto homotopy using the fundamental group invariant. Does anyone know what are those issues?
Thanks, K.
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3
– Soarer Aug 9 '10 at 13:51
6
I think that the OP is alluding to the insolvability of the word problem, and the impact that this has on the feasibility of giving an algorithmic classification of higher dimensional manifolds. See my answer below for more details. – Matt E Aug 9 '10 at 14:35
I also read this in the way Soarer does, maybe the OP could clarify. However Matt E makes a good and interesting point. – BBischof Aug 10 '10 at 2:27
## 5 Answers
This would have belonged as a comment to MattE answer, but I do not have the reputation to leave comments. The reason it is impossible to algorithmically classify topological manifolds (up to homotopy) is, as MattE very clearly explained, that the word problem for groups can be embedded in the question of classifying manifolds. Indeed, suppose that you want to decide if a finitely presented group $G$ is trivial. First, there are several constructions that yield a compact 4-manifold whose fundamental group is isomorphic to $G$; choose your favorite construction and let $M$ be the resulting manifold. Observe that the intersection pairing in homology is computable, so we may just as well assume that we know what this is for our manifold $M$. It turns out that the homotopy type of a simply connected manifold is completely determined by the intersection form on the second integral homology group (and that any intersection form can be realised). So if on the side we construct the simply connected manifold $N$ with intersection form coinciding with the one of $M$, then deciding whether the homotopy types of $M$ and $N$ coincide will decide if the fundamental group of $M$ is trivial or not, i.e. if $G$ is the trivial group or not.
There are some issues on how you compute the intersection pairing and how you find the manifold $N$ given the intersection form, and there are tricks to simplify all these questions, but I will not go into this, in the hope that what I wrote will suffice! Of course, if you want more details, feel free to ask, and I will try to answer!
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Thanks very much for this. – Matt E Aug 10 '10 at 20:34
+1: thanks for this very informative answer. – Pete L. Clark Aug 10 '10 at 21:31
I think that what you are thinking of is the insolvability of the so-called word problem in group theory. What this means is (among other things) that it is not possible to write an algorithm (i.e. a computer program) which can tell whether two groups, each presented by finitely many generators and relations, are or aren't isomorphic.
If you are given a manifold (presented say by gluing together various open balls), you can get such a presentation of its fundamental group, but (at least when the dimension is 4 or higher --- and I think that 4 is the right bound here) in fact you can get an arbitrary presentation in this way, and so the insolvability of the word problem means that, in practice, you don't have any way to figure out what the fundamental group of your manifold is (even whether or not it is trivial!)
As a consequence, you can't find an algorithm to determine whether two manifolds (of dimension 4 or higher) are homotopic (let along homeomorphic or diffeomorphic), since homotopic manifolds will have isomorphic $\pi_1$s, and so we could use such an algorithm to solve the word problem, by encoding two finitely presented groups as the $\pi_1$ of some manifolds, and then applying our hypothetical manifold algorithm. Following Pete Clark's comment below, I'm no longer sure about the correctness of the preceding statement. What is true is that there is no general algorithm to determine the fundamental group of a manifold (in dimension $\geq 4$), or even to determine if a manifold is simply connected. (Whether this actually prohibits a classification is not clear to me; with luck, someone with more expertise will weigh in.)
For this reason, a lot (although not all) investigations of higher dimensional manifolds restrict attention to simply connected manifolds. Then there is no obstruction to classification coming from issues of the uncomputatibility of $\pi_1$. Of course, as others have noted in their answers, there can be many non-diffeomorphic, non-homeomorphic manifolds, and non-homotopic manifolds of a given dimension, so the problem with $\pi_1$ is far from the only problem when it comes to classifying manifolds. But it is certainly one problem, and you are correct that (at least from the point of view I'm explaining here) it is a problem of group theory.
Added: See Damiano's answer to this question for a precise explanation of the relationship between the word problem for groups and the classification problem for manifolds.
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I am a little confused about your third paragraph (although I have heard similar claims many times, so I must just be missing something). How does an algorithm to classify manifolds up to diffeomorphism lead to an algorithm to test for isomorphism of finitely presented groups? You can find two compact $4$-manifolds with corresponding presentations. If they are diffeomorphic, you know that the groups are isomorphic. But maybe they are not diffeomorphic even though the fund. groups are isomorphic. What then? – Pete L. Clark Aug 10 '10 at 1:05
Dear Pete, This is a good question, to which I don't have a good answer. What does follow from insolvability of the word problem (or, more precisely, from unrecognizability of the trivial group), is that one cannot algorithmically determine if a general $n$-manifold (with $n \geq 4$) is simply connected. But now that you raise the question, I don't see immediately why this prohibits a classification (or at least, an algorithmic recognition procedure); it just means that any such classification can't use "is simply connected" as one of its determining attributes. – Matt E Aug 10 '10 at 2:17
Dear Matt,Thanks for your prompt answer, although I was hoping you would explain why I was missing something silly! I started thinking about this sort of thing after a talk that Bjorn Poonen gave at UGA (google "Poonen Undecidability Everywhere") as a possible attack on the open problem of decidability of isomorphism of varieties over Q-bar: I wanted (momentarily, at least) to use the etale fund. group to reduce to some profinite version of the word problem (not that I know that such a thing exists). But I ran into the same problem as above... – Pete L. Clark Aug 10 '10 at 3:28
For a better pair of examples take the once punctured torus and a pair of pants. Both have fundamental groups free of rank 2. They are both homotopy equivalent to a figure 8, but as manifolds they are not homeomorphic since one has 3 boundary components and the other has one boundary component.
There are simply connected closed manifolds of dimension 4 and higher that are not homeomorphic. In even dimensions they can be distinguished by their middle dimensional homology intersection forms.
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Well, there are no reasons to expect that manifolds with equal π1 are homotopy equivalent, and in general, they indeed aren't. Even for closed manifolds of fixed dimension: say, both CP2 and S4 have trivial fundamental group.
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The following (famous) question is perhaps in the spirit of the OP's question:
If $M$ and $N$ are closed aspherical $n$-manifolds with isomorphic fundamental groups, are $M$ and $N$ homeomorphic? (This is known as the Borel Conjecture.)
Scott Carter's example illustrates the need for the assumption that the manifolds are closed. But the question above suggests that perhaps the fundamental group is sufficient to characterize the topological type of a closed aspherical manifold.
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http://stochastix.wordpress.com/category/mathematics/control-theory-mathematics/ | # Rod Carvalho
## Archive for the ‘Control Theory’ Category
### Sailing in state space
April 11, 2011
Consider the following controlled (autonomous) dynamical system [1]
$\dot{x} (t) = f (x (t), u (t))$
where $x : [0, \infty) \to \mathbb{R}^n$ is the state trajectory, $u : [0,\infty) \to \mathcal{U}$ is the control input history, $\mathcal{U} \subseteq \mathbb{R}^m$ is the set of admissible control inputs, and $f : \mathbb{R}^n \times \mathcal{U} \to \mathbb{R}^n$ is a (known) vector field.
If we know the control input $u (t)$ for all $t \geq 0$, then we can integrate the ODE above with initial condition $x(0) = x_0$ and obtain the state trajectory $x : [0, \infty) \to \mathbb{R}^n$, as follows
$x (t) = x_0 + \displaystyle \int_{0}^{t} f \left(x (\tau), u (\tau) \right) \mathrm{d} \tau$
which can be represented pictorially by a single streamline in $\mathbb{R}^n$
I am (quite explicitly) alluding to fluid flow. Just like a cork on the ocean will follow a certain path depending on the velocity field (i.e., “ocean currents”), a point in state space will flow along a certain streamline.
__________
Let us fix the control input, i.e., $u (t) = \bar{u}$ for all $t \geq 0$, and define
$\begin{array}{rl} v^{(\bar{u})} : \mathbb{R}^n & \to \mathbb{R}^n\\ x & \mapsto f(x, \bar{u})\\\end{array}$
For the sake of simplicity, let us consider $\mathcal{U} = \{-1, +1\}$. Since $\bar{u}$ can only take two values, we have two vector fields, $v^{(+1)}(x) := f(x, +1)$ and $v^{(-1)}(x) := f(x, -1)$. Integrating the following ODEs
$\begin{array}{rl} \dot{x} (t) &= v^{(+1)} ( x(t) )\\ \dot{x} (t) &= v^{(-1)} ( x(t) )\\\end{array}$
for various initial conditions, we obtain two families of streamlines, as depicted below
For each fixed control input and collection of initial conditions, we will have a family of streamlines in state space.
Imagine that we have the following control input history
$u (t) = \begin{cases} +1, & t \in [0, t^{\star})\\ -1, & t \geq t^{\star}\\\end{cases}$
where we toggle the control input at time $t^{\star} > 0$, thus interrupting the flow along a “blue” streamline and initiating the flow along a “pink” streamline. Pictorially, we have
If the control input is fixed, say, $\bar{u} = +1$, then given an initial condition $x(0)$, we will flow along the “blue” streamline that passes through $x(0)$. This streamline is $1$-dimensional and does not allow us to “explore” the $n$-dimensional state space. However, if we switch between $\bar{u} = +1$ and $\bar{u} = -1$, then we are able to “travel” to other regions of the state space. Allowing the control input to take values in $\mathcal{U} = [-1,+1]$ would give us even more freedom.
Lastly, we arrive at a most childish idea: in some cases, controller design can be viewed as shaping the streamlines differently in different parts of the state space. This is childish because it is purely conceptual. One obviously cannot design optimal controllers by doodling with crayons.
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References
[1] Hassan K. Khalil, Nonlinear Systems, 3rd edition, Prentice Hall.
Tags:Control Theory, Differential Equations, Dynamical Systems, ODEs, Ordinary Differential Equations
Posted in Control Theory, Dynamical Systems | 2 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": 29, "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": 0.8986668586730957, "perplexity_flag": "head"} |
http://physics.stackexchange.com/questions/15899/why-the-principle-of-least-action/15958 | # Why the Principle of Least Action?
I'll be generous and say it might be reasonable to assume that nature would tend to minimize, or maybe even maximize, the integral over time of $T-V$. Okay, fine. You write down the action functional, require that it be a minimum (or maximum), and arrive at the Euler-Lagrange equations. Great. But now you want these Euler-Lagrange equations to not just be derivable from the Principle of Least Action, but you want it to be equivalent to the Principle of Least Action. After thinking about it for awhile, you realize that this implies that the Principle of Least Action isn't really the Principle of Least Action at all: it's the "Principle of Stationary Action". Maybe this is just me, but as generous as I may be, I will not grant you that it is "natural" to assume that nature tends to choose the path that is stationary point of the action functional. Not to mention, it isn't even obvious that there is such a path, or if there is one, that it is unique.
But the problems don't stop there. Even if you grant the "Principle of Stationary Action" as fundamentally and universally true, you realize that not all the equations of motions that you would like to have are derivable from this if you restrict yourself to a Lagrangian of the form $T-V$. As far as I can tell, from here it's a matter of playing around until you get a Lagrangian that produces the equations of motion you want.
From my (perhaps naive point of view), there is nothing at all particularly natural (although I will admit, it is quite useful) about the formulation of classical mechanics this way. Of course, this wouldn't be such a big deal if these classical ideas stayed with the classical physics, but these ideas are absolutely fundamental to how we think about things as modern as quantum field theory.
Could someone please convince me that there is something natural about the choice of the Lagrangian formulation of classical mechanics (I don't mean in comparison with the Hamiltonian formulation; I mean period), and in fact, that it is so natural that we would not even dare abandon these ideas?
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2
– Mark Eichenlaub Oct 19 '11 at 4:16
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Lagrange's equation was originally discovered without the Principle of Least action, and can be derived directly from the Newtonian formulation of mechanics. Goldstein does it that way (and has a discussion of the history of stationary principles in classical physics). – dmckee♦ Oct 19 '11 at 15:36
2
While I can see how this could be considered a duplicate of the other question, it's well written and its getting a good response so I don't feel particularly compelled to close it at this point. – David Zaslavsky♦ Oct 19 '11 at 22:12
– Luboš Motl Oct 22 '11 at 7:30
## 6 Answers
Could someone please convince me that there is something natural about the choice of the Lagrangian formulation...
If I ask a high school physics student, "I am swinging a ball on a string around my head in a circle. The string is cut. Which way does the ball go?", they will probably tell me that the ball goes straight out - along the direction the string was pointing when it was cut. This is not right; the ball actually goes along a tangent to the circle, not a radius. But the beginning student will probably think this is not natural. How do they lose this instinct? Probably not by one super-awesome explanation. Instead, it's by analyzing more problems, seeing the principles applied in new situations, learning to apply those principles themselves, and gradually, over the course of months or years, building what an undergraduate student considers to be common intuition.
So my guess is no, no one can convince you that the Lagrangian formulation is natural. You will be convinced of that as you continue to study more physics, and if you expect to be convinced of it all at once, you are going to be disappointed. It is enough for now that you understand what you've been taught, and it's good that you're thinking about it. But I doubt anyone can quickly change your mind. You'll have to change it for yourself over time.
That being said, I think the most intuitive way to approach action principles is through the principle of least (i.e. stationary) time in optics. Try Feynman's QED, which gives a good reason to believe that the principle of stationary time is quite natural.
You can go further mathematically by learning the path integral formulation of nonrelativistic quantum mechanics and seeing how it leads to high probability for paths of stationary action.
More importantly, just use Lagrangian mechanics as much as possible, and not just finding equations of motion for twenty different systems. Use it to do interesting things. Learn how to see the relationship between symmetries and conservation laws in the Lagrangian approach. Learn about relativity. Learn how to derive electromagnetism from an action principle - first by studying the Lagrangian for a particle in an electromagnetic field, then by studying the electromagnetic field itself as described by a Lagrange density. Try to explain it to someone - their questions will sharpen your understanding. Check out Leonard Susskind's lectures on YouTube (series 1 and 3 especially). They are the most intuitive source I know for this material.
Read some of the many questions here in the Lagrangian or Noether tags. See if you can figure out their answers, then read the answers people have provided to compare.
If you thought that the Lagrangian approach was wrong, then you might want someone to convince you otherwise. But if you just don't feel comfortable with it yet, you'd be robbing yourself of a great pleasure by not taking the time to learn its intricacies.
Finally, your question is very similar to this one, so check out the answers there as well.
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Beautiful response. It reminds me of the Rambam's response to the request to explain the Torah while standing on one foot. "Do not do to others what is hateful to you. The rest is commentary. Now go study." – AdamRedwine Oct 19 '11 at 13:28
A philosophical answer but a good one. ;-) – Luboš Motl Oct 22 '11 at 7:32
The intuition for the Lagrangian principle comes specific applications of Newton's laws, especially reversible systems with constraints, like nonspherical particles rolling along complicated surfaces. Newton's formulation of Newton's laws was not the end of the story, because there was more structure in the solutions of these types of problems than that which Newton made obvious.
One thing left unsaid by Newton is conservation of energy. Elastic processes are more fundamental than inelastic ones. But energy conservation is only part of the story. Suppose you have a bunch of masses connected by springs, and one of them is attached to a double-pendulum. You could theoretically have energy conservation in such a system by having all the energy leak out of the masses on the springs and go into the double pendulum. Perhaps every frictionless motion of the springs eventually settles all the energy into a single mode.
Your intuition is probably rebelling, telling you "that's infinitely unlikely! How could the pendulum move around and not set the springs vibrating!" But there is nothing in Newton's laws by themselves, even with the principle of conservation of energy, that prevents this sort of concentration of energy. But the solutions do not exhibit such phenomena, and there must be a reason why.
This intuition tells you that a perfect frictionless mechanical system is more than energy-conserving, it must conserve some notion of "motion-volume", so that if you alter the initial state by a certain amount, the final state should alter the same way. It can't concentrate all motion into one mode. This principle is the principle of conservation of phase-space volume, or the conservation of information. If all the motion got concentrated into one mode, the information about where everything was would have to get absurdly compressed into a tiny region of the phase space, the space of all possible motions.
The conservation of information is just about as fundamental as Newton's laws of motion--- it is revealing new facts about nature which are essential for the description of statistical and quantum systems. But it is nowhere to be found in Newton's formulation, because it does not follow from Newton's laws alone, even with the principle of conservation of energy added.
So you need to understand what type of law will give a law of conservation of information. There are two paths to go down, and both lead to the same structure, but from two different points of view, local in time and global in time.
One path is Hamiltonian: you consider formulating the law of motion as a set of symplectic equations for the position and momentum. This formulation clearly separates between reversible and irreversible dynamics, because it only works for reversible. It also explains the fundamental mathematical structure behind reversible classical mechanics, the symplectic geometry. The volume of symplectic geometry gives the precise law of information conservation, and further, the geometrical structure of systems with multiperiodic solutions, the integrable systems, is made clear.
But this point of view is centered on a time-slicing--- it describes things going from one instant of time to another. This is not playing very nice with relativity. So you also want to think about the solution globally, and consider the space of all solutions as the phase space. The initial position and velocities are good coordinates, and intuitive ones, because they determine the future. But if you want a global picture, you want coordinates which are symmetric between the final and initial state, since the dynamics are reversible. An explicit revesible description should treat the initial time and final time symmetrically. So you can use the initial positions and final positions, which also, generically, away from certain bad choices, determine the motion.
For these types of coordinates on phase space, you give the dynamical law as a condition on the trajectory between the intial and final positions. The condition should not be stated as a differential equation, because such a description is unnatural for boundary conditions of this sort. But when you have an action principle, you determine the trajectory by extremizing the action between the end points, you automatically have a notion of phase space volume, which is intuitive--- the phase space volume is defined by the change in the action of extremal trajectories with respect to changes in the initial velocities. This volume is the same as for the changes of the extremal trajectories with respect to changes in the final velocities. This is a straightforward consequence of the equivalence of Lagrangian and Hamiltonian formulation.
The full justification for both principles comes only with quantum mechanics. There you learn that the least action principle is a geometric optics Fermat principle for matter waves, and it is saying that the trajectories are perpendicular to constant-phase lines. But historically, the Lagrangian formulation was recognized to be more fundamental a century before Hamilton conjectured that classical mechanics was a wave mechanics, and this was many decades before Schrodinger. Still, with our modern point of view, it does not hurt to learn the quantum version of these formulations first, and it certainly provides a more solid motivation than the heuristic considerations I gave above.
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OP wrote:
As far as I can tell, from here it's a matter of playing around until you get a Lagrangian that produces the equations of motion you want.
Too often, as a student, one is only shown how to derive Newton's 2nd law from Euler-Lagrange equations by postulating some particular Lagrangian $L$. If one believes that Newton's laws are more natural (within the context of non-relativistic classical mechanics), then perhaps it would be more satisfying to see a derivation in the other direction, i.e. to see Euler-Lagrange equations derived from Newton's laws. This is e.g. done in the first chapter of Herbert Goldstein, Classical Mechanics. (An important element in this derivation is to show that a large class of constraint forces do no virtual work, leading to D'Alembert's principle.)
Throwing a wrench into the works, let me finally mention that there exist equations of motion that have no action principle.
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The main point of Lagrangian formulation of classical mechanics was to get rid of the constraint relations completely so that one does not have to bother about them while calculating anything (see this answer of mine. Remember all the valuable symmetries of the physical situation is automatically inbuilt in this formulation of mechanics.
Now after having such a powerful technique at our disposal, it is natural to ask the simple question. Can the classical theory of electromagnetism i.e. Maxwell's equations, be expressed as Euler-Lagrange equations by suitably defining a Lagrangian of the electromagnetic field, so that we may readily get all those beautiful results of the structure of this formulation (for example avoid annoying field constraints)? The answer is 'yes', provided we suitably define a Lagrangian.
In fact, it is observed that any general classical field equations can be expressed by Euler Lagrange equation of motion and in each case you need to define a Lagrangian (you get a corresponding "Action") and get everything for free.
This Lagrangian approach is so powerful that even quantum field theories exploit them fully and almost all modern theories of physics exploit them in some way.
However, it is not at all clear whether this approach is completely general. Can't it be the case for a future theory that the equations of the theory can't be expressed in Lagrangian formulation? I asked this question here. See some good answers.
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The Lagrangian is just a (special, functional kind of) anti-derivative of an equation of motion. In simple cases, the equation of motion of classical particles is a function $F(x,\dot x,\ddot x,...)$ of the positions, velocities, accelerations, etc. of the particles, for which the equation $F(x,\dot x,\ddot x,...)=0$ determines the trajectory. When a function $F(...)$ admits a Lagrangian anti-derivative, the maxima and minima of the Lagrangian anti-derivative determine where to find the zeros of the function $F(x,\dot x,\ddot x,...)$, the derivative of the Lagrangian.
There is not necessarily anything fundamental or natural about a Lagrangian. The Lagrangian has many formal properties that often make it extremely useful, and quite often make it rather simple or beautiful. We like beautiful things. Appreciating beauty is a tricky thing, to some extent a matter of experience, to some extent a matter of just seeing it. Mark and Anna are both pretty good guides.
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What does "principle" mean? It has many definitions but in the context of this discussion it has the power that "axiom" has in mathematics: a very basic assumption, which, if changed, the whole construct theory shifts or is destroyed.
Physics has adopted this from the geometric observation that : the shortest distance between two points is a straight line which logically led to "minimum time taken" and the search for the shortest distance when unknown.
that it is so natural that we would not even dare abandon these ideas?
If somebody brilliant enough can come out with another principle for the system of mathematical formulation of classical mechanics and subsequently quantum field theory that does not follow least action but incorporates perfectly the large data base / existing equations etc there is no problem. It might even be adopted generally if it could predict new spectacular results. Otherwise, from economy of mental effort( another principle :) ) the system developed under the principle of least action will still prevail.
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http://mathoverflow.net/questions/49946/principal-curvatures-and-curvature-directions-closed | ## Principal curvatures and curvature directions [closed]
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Last week I considered again principal curvature (pc) and principal curvature directions (pcd) of a, for the sake of simplicity, 2-manifold embedded in 3-space. In this simple case, the pc and pcd of at a point are the eigenvalues and eigenvectors of the shape operator. The magnitude of the pc's corresponds to the minimal and maximal normal curvature at the point. My question, however, is:
What does the principal curvatures direction magnitude represents?
In the textbooks I looked up in (Kühnel and do Camro) I couldn't find a reference to the principal curvature direction's magnitude. Is there something known about this? Is it something basic (maybe even from linear algebra)?
Edit 1: A somewhat more general, but related, question is:
What is the geometrical meaning of an eigenvector's magnitude?
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Sorry, this really should be over at math.stackexchange.com. Closing the gate after the horse has bolted, etc, but we've belatedly closed the question. – Scott Morrison♦ Dec 21 2010 at 6:34
## 3 Answers
Unless there's an additional constraint on the defintion of the principal curvature directions, being just defined as eigenvectors means their magnitude is arbitrary and so meaningless.
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That is my understanding as well. However, is it possible that some geometrical meaning of the pcd's magnitude does exist? – Dror Atariah Dec 20 2010 at 10:59
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How could there be any meaning? If $\vec{v}$ is a principal curvature direction then so is $2\vec{v}$. So, in order for there to be a geometrical meaning you would have to modify the definition so that the magnitude isn't allowed to be arbitrary. – Zhen Lin Dec 20 2010 at 11:54
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While all of the above remarks about the principal curvature magnitudes being arbitrary are correct, I think it is fairly customary for differential geometers to think of the eigenvectors of the shape operator as being the principal axes of the the curvature ellipsoid, and so giving them lengths equal to the principal curvatures. Of course this only makes sense on the complement of the umbilic points.(Recall that the umbilics are the points where the principal curvatures are equal, so that at these points the principal directions are not well-defined.)
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In general, the concept of "eigenvector" is slightly misleading. The fundamental concept is eigenspace. However, eigenspaces of dimension greater than 1 are generally considered pathological and rather a nuisance. Mostly we would rather have one-dimensional eigenspaces, and for these any convenient nonzero element---an eigenvector--- will serve as a representative. It would be pedantically correct but encumbering to insist on talking about one-dimensional eigenspaces rather than eigenvectors. The arbitrariness of eigenvectors becomes clearer when we really have to deal with a multidimensional eigenspace.
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http://math.stackexchange.com/questions/142015/is-my-riemann-sum-correct-example-2?answertab=votes | # Is my Riemann Sum correct (Example # 2)? [duplicate]
Possible Duplicate:
Is my Riemann Sum correct?
This is my second attempt, the answer seems rather odd so I thought I would have it checked as well. For the integral: $$\int_{-5}^{2} \left( x^{2} -4 \right) dx$$
My calculations:
$$\begin{align*}\Delta x &= \frac7n\\\\ x_i &= -5 + \frac{7i}n\\\\ f(x_i) &= 21 - \frac{70i}{n} + \frac{49i^2}{n^2} \\\\ A&=-738 \end{align*}$$
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1
The answer is certainly wrong. How did you get $A$? – Arturo Magidin May 7 '12 at 1:30
But +1 for picking up on the $\LaTeX$. – Brian M. Scott May 7 '12 at 1:47
## marked as duplicate by Peter Tamaroff, Thomas, Noah Snyder, Norbert, draks ...Oct 9 '12 at 18:21
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
## 1 Answer
The preliminary computations are fine. That means that the $n$th right hand Riemann sum will be: $$\begin{align*} \text{RHS} &= \sum_{i=1}^n f(x_i)\Delta x\\ &= \sum_{i=1}^n\left(21 - \frac{70i}{n} +\frac{49i^2}{n^2}\right)\frac{7}{n}\\ &= \frac{7(21)}{n}\sum_{i=1}^n1 - \frac{7(70)}{n^2}\sum_{i=1}^n i + \frac{7(49)}{n^3}\sum_{i=1}^ni^2\\ &= \frac{147}{n}(n) - \frac{490}{n^2}\left(\frac{n(n+1)}{2}\right) + \frac{343}{n^3}\left(\frac{n(n+1)(2n+1)}{6}\right)\\ &= 147 - 245\frac{n^2}{n^2+n} + \frac{343}{6}\frac{n(n+1)(2n+1)}{n^3}, \end{align*}$$ using the formulas that say that $$\begin{align*} 1+2+3+\cdots + n &= \frac{n(n+1)}{2}\\ 1^2+2^2+3^2+\cdots+n^2 &= \frac{n(n+1)(2n+1)}{6}. \end{align*}$$ Now, if we take the limit as $n\to\infty$, we have $$\begin{align*} \lim\limits_{n\to\infty}\frac{n^2}{n^2+n} &= 1\\ \lim\limits_{n\to\infty}\frac{n(n+1)(2n+1)}{n^3} &= 2, \end{align*}$$ which means the area should be $$147 -245 +\frac{343}{3} = -98 + 114+\frac{1}{3} = 16+\frac{1}{3} = \frac{49}{3}.$$
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http://www.physicsforums.com/showthread.php?p=1197128 | Physics Forums
## Compton Generator
*I don't know if this belongs to homework section or not.
I just found the description in my book interesting and want to know more. If this is inappropriate, please move it to homework section.
Compton Generator is basically a device invented by Compton which is used to measure the Coriolis Force (a fictitious force in non-inertial frame). As described in my physics book, it works like this:
First, you get a torus filled with water and flat on the ground (the water is initially non-rotating in the frame of the earth), then you flip the device 180 degrees through the east-west axis.
My physics book claims that after the flip, the water in the torus begins to rotate. The book even gives a explicit formula for the speed after the flip, which is only dependent on the radius of the torus and the colatitude (assuming the cross section of the torus is very very small comparing to the torus as a whole).
my understand of how the device works is that when one turns the torus, the water gains velocity in the turning direction. The Coriolis Force then changes the direction of that vector, resulting in a net tangential velocity.
The device seems very interesting... and I've been trying to figure out the mathematics behind it... I tried to assume constant angular velocity when flipping the torus, but that integral turns out quite messy and is dependent on the angular velocity. I couldn't find any work-energy relationship either, since Coriolis Force is absolutely non-conservation.
there might be some kind of parallelism between magnetic flux and this. But that seems to be true only if an initial "current" (flow of water) is presented (similar to how energy is magnetic moment dot magnetic field). I can express the torque in terms of the flow in the torus, radius and such but that doesn't seem to help much (since it is dependent on the flow/speed of the water)
The speed of the water after 180 flip is as following:
$$v=2\Omega R \cos\theta$$
where Omega is the angular velocity of the earth, theta is the colatitude.
I wonder if the equation in my book really works. And what about the equation for speed for turning any amount of degrees (v in terms of the angular displacement). And what happen if the shape is not a torus, but a random closed loop (smooth) of water? And what if you turn the torus the other way around 180 degrees? (I suppose the water will stop rotating)
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
Come on... anyway has any idea? I have been trying to figure this out for a couple of days... it is really frustrating how I cannot understand where the equation comes from.. How can I derive the equation for speed of the water after the 180-degrees flip?
I think the correct equation shouldn't contain the colatitude dependence. The coriolis force has equal magnitude everywhere on a circle (center to circumference), so a measurement at a particular (co)latitude (say the equator) should produce the same result as if it could be repeated with the same orientation closer to (or along) the polar axis. For this particular device a 180 degree flip seems to return it to its starting condition, that is, it doesn't appear to matter whether it is oriented initially flat on the ground (of the earth's sphere) or tilted arbitrarilly (eg. so as to be flat if the earth were a cylinder with axis aligned to rotation). If so then it should produce the same measurement at any latitude.
## Compton Generator
Finally, I seem to have reasoned out the equation. It turns out that it doesn't really have anything to do with Coriolis Force.
when viewed in the inertial that shares the same center as the earth, the equation can be easily derived.
let the center of the torus be at latitude $$\theta$$
then the northern end of the torus would be at latitude
$$\theta+\epsilon$$
and the southern end would be at latitude
$$\theta-\epsilon$$
the torus is rotating as seen in the inertial frame. When the torus is quickly flipped, the speed of the water in either end is not affected. However, when the southern end is brought to the position of the northern end, the speed of the water is not rotating fast enough to remain stationary in the rotating frame. the difference in speed would be the difference in speed between the northern end and the southern end before the flip. That difference is:
$$R_e\Omega\sin(\theta+\epsilon)-R_e\Omega\sin(\theta-\epsilon)$$
(R_e is the radius of the earth)
since,
$$\vec{v}=\vec{\Omega}\times\vec{r}$$
simplify the expression, the difference in speed is:
$$\Delta v=2R_e\Omega\cos\theta\sin\epsilon$$
a little bit of trig reveals that
$$\sin\epsilon=\frac{R}{R_e}$$
so, in the rotating frame, the speed of rotation is exactly the difference in speed, hence:
$$v=2R\Omega\cos\theta$$
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| | Introductory Physics Homework | 4 | | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 9, "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": 0.9393356442451477, "perplexity_flag": "middle"} |
http://math.stackexchange.com/questions/260937/does-this-set-contain-the-given-integer | # Does this set contain the given integer?
Let $a = 31$. Consider the set of integers $T = \{a, 8a, 8^{2}a, 8^{3}a, \cdots \}$. Does $T$ contain the integer:
999999999900000000000090909090000000000000000008?
So far I've deduced that if we work mod 9 that the set $T$ can be reduced to a reduced residue system modulo 9 by using 8 as a root. Additionally, because $(a, 9) = (31, 9) = 1$ we can eliminate $a$ from the elements of $T$.
Continuing, remark that $ord_{9}(2) = \phi(9) = 6$ and $ord_{9}(2^{3}) = \frac{6}{3} = 2$. Taking $8^{k} \mod 9$ for $k \in \mathbb{Z^{+}}$ we see that we get the set of reduced residues $\{8, 1\}$. 8 is an element of this set so $T$ contains $999\ldots 0008$.
That's where I am so far but I have a feeling that I went wrong early on in this one.
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1
Can whoever just downvoted my question explain the downvote? – user39898 Dec 17 '12 at 19:22
I don't know, but maybe is because it looks too exaggerated to be an actual question and you show no own work on it? This can be solved sharing some of your own efforts, insights and self work in this problem and, perhaps, adding some background of where/how did this question come from. – DonAntonio Dec 17 '12 at 19:25
I suppose...it was downvoted like 10 seconds after I uploaded it though so whoever downvoted it didn't even bother to think about it. – user39898 Dec 17 '12 at 19:26
## 2 Answers
Hint
$$8 \equiv -1 \pmod 9$$ $$31 \equiv 4 \pmod 9$$
Thus
$$8^k a \equiv ??? \pmod 9$$
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I'm sure you're on the right track but could you explain where you're coming from? I'm not seeing your thought process here. – user39898 Dec 17 '12 at 19:25
2
@decave Your number is huge, but the "huge" part contains only 9's and 0's, so it is natural to look at it mod 9... Modulo 9, the question you ask implies "Can $8^ka \equiv 8 \pmod 9$?" – N. S. Dec 17 '12 at 19:29
@N.S: Amusing! :) What if the number was 999999999900000000000090909090000000000000000004? – Isomorphism Dec 17 '12 at 19:30
So we know that $9 \mid 99999\ldots00090909\ldots 0000$? I guess you could just factor out a 9....wow...I really need to learn to work better under pressure haha. – user39898 Dec 17 '12 at 19:31
1
@Isomorphism That's why I said implies... If the number was what you said, that one needs to find a differenta approach, because that is a DIFFERENT problem ;) – N. S. Dec 17 '12 at 19:31
show 5 more comments
The largest power of $8$ that divides our number is $8^1$. Our number is not $(8)(31)$.
Note that the only thing that was used is that our number ends in $008$.
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Also, the number is not divisible by $31$. It is, in fact, $8 \times 220654023912941 \times 566497713347021146672084609906661$ where the last two factors are primes. – Robert Israel Dec 17 '12 at 19:49
5
But $31$ is further than I can comfortably count, even after taking my shoes off. – André Nicolas Dec 17 '12 at 19:53 | {"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": 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": 0.9571415185928345, "perplexity_flag": "middle"} |
http://quant.stackexchange.com/questions/519/missing-step-in-stock-price-movement-equations/520 | # Missing step in stock price movement equations
Assuming a naive stochastic process for modelling movements in stock prices we have:
$dS = \mu S dt + \sigma S \sqrt{dt}$
where S = Stock Price, t = time, mu is a drift constant and sigma is a stochastic process.
I'm currently reading Hull and they consider a simple example where volatility is zero, so the change in the stock price is a simple compounding interest formula with a rate of mu.
$\frac{dS}{S} = \mu dt$
The book states that by, "Integrating between time zero and time T, we get"
$S_{T} = S_{0} e^{\mu T}$
i.e. the standard continuously compounding interest formula. I understand all the formulae but not the steps taken to get from the second to the third. This may be a simple request as my calculus is a bit rusty but can anyone fill in the blanks?
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Is this really $\cdots + \sigma S \sqrt{dt}$? Maybe this doesn't matter, but I would assume that it is $\cdots + \sigma S dB_t$. The $\sqrt{dt}$ could come from a discrete simulation of the path of $S_t$ where $\sqrt{t}$ is the volatility of $dB_t$. – Richard Dec 13 '12 at 9:24
## 2 Answers
This is the separable differential equation for simple continuous compounding!
See this very accessible article for a step-by-step derivation (esp. under continuous compounding):
http://plus.maths.org/content/have-we-caught-your-interest
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Do his first step first; integrate both sides:
$$\displaystyle \ \ \int_0^T \frac{dS(t)}{S(t)} = \mu T - 0 \,\,\,\,\,\,\,\,\,\,\,(1)$$
With zero diffusion, we know that $\langle S_.\rangle_t = 0$. Therefore, by applying Ito's lemma (or actually normal calculus):
$$d\ln{S(t)} = \frac{1}{S(t)}dS(t)\,\,\,\,\,\,\,\,\,\,\,(2)$$
Sub this into $(1)$:
$$\displaystyle \ \ \int_0^T d\ln{S(t)} = \mu T$$
$$\therefore \ln{S(T)} = \ln{S(0)} + \mu T$$
$$\therefore e^{\ln{S(T)}} = e^{\ln{S(0)}}e^{\mu T}$$
$$\therefore \boxed{S(T) = S(0)e^{\mu T}}$$
- | {"extraction_info": {"found_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": 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": 0.9271895289421082, "perplexity_flag": "middle"} |
http://math.stackexchange.com/questions/169623/how-complex-exponential-converges-and-sum-of-exponents-rule-holds/169663 | # How complex exponential converges and “sum of exponents” rule holds
1. How is it the complex exponential converges for any value of $z$ in the complex plane? $$e^{z} = 1 + \frac{z}{1!} + \frac{z^2}{2!} \cdots\cdots$$
2. How is it the "sum of exponents" rule holds for complex exponential, that is $e^{w}e^{z} = e^{w+z}$?
...using only the definition of $e^z$?
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Ratio test, binomial theorem. – anon Jul 11 '12 at 20:01
Hint (for the second part): If you multiply the two power series together and collect up terms, and use the binomial theorem, you should be able to prove this. It can be justified by absolute convergence. – John Wordsworth Jul 11 '12 at 20:02
i know 2nd expression can be proved by binomial theorem, but failed to do so. and for the first one, if only the postulates of "complex arithmetic" is given, how it converges? – Aftnix Jul 11 '12 at 20:03
Latex output is not correct in my browser, sometimes its ok, sometimes it not. – Aftnix Jul 11 '12 at 20:05
@Aftnix If you are familiar with the proof of the ratio test in the real case, you should see that it carries over easily for complex series. – Tim Duff Jul 11 '12 at 20:17
## 3 Answers
$$e^z=\sum_{n=0}^\infty\frac{z^n}{n!}$$ Putting $$a_n:=\frac{1}{n!}\Longrightarrow R:=\frac{1}{\lim_{n\to\infty}\sqrt[n]{|a_n|}}=\lim_{n\to\infty}\sqrt[n]{n!}=\infty$$
So the series has infinite convergence radius and is thus absolutely convergent for any $\,z\in\Bbb C\,$ , and from here it follows that $$e^{w+z}=\sum_{n=0}^\infty\frac{(w+z)^n}{n!}=\sum_{n=0}^\infty\sum_{k=0}^n\binom{n}{k}\frac{w^kz^{n-k}}{n!}=\sum_{n=0}^\infty\sum_{k=0}^n\frac{w^kz^{n-k}}{k!(n-k)!}=$$ $$=\sum_{k=0}^\infty\frac{w^k}{k!}\sum_{n=0}^\infty\frac{z^n}{n!}=:e^we^z$$
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Thanks, thats what i was looking for. – Aftnix Jul 12 '12 at 8:51
1. One can use the ratio test or notice that for every $z \in \mathbb{C}$ one has $|z| \in \mathbb{R}$ and $$\sum_{n=0}^\infty\frac{|z|^n}{n!}=e^{|z|},$$ i.e. the series $\sum_{n=0}^\infty z^n/n!$ converges absolutely, and therefore it converges for every $z \in \mathbb{C}$.
2. $$e^we^z=(\sum_{n=0}^\infty\frac{w^n}{n!})(\sum_{n=0}^\infty\frac{z^n}{n!})=\sum_{n=0}^{\infty}A_n(w,z),$$ where with the use of Cauchy product one has \begin{eqnarray} A_n(w,z)&=&\sum_{k=0}^n\frac{w^k}{k!}\frac{z^{n-k}}{(n-k)!}=\sum_{k=0}^n\frac{1}{n!}{n\choose k}w^kz^{n-k}=\frac{1}{n!}\sum_{k=0}^n{n\choose k}w^kz^{n-k}=\frac{(w+z)^n}{n!}. \end{eqnarray}
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The second part can be proved also using the identity theorem: Two analytic functions are the same if they coincide on a set with a limit point. Look at $e^{w+z}$ and $e^w e^z$ with $w$ a fixed real number. The two functions (as functions of $z$) coincide on the real line. Therefore, they are equal for all $z \in \mathbb{C}$. Since $w$ was arbitrary they also coincide as functions of $w$ on the real line. Again by the identity theorem $e^{w+z}=e^w e^z$ for all $w \in \mathbb{C}$.
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http://mathhelpforum.com/advanced-algebra/125559-external-direct-product.html | # Thread:
1. ## External Direct Product
Let $C_n$ denote a cyclic group of order n. Show that if p is a prime number then every abelian group of order $p^2$ is isomorphic to one of the two groups $C_{p^2}, C_p \times C_p$.
OK!
Let G be an abelian group, with $|G|=p^2$. Suppose G is not isomorphic to a cyclic group of order $p^2$. So we need to show that $G \cong C_p \times C_p$.
Let $g \in G$ with $g \not= 1_G$. So |g| = p. |g| cannot be $p^2$ because then g would generate the whole group and so G would be equal and isomorphic to a cyclic subgroup of order $p^2$.
Let $h \in G \setminus <g>$. Then |h| = p.
Define $\theta : <g>\times <h> \rightarrow G$, given by $\theta (g^i,h^j) = g^i g^j$.
This is a homomorphism (need to use abelian-ness to prove this)
But how do I show that this is surjective and injective? It says here that $|<g> \times <h> | = |G| = p^2$ , so it is only required to prove injectivity or surjectivity, but why is $|<g> \times <h> | = |G| = p^2$ true?
In trying to work out the kernel it seems that it is only equal to {1} if $<g> \cap <h> = \{1\}$ but is this true? Any help would be appreciated here. Thanks very much
2. Originally Posted by slevvio
Let $C_n$ denote a cyclic group of order n. Show that if p is a prime number then every abelian group of order $p^2$ is isomorphic to one of the two groups $C_{p^2}, C_p \times C_p$.
OK!
Let G be an abelian group, with $|G|=p^2$. Suppose G is not isomorphic to a cyclic group of order $p^2$. So we need to show that $G \cong C_p \times C_p$.
Let $g \in G$ with $g \not= 1_G$. So |g| = p. |g| cannot be $p^2$ because then g would generate the whole group and so G would be equal and isomorphic to a cyclic subgroup of order $p^2$.
Let $h \in G \setminus <g>$. Then |h| = p.
Define $\theta : <g>\times <h> \rightarrow G$, given by $\theta (g^i,h^j) = g^i g^j$.
This is a homomorphism (need to use abelian-ness to prove this)
But how do I show that this is surjective and injective? It says here that $|<g> \times <h> | = |G| = p^2$ , so it is only required to prove injectivity or surjectivity, but why is $|<g> \times <h> | = |G| = p^2$ true?
In trying to work out the kernel it seems that it is only equal to {1} if $<g> \cap <h> = \{1\}$ but is this true? Any help would be appreciated here. Thanks very much
If $h^a=g^b$ , for some $0\leq a,b< p$ , then as $(a,p)=1$ there exists $k\in\mathbb{Z}$ s.t. $ak=1\!\!\!\pmod p\Longrightarrow h=h^{ak}=g^{bk}\in <g>$ , contradicting your choice of $h$ , and
this simply means $<g>\cap <h>=1$ .
Tonio
3. Originally Posted by slevvio
But how do I show that this is surjective and injective? It says here that $|<g> \times <h> | = |G| = p^2$ , so it is only required to prove injectivity or surjectivity, but why is $|<g> \times <h> | = |G| = p^2$ true?
Because $\left|\left\langle g\right\rangle\right|=\left|\left\langle h\right\rangle\right|=p$. So from basic set theory and combinatorics $\text{card }M=m,\text{card }N=n\implies \text{card }M\times n=mn$. Also, the reason you only have to prove either surjectivity or injectivity is that a surjective function from a set to an equipotent set is injective and vice versa if the sets are finite. | {"extraction_info": {"found_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": 43, "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": 0.9414840936660767, "perplexity_flag": "head"} |
http://www.physicsforums.com/showthread.php?t=616867 | Physics Forums
Equations with integrals,
Hi,
How would one go about solving equations like $∫^{b}_{a}f(s,t)g(s)ds$=g(t),for f(s,t). Could we turn it into a differential equation somehow?
Thanks
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Quote by 6.28318531 Hi, How would one go about solving equations like $∫^{b}_{a}f(s,t)g(s)ds$=g(t),for f(s,t). Could we turn it into a differential equation somehow? Thanks
Hey 2pi.
Have you tried first using the fundamental theorem of calculus and the chain rule to get rid of the integral expression (assuming s and t are unrelated and orthogonal)?
Yeah s and t are independent, but isn't the problem the fact that a and b are constants,and we don't know what f is, and as such the FTC doesn't really get us anywhere useful?
Equations with integrals,
Quote by 6.28318531 Yeah s and t are independent, but isn't the problem the fact that a and b are constants,and we don't know what f is, and as such the FTC doesn't really get us anywhere useful?
You will get a definition for the partial derivative in terms of f(s,t) and from that, you should be able to get something useful.
Remember that the integral is done with respect to ds, so you can consider this partial derivative with respect to the fundamental theorem of calculus and through the chain rule, obtain a partial differential relationship for the integral relationship.
You also take the derivative of the RHS with respect to s (partial derivative) and from this you get 0 (since g(t) will be considered more or less a constant).
This means you will be left with an expression involving partial with respect to s involving the chain rule of f(s,t)g(s) and the RHS will be zero. This should give you a PDE.
More information for this kind of problem, look for integro-differential equations either on the internet or in textbooks.
Quote by 6.28318531 Hi, How would one go about solving equations like $∫^{b}_{a}f(s,t)g(s)ds$=g(t),for f(s,t). Could we turn it into a differential equation somehow? Thanks
Would you mind make clear what is known and what is unknown in the equation.
Is the function g(t) known and then are you searching an unknown function f(s,t) consistant with the équation ?
Or, is the function f(s,t) known and then are you searching an unknown function g(t) consistant with the équation ?
Since there are two parameters a and b involved into the relationship, necesserally they will appear in the function that we are looking for.
So, if f(s,t) is unknown, then a and b will appear in g(t) and, as a matter of fact, the analytical expression of g(t) is g(a,b,t). Rigth or not ?
If g(t) is unknown, then a and b will appear in f(s,t) and, as a matter of fact, the analytical expression of f(s,t) is f(a,b,s,t). Rigth or not ?
@JJacquellin Sorry I should have been a bit clearer, f(s,t) is unknown. We are given g(t) and g(s). So then f is actually f(a,b,s,t), as you said? I think I can see where this is going but would you mind elaborating on how we can use FTC and the chain rule. Im still not 100% sure. Thanks
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http://mathhelpforum.com/geometry/89963-find-angles-triangle.html | # Thread:
1. ## Find the angles of the triangle?
Each of the two triplets of numbers $(a, b, c)\ \& \left(\frac{a}{2b},\ \frac{2b}{3c},\ \frac{3c}{a}\right)$ is a Geometric Progression. Can the numbers $a,b,c$ be the lengths of the sides of a triangle? If they can what triangle is it? Find the angles of the triangle.
2. Originally Posted by fardeen_gen
Each of the two triplets of numbers $(a, b, c)\ \& \left(\frac{a}{2b},\ \frac{2b}{3c},\ \frac{3c}{a}\right)$ is a Geometric Progression. Can the numbers $a,b,c$ be the lengths of the sides of a triangle? If they can what triangle is it? Find the angles of the triangle.
Use the fact that they are in a GP, the two equations ( $b^2 = ac$ and $\frac{4b^2}{9c^2} = \frac{a}{2b}\frac{3c}{a}$)to get $9c = 4a = 6b$. They clearly form a triangle. As for the angles, compute them for the triangle with sides 4,6,9. | {"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": 0, "codecogs_latex": 7, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.8992762565612793, "perplexity_flag": "head"} |
http://mathhelpforum.com/calculus/67061-help-delta-epsilon-proofs-print.html | # Help with delta epsilon proofs...
Printable View
• January 6th 2009, 11:32 AM
bemidjibasser
Help with delta epsilon proofs...
I need to prove that lim (mx+b) as x approaches c = mc +b. Any tips or help would be greatly appreciated.
Also, What exactly is the relationship of the definite integral to the derivative?
Again, help and or tips would be greatly appreciated.
• January 6th 2009, 12:09 PM
Moo
Hello,
Quote:
Originally Posted by bemidjibasser
I need to prove that lim (mx+b) as x approaches c = mc +b. Any tips or help would be greatly appreciated.
Also, What exactly is the relationship of the definite integral to the derivative?
Again, help and or tips would be greatly appreciated.
For the second one, if f is a function and f ' is its derivative, we have :
$\int_a^b f'(x) ~dx=f(b)-f(a)$
Usually, we define an antiderivative of f, F and say $\int_a^b f(x) ~dx=F(b)-F(a)$
Since f is an antiderivative of f ', we have the formula above.
• January 6th 2009, 12:12 PM
galactus
This actually is not that bad. I know that the epsilon-delta thing can be confusing.
$f(x)=mx+b, \;\ m\neq 0$
$\lim_{x\to c}(mx+b)=mc+b$
$|(mx+b)-(mc+b)|<{\epsilon}$
$|m(x-c)|<{\epsilon}$
$|x-c|<\frac{\epsilon}{|m|}={\delta}$
Therefore, for ${\epsilon}>0$, let ${\delta}=\frac{\epsilon}{|m|}$
• January 6th 2009, 03:41 PM
bemidjibasser
Thank you both. Is there a way to expand on the relationship of the derivative and the definite intergal? I think I can follow the formulas above, but how would you go about explaining more in words? Thanks again for help.
• January 6th 2009, 05:02 PM
Mathstud28
An alternative to Moo's post would be that in a sense integration and differentiation are inverses. If $F(x)=\int_{a}^x f~dx$ and $f$ is continuous at some point $x_0$ then $F(x)$ is differentiable at $x_0$ and $F'(x_0)=f(x_0)$
• January 6th 2009, 05:35 PM
bemidjibasser
Does the fundamental theorem have anything to do with this? Or the mean value theorem? Thank you again...
• January 6th 2009, 05:45 PM
Mathstud28
Quote:
Originally Posted by bemidjibasser
Does the fundamental theorem have anything to do with this? Or the mean value theorem? Thank you again...
These are the fundamental theorems of calculus (what me and Moo gave)
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http://mathoverflow.net/questions/38763/is-lp-mathbbr-minus-the-zero-function-contractible/38764 | ## Is $L^p(\mathbb{R})$ minus the zero function contractible?
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Is $L^p(\mathbb{R}) \setminus 0$ contractible? My intuition says that the answer is yes, but I'm afraid that this is based on thinking of this as somehow similar to a limit of $\mathbb{R}^n \setminus 0$ as n approaches $\infty$, which is of course nonsense. In any case, every contraction I've tried ends up making some function pass through $0$.
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Technically, this is a duplicate of: mathoverflow.net/questions/198/… , my answer there works for any space that is "stable" in the sense that $X \oplus \mathbb{R} \cong X$. – Andrew Stacey Sep 15 2010 at 7:18
(Looking again at my answer to the contractiblity of the sphere, I realised one could weaken the conditions so the "stable" in the above is stronger than needed.) – Andrew Stacey Sep 15 2010 at 7:50
Meta discussion on whether to close as duplicate or not: meta.mathoverflow.net/discussion/672/… – Andrew Stacey Sep 15 2010 at 8:09
## 4 Answers
Here is something really cheap and dirty. Let $p<+\infty$. Take $g=\frac{1}{1+x^2}$. Then $f(x,t)=e^{-(1+|x|)t/(1-t)}f(x)$ ($0\le t\le 1$) is a continuous contraction of `$L^p\setminus\{g\}$` to $0$. (the reason is that your only chance to hit $g$ is to start with it because $g(x)e^{s(1+|x|)}$ is not in $L^p$ for $s>0$).
Let's make it more interesting without making it more abstract. Can we find a uniformly continuous (both in space and time, as usual) contraction of the unit ball in $L^p$ without the center to a point?
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That's nice! I have no idea about your more general question. Maybe it's worth asking separately? – Nikita Sep 15 2010 at 4:07
I updated my answer to mathoverflow.net/questions/198/… to note that the homotopy described there is jointly continuous. – Andrew Stacey Sep 15 2010 at 7:49
Fedja, Benyamini and Sternfeld proved in 1983 that the unit sphere of any infinite dimensional Banach space is Lipschitz contractible. A lot of related things are mentioned in a survey talk he gave in 2007: nim.nankai.edu.cn/activites/conferences/hy200707/… As Borcherds mentioned, the topological structure of infinite dimensional spaces is very simple (not that the proofs are easy!). The uniform and Lipschitz structures are complicated and there is a lot of current research on these topics. The book of Benyamini-Lindenstrauss is the standard reference. – Bill Johnson Sep 15 2010 at 12:42
Bill, I read that as a paper by "Fedja, Benyamini and Sternfeld" so was about to search for those three names! Then I wondered why Fedja didn't know about that result ... English is a funny language. – Andrew Stacey Sep 16 2010 at 7:44
Oops! Sorry about that, Andrew. – Bill Johnson Sep 17 2010 at 2:53
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An infinite dimensional Banach space is homeomorphic to itself minus a point. Maybe R.D. Anderson or V. Klee proved this.
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Do you know a reference for this? – Nikita Sep 15 2010 at 2:32
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Start with projecteuclid.org/DPubS/Repository/1.0/… Also look for Bessaga's papers, maybe joint with Klee, in the 1960s in MathSciNet (which I cannot access right now). – Bill Johnson Sep 15 2010 at 2:54
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Wow, that's a pretty serious body of work! I'm going to hold off marking this as the accepted answer for a while in the hope that someone will suggest an easier proof of the fact that I asked for (which is much weaker than these papers prove!). – Nikita Sep 15 2010 at 3:00
According to mathscinet, the results of Anderson given in "Topological properties of the Hilbert cube and the infinite product of open intervals" Trans. Amer. Math. Soc. 126 1967 200--216, and "Hilbert space is homeomorphic to the countable infinite product of lines" Bull. Amer. Math. Soc. 72 1966 515--519 show that all infinite dimensional separable Frechet spaces are homeomorphic, and that removing any countable union of compact sets from such a space leaves the homeomorphism type unchanged.
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Here is a simple proof for case of a Hilbert space $V$. Since $V$ minus the origin deformation retracts onto the unit sphere $S^\infty$, it suffices to show that $S^\infty$ is contractible, and that will follow if we can show that $S^\infty$ is a deformation retract of the unit disk. Below is a simple proof of that fact taken from my book "Critical Point Theory and Submanifold Geometry". (I have an old paper called "On the Homotopy Theory of Infinite Dimensional Manifolds" that proves much more general results of this nature. It appeared in vol.3 of Topology (1966).)
Proposition. If $D^\infty$ is the closed unit disk in an infinite dimensional Hilbert space $V$, and $S^\infty=\partial D^\infty$ is the unit sphere in $V$, then there is a deformation retraction of $D^\infty$ onto $S^\infty$.
Proof. Since $D^\infty$ is convex, it will suffice to show that there is a retraction of $D^\infty$ onto $S^\infty$. Now recall the standard proof of the Brouwer Fixed Point Theorem. If there were a fixed point free map $h:D^n\to D^n$ it would imply the existence of a deformation retraction $r$ of $D^n$ onto $S^{n-1}$; namely $r(x)$ is the point where the ray from $h(x)$ to $x$ meets $S^{n-1}$. If $n<\infty$ this would contradict the fact that $H_n(D^n,S^{n-1})=Z$, so there can be no such retraction and hence no such fix point free map. But when $n=\infty$ we will see that such a fixed point free map does exist, and hence so does the retraction $r$. This will be a consequence of two simple lemmas.
Lemma 1. $D^\infty$ has a closed subspace homeomorphic to $R$.
Proof Let ${e_n}$ be an orthonormal basis for $V$ indexed by $Z$, and define $F:R \to D^\infty$ by $F(t)=\cos({1\over2}(t-n)\pi)e_n +\sin({1\over2}(t-n)\pi)e_{n+1}$ for $n\le t \le n+1$. It is easily checked that $F$ is a homeomorphism of $R$ into $D^\infty$ with closed image. QED
Lemma 2. If a normal space $X$ has a closed subspace $A$ homeomorphic to $R$ then it admits a fixed point free map $H:X\to X$.
Proof. Since $A$ is homeomorphic to $R$ it admits a fixed point free map $h:A\to A$, corresponding to say translation by $1$ in $R$. Since $A$ is closed in $X$ and $X$ is normal, by the Tietze Extension Theorem $h$ can be extended to a continuous map $H:X\to A$, and we may regard $H$ as a map $H:X\to X$. If $x\in A$ then $x\not=h(x)=H(x)$, while if $x\in X\setminus A$ then, since $H(x)\in A$, again $H(x)\not=x$. QED
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I cannot help saying once again my enthusiasm for those papers of yours about Banach manifolds on Topology... they are simply great! – Pietro Majer Sep 15 2010 at 7:51
I first learnt of this sort of fun-and-games in infinite dimensions from those papers and I've been hooked ever since. Finite dimensions is just so dull! – Andrew Stacey Sep 15 2010 at 8:04 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 74, "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": 0.928075909614563, "perplexity_flag": "head"} |
http://nrich.maths.org/6603/index?nomenu=1 | ## 'How Steep Is the Slope?' printed from http://nrich.maths.org/
### Show menu
#### The gradient of a line tells us how far up or down we go when
we take one step to the right:
On a grid like the one below we can draw lines with different gradients.
Check you agree that the black line is one of several that could be drawn with a gradient of 2 and the red line is one of several that could be drawn with a gradient of $-\frac{2}{3}$
Picture some more lines with different gradients. You may want to use this sheet to record your working. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9207726716995239, "perplexity_flag": "head"} |
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