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anonymous
 one year ago
First of a series of trig integrals: Show that
\[\lim_{n\to\infty}\int_{\pi/4}^{3\pi/4}\sum_{k=1}^n\sin(2k1)x\,dx=\ln\left(\cot\frac{\pi}{8}\right)\]
anonymous
 one year ago
First of a series of trig integrals: Show that \[\lim_{n\to\infty}\int_{\pi/4}^{3\pi/4}\sum_{k=1}^n\sin(2k1)x\,dx=\ln\left(\cot\frac{\pi}{8}\right)\]

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0The next one looks a bit tougher, I'll post it once this one's resolved.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0By the way, this problem's inspired by the observation that the sum of sines is bounded by a single cosecant: http://www.wolframalpha.com/input/?i=Plot%5B%7BSum%5BSin%5B%282k1%29x%5D%2C%7Bk%2C1%2C15%7D%5D%2C+Csc%5Bx%5D%7D%2C+%7Bx%2C0%2CPi%7D%5D

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0(I suppose that qualifies as a hint)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0*area under the sum of sines

Empty
 one year ago
Best ResponseYou've already chosen the best response.1Oooh interesting this reminds me of how phase velocity is related to group velocity.

Empty
 one year ago
Best ResponseYou've already chosen the best response.1Oh so wait is this finding the coefficients on the fourier series representation of this function?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Fourier series of which function? I'm not seeing the connection, but that's not to say there isn't one. This is just a problem I came up with after working with the limit, \[\lim_{x\to0}\frac{\sum\limits_{k=1}^n\sin(2k1)x}{\sum\limits_{k=1}^n\sin 2kx}\] (a neat problem on its own).

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0consider \(\sin a\sin b=\frac12\left[\cos(ab)\cos(a+b)\right]\) so we have $$\sin x\sin((2k1)x)=\frac12\left[\cos((2k2)x)\cos(2kx)\right]\\\implies\sin x\sum_{k=1}^n\sin((2k1)x)=\frac12\left[1\cos(2nx)\right]\\\implies\sum_{k=1}^n\sin((2k1)x)=\frac{1\cos(2nx)}{2\sin x}$$

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0now it's tempting to say we could take the limit now, even though strictly that's not true, but intuitively as \(n\to\infty\) the frequency of \(\cos(2nx)\) grows without bound and we get that it oscillates wildly and on a given compact interval (like \((\pi/4,3\pi/4)\)) will 'average out' in the integral to \(0\), giving $$\lim_{n\to\infty}\int_{\pi/4}^{3\pi/4}\frac{1\cos(2nx)}{2\sin x}\, dx=\int_{\pi/4}^{3\pi/4}\frac1{2\sin x}\, dx=\frac12\int_{\pi/4}^{3\pi/4}\csc x\, dx$$

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0where we know $$\int\csc x\, dx=\log\csc x+\cot x+C$$ so $$\log\csc(3\pi/4)+\cot(3\pi/4)+\log\csc(\pi/4)+\cot(\pi/4)\\\qquad=\log\sqrt21+\log\sqrt2+1\\\qquad=\log\frac{\sqrt2+1}{\sqrt21}\\\qquad=\log(3+2\sqrt2)$$ so we find $$\int_{\pi/4}^{3\pi/4}\frac1{2\sin x}\, dx=\frac12\log(3+2\sqrt2)\approx 0.88137$$

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I really like your approach. My attempt was a bit taxing by comparison. Now, I'm not entirely clear on the necessary conditions for being able to do so, but to start with I tried swapping the order of summation/integration. (I'm not fluent in measure theory, but I *think* this function satisfies Fubini's theorem.) I get \[\begin{align*} \lim_{n\to\infty}\int_{\pi/4}^{3\pi/4}\sum_{k=1}^n\sin(2k1)x&=\sum_{k=1}^\infty \int_{\pi/4}^{3\pi/4}\sin(2k1)x\,dx\\[1ex] &=\sum_{k=1}^\infty \frac{\cos\dfrac{(2k1)\pi}{4}\cos\dfrac{(2k1)3\pi}{4}}{2k1}\\[1ex] &=\frac{2}{\sqrt2}\sum_{k=1}^\infty \frac{\cos\dfrac{k\pi}{2}+\sin\dfrac{k\pi}{2}}{2k1} \end{align*}\] Alternating terms in the numerator disappear according to this pattern: \[\begin{array}{cccccccccc} k&1&2&3&4&5&6&7&8&\cdots\\ \hline \cos&0&1&0&1&0&1&0&1&\cdots\\ \sin&1&0&1&0&1&0&1&0&\cdots \end{array}\] so I can write the sum as \[\sqrt2\sum_{k=1}^\infty \left(\frac{1}{8k5}+\frac{1}{8k1}+\frac{1}{8k7}\frac{1}{8k3}\right)\] I'm trying to use to rewrite each term in terms of harmonic numbers, but I'm failing to wrap my mind around the proper manipulation.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0you can write the series in a closed form using digamma functions, if that helps: https://en.wikipedia.org/wiki/Digamma_function#Series_formula

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Ah, another thing to learn, cool. Thanks.
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