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anonymous

  • one year ago

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

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  1. anonymous
    • one year ago
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    The next one looks a bit tougher, I'll post it once this one's resolved.

  2. anonymous
    • one year ago
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    By 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%282k-1%29x%5D%2C%7Bk%2C1%2C15%7D%5D%2C+Csc%5Bx%5D%7D%2C+%7Bx%2C0%2CPi%7D%5D

  3. anonymous
    • one year ago
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    (I suppose that qualifies as a hint)

  4. anonymous
    • one year ago
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    *area under the sum of sines

  5. Empty
    • one year ago
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    Oooh interesting this reminds me of how phase velocity is related to group velocity.

  6. Empty
    • one year ago
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    Oh so wait is this finding the coefficients on the fourier series representation of this function?

  7. anonymous
    • one year ago
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    Fourier 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, \[\lim_{x\to0}\frac{\sum\limits_{k=1}^n\sin(2k-1)x}{\sum\limits_{k=1}^n\sin 2kx}\] (a neat problem on its own).

  8. anonymous
    • one year ago
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    consider \(\sin a\sin b=\frac12\left[\cos(a-b)-\cos(a+b)\right]\) so we have $$\sin x\sin((2k-1)x)=\frac12\left[\cos((2k-2)x)-\cos(2kx)\right]\\\implies\sin x\sum_{k=1}^n\sin((2k-1)x)=\frac12\left[1-\cos(2nx)\right]\\\implies\sum_{k=1}^n\sin((2k-1)x)=\frac{1-\cos(2nx)}{2\sin x}$$

  9. anonymous
    • one year ago
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    now 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 $$\lim_{n\to\infty}\int_{\pi/4}^{3\pi/4}\frac{1-\cos(2nx)}{2\sin x}\, dx=\int_{\pi/4}^{3\pi/4}\frac1{2\sin x}\, dx=\frac12\int_{\pi/4}^{3\pi/4}\csc x\, dx$$

  10. anonymous
    • one year ago
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    where we know $$\int\csc x\, dx=-\log|\csc x+\cot x|+C$$ so $$-\log|\csc(3\pi/4)+\cot(3\pi/4)|+\log|\csc(\pi/4)+\cot(\pi/4)|\\\qquad=-\log|\sqrt2-1|+\log|\sqrt2+1|\\\qquad=\log\frac{\sqrt2+1}{\sqrt2-1}\\\qquad=\log(3+2\sqrt2)$$ so we find $$\int_{\pi/4}^{3\pi/4}\frac1{2\sin x}\, dx=\frac12\log(3+2\sqrt2)\approx 0.88137$$

  11. anonymous
    • one year ago
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    I 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 \[\begin{align*} \lim_{n\to\infty}\int_{\pi/4}^{3\pi/4}\sum_{k=1}^n\sin(2k-1)x&=\sum_{k=1}^\infty \int_{\pi/4}^{3\pi/4}\sin(2k-1)x\,dx\\[1ex] &=\sum_{k=1}^\infty \frac{\cos\dfrac{(2k-1)\pi}{4}-\cos\dfrac{(2k-1)3\pi}{4}}{2k-1}\\[1ex] &=\frac{2}{\sqrt2}\sum_{k=1}^\infty \frac{\cos\dfrac{k\pi}{2}+\sin\dfrac{k\pi}{2}}{2k-1} \end{align*}\] Alternating terms in the numerator disappear according to this pattern: \[\begin{array}{c|cccc|cccc|c} k&1&2&3&4&5&6&7&8&\cdots\\ \hline \cos&0&-1&0&1&0&-1&0&1&\cdots\\ \sin&1&0&-1&0&1&0&-1&0&\cdots \end{array}\] so I can write the sum as \[\sqrt2\sum_{k=1}^\infty \left(-\frac{1}{8k-5}+\frac{1}{8k-1}+\frac{1}{8k-7}-\frac{1}{8k-3}\right)\] 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.

  12. anonymous
    • one year ago
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    you can write the series in a closed form using digamma functions, if that helps: https://en.wikipedia.org/wiki/Digamma_function#Series_formula

  13. anonymous
    • one year ago
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    Ah, another thing to learn, cool. Thanks.

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