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Interesting Trigonometric Series

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Post Sat May 11, 2013 11:58 am
Shobhit Site Admin
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Prove that

\(\displaystyle \sum_{n=1}^\infty \frac{n\sin(nx)}{a^2+n^2}=\frac{\pi}{2}\frac{\sinh(a(\pi-x))}{\sinh(a\pi)}\)

Post Sat May 11, 2013 9:59 pm
galactus User avatar
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Hey E,

There is another series very close to this one that is a nice candidate for residues.

\(\displaystyle \sum_{n=1}^{\infty}\frac{(-1)^{n-1}n\sin(nx)}{n^{2}+a^{2}}=\frac{\pi\sinh(ax)}{2\sinh(a\pi)}\)

One can use this one to show the other by letting \(\displaystyle x\to \pi-x\)

The residue for the above at \(\displaystyle z=ai\) is

\(\displaystyle \frac{\pi\sinh(ax)}{2\sinh(a\pi)}\)

Post Sun May 12, 2013 7:25 am
Shobhit Site Admin
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Location: Jaipur, India

Thanks!

Post Sun May 12, 2013 9:12 am
galactus User avatar
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You know, E, if one wanted to take a non-residue approach, this looks like a classic case of Fourier series.

Post Sun May 12, 2013 4:12 pm
zaidalyafey Global Moderator
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galactus wrote:
You know, E, if one wanted to take a non-residue approach, this looks like a classic case of Fourier series.


The series is the derivative of w.r.t x of the series

\(\displaystyle -\sum^{\infty}_{n=1}\frac{\cos(nx)}{a^2+n^2}\)

which can be solved as you stated ...
Wanna learn what we discuss , see Book

Post Fri Jun 02, 2017 1:34 am

Posts: 5
Shobhit wrote:
Prove that

\(\displaystyle \sum_{n=1}^\infty \frac{n\sin(nx)}{a^2+n^2}=\frac{\pi}{2}\frac{\sinh(a(\pi-x))}{\sinh(a\pi)}\)

usando serie de fourier
$\sinh n\theta = \sum\limits_{k = 1}^{ + \infty } {b_n \sin k\theta }$

donde $b_n = \frac{1}{\pi }\int\limits_{ - \pi }^\pi {\sinh n\theta \sin k\theta d\theta }$
$b_n = \frac{1}{\pi }\int\limits_{ - \pi }^\pi {\sinh n\theta \sin k\theta d\theta } = \frac{1}{\pi }\int\limits_{ - \pi }^\pi {\left( {\frac{{e^{n\theta } - e^{ - n\theta } }}{2}} \right)\left( {\frac{{e^{ik\theta } - e^{ - ik\theta } }}{{2i}}} \right)d\theta } $

$ = \frac{1}{{4i\pi }}\int\limits_{ - \pi }^\pi {\left( {e^{n\theta } - e^{ - n\theta } } \right)\left( {e^{ik\theta } - e^{ - ik\theta } } \right)d\theta } = \frac{1}
{{4i\pi }}\int\limits_{ - \pi }^\pi {\left( {e^{n\theta } e^{ik\theta } - e^{n\theta } e^{ - ik\theta } - e^{ - n\theta } e^{ik\theta } + e^{ - n\theta } e^{ - ik\theta } } \right)d\theta } $

$ = \frac{1}{{4i\pi }}\int\limits_{ - \pi }^\pi {\left( {e^{\left( {n + ik} \right)\theta } + e^{ - \left( {n + ik} \right)\theta } - \left( {e^{\left( {n - ik} \right)\theta } + e^{ - \left( {n - ik} \right)\theta } } \right)} \right)d\theta } = \frac{1}{{2i\pi }}\int\limits_{ - \pi }^\pi {\left( {\frac{{e^{\left( {n + ik} \right)\theta } + e^{ - \left( {n + ik} \right)\theta } }}{2} - \left( {\frac{{e^{\left( {n - ik} \right)\theta } + e^{ - \left( {n - ik} \right)\theta } }}{2}} \right)} \right)d\theta } $

$ = \frac{1}{{2i\pi }}\int\limits_{ - \pi }^\pi {\left( {\cosh \left( {\left( {n + ik} \right)\theta } \right) - \cosh \left( {\left( {n - ik} \right)\theta } \right)} \right)d\theta }
$
pero $\int\limits_{ - \pi }^\pi {\left( {\cosh \left( {\left( {n \pm ik} \right)\theta } \right)} \right)d\theta = } \frac{1}{{n \pm ik}}\int\limits_{ - \pi \left( {n \pm ik} \right)}^{\pi \left( {n \pm ik} \right)} {\cosh ydy} = \frac{2}{{n \pm ik}}\sinh \left( {\pi \left( {n \pm ik} \right)} \right)$

entonces $ = \frac{1}{{i\pi }}\left( {\frac{1}{{n + ik}}\sinh \left( {\pi n + i\pi k} \right) - \frac{1}{{n - ik}}\sinh \left( {\pi n - i\pi k} \right)} \right)$
$ = \frac{1}{{i\pi }}\left( {\frac{1}{{n + ik}}\left( {\sinh \left( {\pi n} \right)\cosh \left( {i\pi k} \right) + \cosh \left( {\pi n} \right)\sinh \left( {i\pi k} \right)} \right) - \frac{1}{{n - ik}}\left( {\sinh \left( {\pi n} \right)\cosh \left( {i\pi k} \right) - \cosh \left( {\pi n} \right)\sinh \left( {i\pi k} \right)} \right)} \right)$

usando la conocida identidad $\sinh ix = i\sin x \wedge \cosh ix = \cos x$

$ = \frac{1}{{i\pi }}\left( {\frac{1}{{n + ik}}\left( {\sinh \left( {\pi n} \right)\cos \left( {\pi k} \right) + i\cosh \left( {\pi n} \right)\sin \left( {\pi k} \right)} \right) - \frac{1}{{n - ik}}\left( {\sinh \left( {\pi n} \right)\cos \left( {\pi k} \right) - i\cosh \left( {\pi n} \right)\sinh \left( {\pi k} \right)} \right)} \right)$

todo se reduce a $ = \frac{{\sinh \left( {\pi n} \right)\cos \left( {\pi k} \right)}}{{i\pi }}\left( {\frac{1}{{n + ik}} - \frac{1}{{n - ik}}} \right) = - \frac{{\sinh \left( {\pi n} \right)\cos \left( {\pi k} \right)}}{\pi }\left( {\frac{{2k}}{{\left( {n + ik} \right)\left( {n - ik} \right)}}} \right)$

$ = - \frac{{\sinh \left( {\pi n} \right)\cos \left( {\pi k} \right)}}{\pi }\left( {\frac{{2k}}{{n^2 + k^2 }}} \right)$

entonces $\sinh n\theta = \sum\limits_{k = 1}^{ + \infty } {b_n \sin k\theta } = \sum\limits_{k = 1}^{ + \infty } { - \frac{{\sinh \left( {\pi n} \right)\cos \left( {\pi k} \right)}}{\pi }\left( {\frac{{2k}}{{n^2 + k^2 }}} \right)\sin k\theta } $

$ = \frac{2}{\pi }\sum\limits_{k = 1}^{ + \infty } {\frac{{\left( { - 1} \right)^{k - 1} k\sin k\theta \sinh \left( {\pi n} \right)}}{{n^2 + k^2 }}}$

donde finalmente se deduce
$\frac{\pi }{2}\frac{{\sinh \left( {n\theta } \right)}}{{\sinh \left( {\pi n} \right)}} = \sum\limits_{k = 1}^{ + \infty } {\frac{{\left( { - 1} \right)^{k - 1} k\sin k\theta }}
{{n^2 + k^2 }}} $


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