Computing a sum involving sin

**4Math** · September 15th 2010, 09:09 AM

I have trouble solving the following sum:

$<br /> \displaystyle\frac{1}{2}\sum_{k=1}^{\infty}4\frac{ sin^{2}{k}}{k^{2}}$

I've tried the following:

$\displaystyle\frac{1}{2}\sum_{k=1}^{\infty}4(\frac {sin{k}}{k})^{2}$

The first term is:

$<br /> \displaystyle4(\frac{sin{1}}{1})^{2}=4(\frac{\pi}{ 2})^{2}=2{\pi}^{2}$

The sum is than:

$\displaystyle\frac{1}{2}\sum_{k=1}^{\infty}4\frac{ sin^{2}{k}}{k^{2}}=\frac{1}{2}2{\pi}^{2}={\pi}^{2}$ but this is incorrect.

Any help to compute this sum would be appriciate it. Thank you

**chisigma** · September 15th 2010, 09:43 AM

In...

http://www.mathhelpforum.com/math-he...on-153971.html

... it has been demonstrated that the 'mirror imaging' of the function...

$\displaystyle f(x)=\left\{\begin{array}{ll}2-x ,\,\,0 < x < 2\\{}\\0 ,\,\, 2 < x< \pi\end{array}\right.$ (1)

... has the following Fourier series expansion...

$\displaystyle f(x)= \frac{2}{\pi} + \frac{4}{\pi}\ \sum_{n=1}^{\infty} \frac{\sin^{2} n}{n^{2}}\ \cos nx$ (2)

Setting $x=0$ in (2) You obtain...

$\displaystyle \frac{2}{\pi} + \frac{4}{\pi}\ \sum_{n=1}^{\infty} \frac{\sin^{2} n}{n^{2}} = 2$ (3)

... so that is...

$\displaystyle \sum_{n=1}^{\infty} \frac{\sin^{2} n}{n^{2}} =\frac{\pi -1}{2} = 1.07079632679...$ (4)

Kind regards

$\chi$ $\sigma$

**4Math** · September 15th 2010, 11:54 AM

... has the following Fourier series expansion...

$\displaystyle f(x)= \frac{2}{\pi} + \frac{4}{\pi}\ \sum_{n=1}^{\infty} \frac{\sin^{2} n}{n^{2}}\ \cos nx$ (2)

Setting $x=0$ in (2) You obtain...

$\displaystyle\frac{2}{\pi} + \frac{4}{\pi}\ \sum_{n=1}^{\infty} \frac{\sin^{2} n}{n^{2}} = 2$ (3)

So to compute the following sum:

$\displaystyle \sum_{n=1}^{\infty} \frac{\sin^{4} n}{n^{4}}$

for this fourier expansion (using Parseval):

$\displaystyle{\frac{1}{2\pi}\int_{0}^{2\pi}|f(x)|^ {2}dx=\frac{4}{\pi^2} + \frac{16}{\pi^2}\ \sum_{n=1}^{\infty} \frac{\sin^{4} n}{n^{4}} = 2$

..and I get the following:

$\displaystyle \sum_{n=1}^{\infty} \frac{\sin^{4} n}{n^{4}}=\frac{\pi^2 -2}{8}=\frac{\pi^2}{8}-\frac{1}{4}$

But this is incorrect answer. What am I doing wrong here?

By the way, thank you for your response.

Kind regards

**4Math** · September 16th 2010, 07:48 AM

I think I found my mistake for computing the following sum:

$\displaystyle \sum_{n=1}^{\infty} \frac{\sin^{4} n}{n^{4}}$

using the following Fourier series

$\displaystyle \frac{2}{\pi} + \frac{4}{\pi}\ \sum_{n=1}^{\infty} \frac{\sin^{2} n}{n^{2}}\ \cos nx$

$\displaystyle a_{0} = \frac{2}{\pi}$

by using Parseval's theorem:

$\displaystyle{\frac{2}{\pi}= {\frac{1}{2\pi}\int_{0}^{2\pi}|f(x)|^{2}dx=\frac{4 }{\pi^2} + {\frac{1}{2}\ \sum_{n=1}^{\infty} \frac{16}{\pi^2}\frac{\sin^{4} n}{n^{4}}$

$\displaystyle\implies$ $\displaystyle{\frac{2}{\pi}= \frac{4}{\pi^2} + \frac{8}{\pi^2}\ \sum_{n=1}^{\infty}\frac{\sin^{4} n}{n^{4}}$ $\displaystyle\implies$

$\displaystyle\implies$ $\displaystyle\sum_{n=1}^{\infty}\frac{\sin^{4} n}{n^{4}}=\frac{2\pi -4}{8}=\frac{\pi}{4}-\frac{1}{2}$

Please let me know if this is not correct. Thank you

**chisigma** · September 16th 2010, 11:33 AM

The Parseval's identity is...

$\displaystyle \frac{1}{\pi} \int_{-\pi}^{\pi} f^{2} (x)\ dx= \frac{a_{0}^{2}}{2} + \sum_{n=1}^{\infty} (a^{2}_{n} + b^{2}_{n})$ (1)

... and in Your case is...

$a_{0}=\frac{4}{\pi}$ , $a_{n}=\frac{4}{\pi}\ \frac{\sin^{2} n}{n^{2}}$ , $b_{n}=0$ (2)

... and ...

$\displaystyle \int_{-\pi}^{\pi} f^{2} (x)\ dx = 2\ \int_{0}^{2} (4 - 4\ x + x^{2})\ dx = \frac{16}{3}$ (3)

... so that is...

$\displaystyle \frac{2}{3\ \pi} =\frac{1}{\pi^{2}} + \frac{2}{\pi^{2}} \ \sum_{n=1}^{\infty} \frac{\sin^{4} n}{n^{4}} \implies \sum_{n=1}^{\infty} \frac{\sin^{4} n}{n^{4}} = \frac{\pi}{3} - \frac{1}{2} = .547197551196...$ (4)

Kind regards

$\chi$ $\sigma$

**4Math** · September 17th 2010, 07:05 AM

Originally Posted by chisigma

The Parseval's identity is...

$\displaystyle \frac{1}{\pi} \int_{-\pi}^{\pi} f^{2} (x)\ dx= \frac{a_{0}}{2} + \sum_{n=1}^{\infty} (a^{2}_{n} + b^{2}_{n})$ (1)

Isn't it: $\displaystyle \frac{a_{0}^{2}}{2}$ in the right hand side?

$\displaystyle \int_{-\pi}^{\pi} f^{2} (x)\ dx = 2\ \int_{0}^{2} (4 - 4\ x + x^{2})\ dx = \frac{16}{3}$ (3)

Why do you double the integral? Is it because of that the given function is even?

Kind regards

**chisigma** · September 17th 2010, 07:13 AM

Originally Posted by 4Math

Isn't it: $\displaystyle \frac{a_{0}^{2}}{2}$ in the right hand side? Yes!... thank You very much!... typical transcription error!... [/COLOR]

Why do you double the integral? Is it because of that the given function is even? Yes again!...

Kind regards

Kind regards

$\chi$ $\sigma$

**4Math** · September 18th 2010, 03:02 AM

Originally Posted by chisigma

$a_{0}=\frac{4}{\pi}$

Since the given function is even, continuous and its periodic with period ${2\pi}$ , I get the following when I compute $a_{0}$ ...

$\displaystyle a_{0}= {\frac{2}{T}\int_{0}^{{\frac{T}{2}}} f(x) dx \displaystyle\implies<br />$

in this case $T={2\pi}$ ...

$\displaystyle\implies a_{0}= {\frac{2}{2\pi}(2\int_{0}^{2} (2-x) dx)=\frac{4}{\pi}$

Am I using a correct formula to compute $a_{0}$ ?
------------------------------------------------------------------------------------------------------

The Parseval's identity is...

$\displaystyle \frac{1}{\pi} \int_{-\pi}^{\pi} f^{2} (x)\ dx= \frac{a_{0}^{2}}{2} + \sum_{n=1}^{\infty} (a^{2}_{n} + b^{2}_{n})$ (1)

In our course book says that the Parseval's identity is:
$\displaystyle \frac{1}{T} \int_{P} f^{2} (x)\ dx= {a_{0}^{2}} + \frac{1}{2} \sum_{n=1}^{\infty} (a^{2}_{n} + b^{2}_{n})$ (4)

But when I apply (4) to the the same problem above I get a different result. Why is that? But your result $\displaystyle \sum_{n=1}^{\infty} \frac{\sin^{4} n}{n^{4}} = \frac{\pi}{3}- \frac{1}{2}$ is the correct result according to the answer in the problem book.

I used $T=\pi$ and the limits for the integral $[0,2]$ . I also tried with $T=2\pi$ , but still get different result.
---------------------------------------------------------------------------------------------------------------

$\displaystyle \frac{2}{3\ \pi} =\frac{1}{\pi^{2}} + \frac{2}{\pi^{2}} \ \sum_{n=1}^{\infty} \frac{\sin^{4} n}{n^{4}}$

I would like to ask you kindly to write out all the terms without simplifications in other words:

$\displaystyle \frac{16}{3\ \pi} =... + ... \ \sum_{n=1}^{\infty} \frac{\sin^{4} n}{n^{4}}<br />$

Thank you
Kind regards

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