# Math Help - Computing a sum using Parseval's theorem

1. ## Computing a sum using Parseval's theorem

Hi,

I have the following $2\pi$-periodic function:

$\displaystyle g(t)=e^{-t} ,\,\,0 < t < 2\pi$

I've computed the Exponential Fourier Series for the given function:

$\displaystyle{g(t)=\sum_{k=-\infty}^{\infty}c_{k}e^{ik\Omega t}}$

$\displaystyle c_{k}={\frac{1-e^{-2\pi}}{{2\pi}(1+ik)}$

$\displaystyle\implies{g(t)=\sum_{k=-\infty}^{\infty}({\frac{1-e^{-2\pi}}{{2\pi}(1+ik)})e^{ikt}}$ (1)
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I need to compute the following sum:

$\displaystyle \sum_{k=0}^{\infty} \frac{1}{1+k^{2}}$

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The Parseval's identity is:

$\displaystyle \frac{1}{2\pi} \int_{0}^{2\pi}|f(t)|^{2}dt=\sum_{k=-\infty}^{\infty}|c_{k}(f)|^{2}$

$\displaystyle \frac{1}{2\pi} \int_{0}^{2\pi}|f(t)|^{2}dt=\frac{1}{2\pi} \int_{0}^{2\pi}e^{-2t}dt={\frac{1}{4\pi}(1-e^{-4\pi})$ (2)

$\displaystyle |c_{k}(f)|={\frac{1-e^{-2\pi}}{{2\pi}(1+k)}$(3)

$\displaystyle |c_{k}(f)|^{2}=({\frac{1-e^{-2\pi}}{{2\pi}(1+k)})^{2}={\frac{(1-e^{-2\pi})^{2}}{{4\pi^{2}}(1+k)^{2}}={\frac{(1-e^{-2\pi})^{2}}{{4\pi^{2}}(1+k)(1+k)}={\frac{(1-e^{-2\pi})^{2}}{{4\pi^{2}}(1+2k+k^{2})}$(4)
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What is:

$\displaystyle |{{2\pi}(1+ik)}|$

Isn't it:

$\displaystyle {{2\pi}(1+k)}$

Is (3) and (4) correct?

I would greatly appriciate guidance to compute the sum.

Thank you

2. If $\sigma= \alpha + i\ \beta$ is any complex number, then $|\sigma|^{2} = \sigma\ \sigma^{*} = \alpha^{2} + \beta^{2}$. If...

$\displaystyle \sigma= \frac{1-e^{-2 \pi}}{2\ \pi\ (1+ i\ k)}$ (1)

... what is $\sigma^{*}$ and $|\sigma|^{2}$ ?...

Kind regards

$\chi$ $\sigma$

3. Originally Posted by chisigma

... what is $\sigma^{*}$ and $|\sigma|^{2}$ ?...
$\sigma^{*}=(1-ik)$

and

$\displaystyle|(1-ik)|^{2}=\sqrt{1^{2}+k^{2}}=(1+k)^{2}$

therefore:

$|\displaystyle\sigma|^{2}={\frac{(1-e^{-2\pi})^{2}(1+k)^{2}}{{4\pi^{2}}(1+k^{2})^{2}}$

Kind regards

4. Originally Posted by chisigma
If $\sigma= \alpha + i\ \beta$ is any complex number, then $|\sigma|^{2} = \sigma\ \sigma^{*} = \alpha^{2} + \beta^{2}$. If...

$\displaystyle \sigma= \frac{1-e^{-2 \pi}}{2\ \pi\ (1+ i\ k)}$ (1)

... what is $\sigma^{*}$ and $|\sigma|^{2}$ ?...

Kind regards

$\chi$ $\sigma$
$\displaystyle \sigma= \frac{1-e^{-2 \pi}}{2\ \pi\ (1+ i\ k)} \implies \sigma^{*}= \frac{1-e^{-2 \pi}}{2\ \pi\ (1- i\ k)} \implies |\sigma|^{2} = \sigma\ \sigma^{*} = \frac{(1-e^{-2\ \pi})^{2}}{4\ \pi^{2}\ (1+k^{2})}$

Kind regards

$\chi$ $\sigma$

5. Originally Posted by chisigma
$\displaystyle \sigma= \frac{1-e^{-2 \pi}}{2\ \pi\ (1+ i\ k)} \implies \sigma^{*}= \frac{1-e^{-2 \pi}}{2\ \pi\ (1- i\ k)} \implies |\sigma|^{2} = \sigma\ \sigma^{*} = \frac{(1-e^{-2\ \pi})^{2}}{4\ \pi^{2}\ (1+k^{2})}$
If the denominator is multiplied by $\ (1- i\ k)$ don't I need to also multiply the numerator with $\ (1- i\ k)$ ??...

Kind regards

6. Once You have $\sigma$ and $\sigma^{*}$ the only You have to do is multiplying them, 'numerator x numerator' and 'denominator x denominator' independently...

Kind regards

$\chi$ $\sigma$

7. Originally Posted by chisigma
Once You have $\sigma$ and $\sigma^{*}$ the only You have to do is multiplying them, 'numerator x numerator' and 'denominator x denominator' independently...
Ok, thank you kindly for clarifying that.

I get the following resulat:

$\displaystyle \sum_{k=-\infty}^{\infty}\frac{1}{1+k^{2}}={\frac{{\pi}(1+e ^{-2\pi})}{1-e^{-2\pi}}$ (5)

In (5) the sum goes from ${k=-\infty}$ to ${k=\infty}$ according to Parseval's identity.

...but the given sum goes from ${k=0$ to ${k=\infty}$.

How do I compute:

$\displaystyle \sum_{k=0}^{\infty}\frac{1}{1+k^{2}}$

Kind regards

8. Originally Posted by 4Math
Ok, thank you kindly for clarifying that.

I get the following resulat:

$\displaystyle \sum_{k=-\infty}^{\infty}\frac{1}{1+k^{2}}={\frac{{\pi}(1+e ^{-2\pi})}{1-e^{-2\pi}}$ (5)

In (5) the sum goes from ${k=-\infty}$ to ${k=\infty}$ according to Parseval's identity.

...but the given sum goes from ${k=0$ to ${k=\infty}$.

How do I compute:

$\displaystyle \sum_{k=0}^{\infty}\frac{1}{1+k^{2}}$

Kind regards
If the result You have found...

$\displaystyle \sum_{k= -\infty}^{+ \infty} \frac{1}{1+k^{2}}= \frac{{\pi}(1+e^{-2\pi})}{1-e^{-2\pi}} = \alpha$ (1)

... is correct, the procedure to obtain the final result is...

$\displaystyle \sum_{k=0}^{\infty} \frac{1}{1+k^{2}} = \frac{\alpha+1}{2}$ (2)

Kind regards

$\chi$ $\sigma$

9. Originally Posted by chisigma
If the result You have found...

$\displaystyle \sum_{k= -\infty}^{+ \infty} \frac{1}{1+k^{2}}= \frac{{\pi}(1+e^{-2\pi})}{1-e^{-2\pi}} = \alpha$ (1)

... is correct, the procedure to obtain the final result is...

$\displaystyle \sum_{k=0}^{\infty} \frac{1}{1+k^{2}} = \frac{\alpha+1}{2}$ (2)
Is it:

$\displaystyle \sum_{k= -\infty}^{+ \infty} \frac{1}{1+k^{2}}=\sum_{k= -\infty}^{-1} \frac{1}{1+k^{2}}+ \sum_{k= 0}^{+ \infty} \frac{1}{1+k^{2}}=\sum_{n= 1}^{+ \infty} \frac{1}{1+n^{2}}+\sum_{k=0}^{\infty} \frac{1}{1+k^{2}} =\sum_{n= 0}^{+ \infty} \frac{1}{1+n^{2}}+\frac{1}{2}+\sum_{k=0}^{\infty} \frac{1}{1+k^{2}}$

I am just trying to understand how to get to:

$\displaystyle\frac{\alpha+1}{2}$

Kind regards

10. One and half year ago I opened this thread...

http://www.mathhelpforum.com/math-he...ies-95343.html

... with the task of verify the result I obtained in the Fourier series expansion of the function $\cosh t$ in $-\pi < t < \pi$. Opalg and DeMath confirmed the following result...

$\displaystyle \cosh t = \frac{2\ \sinh \pi}{\pi}\ \{\frac{1}{2} + \sum_{n=1}^{\infty} \frac{(-1)^{n}}{1+n^{2}}\ \cos n t \}$ (1)

Now if You set in (1) $t= \pi$ with some easy steps You obtain...

$\displaystyle \sum_{n=0}^{\infty} \frac{1}{1+n^{2}} = \frac{1 + \pi\ \coth \pi}{2} = 2.07667404746858...$ (2)

A simple computer program can verify the convergence of (2) to that value. It seems to me however that with the approach used by 4Math we arrive at a different result , and I'm not able to explain why. Regarding the series (2) it is remarkable the fact that in F. Sheid, Numerical Analysis, McGraw-Hill 1968 in the prob. 17.15 [pag. 162] You can read regarding series (2)...

... Stegun & Abramowitz [Journal of SIAM, 1956] demonstrated that 5745 terms are needed for three exact decimals...

Well!... my computer for three exact decimals needs 'only' 1484 terms... did something change in the last 55 years? ...

Kind regards

$\chi$ $\sigma$

11. Originally Posted by chisigma
If the result You have found...

$\displaystyle \sum_{k= -\infty}^{+ \infty} \frac{1}{1+k^{2}}= \frac{{\pi}(1+e^{-2\pi})}{1-e^{-2\pi}} = \alpha$ (1)

... is correct, the procedure to obtain the final result is...

$\displaystyle \sum_{k=0}^{\infty} \frac{1}{1+k^{2}} = \frac{\alpha+1}{2}$ (2)

Kind regards

$\chi$ $\sigma$
In detail...

$\displaystyle \sum_{k= -\infty}^{+ \infty} \frac{1}{1+k^{2}} = 1 + 2\ \sum_{n=1}^{\infty} \frac{1}{1+k^{2}}= \alpha \implies$

$\displaystyle \sum_{n=1}^{\infty} \frac{1}{1+k^{2}} = \frac{\alpha-1}{2} \implies \sum_{k=0}^{\infty} \frac{1}{1+k^{2}} = \frac{\alpha+1}{2}$

Kind regards

$\chi$ $\sigma$

12. Originally Posted by chisigma
In detail...

$\displaystyle \sum_{k= -\infty}^{+ \infty} \frac{1}{1+k^{2}} = 1 + 2\ \sum_{n=1}^{\infty} \frac{1}{1+k^{2}}= \alpha$
I don't follow...

Where does 1 in the right hand side comes in and why do we double the sum: $\displaystyle2\ \sum_{n=1}^{\infty} \frac{1}{1+k^{2}}$

I greatly appriciate for an explanation.

Thank you
Kind regards

13. ... because $k$ and $- k$ both generate the same contribute $\displaystyle \frac{1}{1+k^{2}}$ ...

Kind regards

$\chi$ $\sigma$

14. Originally Posted by chisigma

$\displaystyle \sum_{k= -\infty}^{+ \infty} \frac{1}{1+k^{2}} = 1 + 2\ \sum_{n=1}^{\infty} \frac{1}{1+k^{2}}= \alpha$
I would like to ask you kindly if you could please write out the sums that results in $1+2\sum_{n=1}^{\infty} \frac{1}{1+k^{2}}$ in detail:

$\displaystyle \sum_{k= -\infty}^{+ \infty} \frac{1}{1+k^{2}} = .........=1 + 2\ \sum_{n=1}^{\infty} \frac{1}{1+k^{2}} = \alpha$

Kind regards

15. Originally Posted by 4Math
I would like to ask you kindly if you could please write out the sums that results in $1+2\sum_{n=1}^{\infty} \frac{1}{1+k^{2}}$ in detail:

$\displaystyle \sum_{k= -\infty}^{+ \infty} \frac{1}{1+k^{2}} = .........=1 + 2\ \sum_{n=1}^{\infty} \frac{1}{1+k^{2}} = \alpha$

Kind regards
$\displaystyle \sum_{k= -\infty}^{+ \infty} \frac{1}{1+k^{2}} =\sum_{k= -\infty}^{-1} \frac{1}{1+k^{2}} + 1 + \sum_{1}^{+ \infty} \frac{1}{1+k^{2}}$

... but is...

$\displaystyle \sum_{k= -\infty}^{-1} \frac{1}{1+k^{2}} = \sum_{k= 1}^{+ \infty} \frac{1}{1+k^{2}}$

... so that...

$\displaystyle \sum_{k= -\infty}^{+ \infty} \frac{1}{1+k^{2}} = 1 + 2 \sum_{k= 1}^{\infty} \frac{1}{1+k^{2}}$

Kind regards

$\chi$ $\sigma$