Fourier Transform & Regularity

**davidmccormick** · May 3rd 2010, 10:55 AM

Here is a few transform that I'm trying to calculate:

1. Calculate the fourier coefficient of $e^x$ on $[-\pi, \pi]$ and thus determine the sum $\sum_{k=-\infty}^{\infty} 1/(1+k^2)$ . I get $F_k = (-1)^k (e^{\pi}-e^{-\pi})(1+ki) /(1+k^2)$ . I'm not sure if this is correct and can't see how using any identities (e.g. Parseval) I can calculate the sum.

2. Suppose that $f \in \{C(\mathbb{R}) \cap L^1 (\mathbb{R})\}$ and $p \in \mathbb{N}$ . Show that if $|k|^p F_k$ is integrable then $f$ has $p$ continuous bounded derivatives.

**Opalg** · May 4th 2010, 12:34 AM

Originally Posted by davidmccormick

Here is a few transform that I'm trying to calculate:

1. Calculate the fourier coefficient of $e^x$ on $[-\pi, \pi]$ and thus determine the sum $\sum_{k=-\infty}^{\infty} 1/(1+k^2)$ . I get $F_k = (-1)^k (e^{\pi}-e^{-\pi})(1+ki) /(1+k^2)$ . I'm not sure if this is correct and can't see how using any identities (e.g. Parseval) I can calculate the sum.

The Fourier series converges to the function at points where the function is continuous. At points where there is a jump discontinuity, it converges to the midpoint of the jump. So you know that $e^x = \frac1{2\pi}\sum_{k=-\infty}^\infty \frac{(-1)^k (e^{\pi}-e^{-\pi})(1+ki)}{1+k^2}e^{ikx}$ when $|x|<\pi$ . But what happens when $x=\pi$ ?

**chisigma** · May 4th 2010, 04:28 AM

Proceeding as in...

http://www.mathhelpforum.com/math-help/advanced-applied-math/141470-fourier-series.html

... we obtain...

$a_{0} = \frac{1}{\pi} \int_{-\pi}^{+ \pi} e^{x}\cdot dx = \frac{2}{\pi}\cdot \sinh \pi$

$a_{n} = \frac{1}{\pi} \int_{-\pi}^{+ \pi} e^{x}\cdot \cos nx \cdot dx = (-1)^{n} \frac{2\cdot \sinh \pi}{\pi\cdot (1+n^{2})}$

$b_{n} = \frac{1}{\pi} \int_{-\pi}^{+ \pi} e^{x}\cdot \sin nx \cdot dx = (-1)^{n-1} \frac{2\cdot n\cdot \sinh \pi}{\pi\cdot (1+n^{2})}$ (1)

... so that for $-\pi < x < \pi$ is...

$e^{x} = \frac{\sinh \pi}{\pi} \cdot \{1 + 2 \sum_{n=1}^{\infty} \frac{(-1)^{n}}{1 + n^{2}}\cdot (\cos nx - n\cdot \sin nx)\}$ (2)

Now if we remember that the series (2) for $x= \pm \pi$ converges to $\frac{e^{\pi} + e^{- \pi}}{2} = \cosh \pi$ we obtain...

$1 + 2 \sum_{n=1}^{\infty} \frac{1}{1 + n^{2}} = \sum_{n= -\infty}^{+\infty} \frac{1}{1+n^{2}} = \frac{\pi}{\tanh \pi} = 3.153348094937...$ (3)

... and from (3) ...

$\sum_{n=1}^{\infty} \frac{1}{1 + n^{2}} = \frac{\pi}{2\cdot \tanh \pi} - \frac{1}{2} = 1.076674047468...$ (4)

Kind regards

$\chi$ $\sigma$

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