Hello,

I can't solve the following problem involving the Fourier series of a function and need som guidance. I appriciate any suggestion. Thank you kindly.

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- Aug 18th 2010, 12:50 AM4MathShow the Fourier series of a function
Hello,

I can't solve the following problem involving the Fourier series of a function and need som guidance. I appriciate any suggestion. Thank you kindly. - Aug 18th 2010, 05:36 AMchisigma
The even function in $\displaystyle [-\pi,\pi]$ is...

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

... and its cosine Fourier coefficients are...

$\displaystyle \displaystyle a_{n} = \frac{2}{\pi} \int_{0}^{2} (2-x)\ \cos nx \ dx = $

$\displaystyle \displaystyle = \frac{4}{\pi} |\frac{\sin n x }{n}|_{0}^{2} - \frac{2}{\pi} |\frac{n x \sin nx + \cos nx}{n^{2}}|_{0}^{2} = $

$\displaystyle \displaystyle = \frac{2}{\pi} \frac{1- \cos 2n}{n^{2}} = \frac{4}{\pi} \frac{\sin^{2} n}{n^{2}}$ (2)

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$ - Aug 18th 2010, 08:07 AM4Math
So when can the rule: odd function times an even function becomes zero, applies when solving for Fourier coefficients??

- Aug 19th 2010, 09:56 PMchisigma
- Aug 20th 2010, 03:02 AM4Math
When solving for $\displaystyle a_{n}$ :

$\displaystyle \displaystyle a_{n} = \frac{2}{\pi} (({2}\int_{0}^{2} cos nx \ dx)-(\int_{0}^{2} x\ \cos nx \ dx))$

the second integral:

$\displaystyle -(\int_{0}^{2} x\ \cos nx \ dx)$

$\displaystyle x$ is an odd function and $\displaystyle cos nx $ is an even function. And the result of that is an odd function times an even function becomes zero.

Why can this not be applied here?

By the way, thank you kindly for your response. - Aug 20th 2010, 03:34 AMchisigma