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.
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
$\displaystyle \chi$ $\displaystyle \sigma$
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.