Given that f(x)= sinx from 0<x< pi, why is (f- sinx) an even function and hence perform fourier cosin series?
I thought f(x) is an odd function hencd fourier sine?
photo 1.pdfphoto 2.pdf
here is the question
You question is:
Show that the Fourier series S(x) of:
$\displaystyle f(x)=\left\{ \begin{array}{ll}0, & -\pi \le x \le 0 \\sin(x), & 0<x\le \pi \end{array} \right.$
is:
$\displaystyle S(x)=\frac{2}{\pi}\left(\frac{1}{2}+\sum_{k=1}^\in fty \frac{\cos(2\pi x)}{1-4k^2} \right )+\frac{1}{2}\sin(x)$
The answer to your original quaetion is because:
$\displaystyle f(x)=\left|\frac{\sin(x)}{2} \right|+\frac{\sin(x)}{2} \ \ -\pi \le x \le \pi$
The first term on the right is even and so has a Fourier series with cosine terms only and the second is already in Fourier form and indeed is the right most term in the given series.
(there may be errors in the LaTeX above, but I can't read the rendered images on this machine, but they look OK on CodeCogs)
CB
f(x) is neither odd nor even. A general function on an interval symettric about the origin can always be split into the sum of an even and an odd function (there is even a systematic method of doing so).
All I have done is split your function into the sum of an odd and even fuction. It is convienient that the odd part is already in the form of a Fourier series and so needs no further work.
CB