0
$\begingroup$

I've been told to construct a Fourier Series for the odd function that has period $2\pi$ and is equal to $\cos(x)$ for $x \in (0,\pi]$. For $f$ that is $2\pi$ period I have a formula
$$b_n=\int_{-\pi}^\pi f(x)\sin(nx) \, dx.$$
I don't know what this question is asking, I thought it wanted $f(x)=\cos(x)$ but when substituting in the equation, that would give an odd integrand so it would just be $0$.

$\endgroup$
  • $\begingroup$ First define $f(x) := \cos(x)$ for $x \in (0,\pi]$, then extend it to $[-\pi,0]$ in such a way that $f(x)$ is an odd function. Next extend it to a periodic function with period $ 2 \pi$. Now find the Fourier series of this $f$. $\endgroup$ – MichaelNgelo Jan 23 '13 at 0:36
  • $\begingroup$ This makes sense, but how would you extend $\cos(x)$ to $[-\pi,0]$ so that f(x) is an odd function since at $x=0$, $\cos(x)$ is not $0$? Don't odd functions have to pass through $(0,0)$ $\endgroup$ – user51327 Jan 23 '13 at 0:52
  • 1
    $\begingroup$ The extension is discontinuous: $f(x)=\cos x$ on $(0,\pi]$, $f(0)=0$, $f(x)=-\cos x$ on $[-\pi,0)$. $\endgroup$ – David Moews Jan 23 '13 at 1:23
0
$\begingroup$

Since $f(x)$ is odd and equals $\cos x$ on $(0,\pi]$, it equals $-\cos (-x)=-\cos x$ on $[-\pi,0)$. So, $$ b_n:=\int_{-\pi}^\pi f(x) \sin nx \, dx $$ $$ = \int_{-\pi}^0 (-\cos x) \sin nx \, dx + \int_{0}^\pi (\cos x) \sin nx \, dx. $$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.