Given that $f:[-\pi,\pi] \rightarrow \mathbb{R}$ satisfies
$$f(-x) = -f(x), \quad f(\pi) = 0, \quad f'''(x) = -6.$$
Prove that $f''(x) = -6x$ and hence, find the fourier series of $f$ by using integration by parts (repeatedly).
So I know how to find fourier series when explicitly given a function (integrate with respective bounds and equate coefficients), but this one requires us to find it without am explicit function given.
Can someone show me how this would be done $\textbf{without}$ using an explicit function (i.e. just using the given information and that second derivative property)?
This is what I did so far:
Since $f(-x) = -f(x)$ then $-f'(-x) = -f'(x)$ and $f''(-x) = -f''(x)$.
Hence, the second derivative is also an odd function and so $\int f'''(x)dx = \int -6 \ dx \implies f''(x) = -6x$ since $f''$ is odd (no constant term).
Then now I'm stuck, because I'm not sure what to even integrate to get the fourier series in the form:
$$Sf(x) = \sum_{k=1}^\infty b_k \sin kx.$$
Note that $f(0) = 0$ since $f$ is odd and continuous.
\begin{eqnarray} \int_{-\pi}^\pi e^{-inx} f(x) dx &=& -\int_{-\pi}^\pi \sin (nx) f(x) dx \\ &=& -2\int_{0}^\pi \sin (nx) f(x) dx \\ &=& -2 \left ( -[{1 \over n} \cos (nx) f(x) ]_0^\pi + {1 \over n} \int_{0}^\pi \cos (nx) f'(x) dx \right) \\ &=& - {2 \over n} \left ( \int_{0}^\pi \cos (nx) f'(x) dx \right) \\ &=& - {2 \over n} \left ( [{1 \over n} \sin (nx)]_0^\pi + {1 \over n} \int_{0}^\pi \sin (nx) f''(x) dx \right) \\ &=& {2 \over n^2} \int_{0}^\pi \sin (nx) f''(x) dx \\ \end{eqnarray} (Assuming I have made no mistakes, which is unlikely.)