Hi i was wondering if anyone could help me with the following problem.
In particular part c i have obtained the series i just do not know how to get to the result is it a simple case of just substituting the value of x into the Fourier series. also i am struggling with part a i don't understand how to sketch a diagram over three periods
if anyone could provide help or a website in which i could use to learn how to do these parts of the problem i would be extremely happy, Thanks
My Fourier series is $F(x)=\frac{\pi}{2}+\sum_{i=1}^n \frac{1-\cos(\pi n)}{n^2}\cos(nx)$
(a) You are given $f(x)$ for $-\pi\leq x\leq \pi$. Hence you can sketch the function over $[-\pi,\pi]$. The condition $f(x+2\pi)=f(x)$ means that $f$ is $2\pi$-periodic. So to sketch $f$ over three periods means to sketch $f$ on an interval of length $3\cdot2\pi=6\pi$ and you can do so by replicating the sketch on $[-\pi,\pi]$. What you get is:
(b) To obtain the Fourier series of $f$ we must calculate its Fourier coefficients. In complex exponential form, the $k$-th coefficient is $$ \hat{f}(k) = \frac{1}{2\pi} \int_{-\pi}^{\pi}f(t)e^{-ikt} \,dt $$ Here because $f$ is given piecewise we split the integral in two: \begin{align} \hat{f}(k) &= \frac{1}{2\pi} \int_{-\pi}^{\pi}f(t)e^{-ikt} \,dt \\ &= \frac{1}{2\pi}\left( \int_{-\pi}^{0} (t-\pi)e^{-ikt} \,dt+\int_{0}^{\pi} (\pi-t)e^{-ikt} \,dt \right) \end{align} It is a simple calculus exercise to evaluate these integrals (do it by integration by parts for the $te^{-ikt}$ components). Don't forget to treat the cases $k=0$ and $k\neq0$ separately. The result is: \begin{align} \hat{f}(k) &= \begin{cases}\frac{\pi}{2}&&k=0\\ \frac{1+e^{-i\pi k}}{\pi k^2} && k\neq0\end{cases} \\ &=\begin{cases}\frac{\pi}{2}&&k=0\\ \frac{1+(-1)^{k+1}}{\pi k^2} && k\neq0\end{cases} \end{align} The Fourier series of $f$ is then $$ \sum_{k=-\infty}^{\infty}\hat{f}(k)e^{ikt} $$
(c) Since $f$ is continuous at $0$ and $C^1$ in a punctured neighborhood of $0$, we know the symmetric partial sums of the Fourier series of $f$ converge to $f(0)$. That is, $$ \lim_{n\to\infty}\sum_{k=n}^{n}\hat{f}(k) = f(0) $$ which, in this case, becomes $$ \frac{\pi}{2} + \lim_{n\to\infty}\sum_{\substack{-n\leq k \leq n \\ k\neq0}}\frac{1+(-1)^{k+1}}{\pi k^2} = \pi\tag{1} $$ Since $\displaystyle\frac{1+(-1)^{k+1}}{\pi k^2}$ is even, $(1)$ reduces to $$ \frac{\pi}{2} + 2 \lim_{n\to\infty}\sum_{k=1}^{n}\frac{1+(-1)^{k+1}}{\pi k^2} = \pi\tag{2} $$ Finally, we note that the numerator $1+(-1)^{k+1}$ equals $0$ when $k$ is even and $2$ when $k$ is odd. So $(2)$ reduces to $$ \frac{\pi}{2} + \frac{4}{\pi} \sum_{k=1}^{\infty}\frac{1}{(2k-1)^2} = \pi\tag{3} $$ The series in $(3)$ is precisely the one we are asked to evaluate. From $(3)$ we obtain $$ \sum_{k=1}^{\infty}\frac{1}{(2k-1)^2} = \frac{\pi^2}{8} $$ as was to be shown.