<s>[INST] Fourier series applied I have a question concerning the Fourier series:
I started with the following:
$$\cos (\alpha x)= \frac{1}{2}a_0+ \sum_{k=1}^{\infty}a_k\cos(kx).$$
I proved that this series with Fourier coefficients is equal to:
$$\cos \alpha x= \frac {\sin \alpha \pi}{\alpha \pi}+ \frac{2 \alpha}{\pi}\sin \alpha \pi \sum_{k=1}^{\infty}(-1)^{k-1}\frac{\cos kx}{k^2-\alpha^2},$$
and $\alpha$ is NOT an integer.
But now I am stuck:
Suppose that $F_N(x)$ is the above Fourier series truncated at the $N$-th term. Can we show that:
$$F_N(\pi) = \cos \alpha \pi+ \frac{2 \alpha}{\pi}\sin \alpha \pi \sum_{k=N+1}^{\infty}\frac{1}{k^2-\alpha^2},$$
and hence: $$F_N(\pi) \simeq \cos \alpha \pi + \frac{2\alpha}{N \pi} \sin \alpha \pi?$$
I started by writing:
$$\cos \alpha \pi = \frac {\sin \alpha \pi}{\alpha \pi}+ \frac{2 \alpha}{\pi}\sin \alpha \pi \sum_{k=1}^{\infty}(-1)^{k-1}\frac{\cos k\pi}{k^2-\alpha^2}$$
and $\cos k \pi = (-1)^k$. But now I am completely lost in what to do.. Can someone help me?
 [/INST] Because the series $\displaystyle \sum\limits_{k = 1}^\infty (-1)^{k - 1} \frac{\cos kπ}{k^2 - α^2} = -\sum\limits_{k = 1}^\infty \frac{1}{k^2 - α^2}$ converges, then\begin{align*}
\cos απ &= \frac{\sin απ}{απ} + \frac{2α}{π} \sin απ \sum_{k = 1}^\infty (-1)^{k - 1} \frac{\cos kπ}{k^2 - α^2}\\
&= \frac{\sin απ}{απ} - \frac{2α}{π} \sin απ \sum_{k = 1}^\infty \frac{1}{k^2 - α^2}\\
&= \frac{\sin απ}{απ} - \frac{2α}{π} \sin απ \sum_{k = 1}^N \frac{1}{k^2 - α^2} - \frac{2α}{π} \sin απ \sum_{k =  N + 1}^\infty \frac{1}{k^2 - α^2}\\
&= F_N(π) - \frac{2α}{π} \sin απ \sum_{k =  N + 1}^\infty \frac{1}{k^2 - α^2},
\end{align*}
which implies\begin{align*}
F_N(π) &= \cos απ + \frac{2α}{π} \sin απ \sum_{k =  N + 1}^\infty \frac{1}{k^2 - α^2}\\
&\approx \cos απ + \frac{2α}{π} \sin απ \sum_{k =  N + 1}^\infty \frac{1}{k^2 - k}\\
&= \cos απ + \frac{2α}{π} \sin απ \sum_{k =  N + 1}^\infty \left( \frac{1}{k - 1} - \frac{1}{k} \right)\\
&= \cos απ + \frac{2α}{π} \sin απ · \frac{1}{N} = \cos απ + \frac{2α}{Nπ} \sin απ.
\end{align*}
</s>