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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?

Ѕᴀᴀᴅ
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Joe Goldiamond
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1 Answers1

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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*}

Ѕᴀᴀᴅ
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  • Alex, thanks for your clear explanation, just 2 minor questions. In the second line after which implies: you write $k^2 - k$, I don't understand what you are doing and why. And in the last line 'suddenly' $\frac{1}{N}$ appears. Is this a standard result? – Joe Goldiamond Mar 21 '18 at 08:01
  • @JoeGoldiamond For the first question, since it's intended to make approximation, so for large $k$, there is $\frac1{k^2-α^2}≈\frac1{k^2-k}$. For the second question, it's a telescoping sum, i.e.$$\sum_{k=N+1}^∞\left(\frac1{k-1}-\frac1k\right)=\frac1N-\frac1{N+1}+\frac1{N+1}-\frac1{N+2}+\cdots=\frac1N.$$ – Ѕᴀᴀᴅ Mar 21 '18 at 08:18
  • That makes sense! Thanks for the link to the telescoping sum. – Joe Goldiamond Mar 21 '18 at 08:50