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I'm trying to find Fourier cosine series for function $$f(x) = \begin{cases} \frac{\pi}{2}-x,\quad &\text{if } x\in\left[0,\frac{\pi}{2}\right) \\ \pi,\quad &\text{if } x\in\left[\frac{\pi}{2},\pi\right] \end{cases}$$ in $[0,\pi]$.

I found coefficients $$b_k=0,\quad a_0=\frac{5\pi}{4}\quad \text{and}\quad a_k=\frac{2}{\pi}\cdot\frac{1-\cos\frac{k\pi}{2} + k\pi\sin k\pi - k\pi\sin\frac{k\pi}{2}}{k^2}$$ where $k\in\mathbb{N}$, but I don't know how to simplify $a_k$ to write cosine series as $$\frac{5\pi}{4}+\sum_{k=1}^{\infty}\ ...\ \cos kx$$ and investigate pointwise convergence in $[0,\pi]$.

Can anyone help me with this?

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  • $\begingroup$ Finding the coefficients amounts to finding the series. For convergence use the Dirichlet condition $\endgroup$
    – Shubham Johri
    Dec 7, 2020 at 21:50

1 Answer 1

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$a_k=0$ if $k$ is a multiple of $4$. $a_k=\frac 1 {\pi j^{2}}$ if $k=2j$ with $j$ odd. $a_k=\frac 1 {\pi} \frac { 1-(2j-1)\pi(-1)^{j+1}} {(2j-1)^{2}}$ if $k=2j-1$.

In all cases $|a_k| \leq \frac C {k^{2}}$ for some $C$ so $\sum |a_k| <\infty$. This imlies that the Fourier series converges to $f$ at very point.

I have just used basic properties sine and cosine functions.

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