It is asked to find the Fourier Cosine Series for the function defined by
$$f(x) = \cos \frac{\pi x}{l}, x \in [0, l/2]$$ $$f(x) = 0, (l/2, l]$$
I thought it should be
$$\frac{a_o}{2} + \sum a_n \cos(\frac{n \pi x}{l/2})$$
where $a_n = \frac{2}{l/2} \int_0^{l/2} f(x) \cos ( \frac{n \pi x}{l/2})$
But this is not the answer (this function is 0 for [0, l/2]!).
What should I do? And how do I find the correct formula for it? If someone could give a brief explain of how to obtain this, I would be really glad!
Thanks in advance
Use a Fourier Sine series for which \begin{align} f(x) = \sum_{n=1}^{\infty} B_{n} \, \sin\left(\frac{n \pi x}{L} \right) \end{align} where \begin{align} B_{n} = \frac{2}{L} \, \int_{0}^{L} f(x) \, \sin\left(\frac{n \pi x}{L} \right) dx. \end{align}
For the given $f(x)$ it is seen that \begin{align} B_{n} &= \frac{2}{L} \, \int_{0}^{L} \cos\left(\frac{\pi x}{L} \right) \, \sin\left(\frac{n \pi x}{L} \right) dx + \frac{2}{L} \, \int_{0}^{L}(0) \, \sin\left(\frac{n \pi x}{L} \right) dx \\ &= \frac{2 (n- \sin\left( \frac{\pi n}{2} \right))}{\pi (n^2-1)} \end{align} this is valid for $n \geq 2$, whereas $B_{1} = \frac{1}{2}$.
This leads to \begin{align} f(x) = \left\{ \begin{array}{cc} \cos\left(\frac{\pi x}{L} \right) & 0 \leq x \leq \frac{L}{2} \\ 0 & \frac{L}{2} \leq x \leq L \end{array} \right. \hspace{5mm} = \frac{1}{\pi} \, \sin\left(\frac{\pi x}{L} \right) + \frac{2}{\pi} \, \sum_{n=2}^{\infty} \frac{ (n- \sin\left( \frac{\pi n}{2} \right))}{n^2-1} \, \sin\left(\frac{n \pi x}{L} \right) \end{align}
Fourier Cosine series
\begin{align} f(x) = \frac{A_{0}}{2} + \sum_{n=1}^{\infty} A_{n} \, \cos\left( \frac{n \pi x}{L} \right) \end{align} where \begin{align} A_{n} = \frac{2}{L} \, \int_{0}^{L} f(x) \, \cos\left(\frac{n \pi x}{L} \right) dx. \end{align} Now, \begin{align} A_{0} &= \frac{2}{L} \, \int_{0}^{L} f(x) dx = 0 \\ A_{1} &= \frac{2}{L} \, \int_{0}^{L/2} \cos^{2}\left(\frac{\pi x}{L}\right) dx = 1 \\ A_{n \geq 2} &= \frac{2}{L} \, \int_{0}^{L} f(x) \, \cos\left(\frac{n \pi x}{L} \right) dx = 0 \end{align} The Fourier Cosine series of $f(x)$ is $f(x)$.