Fourier series for function $f(x)=c^x$, $c\in\mathbb Z$, $c>1$ on interval $(a,b)$, where $a,b\in\mathbb R$, $a<b$.
Can I use the next formulas for this case?:
$$f(x)=\frac{a_0}{2}+\sum\limits_{n=1}^{+\infty}\left[a_n\cos\left(\frac{\pi nx}{l}\right)+b_n\sin\left(\frac{\pi nx}{l}\right)\right],$$
$$a_n=\frac{1}{l}\int\limits_a^bf(x)\cos\left(\frac{\pi nx}{l}\right)dx,$$
$$b_n=\frac{1}{l}\int\limits_a^bf(x)\sin\left(\frac{\pi nx}{l}\right)dx,$$
where $l=(b-a)/2$.
Particularly I need a Fourier series for function $f(x)=2^x$ on interval $(0;1)$.
Best Answer
For the specific case of interest where
$$f(x)=2^x\tag{1}$$
the corresponding Fourier series is
$$\tilde{f}(x)=\frac{1}{\log (2)}+\underset{K\to \infty }{\text{lim}}\left(\sum\limits_{n=1}^K \left(\frac{\log(4)}{4 \pi^2 n^2+\log^2(2)}\, \cos(2 \pi n x)-\frac{4 \pi n}{4 \pi^2 n^2+\log^2(2)}\, \sin(2 \pi n x)\right)\right)\tag{2}$$
which is valid on the interval $0<x<1$.
Figure (1) below illustrates formula (2) for $\tilde{f}(x)$ in orange overlaid on the blue reference function $f(x)=2^x$ where formula (2) is evaluated at $K=100$.
Figure (1): Illustration of formula (2) for $\tilde{f}(x)$ in orange overlaid on $f(x)=2^x$ in blue