Fourier series for exponential function

Question

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 $(0;1)$.

Answer 1

In the most general case you proposed, you can perfectly use the written formulas. But, for your particular case (2^x, 0<x<1), since the representation can possibly be odd, I'd recommend you to use the formulas that just involve the sine (they're the easiest ones to calculate).

Answer 2

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 $0<x<1$.

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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