# Fourier series for exponential function

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.

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 $$(0;1)$$.

• Find this theorem in your textbook. See what properties of $f$ are required for the formula. Does $c^x$ satisfy those properties? (In particular, the formula fails at the endpoints $a,b$ of your interval.) Jun 21, 2022 at 14:52
• Thanks. I've found the properties, the main difficulty for me was non-periodicity of my function and the interval other than $(-\pi;\pi)$. Jun 21, 2022 at 20:46

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.

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

• Thanks! I've obtained the same result using my formulas, so those formulas seem to be correct. I just wasn't sure about them. Jun 21, 2022 at 20:33

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).