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 $(0;1)$.

  • 1
    $\begingroup$ 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.) $\endgroup$
    – GEdgar
    Jun 21, 2022 at 14:52
  • $\begingroup$ Thanks. I've found the properties, the main difficulty for me was non-periodicity of my function and the interval other than $(-\pi;\pi)$. $\endgroup$ Jun 21, 2022 at 20:46

2 Answers 2


For the specific case of interest where


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 $K=100$.

Illustration of formula (2) in orange overlaid on formula (1) in blue

Figure (1): Illustration of formula (2) for $\tilde{f}(x)$ in orange overlaid on $f(x)=2^x$ in blue

  • $\begingroup$ Thanks! I've obtained the same result using my formulas, so those formulas seem to be correct. I just wasn't sure about them. $\endgroup$ 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).

  • $\begingroup$ As it’s currently written, your answer is unclear. Please edit to add additional details that will help others understand how this addresses the question asked. You can find more information on how to write good answers in the help center. $\endgroup$
    – Community Bot
    Jun 21, 2022 at 15:12
  • $\begingroup$ Thanks, I've tried to use the cosine, obtained the Fourier series and plotted the graphs, it fits. I'll try the sine, too. $\endgroup$ Jun 21, 2022 at 20:50
  • $\begingroup$ The "sine method" works well too! $\endgroup$ Jun 21, 2022 at 21:13

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