# Fourier series of non-periodic function $f(x)=e^{-\frac{ax}{L}}$

The definition of Fourier series states that

It decomposes any periodic function or periodic signal into the weighted sum of a (possibly infinite) set of simple oscillating functions, namely sines and cosines (or, equivalently, complex exponentials)

I was a little confused as to how then, non-periodic functions like $$f(x) = e^{\frac{-ax}{L}}$$ defined over an interval $$[0,L]$$ can have a Fourier expansion ? We know that $$f(x) = \frac{a_0}{2} + \sum_{n=1}^{\infty}a_n\cos(nx) + \sum_{n=1}^{\infty}b_n\sin(nx)$$

Note $$\rightarrow$$ $$a,L$$ are constants $$>0$$

What I fail to understand is, how is this possible for a non-periodic function like an exponential ? Somewhere on another question in MSE, I learned that The Fourier series is described for the periodic extension of a non-periodic function, which failed to clarify my doubts.

The motivation to ask this question comes from my attempts to solve

Evaluating Coefficients for a Fourier Series when Exponential terms are present [Approach needed]

• What are the "standard coefficient finding formulae"? The formula of which I am aware involves integrating over a single period of a periodic formula. How do you define the $n$-th Fourier coefficient of a non-periodic function? Feb 27, 2019 at 5:03
• If you write down the formula for the Fourier coefficients of a non-periodic function, you get a representation of the Fourier series of its periodic extension (which is a discontinuous function even if the original function was continuous).
– Ian
Feb 27, 2019 at 5:31
• @Ian My recollection is that the periodic extension of a function refers to the extension of a function which is defined on a finite interval to a function defined on $\mathbb{R}$. For example, the function $$f : (-\frac{1}{2},\frac{1}{2})\to\mathbb{R} : x \mapsto x$$ is non-periodic, but has a sawtoolh function as its periodic extension. This is neither here nor there with respect to the question of a function defined on all of $\mathbb{R}$. Feb 27, 2019 at 5:41
• @XanderHenderson I have added the relations I used to find the coefficients Feb 27, 2019 at 5:41
• @IndrasisMitra In those formulae, $L$ the the period of the function. How are you choosing $L$ when $f$ is a non-periodic function defined on $\mathbb{R}$? Feb 27, 2019 at 5:42

Thanks to Ian and Xander Henderson for their suggestions.

We can consider any function defined on a finite interval $$(a,b)$$ or $$[a,b]$$ as a periodic function defined on $$R$$ by thinking that the function is extended to $$R$$ by repeating the values in $$[a,b]$$ to the remaining part of $$R$$.

Thus for $$f(x)$$ defined on $$[a,b]$$ where $$(\frac{b-a}{2}) = l$$, we have

$$f(x) = \frac{a_0}{2} + \sum_{n=1}^{\infty} \bigg[ a_n \cos\bigg(\frac{n\pi x}{l}\bigg) + b_n \sin\bigg(\frac{m\pi x}{l}\bigg)\bigg]$$

where,

$$a_0 = \frac{1}{l} \int_a^b f(x) \mathrm{d}x$$

$$a_n = \frac{1}{l} \int_a^b f(x)\cos\bigg(\frac{n\pi x}{l}\bigg)\mathrm{d}x$$

$$b_n = \frac{1}{l} \int_a^b f(x)\sin\bigg(\frac{n\pi x}{l}\bigg)\mathrm{d}x$$

Applying these for our function $$f(x) = e^{-\frac{ax}{L}}$$ defined on $$x \in [0,L]$$.

Hence, $$a = 0$$,$$b = L$$ and $$l=\frac{L}{2}$$, leads us to:

$$e^{\frac{-a x}{L}} = \frac{(1-e^{-a})}{a} + \sum_{n=1}^{\infty}\bigg[\frac{2a(1-e^{-a})}{(a)^2 + (2n\pi)^2}\cos\bigg(\frac{2n\pi x}{L}\bigg) + \frac{4n\pi(1-e^{-a})}{(a)^2 + (2n\pi)^2}\sin\bigg(\frac{2n\pi x}{L}\bigg)\bigg]$$