Extending the calculation of fourier coefficients to general interval

Question

While I was learning about Fourier Series I stumbled upon the following Let $f:[0, 2 \pi] \rightarrow \mathbb{C}$. One can calculate the coefficients $c_n$ of the Fourier series of $f$, as follows: (1) $c_k=\frac{1}{2 \pi} \int_0^{2\pi}f(x)e^{-ikx}dx$ (by an orthogonality argument).

In the book I am reading it is mentioned that the interval $[0,2\pi]$ can be easily extended to an interval $(a,b)$, where the coefficients can be calculated as follows: (2) $c_k=\frac{1}{b-a}\int_a^b f(x)e^{-2\pi ik \frac{x}{b-a}} dx$

I don't understand why the formula changed in that way (or how). Would be thankful if someone could show me how this transition works.

Answer 1

Here’s a general definition:

Definition (Fourier Coefficient):

Let $H$ be a Hilbert space, with inner product denoted $\langle\cdot,\cdot\rangle$ (in the case of complex Hilbert spaces, suppose the inner product is linear in the first slot). Let $(u_{\alpha})_{\alpha\in A}$ be an orthonormal basis (in the sense of Hilbert spaces). For any element $f\in H$, we define the Fourier coefficients of $f$ relative to $(u_{\alpha})_{\alpha\in A}$ to be the collection of numbers $(f_{\alpha})_{\alpha\in A}$, where $f_{\alpha}:=\langle f,u_{\alpha}\rangle$.

So, more generally, we see that the concept of “Fourier coefficients” relies on having an inner product space, and a specific choice of orthonormal basis. These $f_{\alpha}$ are simply “the components of the vector $f$ along the vector $u_{\alpha}$”, so it is a simple extension of an idea which we’re all familiar with in 3-dimensions, to arbitrary dimensions. Now, with this preliminary general definition out of the way, let’s talk about your example.

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First way of describing things (your book)

This is the way your book seems to be doing things.

We shall take $H= L^2([a,b],\frac{1}{b-a}m_{\text{Lebesgue}})$ the space of complex-valued $L^2$ functions with respect to the normalized Lebesgue measure on the interval. The inner product on this space is \begin{align} \langle f,g\rangle&:=\frac{1}{b-a}\int_a^bf(x)\overline{g(x)}\,dx. \end{align} Then, consider the sequence of functions $(u_k)_{k\in\Bbb{Z}}$ given by \begin{align} u_k(x)&=e^{\frac{2\pi i k x}{b-a}}. \end{align} Now, here are some facts:

• $H$ with the inner product defined above is a Hilbert space. I won’t prove this; this is a standard fact you can look up. Elements of this space are (equivalence classes of) functions, so this is also referred to as a “function space” (of (complex-valued) square-integrable functions on the interval $[a,b]$), but keeping in line with our general terminology above, we can also reasonably refer to an element $f\in H$ as a “vector” in this space.
• The sequence $(u_k)_{k\in\Bbb{Z}}$ defined above is an orthonormal basis. The fact that it is orthonormal is a direct computation: \begin{align} \langle u_k,u_l\rangle&=\frac{1}{b-a}\int_a^be^{\frac{2\pi ikx}{b-a}}\cdot\overline{e^{\frac{2\pi i lx}{b—a}}}\,dx\\ &=\frac{1}{b-a}\int_a^be^{\frac{2\pi i (k-l)x}{b-a}}\,dx\\ &=e^{\frac{2\pi i (k-l)a}{b-a}}\int_0^1e^{2\pi i (k-l)t}\,dt\tag{put $x=a+(b-a)t$}\\ &=\begin{cases} 1&\text{if $k=l$}\\ 0&\text{else} \end{cases}. \end{align} The fact that it forms an orthonormal basis is slightly harder to prove, but it’s also a standard fact you can find in books or online.

So, with this, we are perfectly in the situation of the definition above. We can thus calculate, for any $f\in H$, the Fourier coefficients of $f$, relative to this exponential basis, and we get \begin{align} f_k&=\langle f,u_k\rangle=\frac{1}{b-a}\int_a^bf(x)e^{-\frac{2\pi i kx}{b-a}}\,dx. \end{align}

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Alternative approach:

In some other texts, you may find things done this way, so I’ll describe it as well. You see, the constant factors are always a pain, and it’s a question of where you want to place the normalization factors.

This time, let’s define $H_1=L^2([a,b])$, with its usual inner product (which one can show forms a Hilbert space, but I won’t do it) \begin{align} \langle f,g\rangle_1&=\int_a^bf(x)\overline{g(x)}\,dx. \end{align} So, we no longer normalize the measure. That has to be taken into account elsewhere, and indeed, now we consider the sequence $(v_k)_{k\in \Bbb{Z}}$, \begin{align} v_k(x)&=\frac{1}{\sqrt{b-a}}e^{\frac{2\pi i k x}{b-a}}. \end{align} Then, you can verify that $(v_k)_{k\in\Bbb{Z}}$ is an orthonormal basis in $H_1$ (i.e rather than having the normalization factor in absorbed in the measure, we put absorb it into the basis).

We once again have a Hilbert space and an orthonormal basis, so we can consider Fourier coefficients: \begin{align} \langle f,v_k\rangle_1&=\int_a^bf(x)\overline{\frac{1}{\sqrt{b-a}}e^{\frac{2\pi i kx}{b-a}}}\,dx=\frac{1}{\sqrt{b-a}}\int_a^bf(x)e^{-\frac{2\pi i kx}{b-a}}\,dx. \end{align}

This is also a perfectly valid way of doing things; it’s a matter of preference.

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So, tldr: get yourself a Hilbert space, and an orthonormal basis, and simply look at the “components relative to that basis”, because that is exactly the purpose of Fourier coefficients.

Answer 2

The mapping $f\mapsto F$ defined by $$F(t)=f\left (a+{b-a\over 2\pi}t\right )$$ is an isometric isomorphism from $L^2((a,b);dx/(b-a))$ onto $L^2((0,2\pi);dt/2\pi).$ The $k$th Fourier coefficient of $F$ is equal $$ \widehat{F}(k)={1\over 2\pi}\int\limits_0^{2\pi}F(t)e^{-ikt}\,dt$$ Substitution $x=a+{b-a\over 2\pi}t$ leads to $$\widehat{F}(k)={1\over b-a}\int\limits_a^b f(x)e^{-2\pi i k(x-a)/(b-a)}\,dx\\ = e^{2\pi i ka/(b-a)}{1\over b-a}\,\int\limits_a^b f(x)e^{-2\pi i kx/(b-a)}\,dx $$ Thus it is natural to define the Fourier coefficient of $f$ as in OP. For simplicity the coefficients $a_k:=e^{2\pi i ka/(b-a)}$ of absolute value $1$ are ignored as $$\langle f,a_k\varphi_k\rangle (a_k\varphi_k)=\langle f,\varphi_k\rangle \varphi_k$$ where $f\in L^2$ and $\varphi_k(x)=^{-2\pi i kx/(b-a)}$