# Using Inner Product Spaces to Derive Fourier Series: Require Clarification.

I'm going through a chapter on Fourier series, and I've encountered some confusion.

The textbook states the following:

In the context of linear spaces, we can immediately write $f(x) = (a_0, a_1, b_1, ..., a_n, b_n, ...)$ is a vector express in terms of the orthonormal basis $e = (e_0, e_{\alpha_1}, e_{\beta_1}, ..., e_{\alpha_n}, e_{\beta_n}, ...)$

$e_0 = \dfrac{1}{\sqrt{2}}, e_{\alpha_n} = \cos(nx), e_{\beta_n} = \sin(nx)$

The series $f = \sum_{k = 0}^\infty \langle f,e_k\rangle e_k$, where $e_k, k = 0, ....$ is a renumbering of the basis vectors. This is the standard expansion of $f$ in terms of the orthonormal basis and is the Fourier series for $f$. Invoking the linear space theory therefore helps us understand how it is possible to express any function piecewise continuous in $[-\pi, \pi]$ as the series expansion

$f(x) \sim \dfrac{a_0}{\sqrt{2}} + \sum_{n = 1}^\infty (a_n \cos(nx) + b_n \sin(nx)), -\pi < x < \pi$,

where $a_n = \dfrac{1}{\pi} \int_{-\pi}^\pi f(x)\cos(nx) \ dx$, and

$b_n = \dfrac{1}{\pi} \int_{-\pi}^\pi f(x)\sin(nx) \ dx, n = 0, 1, 2, ...$

The orthonormal basis is $\{ 1/\sqrt{2}, \sin(x), \cos(x), \sin(2x), \cos(2x), ... \}$.

I am also told the following:

The sequence of functions $\{ 1/\sqrt{2}, \sin(x), \cos(x), \sin(2x), \cos(2x), ... \}$ form an infinite orthonormal sequence in the space of all piecewise continuous functions on the interval $[-\pi, \pi]$, where the inner product $\langle f, g\rangle$ is defined by $\langle f, g\rangle = \dfrac{1}{\pi} \int^\pi_{-\pi} f \ \bar{g} \ dx$, where the overbar denotes the complex conjugate.

However, I am struggling to expand $f = \sum_{k = 0}^\infty \langle f,e_k\rangle e_k$ to get $f(x) \sim \dfrac{a_0}{\sqrt{2}} + \sum_{n = 1}^\infty (a_n \cos(nx) + b_n \sin(nx)), -\pi < x < \pi$. I would greatly appreciate it if people could please take the time to demonstrate this step-by-step.

• Can you say more about what you're confused about? Some of the $e_k$ are equal to $\cos nx$ for some $n$, and that's where the cosine terms come from. Some of the $e_k$ are equal to $\sin nx$ for some $n$, and that's where the sine terms come from. And one of them is equal to $\frac{1}{\sqrt{2}}$, and that's where the $\frac{a_0}{\sqrt{2}}$ term comes from. – Qiaochu Yuan Sep 26 '17 at 22:41
• @QiaochuYuan Thanks for the response. I'm primarily concerned with how we expand $f = \sum_{k = 0}^\infty \langle f,e_k\rangle e_k$ to get, as the textbook says, $f(x) \sim \dfrac{a_0}{\sqrt{2}} + \sum_{n = 1}^\infty (a_n \cos(nx) + b_n \sin(nx)), -\pi < x < \pi$. – The Pointer Sep 26 '17 at 22:47
• Yes, I know, but what are you confused about when you try to do that? – Qiaochu Yuan Sep 26 '17 at 22:48
• @QiaochuYuan The term $f = \sum_{k = 0}^\infty \langle f,e_k\rangle e_k$. I've never dealt with abstract inner product spaces, so I think this is where the confusion lies. – The Pointer Sep 26 '17 at 22:48
• There are various steps involved in converting that expression into a Fourier series, and you're stuck on at least one of them; which one is it? Can you write down the integral $\langle f, e_k \rangle$, using the definition of the inner product provided? – Qiaochu Yuan Sep 26 '17 at 22:50

It boils down to calculate the product

$$\langle f, e_k\rangle = \frac{1}{\pi}\int_{-\pi}^{\pi}{\rm dx}~f(x)e_k(x)$$

Note that

$k = 0$

$$\langle f, e_0\rangle = \frac{1}{\pi}\int_{-\pi}^{\pi}{\rm dx}~f(x)e_0(x) = \frac{1}{\sqrt{2}} \left[\frac{1}{\pi}\int_{-\pi}^{\pi}{\rm dx}~f(x)\right] = \frac{a_0}{\sqrt{2}}$$

$k = \alpha_n$

$$\langle f, e_{\alpha_n}\rangle = \frac{1}{\pi}\int_{-\pi}^{\pi}{\rm dx}~f(x)e_{\alpha_n}(x) = \frac{1}{\pi}\int_{-\pi}^{\pi}{\rm dx}~f(x)\cos(nx) = a_n$$

$k = \beta_n$

$$\langle f, e_{\beta_n}\rangle = \frac{1}{\pi}\int_{-\pi}^{\pi}{\rm dx}~f(x)e_{\beta_n}(x) = \frac{1}{\pi}\int_{-\pi}^{\pi}{\rm dx}~f(x)\sin(nx) = b_n$$

Putting everything together you get

$$f(x) = \sum_{k=0}^{+\infty}\langle f, e_k\rangle e_k(x) = \frac{a_0}{\sqrt{2}} + \sum_{k=1}^{+\infty}a_n\cos(nx) + b_n\sin(nx)$$

• Thanks for the comprehensive response! I understand now. :) – The Pointer Sep 26 '17 at 22:54