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 $f(x) = (a_0, a_1, b_1, \dots, a_n, b_n, \dots)$ is a vector expressed in terms of the orthonormal basis $e = (e_0, e_{\alpha_1}, e_{\beta_1}, \dots, e_{\alpha_n}, e_{\beta_n}, \dots)$, where

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

The series $f = \sum_{k = 0}^\infty \langle f,e_k\rangle e_k$, where $e_k, k = 0, \dots$ 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 $$b_n = \dfrac{1}{\pi} \int_{-\pi}^\pi f(x)\sin(nx) \ dx, \ n = 0, 1, 2, \dots$$

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

I am also told the following:

The sequence of functions $\{ 1/\sqrt{2}, \sin(x), \cos(x), \sin(2x), \cos(2x), \dots \}$ 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.

  • $\begingroup$ 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. $\endgroup$ Sep 26, 2017 at 22:41
  • $\begingroup$ @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, $f(x) \sim \dfrac{a_0}{\sqrt{2}} + \sum_{n = 1}^\infty (a_n \cos(nx) + b_n \sin(nx)), -\pi < x < \pi$. $\endgroup$ Sep 26, 2017 at 22:47
  • $\begingroup$ Yes, I know, but what are you confused about when you try to do that? $\endgroup$ Sep 26, 2017 at 22:48
  • $\begingroup$ @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. $\endgroup$ Sep 26, 2017 at 22:48
  • $\begingroup$ 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? $\endgroup$ Sep 26, 2017 at 22:50

1 Answer 1


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

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

  • $\begingroup$ Thanks for the comprehensive response! I understand now. :) $\endgroup$ Sep 26, 2017 at 22:54

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .