For Fourier series, where does the $\pi$ and the $2\pi$ come from?

Question

If a signal is given on an interval $[0, 2\pi]$, the Fourier series can be written as $$ f(t) = c_0 + \sum_{n = 1}^\infty a_n\cos nt + b_n\sin nt $$ with coefficients $$ c_0 = \frac1{2\pi}\int_0^{2\pi}f(t)\,dt\\ a_n = \frac1{\pi}\int_0^{2\pi}f(t)\cos nt\,dt\\ b_n = \frac1{\pi}\int_0^{2\pi}f(t)\sin nt\,dt $$

Specifically, for the coefficients, where does the $1/\pi$ and $1/2\pi$ come from?

Answer 1

They come from the periodicity of $2\pi$ (The length of the interval $[0,2\pi]$ is $2\pi-0 = 2\pi$). If the period is something else, say $T$, then the values would change.

Note that when $T = 2\pi$, then $\frac{1}{\pi} = \frac{2}{T}$ and $\frac{1}{2\pi} = \frac1T$. Similarly, the integration would be $\int_0^{2\pi} f(t) dt = \int_0^T f(t) dt.$

Answer 2

It's for niceness. For instance, if $f(x) = 1$, then all the $a_i$ and $b_i$ integrals turn out to be $0$, and you want to recover $c_0 = 1$. But it turns out that $\int_0^{2\pi} 1\,dt = 2\pi$ is too large, so you scale it back down.

Similarily, if $f(x) = \cos kx$, for some natural number $k>0$, then $a_k$ is the only integral that becomes non-zero, and we want to recover $a_k = 1$. However, it turns out that $\int_0^{2\pi}\cos kt\cdot \cos kt\,dt = \pi$, which is too large, so we need to scale it down. (And the argument for $b_k$ is exactly the same.)

Alternatively, we could do the definitions of $a_k, b_k$ and $c_0$ without the $\frac1{2\pi}$ and $\frac1\pi$ factors, but then we would have to make up for that someplace else, like with

$$f(x) = \frac{c_0}{2\pi} + \frac1\pi\sum_{n = 0}^\infty a_n\cos nx + b_n\sin nx$$

It's a matter of personal preference where you would best like to put these factors, but they must appear somewhere.

Answer 3

More abstractly, with an orthonormal basis $\{u_1, u_2, \dots\}$ in a (real) inner product space, we have $$ f = \sum a_n u_n $$ where the coefficients are the inner products $a_n = \langle f,u_n\rangle$. In our traditional case, if the inner product is $$ \langle f,g \rangle = \int_0^{2\pi} f(x)\;g(x)\;dx $$ we need to do integrals $$ \int_0^{2\pi} 1^2\;dx = 2\pi,\\ \int_0^{2\pi} \cos(nx)^2\;dx = \pi,\\ \int_0^{2\pi} \sin(nx)^2\;dx = \pi, $$ to find the denominators to use so that we have an orthonormal basis.

Answer 4

Please look at the derivation of these coefficients, Suppose you have the formula and you want to calculate the coefficients then for example i will tell you for $c_0$ ,

Take integration on both sides w.r.t to $t$ with limit from $0$ to $2\pi$, $$ \int_0^{2\pi}f(t)dt=\int_0^{2\pi}c_0dt+\sum_{n=1} \left( \int_0^{2\pi}a_ncos(nt)dt+\int_0^{2\pi}b_nsin(nt)dt\right) $$ All coefficients are constant w.r.t. $t$ so, $$ \int_0^{2\pi}f(t)dt=c_0\int_0^{2\pi}dt+\sum_{n=1} \left( a_n\int_0^{2\pi}cos(nt)dt+b_n\int_0^{2\pi}sin(nt)dt\right) $$ But $\int_0^{2\pi}cos(nt)dt$ and $\int_0^{2\pi}sin(nt)dt$ are $0$ so you get, $$ \int_0^{2\pi}f(t)dt=c_0\int_0^{2\pi}dt $$ Which is eqaul to, $$ \int_0^{2\pi}f(t)dt=c_02\pi $$ Thus you get, $$ c_0=\frac{1}{2\pi}\int_0^{2\pi}f(t)dt $$