Infinite Addition

Big Idea

We can approximate any continuous $T$ -periodic function (that is differentiable on $(a, a+T)$ for some $a \in \mathbb R$ ) using its Fourier series given by

$\displaystyle f(t) = \frac{a_0}{2} + \sum_{n=1}^\infty a_n \cos\left( \frac{2\pi n t}{T} \right) + b_n \cos\left( \frac{2\pi n t}{T} \right),$

where

$\displaystyle a_n = \frac{2}{T} \int_{-T/2}^{T/2} f(t)\cos\left( \frac{2\pi n t}{T} \right)\, \mathrm dt,\quad \displaystyle b_n = \frac{2}{T} \int_{-T/2}^{T/2} f(t)\sin\left( \frac{2\pi n t}{T} \right)\, \mathrm dt.$

To simplify computations, it helps to exploit the symmetric properties of odd and even functions. Furthermore, we can use Fourier series to compute infinite sums.

A derivation of the Fourier series (and even a proof that it converges) will require us to venture into abstract linear algebra and even advanced real analysis, far beyond the scope of pre-university mathematics.

Questions

Question 1. Given $f(t) = |t|$ for $-2 \leq t \leq 2$ with periodic extension $f(t) = f(t+4)$ , evaluate the Fourier series of $f(t)$ .

(Click for Solution)

Solution. We contextualise the Fourier series to $T = 4$ :

$\displaystyle f(t) = \frac{a_0}{2} + \sum_{n=1}^\infty \left(a_n \cos \left( \frac{\pi n t}{2} \right) + b_n \sin \left( \frac{\pi n t}{2} \right) \right),$

where

$\displaystyle a_n = \frac 12 \int_{-2}^2 f(t) \cos \left( \frac{\pi n t}{2} \right)\, \mathrm dt,\quad b_n = \frac 12 \int_{-2}^2 f(t) \sin \left( \frac{\pi n t}{2} \right)\, \mathrm dt.$

We observe that $f(t) = |t| = |{-t}| = f(-t)$ so that $f$ is even on $[-2, 2]$ . Since $f(\cdot) \sin(\cdot)$ is odd on $[-2, 2]$ , $b_n = 0$ . For the even term, since $f(\cdot) \cos(\cdot)$ is even on $[-2 ,2]$ ,

$\displaystyle a_n = \frac 12 \int_{-2}^2 f(t) \cos \left( \frac{\pi n t}{2} \right)\, \mathrm dt = \int_0^2 t \cos \left( \frac{\pi n t}{2} \right)\, \mathrm dt.$

For the case $n = 0$ ,

$\displaystyle a_0 = \int_0^2 t \cos (0)\, \mathrm dt = \int_0^2 t\, \mathrm dt = \frac 12 \cdot 2 \cdot 2 = 2.$

For the case $n \neq 0$ , we first make the substitutions

$\displaystyle u = (\pi n t)/2 \quad \iff \quad \mathrm dt = \frac 2{\pi n}\, \mathrm du$

so that $u(2) = n \pi$ and

$\displaystyle \begin{aligned} a_n &= \int_0^2 t \cos \left( \frac{\pi n t}{2} \right)\, \mathrm dt \\ &= \frac 2{n \pi} \int_0^{n \pi} u \cos (u) \cdot \frac 2{\pi n}\, \mathrm du \\ &= \frac 4{ n^2 \pi^2 } \int_0^{n \pi} u \cos (u) \, \mathrm du. \end{aligned}$

Integrating by parts,

$\displaystyle \begin{aligned} \int u \cos(u)\, \mathrm du &= \underbrace{\sin(u)}_{\mathrm I} \underbrace{u}_{\mathrm S} - \int \underbrace{\sin(u)}_{\mathrm I} \underbrace{1}_{\mathrm D}\, \mathrm du \\ &= u \sin(u) - \int \sin(u)\, \mathrm du\\ &= u \sin(u) - (-\cos(u)) + C \\ &= u \sin(u) + \cos(u) + C. \end{aligned}$

Therefore,

$\begin{aligned} a_n &= \frac 4{ n^2 \pi^2 } \left[u \sin(u) + \cos(u) \right]_0^{n \pi} \\ &= \frac 4{ n^2 \pi^2 } ((0 + \cos(n \pi)) - (0 + 1)) \\ &= \frac 4{ n^2 \pi^2 } ( (-1)^n - 1) \\ &= \begin{cases} -\frac{8}{ n^2 \pi^2 }, & n\ \text{odd}, \\ 0, & n\ \text{even}. \end{cases} \end{aligned}$

Substituting the values of $a_n$ and $b_n$ , the Fourier series of $f(t)$ is given by

$\displaystyle f(t) = 1 - \frac {8}{\pi^2} \sum_{k=1}^\infty \frac{1}{(2k-1)^2} \cos \left( \frac{\pi (2k-1) t}{2} \right).$

Question 2. Given $f(t) = |{ \sin(t) }|$ , use the Fourier series of $f(t)$ to evaluate exactly the infinite sum

$\displaystyle \frac 1{2^2 - 1} + \frac 1{4^2 - 1} + \frac 1{6^2 - 1} + \cdots.$

(Click for Solution)

Solution. We contextualise the Fourier series to $T = \pi$ :

$\displaystyle f(t) = \frac{a_0}{2} + \sum_{n=1}^\infty \left(a_n \cos (2n t) + b_n \sin (2n t) \right),$

where

$\displaystyle a_n = \frac 2{\pi} \int_{-\pi/2}^{\pi/2} f(t) \cos (2nt)\, \mathrm dt,\quad b_n = \frac 2{\pi} \int_{-\pi/2}^{\pi/2} f(t) \sin (2nt)\, \mathrm dt.$

We observe that $f(t) = |{\sin(t)}| = |{\sin(-t)}| = f(-t)$ so that $f$ is even on $[-2, 2]$ . Since $f(\cdot) \sin(\cdot)$ is odd on $[-2, 2]$ , $b_n = 0$ . For the even term, since $f(\cdot) \cos(\cdot)$ is even on $[-2 ,2]$ ,

$\displaystyle a_n = \frac 2{\pi} \int_{-\pi/2}^{\pi/2} f(t) \cos (2nt)\, \mathrm dt = \frac 4{\pi} \int_0^{\pi/2} f(t) \cos (2nt)\, \mathrm dt.$

For the case $n = 0$ ,

$\displaystyle a_0 = \frac 4{\pi} \int_0^{\pi/2} \sin(t)\, \mathrm dt = \frac 4{\pi} \left[-{\cos(t)}\right]_0^{\pi/2} = \frac 4{\pi}(0 - (-1)) = \frac 4{\pi}.$

For the case $n \neq 0$ , we integrate by parts to get

$\displaystyle \begin{aligned} & \int \sin(t) \cos (2nt)\, \mathrm dt \\ &= \underbrace{-\cos(t)}_{\mathrm I} \underbrace{\cos (2nt)}_{\mathrm S} - \int \underbrace{-\cos(t)}_{\mathrm I} \underbrace{-2n \sin (2nt)}_{\mathrm D}\, \mathrm dt \\ &= -\cos(t) \cos(2nt) - 2n \int \cos(t) \sin(2nt)\, \mathrm dt \\ &= -\cos(t) \cos(2nt) - 2n \left( \underbrace{\sin(t)}_{\mathrm I} \underbrace{\sin (2nt)}_{\mathrm S} - \int \underbrace{\sin(t)}_{\mathrm I} \underbrace{2n \cos (2nt)}_{\mathrm D}\, \mathrm dt \right) \\ &= -\cos(t) \cos(2nt) -2n \sin(t) \sin(2nt) + 4n^2 \int \sin(t) \cos (2nt)\, \mathrm dt. \end{aligned}$

By algebruh,

$\displaystyle \int \sin(t) \cos (2nt)\, \mathrm dt = \frac{ \cos(t) \cos(2nt) + 2n \sin(t) \sin(2nt)}{4n^2 - 1} + C$

Therefore,

$\begin{aligned} a_n &= \frac 4{\pi} \left[\frac{ \cos(t) \cos(2nt) + 2n \sin(t) \sin(2nt)}{4n^2 - 1} \right]_0^{\pi/2} \\ &= \frac 4{\pi} \left( 0 - \frac 1{4n^2 - 1}\right) = -\frac{4}{\pi(4n^2 - 1)}. \end{aligned}$