Fourier Series Proof

Math Calculus

Starting the Proof Off

The Taylor Series uses $x^n$ as building blocks for a function: Taylor Series Proof

However, we can use $\sin (nx)$ and $\cos (nx)$ as well. This will be our starting point to derive the Fourier Series:

$$\begin{gathered} f(x)=a_0\cos (0x)+b_0\sin (0x)+a_1\cos (x)+b_1\sin (x)+a_2\cos (2x)+b_2\sin (2x)... \\ f(x)=a_0+\sum _{n=1}^{\infty }(a_n\sin (nx)+b_n\cos (nx)) \end{gathered}$$

This will be the basic equation we will use.

Finding $a_0$

Let’s integrate the equation on both sides, and bound by $[-\pi ,\pi ]$:

$${\int }_{-\pi }^{\pi }f(x)dx={\int }_{-\pi }^{\pi }{a}_{0}dx+\sum _{n=1}^{\infty }{\int }_{-\pi }^{\pi }{a}_n\cos (nx)dx+\sum _{n=1}^{\infty }{\int }_{-\pi }^{\pi }{b}_n\sin (nx)dx$$

The first integral evaluates to $2\pi a_0$. Since the third integral is an odd function, it evaluates to $0$. The second integral can be expressed as:

$$\begin{gathered} a_n{\int }_{-\pi }^{\pi }\cos (nx)dx \\ =\frac{a_n}{n}(\sin (n\pi )-\sin (-n\pi )) \\ =0 \end{gathered}$$

So now we have:

$$\begin{gathered} 2\pi a_0={\int }_{-\pi }^{\pi }f(x)dx \\ {a}_0=\frac{1}{2\pi }{\int }_{-\pi }^{\pi }f(x)dx \end{gathered}$$

Finding $a_n$

Let’s multiply the entire equation by $\cos (mx)$, where $m\in {\mathbb{Z}}^{+}$ ($m$ is a positive integer):

$$f(x)\cos (mx)=a_0\cos (mx)+\sum _{n=1}^{\infty }{a}_n\cos (nx)\cos (mx)+b_n\sin (nx)\cos (mx)$$

Now integrate on both sides, and bound by $[-\pi ,\pi ]$:

$${\int }_{-\pi }^{\pi }f(x)\cos (mx)dx={\int }_{-\pi }^{\pi }{a}_0\cos (mx)dx+\sum _{n=1}^{\infty }{\int }_{-\pi }^{\pi }{a}_n\cos (nx)\cos (mx)dx+\sum _{n=1}^{\infty }{\int }_{-\pi }^{\pi }{b}_n\sin (nx)\cos (mx)dx$$

We have three integrals on the right hand side to evaluate:

First Integral

$$\begin{gathered} {\int }_{-\pi }^{\pi }{a}_0\cos (mx)dx \\ =\frac{a_0}{m}\sin (m\pi )-\frac{a_0}{m}\sin (-m\pi ) \end{gathered}$$

Since $m\pi$ is always a multiple of $\pi$:

$$=0$$

Second Integral

$${\int }_{-\pi }^{\pi }{a}_n\cos (nx)\cos (mx)dx$$

Using $\cos$ addition formula:

$$\begin{gathered} =\frac{a_0}{2}{\int }_{-\pi }^{\pi }\cos (nx+mx)+\cos (nx-mx)dx \\ =\frac{a_0}{2}({\int }_{-\pi }^{\pi }\cos (nx+mx)dx+{\int }_{-\pi }^{\pi }\cos (nx-mx)dx) \\ =[\frac{a_0}{2}(\frac{\sin (nx+mx)}{n+m}+\frac{\sin (nx-mx)}{n-m}){]}_{-\pi }^{\pi } \end{gathered}$$

Here you will notice that this integral doesn’t work for $n=m$. We’ll circle back to that later. For now, this is two odd functions being added together. Since the bounds are the negatives of one another:

$$=0$$

Now, circling back to the extra case, where $n=m$:

$$\begin{gathered} a_m{\int }_{-\pi }^{\pi }{\cos }^2(nx)dx \\ =a_m{\int }_{-\pi }^{\pi }\frac{1+\cos (2x)}{2}dx \\ =a_m[\frac{x}{2}+\frac{\sin 2x}{4}{]}_{-\pi }^{\pi } \\ =a_m[(\frac{\pi }{2}+\frac{\sin 2\pi }{4})-(\frac{-\pi }{2}+\frac{\sin -2\pi }{4})] \\ =a_m\pi \end{gathered}$$

So, the second term in the right hand side evaluates to $a_m\pi$.

Third Integral

$$\begin{gathered} {\int }_{-\pi }^{\pi }\sin (nx)\cos (mx)dx \\ =\frac{1}{2}{\int }_{-\pi }^{\pi }\sin (nx+mx)dx+\frac{1}{2}{\int }_{-\pi }^{\pi }\sin (nx-mx)dx \\ =[-\frac{1}{2}(\frac{\cos (nx+mx)}{n+m}+\frac{\cos (nx-mx)}{n-m}){]}_{-\pi }^{\pi } \end{gathered}$$

Remember that $\cos x=-cos(x+\pi )$:

$$=0$$

Putting it Together

Now we have:

$$\begin{gathered} {\int }_{-\pi }^{\pi }f(x)\cos (mx)dx=a_m\pi \\ \frac{1}{\pi }{\int }_{-\pi }^{\pi }f(x)\cos (mx)dx=a_m \end{gathered}$$

Note in this case $m$ and $n$ both represent any positive integer, and are therefore interchangeable:

$$a_n=\frac{1}{\pi }{\int }_{-\pi }^{\pi }f(x)\cos (nx)dx$$

Finding $b_n$

Multiply the equation by $\sin mx$, where $m\in {\mathbb{Z}}^{+}$,integrate, and bound between $[-\pi ,\pi ]$:

$${\int }_{-\pi }^{\pi }f(x)\sin (mx)dx={\int }_{-\pi }^{\pi }{a}_0\sin (mx)dx+\sum _{n=1}^{\infty }{\int }_{-\pi }^{\pi }{a}_n\cos (nx)\sin (mx)dx+\sum _{n=1}^{\infty }{\int }_{-\pi }^{\pi }{b}_n\sin (nx)\sin (mx)dx$$

The first two terms are already covered, so let’s focus on the final term.

Last Integral

$$\begin{gathered} {\int }_{-\pi }^{\pi }{b}_n\sin (nx)\sin (mx)dx \\ =b_n{\int }_{-\pi }^{\pi }\cos (nx-mx)-\cos (nx+mx)dx \\ =b_n[\frac{\sin (nx-mx)}{n-m}-\frac{\sin (nx+mx)}{n+m}{]}_{-\pi }^{\pi } \end{gathered}$$

Again, there is a special case where $n=m$. Remember $\sin \pi =0$, so:

$$=0$$

With the special case:

$$\begin{gathered} b_m{\int }_{-\pi }^{\pi }{\sin }^2(mx)dx \\ =b_m{\int }_{-\pi }^{\pi }\frac{-\cos (2mx)+1}{2}dx \\ =b_m[\frac{1}{2}(x-\frac{\sin (2mx)}{2m}){]}_{-\pi }^{\pi } \\ =b_m\pi \end{gathered}$$

Putting it Together

$$\begin{gathered} b_m\pi ={\int }_{-\pi }^{\pi }f(x)\sin (mx)dx \\ {b}_m=\frac{1}{\pi }{\int }_{-\pi }^{\pi }f(x)\sin (mx)dx \\ {b}_n=\frac{1}{\pi }{\int }_{-\pi }^{\pi }f(x)\sin (nx)dx \end{gathered}$$

Fourier Series

Using the above, let’s express $f(x)$ as a Fourier Series:

$$f(x)=\frac{1}{2\pi }{\int }_{-\pi }^{\pi }f(x)dx+\sum _{n=1}^{\infty }\frac{\cos (nx)}{\pi }{\int }_{-\pi }^{\pi }f(x)\cos (nx)dx+\sum _{n=1}^{\infty }\frac{\sin (nx)}{\pi }{\int }_{-\pi }^{\pi }f(x)\sin (nx)dx$$

Note that this representation only works for when the function repeats from $[0,2\pi ]$. Using a similar proof, we can get:

$$f(x)=\frac{1}{P}{\int }_{-\frac{P}{2}}^{\frac{P}{2}}f(x)dx+\sum _{n=1}^{\infty }\frac{2\cos (\frac{2\pi nx}{P})}{P}{\int }_{-\frac{P}{2}}^{\frac{P}{2}}f(x)\cos (\frac{2\pi nx}{P})dx+\sum _{n=1}^{\infty }\frac{2\sin (\frac{2\pi nx}{P})}{P}{\int }_{-\frac{P}{2}}^{\frac{P}{2}}f(x)\sin (\frac{2\pi nx}{P})dx$$