How did Fourier arrive at the following regarding his series and coefficients?

Question

I am reading Karen Saxe's "Beginning Functional Analysis." Perhaps it is poor exposition on her part, but she states:

...Fourier begins with an arbitrary function $f$ on the interval from $-\pi$ to $\pi$ and states that if we can write $$f(x) = \frac{a_0}{2} + \sum_{k = 1}^\infty a_k\cos(kx) + b_k\sin(kx),$$ then it must be the case that the coefficients $a_k$ and $b_k$ are given by the formulas...

and then she states the formulas for $a_k$ and $b_k$. However she does not elaborate further on why Fourier concluded this. This is unsatisfactory to me because I would like to know the motivation behind this conclusion. How did he know that it must be the case that this is true?

Also, it is unclear why we don't rewrite $\frac{a_0}{2}$ as $b_0$ since it would seem that $a_0$ is just some constant.

Answer 1

I think Fourier and his contemporaries were looking at the heat equation $u_t = \Delta u$ where $u$ describes distribution of heat on some domain. In a one dimensional case, it would be $u_t = u_{xx}$ on, say, $[-\pi, \pi]$. Physically, we are heating up the wire $[-\pi,\pi]$ from its two ends and the two ends are held at some temperature determined by an external heating device. For simplicity, we may assume they are at $0$ temperature.

A standard procedure to solve this PDE would be to consider $u(t,x) = f(x)g(t)$ for some functions $f$ and $g$. If you plug this back to the heat equation, you arrive at

$ f''/f = g'/g$

But since $f$ depends only on $x$ and $g$ depends only on $t$, the above is a constant, call it $k$. Then we can solve to get $g = Ce^kt$. The other equation is $f'' = k f$, an eigenvalue problem for the operator $\frac{\partial^2}{\partial x^2}$ on $H:=L^2([-\pi, \pi])$. Recall that the inner product on $H$ is

$ (f,g) = \int_{-\pi}^\pi f \bar{g} dx $

Now note that by an integration by part

$ (f'',f) = \int_{-\pi}^\pi f'' \bar{g} dx = f'f \mid_{-\pi}^\pi - \int_{-\pi}^\pi f' \bar{f'} dx $

If we demand $f$ to satisfy $f(-\pi)=f(\pi)$, then we have $(f'',f) = (f',f') \geq 0 $. This tells you that $\frac{\partial^2}{\partial x^2}$ is a positive operator, so its eigenvalues are non-negative.

If $k = 0$, the solutions to $f'' = kf$ are linear polynomials. But they don't satisfy boundary conditions. So $k > 0$. And the solutions to $f'' = k f$ are precisely $A\cos(nt) + B\sin(nt)$ with given boundary condition.

Now to get a general solution to the heat equation, due to linearity of the equation, we expect them to be linear combinations of these trig functions. Thus Fourier write down

$f(x) = \sum A_n \cos(nx) + B_n \sin(nx) $

If we are lucky, and surely we are, all (reasonable) functions can be expressed in this form, so we can solve the heat equation.

Answer 2

If $$ f(x) = \frac{a_0}{2} + \sum_{k = 1}^\infty a_k\cos(kx) + b_k\sin(kx) $$ then $$ \int_{-\pi}^\pi f(x)\cos (nx)\,dx = \int_{-\pi}^\pi \quad \underbrace{\left\{ \frac{a_0}{2} + \sum_{k = 1}^\infty a_k\cos(kx) + b_k\sin(kx) \right\}} \quad \cos(nx)\,dx $$ because the part over the $\underbrace{\text{underbrace}}$ is the same as $f(x)$.

So now notice that one of the values of $k$ is equal to $n$ and all the others are not. For the terms in which $k\ne n$, the integral is zero. Hence you have $$ \int_{-\pi}^\pi f(x)\cos (nx)\,dx = \int_{-\pi}^\pi a_n \cos^2(nx)\,dx = a_n \,\pi. $$

Hence $$ a_n = \frac 1 \pi \int_{-\pi}^\pi f(x)\cos(nx)\,dx. $$