# Fourier Series Basics

### Basic Concepts and Principles

In many physics and mathematics problems it appears trigonometric series, called Fourier Series in the form
$$f(x)=\frac{A_{0}}{2} + \sum_{n=1}^{\infty}(A_n cos nx + B_nsin nx )$$
Where

$A_i, B_i \in \mathbb{R}$

and

$A_0= \frac{1}{\pi} \int_{-\pi}^{\pi}f(x) dx$

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

$B_n= \frac{1}{\pi} \int_{-\pi}^{\pi}f(x) sin\,nx\,dx$
For what kind of functions f it is possible to obtain the Fourier series like this? Do it actually exists? Attempts to answer these questions have advanced various branches of mathematical analysis over the last two centuries.

We anticipate that the class of functions for which Fourier series makes sense, is quite large, much more than in the power series case, in which a function supports such development if has continuous derivatives of all orders. In the trigonometric series case, as we will see there is not need the differentiability condition and nor even continuty . Even though Fourier series converge if the function f is continuous and has first derivate continuous, this is all differentiable function admits a Fourier series.

### Extended Theory

Fourier series appears naturally in many physics problems, for example, in attempting to solve boundary value border problems. Lets go: consider the one-dimensional heat equation

$$a^2\frac{\partial^2 u}{\partial x^2} = \frac{\partial u}{\partial y}$$

Consider metal bar with length pi = 3.1415926 ... whose extreme ponits are maintained at constant temperature, for example 0.

Let's Suppose further that in an initial time t=0, the temperature value is given by the function f(x).
Let's Suppose that at the initial instant there is no change in temperature.
If we put these conditions all together, resulting in

$\left\{\begin{matrix} u(x,0)= f(x) \\ u(x,t)=0 \\ u(\pi,t)=0 \\ \frac{\partial u}{\partial t}(0) = 0 \end{matrix}\right.$

The heat equation is as stated $$a^2\frac{\partial^2 u}{\partial x^2} = \frac{\partial u}{\partial y}$$

We solve this equation by the method separation of variables, where is assumed that solution is of the form

$u(x, t)=v(x)w(t)$

and so heat equation becomes

$v'' w = w' v \Rightarrow \frac{v''}{v} = \frac{w'}{w} = -k$

therefore we have two Ordinary Differential Equations (ODEs) to solve:

v'' + kv = 0    (1)

w' + a2 kv = 0    (2)

We solve first (1) based on the value k, both cases k = 0 and k < 0 are reduced to the trivial solution u = 0
So we take the case k > 0, where the solution is of the form

$v(x)=\sum_{n=1}^{\infty}(A_n cos \sqrt{k}x + B_nsin \sqrt{k} x )$

Applying the initial conditions, we obtain

$A_n = 0, k=n^2, \forall n$ Then the solution of (1) is

$v(x) = \sum_{n=1}^{\infty} B_n sin nx$

with

$B_n = \frac{2}{\pi} \int_{- \pi}^{\pi} f(x) sin\,nx dx$

If we now solve (2), we obtain

$w(t) = e^{(- a^2 n^2 t)}$

Thus the general solution of the one-dimensional heat equation is
$u(x, t)= \sum_{n=1}^{\infty} e^{(- a^2 n^2 t)} B_n sin\,nx$
As shown, the solution displayed the Fourier coefficients. Knowing if (3) makes sense depends if on t = 0 make sense

$f(x)=\sum_{n=1}^{\infty} B_n sin\,nx$

At the beginning, we did not talk about conditions on f or on its derivatives. The questions now are ... Under what conditions a f series converges? In what sense? and if it does ... does it converge necessarily to function f?

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