In attempting to solve many physics and mathematics problems appear
trigonometric series, called Fourier Series in the form

f(x) = A0/2 + ∑ ∞n=1 ( An cos nx + Bn sin nx) (1)

with

Ai, Bi ∈ R

and

A0 = 1/π ∫ π-π f(x) dx.

An = 1/π ∫ π-π f(x) cos nx dx.

Bn = 1/π ∫ π-π f(x) sin nx dx.

For what kind of functions f it is possible to do the Fourier series like (1)? Really it exists? Attempts to answer these questions have advanced various branches of mathematical analysis over
the last two centuries.

We anticipate that functions class for which Fourier series given at (1) 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 feature and nor even be continuing to support development like on type (1), however, this does not mean that the Fourier series (1) will converge if the function f is continuous but as we see, every

differentiable function admits a Fourier series,

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∂

2u/∂x

2 = ∂u/∂t

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

Suppose further that in an initial time t=0, the temperature value is given by the function f(x)

Suppose that at the initial instant there is no change in temperature

If we put these conditions all together, resulting in

u(x, 0) = f(x)

u(x, t) = 0

u(π, t) = 0

∂u/∂t |

t=0 = 0

The heat equation to be solved is, as stated a

2∂

2u/∂x

2 = ∂u/∂t

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

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

so the heat equation becomes

v'' w = w' v => v''/v = w'/w = -k

therefore we have two ODEs to solve:

v'' + kv = 0 (1)

w' + a

2 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) = ∑

∞n=1 ( A

n cos √k x + B

n sen √k x)

Applying the initial conditions, we obtain

A

n = 0, ∀n, k = n

2
Then the solution of (1) is

v(x) = ∑

∞n=1 B

n sin nx

with

B

n = 2/π ∫

π0 f(x) sen nx dx.

If we now solve (2), we obtain

w(t) = exp(-a

2n

2t)

where exp(x) = e

x
Thus the general solution of the one-dimensional heat equation is

v(x) = ∑ ∞n=1 Bn exp(-a2n2t) sin nx (3)

As shown, the solution displayed the Fourier coefficients. Knowing if (3) makes sense depends if on t = 0 make sense

f(x) = ∑

∞n=1 B

n sin nx

From the beginning, we didn't talk about conditions on f or on its derivatives. The questions now are ... Under what conditions f Series converges? In what sense? and if it does, converges necessarily to function f?