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## Monday, 25 March 2013

### Operations on Fourier Series

The results obtained in this page may easily be extended to function defined on any interval [a,b]. So without loss of generality, we will assume that the functions involved are -periodic and defined on .

Let f(x) be a -periodic piecewise continuous function. Then the function

is continuous and is -periodic if and only if , i.e. the Fourier coefficient a0 = 0. It is also quite easy to show that if f(x) is piecewise smooth, then also is F(x). An interesting question will be to find out if a simple relationship between the Fourier coefficients of f(x) and F(x) exist. Denote by An and Bn the Fourier coefficients of F(x). We have

Integration by parts will give

for . Hence

A similar calculation gives

and

This shows the following:

Theorem. Integration of Fourier series
Let f(x) be -periodic piecewise continuous function such that a0 = 0. If

then

where .

Since the function F(x) is continuous, we have for any

because of the main convergence Theorem relative to Fourier series.

Example. Consider the function

We have

Since, for any , we have

then

Simple calculations give

Hence

Let f(x) be -periodic piecewise continuous function such that . Set . Then h(x) is -periodic piecewise continuous and satisfies the condition

Since

the result above implies

which completes the proof.

Theorem. Let f(x) be -periodic piecewise continuous function. Then for any x and y, the integral

may be evaluated by integrating term-by-term the Fourier series of f(x).

Example. In the example above, we showed that

Hence

This implies the formula

This kind of formulas are quite interesting. Indeed, they enable us to find approximations to the irrational number .

Example. Show that the trigonometric series

is not the Fourier series of any function.
Answer. It is easy to see that this series converges for any . Assume there exists a function f(x) such that this series is its Fourier series. Then

must be convergent everywhere since it is going to be the Fourier series of the antiderivative of f(x). But this series fails to be convergent when x=0. Contradiction.

After we discussed the relationship between the Fourier series of a function and its antiderivative, it is natural to ask if a similar relationship exists between a function and its derivative. The answer to this is more complicated. But we do have the following result:

Theoreme. Let f(x) be -periodic continuous and piecewise smooth function. Then, for any , we have

In other words, we obtain the Fourier series of f'(x) by differentiating term-by-term the Fourier series of f(x).

Application: Isoperimetric Inequality

Theoreme. Consider a smooth closed curve in the plane xy. Denote by P its perimeter (total arclength) and by A the area of the region enclosed by the curve. Then we have

The equality holds if and only if the curve is a circle.

Proof. A parametric representation of the curve may be given by

with  and . The formulas giving P and A are

Set

Then . Consider the new variable . If we rewrite the parametric representation in terms of , we get

Easy calculations give

i.e. the new variable enables us to reparametrize the curve while assuming the quantity

constant. Hence

Since the curve is smooth, we get

Previous result, on the relationship between the Fourier coefficients of the function and its derivative, gives

and

Parseval formula implies

On the other hand, we have

Hence

Algebraic manipulations imply

Since the second term of this equality is positive, we deduce the first part of the result above. On the other hand, we will have  if and only if

This implies

for . Therefore the curve is a circle centered at (a0,c0) with radius , which completes the proof of the theorem.