# 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. [Geometry] [Algebra] [Trigonometry ]
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