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 $2\pi$-periodic and defined on $[-\pi,\pi]$.

Let f(x) be a $2\pi$-periodic piecewise continuous function. Then the function

$$F(x) = \int_{-\pi}^{x}f(t)dt$$

is continuous and is $2\pi$-periodic if and only if $F(\pi) = F(-\pi) = 0$, 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 A_n and B_n the Fourier coefficients of F(x). We have

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

Integration by parts will give

$$A_n = \frac{1}{n\pi}[F(x)\sin(nx)]^{\pi}_{-\pi} - \frac{1}{n\pi}\int_{-\pi}^{\pi} F'(x)\sin(nx)dx,$$

for $n \geq 1$. Hence

$$A_n = - \frac{1}{n\pi} \int_{-\pi}^{\pi} f(x)\sin(nx)dx.$$

A similar calculation gives

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

and

$$A_0 = - \frac{1}{2\pi} \int_{-\pi}^{\pi} xf(x)dx.$$

This shows the following:

Theorem. Integration of Fourier series
Let f(x) be $2\pi$-periodic piecewise continuous function such that a₀ = 0. If

$$f(x) \sim \sum_{n=1}^{\infty} (a_n\cos(nx) + b_n\sin(nx)),$$

then

$$\int_{-\pi}^{x}f(t)dt \sim A_0 + \sum_{n=1}^{\infty}\frac{1}{n} (a_n\sin(nx) - b_n\cos(nx)),$$

where $A_0 = - \displaystyle \frac{1}{2\pi} \int_{-\pi}^{\pi} xf(x)dx$.

Since the function F(x) is continuous, we have for any $x \in [-\pi,\pi]$

$$\int_{-\pi}^{x}f(t)dt = A_0 + \sum_{n=1}^{\infty}\frac{1}{n} \Big(a_n\sin(nx) - b_n\cos(nx)\Big),$$

because of the main convergence Theorem relative to Fourier series.

Example. Consider the function

$$f(x) = \left\{ \begin{array}{lll} 1 & 0 \leq x \leq \pi \\ -1 & -\pi \leq x < 0. \end{array} \right.$$

We have

$$f(x) \sim \frac{4}{\pi}(\sin(x) + \frac{\sin(3x)}{3} + \frac{\sin(5x)}{5} +\ldots).$$

Since, for any $x \in [-\pi,\pi]$, we have

$$\int_{-\pi}^{x} f(t)dt = \vert x\vert - \pi,$$

then

$$\vert x\vert - \pi \sim A_0 + \frac{4}{\pi}\left(- \cos(x) - \frac{\cos(3x)}{9} - \frac{\cos(5x)}{25} - \cdots\right).$$

Simple calculations give

$$A_0 = - \frac{1}{2\pi}\int_{-\pi}^{\pi} xf(x)dx = - \frac{\pi}{2}.$$

Hence

$$\vert x\vert = \frac{\pi}{2} - \frac{4}{\pi}\left(\cos(x) + \frac{\cos(3x)}{9} + \frac{\cos(5x)}{25} +\ldots\right).$$

Let f(x) be $2\pi$-periodic piecewise continuous function such that $a_0 \neq 0$. Set $h(x) = f(x) - \displaystyle a_0$. Then h(x) is $2\pi$-periodic piecewise continuous and satisfies the condition

$$\int_{-\pi}^{\pi} h(x)dx = 0.$$

Since

$$\int_{x}^{y} f(t)dt = \int_{-\pi}^{y}f(t)dt - \int_{-\pi}^{x}... ..._{-\pi}^{y}f(t)dt - \int_{-\pi}^{x}f(t)dt + \int_{x}^{y}a_0dt,$$

the result above implies

$$\int_{x}^{y} f(t)dt = a_0 (y-x) + \sum_{n=1}^{\infty} \int_{x}^{y}\Big[a_n\cos(nt) + b_n \sin(nt)\Big]dt$$

which completes the proof.

Theorem. Let f(x) be $2\pi$-periodic piecewise continuous function. Then for any x and y, the integral

$$\int_{x}^{y}f(t)dt$$

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

Example. In the example above, we showed that

$$\vert x\vert = \frac{\pi}{2} - \frac{4}{\pi}\left(\cos(x) + \frac{\cos(3x)}{9} + \frac{\cos(5x)}{25} +\ldots\right).$$

Hence

$$\int_{0}^{\frac{\pi}{2}} \vert x\vert dx = \frac{\pi^2}{4} - \frac{4}{\pi}\left(1-\frac{1}{27} + \frac{1}{5^3}-\ldots \right).$$

This implies the formula

$$\frac{\pi^3}{32} = \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{(2n-1)^3}.$$

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

Example. Show that the trigonometric series

$$\sum_{n=1}^{\infty} \frac{\sin(nx)}{\ln(1+n)},$$

is not the Fourier series of any function.
Answer. It is easy to see that this series converges for any $x \in \mbox{\bf R}$. Assume there exists a function f(x) such that this series is its Fourier series. Then

$$- \sum_{n=1}^{\infty} \frac{\cos(nx)}{n\ln(1+n)},$$

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 $2\pi$-periodic continuous and piecewise smooth function. Then, for any $x \in [-\pi,\pi]$, we have

$$\frac{(f'(x+) + f'(x-))}{2} = \sum_{n=1}^{\infty}n\Big(-a_n \sin(nx) +b_n\cos(nx)\Big).$$

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

$$4\pi A \leq P^2 .$$

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

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

$$(x(t), y(t)) \;\;\;\; t \in [-\pi,\pi]$$

with $x(-\pi) = x(\pi)$ and $y(-\pi) = y(\pi)$. The formulas giving P and A are

$$P = \int_{-\pi}^{\pi}\sqrt{\left(\frac{dx}{dt}\right)^2 + \le... ...;\;\;\;\mbox{and}\;\; A = \int_{-\pi}^{\pi} x \frac{dy}{dt} dt.$$

Set

$$s(t) = \int_{-\pi}^{t} \sqrt{(x'(u))^2 + (y'(u))^2} du .$$

Then $P = s(\pi)$. Consider the new variable $\tilde{t} = -\pi + \displaystyle \frac{2\pi s(t)}{P}$. If we rewrite the parametric representation in terms of $\tilde{t}$, we get

$$(\tilde{x}(\tilde{t}),\tilde{y}(\tilde{t})) = (x(t),y(t)).$$

Easy calculations give

$$\left(\frac{d\tilde{x}}{d\tilde{t}}\right)^2 + \left(\frac{d\tilde{y}}{d\tilde{t}}\right)^2 = \frac{P^2}{4\pi^2},$$

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

$$\left(\displaystyle \frac{dx}{dt}\right)^2 + \left(\displaystyle \frac{dy}{dt}\right)^2$$

constant. Hence

$$\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2 = \frac{P^2}{4 \pi^2}.$$

Since the curve is smooth, we get

$$\left\{\begin{array}{lcll} x(t) = a_0 + \sum_{n=1}^{\infty}\B... ...nfty}\Big(c_n\cos(nx) + d_n\sin(nx)\Big).\\ \end{array}\right.$$

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

$$x'(t) = \sum_{n=1}^{\infty}n\Big(-a_n\sin(nt) + b_n\cos(nt)\Big),$$

and

$$y'(t) = \sum_{n=1}^{\infty}n\Big(-c_n\sin(nx) + d_n\cos(nx)\Big).$$

Parseval formula implies

$$\frac{P^2}{2\pi} = \int_{-\pi}^{\pi} ((x')^2 + (y')^2)dt = \pi \sum_{n=1}^{\infty}n^2\Big(a_n^2 + b_n^2 + c_n^2 + d_n^2\Big).$$

On the other hand, we have

$$A = \int_{-\pi}^{\pi} x(t)y'(t)dt = \frac{1}{4}\int_{-\pi}^{\pi} \Big([x(t) + y'(t)]^2 - [x(t) + y'(t)]^2\Big)dt.$$

Hence

$$A = \pi \sum_{n=1}^{\infty}n \Big(a_nd_n - b_nc_n\Big).$$

Algebraic manipulations imply

$$\frac{P^2}{2\pi} - 2A = \pi \sum_{n=1}^{\infty} \Big[n(a_n-d... ...+ n(b_n + c_n)^2 + n(n-1)(a_n^2 + b_n^2 + c_n^2 + d_n^2)\Big].$$

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 $$\frac{P^2}{2\pi} - 2A = 0$$ if and only if

$$\left\{\begin{array}{lcll} a_n^2 + b_n^2 + c_n^2 + d_n^2 &=& ... ...\\ a_1 - d_1 &=& 0,&\\ b_1 + c_1 &=& 0.\\ \end{array}\right.$$

This implies

$$\left\{\begin{array}{lcll} x(t) &=& a_0 + a_1 \cos(t) - c_1 \... ... y(t) &=& c_0 + c_1 \cos(t) + a_1 \sin(t)\\ \end{array}\right.$$

for $t \in [-\pi,\pi]$. Therefore the curve is a circle centered at (a₀,c₀) with radius $\sqrt{a_1^2 + c_1^2}$, which completes the proof of the theorem.

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Author: M.A. Khamsi

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