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

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

\begin{displaymath}F(x) = \int_{-\pi}^{x}f(t)dt\end{displaymath}

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 An and Bn the Fourier coefficients of F(x). We have

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

Integration by parts will give

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

for $n \geq 1$. Hence

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

A similar calculation gives

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


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

This shows the following:

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

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


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

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 $x \in [-\pi,\pi]$

\begin{displaymath}\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),\end{displaymath}

because of the main convergence Theorem relative to Fourier series.

Example. Consider the function

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

We have

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

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

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


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

Simple calculations give

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


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

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

\begin{displaymath}\int_{-\pi}^{\pi} h(x)dx = 0.\end{displaymath}


\begin{displaymath}\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 +

the result above implies

\begin{displaymath}\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\end{displaymath}

which completes the proof.

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

\begin{displaymath}\int_{x}^{y}f(t)dt \end{displaymath}

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

Example. In the example above, we showed that

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


\begin{displaymath}\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).\end{displaymath}

This implies the formula

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

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

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

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

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

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

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

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

\begin{displaymath}4\pi A \leq P^2 .\end{displaymath}

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

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

\begin{displaymath}(x(t), y(t)) \;\;\;\; t \in [-\pi,\pi]\end{displaymath}

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

\begin{displaymath}P = \int_{-\pi}^{\pi}\sqrt{\left(\frac{dx}{dt}\right)^2 + \le...
A = \int_{-\pi}^{\pi} x \frac{dy}{dt} dt.\end{displaymath}


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

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

\begin{displaymath}(\tilde{x}(\tilde{t}),\tilde{y}(\tilde{t})) = (x(t),y(t)).\end{displaymath}

Easy calculations give

\begin{displaymath}\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},\end{displaymath}

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

\begin{displaymath}\left(\displaystyle \frac{dx}{dt}\right)^2 + \left(\displaystyle \frac{dy}{dt}\right)^2\end{displaymath}

constant. Hence

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

Since the curve is smooth, we get

x(t) = a_0 + \sum_{n=1}^{\infty}\B...
...nfty}\Big(c_n\cos(nx) + d_n\sin(nx)\Big).\\

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

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


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

Parseval formula implies

\begin{displaymath}\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).\end{displaymath}

On the other hand, we have

\begin{displaymath}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.\end{displaymath}


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

Algebraic manipulations imply

\begin{displaymath}\frac{P^2}{2\pi} - 2A = \pi \sum_{n=1}^{\infty}
...+ n(b_n + c_n)^2 + n(n-1)(a_n^2 + b_n^2 + c_n^2 +

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

a_n^2 + b_n^2 + c_n^2 + d_n^2 &=& ...
a_1 - d_1 &=& 0,&\\
b_1 + c_1 &=& 0.\\

This implies

x(t) &=& a_0 + a_1 \cos(t) - c_1 \...
y(t) &=& c_0 + c_1 \cos(t) + a_1 \sin(t)\\

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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