Monday, 25 March 2013

Bessel's Inequality and Parseval's Formula: The Energy Theorem

We have seen some types of approximations, such as Taylor and Fourier approximations. The type of convergence used changes depending on the nature of the approximation. One of the most useful for Fourier approximations is L2-convergence:
Let f (x) be an integrable function on the interval [- $ \pi$,$ \pi$], such that

$\displaystyle \int_{-\pi}^{\pi}$f2(x)dx < $\displaystyle \infty$.

We will say that f (x) is square integrable. Consider the associated Fourier series
f (x$\displaystyle \sim$ a0 + $\displaystyle \sum_{n=1}^{\infty}$$\displaystyle \Big($ancos(nx) + bnsin(nx)$\displaystyle \Big)$.

fN(x) = a0 + $\displaystyle \sum_{n=1}^{n=N}$$\displaystyle \Big($ancos(nx) + bnsin(nx)$\displaystyle \Big)$.

$\displaystyle \int_{-\pi}^{\pi}$(f (x) - fN(x))2dx = $\displaystyle \int_{-\pi}^{\pi}$(f2(x) - 2f (x)fN(x) + f2N(x))dx.

Easy calculations give
$\displaystyle \int_{-\pi}^{\pi}$f2N(x)dx = $\displaystyle \pi$$\displaystyle \left(\vphantom{2 a^2_0 + \sum_{n=1}^{n=N}(a^2_n + b^2_n)}\right.$2a20 + $\displaystyle \sum_{n=1}^{n=N}$(a2n + b2n)$\displaystyle \left.\vphantom{2 a^2_0 + \sum_{n=1}^{n=N}(a^2_n + b^2_n)}\right)$.

We also have
$\displaystyle \int_{-\pi}^{\pi}$f (x)fN(x)dx = $\displaystyle \pi$$\displaystyle \left(\vphantom{2 a^2_0 + \sum_{n=1}^{n=N}(a^2_n + b^2_n)}\right.$2a20 + $\displaystyle \sum_{n=1}^{n=N}$(a2n + b2n)$\displaystyle \left.\vphantom{2 a^2_0 + \sum_{n=1}^{n=N}(a^2_n + b^2_n)}\right)$.

$\displaystyle \int_{-\pi}^{\pi}$(f (x) - fN(x))2dx = $\displaystyle \int_{-\pi}^{\pi}$f2(x)dx - $\displaystyle \pi$$\displaystyle \left(\vphantom{2 a^2_0 + \sum_{n=1}^{n=N}(a^2_n + b^2_n)}\right.$2a20 + $\displaystyle \sum_{n=1}^{n=N}$(a2n + b2n)$\displaystyle \left.\vphantom{2 a^2_0 + \sum_{n=1}^{n=N}(a^2_n + b^2_n)}\right)$.

Since $ \int_{-\pi}^{\pi}$(f (x) - fN(x))2dx $ \geq$ 0, then
$\displaystyle \pi$$\displaystyle \left(\vphantom{2 a^2_0 + \sum_{n=1}^{n=N}(a^2_n + b^2_n)}\right.$2a20 + $\displaystyle \sum_{n=1}^{n=N}$(a2n + b2n)$\displaystyle \left.\vphantom{2 a^2_0 + \sum_{n=1}^{n=N}(a^2_n + b^2_n)}\right)$ $\displaystyle \leq$ $\displaystyle \int_{-\pi}^{\pi}$f2(x)dx,

for any N $ \geq$ 1. This clearly implies the following result:
Theorem. Bessel's Inequality Let f (x) be a function defined on [- $ \pi$,$ \pi$] such that f2(x) has a finite integral on [- $ \pi$,$ \pi$]. If an and bn are the Fourier coefficients of the function f (x), then we have

$\displaystyle \pi$$\displaystyle \left(\vphantom{2a^2_0 + \sum_{n=1}^{\infty}(a^2_n + b^2_n)}\right.$2a20 + $\displaystyle \sum_{n=1}^{\infty}$(a2n + b2n)$\displaystyle \left.\vphantom{2a^2_0 + \sum_{n=1}^{\infty}(a^2_n + b^2_n)}\right)$ $\displaystyle \leq$ $\displaystyle \int_{-\pi}^{\pi}$f2(x)dx.

In particular, the series $ \sum_{n=1}^{\infty}$(a2n + b2n)) is convergent.
Remark. The quantity An = $ \sqrt{a_n^2 + b^2_n}$ is called the amplitude of the nth harmonic. The square of the amplitude has a useful interpretation. Indeed, borrowing terminology from the study of periodic waves, we define theenergy E of a 2$ \pi$-periodic function f (x) to be the number

E = $\displaystyle {\frac{1}{\pi}}$$\displaystyle \int_{-\pi}^{\pi}$f2(x)dx.

So Bessel's Inequality translates into:
$\displaystyle \left(\vphantom{2a^2_0 + \sum_{n=1}^{\infty}(a^2_n + b^2_n)}\right.$2a20 + $\displaystyle \sum_{n=1}^{\infty}$(a2n + b2n)$\displaystyle \left.\vphantom{2a^2_0 + \sum_{n=1}^{\infty}(a^2_n + b^2_n)}\right)$ $\displaystyle \leq$ E.

You may ask the following question: when does the inequality become an equality?
Note that for Fourier polynomials, the inequality does indeed become an equality. Using this, one can show that the answer to the question is in the affirmative if and only if

$\displaystyle \lim_{N \rightarrow \infty}^{}$$\displaystyle \int_{-\pi}^{\pi}$(f (x) - fN(x))2dx = 0.

In this case, we have
$\displaystyle \pi$(2a20 + $\displaystyle \sum_{n=1}^{\infty}$(a2n + b2n)) = $\displaystyle \int_{-\pi}^{\pi}$f2(x)dx.

Theorem. Parseval's Formula or the Energy Theorem. Let f (x) be a function defined on [- $ \pi$,$ \pi$] such that f2(x) has a finite integral on [- $ \pi$,$ \pi$]. If an and bn are the Fourier coefficients of f (x), then we have

2a20 + $\displaystyle \sum_{n=1}^{\infty}$(a2n + b2n) = $\displaystyle {\frac{1}{\pi}}$$\displaystyle \int_{-\pi}^{\pi}$f2(x)dx = E

if and only if
$\displaystyle \lim_{N \rightarrow \infty}^{}$$\displaystyle \int_{-\pi}^{\pi}$(f (x) - fN(x))2dx = 0.

Remark. One may wonder when does Parseval's Formula hold? This is the case, for example, for piecewise smooth functions. The reason behind is the uniform convergence of the Fourier partial sums to f (x), ie.

$\displaystyle \lim_{N \rightarrow \infty}^{}$$\displaystyle \sup_{x \in [-\pi,\pi]}^{}$f (x) - fN(x)| = 0.

The proof in this case is quite easy. Indeed, since f (x) is continuous then
f (x) = a0 + $\displaystyle \sum_{n=1}^{\infty}$(ancos(nx) + bnsin(nx)).

f2(x) = a0f (x) + $\displaystyle \sum_{n=1}^{\infty}$(ancos(nx)f (x) + bnsin(nx)f (x)).

The uniform convergence of this series enables us to integrate term-by-term to obtain
$\displaystyle \int_{-\pi}^{\pi}$f2(x)dx = a0$\displaystyle \int_{-\pi}^{\pi}$f (x)dx + $\displaystyle \sum_{n=1}^{\infty}$$\displaystyle \int_{-\pi}^{\pi}$(ancos(nx)f (x) + bnsin(nx)f (x))dx.

By using the definition of the Fourier coefficients, we get the desired conclusion.
Application: Least Square Error.
One application of Parseval's Formula is the measure of the least square error $ \sigma^{2}_{N}$ defined by

$\displaystyle \sigma^{2}_{N}$ = $\displaystyle {\frac{1}{2\pi}}$$\displaystyle \int_{-\pi}^{\pi}$(f (x) - fN(x))2dx.

If the function f (x) satisfies the assumptions of the Energy Theorem, then we have
$\displaystyle \sigma^{2}_{N}$ = $\displaystyle {\textstyle\frac{1}{2}}$$\displaystyle \sum_{n=N+1}^{\infty}$(a2n + b2n).

Example. Let f (x) = | x| be defined on [- $ \pi$,$ \pi$]. Find $ \sigma^{2}_{N}$ and its asymptotic behavior when N gets large.
Answer. Since f (x) is even, we have bn = 0. On the other hand, easy calculations give

a2n = 0,    and    a2n - 1 = - $\displaystyle {\frac{4}{\pi (2n-1)^2}}$.

$\displaystyle \sigma^{2}_{2N-1}$ = $\displaystyle \sigma^{2}_{2N}$ = $\displaystyle {\textstyle\frac{1}{2}}$$\displaystyle \sum_{n=2N+1}^{\infty}$$\displaystyle \left(\vphantom{\frac{4}{\pi (2n-1)^2}}\right.$$\displaystyle {\frac{4}{\pi (2n-1)^2}}$$\displaystyle \left.\vphantom{\frac{4}{\pi (2n-1)^2}}\right)^{2}_{}$ = $\displaystyle {\frac{8}{\pi^2}}$$\displaystyle \sum_{n=2N+1}^{\infty}$$\displaystyle {\frac{1}{(2n-1)^4}}$.

Using the equality
$\displaystyle \int_{N}^{\infty}$$\displaystyle {\frac{dx}{(2x-1)^4}}$ = $\displaystyle {\textstyle\frac{1}{6}}$$\displaystyle {\frac{1}{(2N-1)^3}}$,

we get
$\displaystyle \sigma^{2}_{N}$ = O$\displaystyle \left(\vphantom{\frac{1}{N^3}}\right.$$\displaystyle {\frac{1}{N^3}}$$\displaystyle \left.\vphantom{\frac{1}{N^3}}\right)$,      as    N $\displaystyle \rightarrow$ $\displaystyle \infty$.

Recall that the sequences {un} and {vn} satisfy

un = O(vn)

if $\displaystyle \left\{\vphantom{\frac{u_n}{v_n}}\right.$$\displaystyle {\frac{u_n}{v_n}}$$\displaystyle \left.\vphantom{\frac{u_n}{v_n}}\right\}$ is a bounded sequence.
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