Approximation by Trigonometric Polynomials

Section 6.4 Approximation by Trigonometric Polynomials

Objectives

A Fourier series with only finite nubmer of terms is called a Trigonometric Polynomial.
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The Trigonometric polynomial for a given $N$ is the best approximation to a periodic function $f(x)$ in the sum of squares error sense. The Fourier Coefficients seen previous result in this trigonometric polynomila.
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Bessel’s Inequality give a bound on the sum of squares of the Fourier Coefficients.
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Parseval’s Theorem shows that the sum of squares of coefficients is proportional to the definite integral of the square of the function.
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Consider a periodic function $f$ of period $2\pi$ on the interval $[-\pi,\pi]\text{.}$ The $N$th partial sum of the Fourier Series of $f$ is denoted $f_N\text{,}$

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\begin{equation*} f_N(x) = a_0 + \sum_{n=1}^N (a_n \cos nx + b_n \sin nx) \end{equation*}

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where $a_0, a_n$ and $b_n$ are the Fourier Coefficients as before. The function $f_N$ is also called the Trigonometric Polynomial of degree $n$.

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Let’s ask a question about approximation. Consider a function of the form:

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\begin{equation*} F_N(x) =A_0 + \sum_{n=1}^N (A_n \cos nx + B_n \sin nx) \end{equation*}

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What is the best approximate for a trigonometric polynomial to another function $F(x)\text{.}$ That is, what coefficients can be chosen $A_0, A_n, B_n\text{?}$

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To answer this question, we will need to know what error we are taking about. Typically the error will be some function of the two functions, called $E(f,g)$ that outputs a number. We would like the error to have the following properties:

\begin{align*} E(f,g) \amp = E(g,f) \\ E(f,g) \amp \geq 0 \\ E(f,f) \amp = 0 \end{align*}

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one such function that we know is the function norm or square of it:

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\begin{align*} E(F,F_N) \amp = \langle F-F_N,F-F_N \rangle = ||F-F_N||^2 \\ \amp = \int_{-\pi}^{\pi} (F(x)-F_N(x))^2 \, dx \\ \amp = \int_{-\pi}^{\pi}\bigl(F(x) - A_0 - \sum_{n=1}^{N} (A_n \cos nx + B_n \sin nx) \bigr)^2 \end{align*}

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To find the minimum of this, we will take the derivatives of $E$ with respect to $A_0, A_n$ and $B_n$ and solve for where the derivative is 0.

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\begin{align*} \frac{\partial E}{\partial A_0} \amp = -2 \int_{-\pi}^{\pi} \bigl(F(x) - A_0 - \sum_{n=1}^{N} (A_n \cos nx + B_n \sin nx ) \bigr) \\ 0 \amp = \langle F, 1\rangle - \langle A_0, 1 \rangle - \sum_{n=1}^{N} (\langle A_n \cos n x, 1 \rangle + \langle B_n \sin nx, 1 \rangle ) \\ 0 \amp = \langle F, 1\rangle - A_0\langle 1, 1 \rangle \end{align*}

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\begin{equation} A_0 = \frac{\langle F,1\rangle}{\langle 1,1 \rangle} = \frac{1}{2\pi} \int_{-\pi}^{\pi} F(x)\,dx\tag{6.4.1} \end{equation}

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Similarly take the derivative with respect to $A_m\text{:}$

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\begin{align*} \frac{\partial E}{\partial A_m} \amp = -2 \int_{-\pi}^{\pi} \cos mx \bigl(F(x) - A_0 - \sum_{n=1}^{N} (A_n \cos nx + B_n \sin nx ) \bigr)\\ \amp \qquad \text{set the derivative to 0} \\ 0 \amp = \langle F, \cos mx \rangle - \langle A_0, \cos mx \rangle \\ \amp \qquad - \sum_{n=1}^{N} (\langle A_n \cos n x, \cos mx \rangle + \langle B_n \sin nx, \cos mx \rangle ) \end{align*}

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Note that $\langle \cos nx,\cos mx\rangle =0$ unless $m=n$ and $\langle 1,\cos mx\rangle=0$

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\begin{align} 0 \amp = \langle F, \cos mx \rangle - A_m \langle \cos mx, \cos mx \rangle \notag\\ \amp \qquad \text{or}\notag\\ A_m \amp = \frac{\langle F, \cos mx\rangle }{\langle \cos mx, \cos mx \rangle } = \frac{1}{\pi} \int_{-\pi}^{\pi} F(x) \cos m x \, dx\tag{6.4.2} \end{align}

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And similarly it can be shown that

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\begin{equation} B_m = \frac{1}{\pi} \int_{-\pi}^{\pi} F(x) \sin mx \, dx\tag{6.4.3} \end{equation}

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Note that the coefficients $A_0, A_m$ and $B_m$ in (6.4.1), (6.4.2) and (6.4.3) are the Fourier Coefficients, seen in Section 6.2

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Remark 6.4.1.

Let $f(x)$ be a piecewise continuous function on $[-\pi,\pi]\text{,}$ $N \gt 0$ and

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\begin{equation*} f_N (x) = a_0 + \sum_{n=1}^N (a_n \cos nx + b_n \sin nx) \end{equation*}

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The values of $a_n$ and $b_n$ that minimize $||f-f_N||$ (or $||f-f_N||^2$) is

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\begin{align*} a_0 \amp = \frac{1}{2\pi} \int_{-\pi}^{\pi} f(x) \, dx \\ a_n \amp = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \cos n x \, dx \\ b_n \amp = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \sin n x \, dx \end{align*}

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that is they are the Fourier Coefficients.

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Example 6.4.2.

In Example Example 6.2.5, the Fourier series of the sawtooth function was found the Fourier Coefficients are:

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\begin{align*} a_0 \amp = \frac{\pi}{2} \\ a_n \amp = \frac{2}{\pi n^2} ((-1)^n-1) \qquad \text{$n \geq 1$} \\ b_n \amp = 0 \end{align*}

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Graph the sawtooth function and $f_{9}(x)\text{,}$ the 9th degree trigonometric polynomial.

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Solution.

Figure 6.4.3. Plot of the sawtooth wave and it’s 9th degree trigonometric approximation.
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and the two plots are indistinguishable on this scale.

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Example 6.4.4.

In Example 6.2.3, the Fourier series of the square wave function was found and the Fourier Coefficients are:

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\begin{align*} a_0 \amp = 0 \\ a_n \amp = 0 \\ b_n \amp = \frac{2}{n\pi} (1- (-1)^n) \end{align*}

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Graph the square wave function and $f_{9}(x)\text{,}$ the $9$th degree trigonometric polynomial.

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Solution.

Figure 6.4.5. Graph of the square wave function and the $9$th degree trigonometric polynomial.
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And in contrast to the previous example, the trigonometric polynomial $f_9(x)$ and the original function $f$ are quite different. This is mainly due to the discontinuities in the original function.

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We will explore this example in a bit more detail after seeing some important theorems.

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Subsection 6.4.1 Theorems Related to Fourier Series

Theorem 6.4.6.

The quantity $||F-F_N||^2$ on the interval $[-\pi,\pi]$ is the minimum if and only if the coefficients of $F_N$ in (2) ar the Fourier coefficients of $F\text{.}$ This minimum value is

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\begin{equation} E^{\star} = \int_{-\pi}^{\pi} f^2 \, dx - 2A_0\pi - \pi \sum_{n=1}^{N} (A_n^2 + B_n^2)\tag{6.4.4} \end{equation}

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Theorem 6.4.7. Bessel’s Inequality.

Let $a_0, a_n$ and $b_n$ be the fourier coefficients related to the function $f$ on $[-\pi,\pi]\text{.}$ Then

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\begin{equation*} 2a_0^2 + \sum_{n=1}^N (a_n^2+b_n^2) \leq \frac{1}{\pi} \int_{-\pi}^{\pi} f^2 \, dx \end{equation*}

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Theorem 6.4.8. Parseval’s Theorem.

Let $a_0, a_n$ and $b_n$ be the fourier coefficients related to the function $f$ on $[-\pi,\pi]\text{.}$ Then

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\begin{equation*} 2a_0^2 + \sum_{n=1}^{\infty} (a_n^2+b_n^2) = \frac{1}{\pi} \int_{-\pi}^{\pi} f^2 \, dx \end{equation*}

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There are two important consequences of this theorem:

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If the integral on the right side is finite, then the series on the left converges. Functions in which the right side is finite are piecewise continuous functions.
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The error, $E^{\star}$ in ((6.4.4)) goes to zero. That is Fourier Series converge to $f$ (using the square error).
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Example 6.4.9.

Calculate $E^{\star}$ for the function:

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\begin{equation*} f(x) = \begin{cases} \pi+x \amp x \in [-\pi,0) \\ \pi-x \amp x \in [0,\pi] \end{cases} \end{equation*}

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and extended periodically and let $N=5,10,25,50,100,250,500,1000\text{.}$

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Solution.

The Fourier Series of this function is

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\begin{equation*} f(x) = \frac{\pi}{2} + \sum_{n=1}^{\infty} \frac{2}{\pi n^2} (1-(-1)^n) \cos nx \end{equation*}

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or $a_0 = \pi/2$ and $a_n = (2/\pi n^2)(1-(-1)^n) $

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\begin{align*} E^{\star} \amp = \int_{-\pi}^{\pi} f^2 \, dx - 2A_0\pi - \pi \sum_{n=1}^{N} (A_n^2 + B_n^2) \\ \amp = 2\int_{0}^{\pi} (\pi-x)^2 \, dx - 2\biggl(\frac{\pi}{2}\biggr) - \pi \sum_{n=1}^{N}\biggl(\frac{2}{\pi n^2} (1-(-1)^n) \biggr) \end{align*}

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Table 6.4.10.

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$n$	$E^{\star}$
5	0.00372
10	0.000832
25	0.0000482
50	$6.78 \times 10^{-6} $
100	$8.49 \times 10^{-8} $
250	$5.43 \times 10^{-8} $
500	$6.79 \times 10^{-9} $
1000	$8.48 \times 10^{-10} $

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A consequence of Parseval’s Theorem is that for piecewise continuous functions, the Fourier Series converges as $n \rightarrow \infty\text{.}$ So in light of the plot in Example Example 6.4.4, that it would appear that the plot of $f_N$ would approach the square wave as $N \rightarrow \infty\text{.}$ However the plots of $f_{25}$ and $f_{100}$ are shown below (with $n=25$ on top):

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Figure 6.4.12. The Trigonmetric Polynomials of degree $N=25$ of the square wave function from above.
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And despite the larger value of $N\text{,}$ $f_N$ does not appear to be approaching the square wave function. The difference is pronounced near the discontinuities in the function. This is called \emph{Gibbs Phenomena} and it can be shown in this situation that the local max near $x=0$ in fact grows without bound as $N \rightarrow \infty\text{,}$ despite the fact that $||f_N-f|| \rightarrow 0\text{.}$

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Subsection 6.4.2 Why is finite Fourier Series called a Polynomial?

You may be scratching your head about why the sum of sines and cosines is called a polynomial. You do recall correctly that polynomials are generally of the form

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\begin{equation*} p(x) = \sum_{i=0}^n a_n x^n \end{equation*}

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that is a linear combination of powers of $x\text{.}$

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However, you may also recall some trigonmetric identities. For example,

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\begin{align*} \sin 2x \amp = 2 \sin x \cos x\\ \cos 2x \amp = 1- 2 \sin^2 x \end{align*}

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And a more complicated set of identities lead to

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\begin{align*} \sin 3x \amp = 3 \sin x - 4 \sin^3 x \\ \cos 3x \amp = 4 \cos^3 x - 3 \cos x \end{align*}

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Note that in these examples, functions of the form $\sin kx$ and $\cos kx$ can be written in terms (for $k=2,3$) of products and powers of $\sin x$ and $\cos x\text{.}$ This continues for larger values of $k$ as well.

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If you continue with similar identities, you can show that the trigonometric polynomial of the form

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\begin{equation*} a_0 + \sum_{n=1}^N (a_n \cos nx + b_n \sin nx) \end{equation*}

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Can be written as the powers of $\sin x$ and $\cos x\text{.}$ This explains why this is called a polynomial.

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\(n\)	\(E^{\star}\)
5	0.00372
10	0.000832
25	0.0000482
50	\(6.78 \times 10^{-6} \)
100	\(8.49 \times 10^{-8} \)
250	\(5.43 \times 10^{-8} \)
500	\(6.79 \times 10^{-9} \)
1000	\(8.48 \times 10^{-10} \)