University of Florida/Eml5526/s11.team2.reiss.HW

Given

Consider the family of functions

$\mathrm{\Im }=\left\{1,\cos (\omega x),\cos (2\omega x),\sin (\omega x),\sin (2\omega x)\right\}$

(4.1)

on the interval [0,T], where T=$2\mathrm{\Pi }/\omega$

Find

A) Construct $\mathrm{\Gamma }(\mathrm{\Im })$ and observe its properties
B) Find $\det [\mathrm{\Gamma }(\mathrm{\Im })]$
C) Is $\mathrm{\Im }$ an orthogonal basis

Solution

Construct $\mathrm{\Gamma }(\mathrm{\Im })$:

$\mathrm{\Gamma }(\mathrm{\Im })=\left(\begin{array}{ccc}<b_1,b_1>& \dots & <b_1,b_n>\\ ⋮& \ddots & ⋮\\ <b_n,b_1>& \cdots & <b_n,b_n>\end{array}\right)$

(4.2)

where

$\begin{array}{c}{b}_1=1\\ b_2=\cos (\omega x)\\ b_3=\cos (2\omega x)\\ b_4=\sin (\omega x)\\ b_5=\sin (2\omega x)\end{array}$

In order to construct the matrix we must first define $<b_i,b_j>$

$<b_i,b_j>=\underset{x_0}{\overset{x}{\int }}{b}_i(x)b_j(x)dx$

(4.3)

Because multiplication of continuous functions is communicative it can be shown from equation 4.3 that

$<b_i,b_j>=<b_j,b_i>$

(4.4)

And therefore $\mathrm{\Gamma }(\mathrm{\Im })$ is a symmetric matrix

We must now evaluate the terms of the matrix

$\begin{array}{c}<b_1,b_1>=<b_1,b_1>=\underset{0}{\overset{T}{\int }}1\cdot 1dx=x{|}_0^T=T\\ <b_1,b_2>=<b_2,b_1>=\underset{0}{\overset{T}{\int }}1\cdot \cos (\omega x)dx=\frac{1}{\omega }\sin (\omega x){|}_0^T=0\\ <b_1,b_3>=<b_3,b_1>=\underset{0}{\overset{T}{\int }}1\cdot \cos (2\omega x)dx=\frac{1}{2\omega }\sin (2\omega x){|}_0^T=0\\ <b_1,b_4>=<b_4,b_1>=\underset{0}{\overset{T}{\int }}1\cdot \sin (\omega x)dx=-\frac{1}{\omega }\cos (\omega x){|}_0^T=0\\ <b_1,b_5>=<b_5,b_1>=\underset{0}{\overset{T}{\int }}1\cdot \sin (2\omega x)dx=-\frac{1}{2\omega }\cos (2\omega x){|}_0^T=0\\ <b_2,b_2>=<b_2,b_2>=\underset{0}{\overset{T}{\int }}\cos (\omega x)\cdot \cos (\omega x)dx=\left(\frac{\sin (2\omega x)}{4\omega }+\frac{x}{2}\right){|}_0^T=\frac{T}{2}\\ <b_2,b_3>=<b_3,b_2>=\underset{0}{\overset{T}{\int }}\cos (\omega x)\cdot \cos (2\omega x)dx=\left(\frac{3\sin (\omega x)+\sin (3\omega x)}{6\omega }\right){|}_0^T=0\\ <b_2,b_4>=<b_4,b_2>=\underset{0}{\overset{T}{\int }}\cos (\omega x)\cdot \sin (\omega x)dx=-\frac{{\cos }^2(\omega x)}{2\omega }{|}_0^T=0\\ <b_2,b_5>=<b_5,b_2>=\underset{0}{\overset{T}{\int }}\cos (\omega x)\cdot \sin (2\omega x)dx=-\frac{2{\cos }^3(\omega x)}{3\omega }{|}_0^T=0\\ <b_3,b_3>=<b_3,b_3>=\underset{0}{\overset{T}{\int }}\cos (2\omega x)\cdot \cos (2\omega x)dx=\left(\frac{\sin (4\omega x)}{8\omega }+\frac{x}{2}\right){|}_0^T=\frac{T}{2}\\ <b_3,b_4>=<b_4,b_3>=\underset{0}{\overset{T}{\int }}\cos (2\omega x)\cdot \sin (\omega x)dx=\frac{\cos (3\omega x)-3\cos (\omega x)}{6\omega }{|}_0^T=0\\ <b_3,b_5>=<b_5,b_3>=\underset{0}{\overset{T}{\int }}\cos (2\omega x)\cdot \sin (2\omega x)dx=-\frac{\cos (4\omega x)}{8\omega }{|}_0^T=0\\ <b_4,b_4>=<b_4,b_4>=\underset{0}{\overset{T}{\int }}\sin (\omega x)\cdot \sin (\omega x)dx=\left(-\frac{\sin (2\omega x)}{4\omega }+\frac{x}{2}\right){|}_0^T=\frac{T}{2}\\ <b_4,b_5>=<b_5,b_4>=\underset{0}{\overset{T}{\int }}\sin (\omega x)\cdot \sin (2\omega x)dx=\left(\frac{2{\sin }^3(\omega x)}{3\omega }\right){|}_0^T=0\\ <b_5,b_5>=<b_5,b_5>=\underset{0}{\overset{T}{\int }}\sin (2\omega x)\cdot \sin (2\omega x)dx=\left(-\frac{\sin (4\omega x)}{8\omega }+\frac{x}{2}\right){|}_0^T=\frac{T}{2}\end{array}$

All values were checked with Wolframalpha

The Gram matrix then becomes

$\mathrm{\Gamma }(\mathrm{\Im })=\left(\begin{array}{ccccc}T& 0& 0& 0& 0\\ 0& \frac{T}{2}& 0& 0& 0\\ 0& 0& \frac{T}{2}& 0& 0\\ 0& 0& 0& \frac{T}{2}& 0\\ 0& 0& 0& 0& \frac{T}{2}\end{array}\right)$

(4.5)

As we can see the Gram matrix based constructed from this set of functions is a diagonal matrix

Finding $\det [\mathrm{\Gamma }(\mathrm{\Im })]$

The determinant of a diagonal matrix is

$\begin{array}{c}\det (A)=\prod _{i=1}^n{a}_{ii}\end{array}$

(4.6)

Where

$A_{nxn}=\left(\begin{array}{ccc}{a}_{11}& & 0\\ & \ddots & \\ 0& & a_{nn}\end{array}\right)$

Based on equation 4.6

$\det [\mathrm{\Gamma }(\mathrm{\Im })]=T\cdot \frac{T}{2}\cdot \frac{T}{2}\cdot \frac{T}{2}\cdot \frac{T}{2}=\frac{T^5}{16}$

(4.7)

For the set to be an orthogonal basis the Gram matrix must be a diagonal matrix with a non-zero determinant. As we can see from equations 4.5 and 4.7 both of these criteria are satisfied. Thus the set of functions is an orthogonal basis.