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

Problem 2.6: Determination of orthogonal functions

## Given

Consider the family of functions
| $\Im =\left\{{1,\cos(\omega x),\cos(2\omega x),\sin(\omega x),\sin(2\omega x)}\right\}$ | (4.1) |
on the interval [0,T], where T=$2\Pi /\omega$

## Find

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

## Solution

Construct $\Gamma (\Im )$:
| $\Gamma (\Im )=\left({\begin{array}{*{20}{c}}{<{b_{1}},{b_{1}}>}&\ldots &{<{b_{1}},{b_{n}}>}\\\vdots &\ddots &\vdots \\{<{b_{n}},{b_{1}}>}&\cdots &{<{b_{n}},{b_{n}}>}\end{array}}\right)$ | (4.2) |
where
* ${\begin{array}{l}{b_{1}}=1\\{b_{2}}=\cos(\omega x)\\{b_{3}}=\cos(2\omega x)\\{b_{4}}=\sin(\omega x)\\{b_{5}}=\sin(2\omega x)\end{array}}$
In order to construct the matrix we must first define $<{b_{i}},{b_{j}}>$
| $<{b_{i}},{b_{j}}>=\int _{x_{0}}^{x}{{b_{i}}(x){b_{j}}(x)dx}$ | (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 $\Gamma (\Im )$ is a symmetric matrix
We must now evaluate the terms of the matrix
* ${\begin{array}{l}<{b_{1}},{b_{1}}>=<{b_{1}},{b_{1}}>=\int _{0}^{T}{1\cdot 1dx=x}|_{0}^{T}=T\\<{b_{1}},{b_{2}}>=<{b_{2}},{b_{1}}>=\int _{0}^{T}{1\cdot \cos(\omega x)dx={\frac {1}{\omega }}\sin(\omega x)|_{0}^{T}=0}\\<{b_{1}},{b_{3}}>=<{b_{3}},{b_{1}}>=\int _{0}^{T}{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}}>=\int _{0}^{T}{1\cdot \sin(\omega x)dx=-{\frac {1}{\omega }}\cos(\omega x)|_{0}^{T}=0}\\<{b_{1}},{b_{5}}>=<{b_{5}},{b_{1}}>=\int _{0}^{T}{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}}>=\int _{0}^{T}{\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}}>=\int _{0}^{T}{\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}}>=\int _{0}^{T}{\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}}>=\int _{0}^{T}{\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}}>=\int _{0}^{T}{\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}}>=\int _{0}^{T}{\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}}>=\int _{0}^{T}{\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}}>=\int _{0}^{T}{\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}}>=\int _{0}^{T}{\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}}>=\int _{0}^{T}{\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
| $\Gamma (\Im )=\left({\begin{array}{*{20}{c}}T&0&0&0&0\\0&{\frac {T}{2}}&0&0&0\\0&0&{\frac {T}{2}}&0&0\\0&0&0&{\frac {T}{2}}&0\\0&0&0&0&{\frac {T}{2}}\end{array}}\right)$ | (4.5) |
As we can see the Gram matrix based constructed from this set of functions is a diagonal matrix
Finding $\det[\Gamma (\Im )]$
The determinant of a diagonal matrix is
| ${\begin{array}{l}\det(A)=\prod \limits _{i=1}^{n}{a_{ii}}\\\end{array}}$ | (4.6) |
Where
* ${A_{nxn}}=\left({\begin{array}{*{20}{c}}{a_{11}}&&0\\&\ddots &\\0&&{a_{nn}}\end{array}}\right)$
Based on equation 4.6
| $\det[\Gamma (\Im )]=T\cdot {\frac {T}{2}}\cdot {\frac {T}{2}}\cdot {\frac {T}{2}}\cdot {\frac {T}{2}}={\frac {T^{5}}{16}}$ | (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.