View source for University of Florida/Eml5526/s11.team2.reiss.HW

{{Big3|Problem 2.6: Determination of orthogonal functions}}


==Given==
Consider the family of functions<br>
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<div style="text-align: center;">
<math>\Im  = \left\{ {1,\cos (\omega x),\cos (2\omega x),\sin (\omega x),\sin (2\omega x)} \right\}</math></div>
|| <p style="text-align:right">
      (4.1)</p>
|}
on the interval [0,T], where T=<math>2\Pi/\omega</math>

==Find==
A) Construct <math>\Gamma (\Im )</math>  and observe its properties<br>
B) Find <math>\det [\Gamma (\Im )]</math><br>
C) Is <math>\Im</math> an orthogonal basis

==Solution==
Construct <math>\Gamma (\Im )</math>:

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<math>\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)</math></div>
|| <p style="text-align:right">
      (4.2)</p>
|}

where
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<math>\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}</math></div>
|}

In order to construct the matrix we must first define <math> < {b_i} , {b_j} > </math>

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<math> < {b_i} , {b_j} > =\int_{{x_0}}^x {{b_i}(x){b_j}(x)dx} </math>
</div>
|| <p style="text-align:right">
      (4.3)</p>
|}

Because multiplication of continuous functions is communicative it can be shown from equation 4.3 that
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<math> < {b_i} , {b_j} > =< {b_j} , {b_i} ></math>
</div>
|| <p style="text-align:right">
      (4.4)</p>
|}

And therefore <math>\Gamma (\Im )</math> is a symmetric matrix
<br>

We must now evaluate the terms of the matrix 
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<math>\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}</math>
</div>
|}
All values were checked with [http://www.wolframalpha.com Wolframalpha]
<br>
<br>
The Gram matrix then becomes

{| style="width:100%" border="0"
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<math>\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)</math>
</div>
|| <p style="text-align:right">
      (4.5)</p>
|}
As we can see the Gram matrix based constructed from this set of functions is a diagonal matrix<br>

Finding <math>\det [\Gamma (\Im )]</math>

The determinant of a diagonal matrix is 
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<math>\begin{array}{l}
\det (A) = \prod\limits_{i = 1}^n {{a_{ii}}} \\
\end{array}</math>
</div>
|| <p style="text-align:right">
      (4.6)</p>
|}
Where 
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<math>
{A_{nxn}} = \left( {\begin{array}{*{20}{c}}
{{a_{11}}}&{}&0\\
{}& \ddots &{}\\
0&{}&{{a_{nn}}}
\end{array}} \right)</math>
</div>
|}
<br>
Based on equation 4.6 
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<math>\det [\Gamma (\Im )]=T \cdot \frac{T}{2} \cdot \frac{T}{2} \cdot \frac{T}{2} \cdot \frac{T}{2} = \frac{{{T^5}}}{{16}}</math>
</div>
|| <p style="text-align:right">
      (4.7)</p>
|}

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.

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