User:Egm6321.f10.team2.oztekin/SandBox

=Problem 2.6=

Given
Consider family of functions $$\displaystyle F=\left \{ 1,cosiwx,siniwx \right \} $$ on interval $$\displaystyle \Omega \left [ 0,T \right ]$$ for i =1,2

So


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$$  \displaystyle F=\left \{ b_{1}(x),b_{2}(x),b_{3}(x),b_{4}(x),b_{5}(x) \right \} $$     (Eq 6.1)
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$$  \displaystyle F=\left \{ 1,coswx,cos2wx,sinwx,sin2wx \right \} $$     (Eq 6.2)
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Find
1)Construct $$\displaystyle \underline{\Gamma }(F)$$ and observe properties of $$\displaystyle \underline{\Gamma }(F)$$

2)Find the determinant of  $$\displaystyle \underline{\Gamma }(F)$$

3)Conclude F is orthogonal basis.

Solution
1)
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$$  \displaystyle \underline{\Gamma }={{[{{\Gamma }_{ij}}]}_{5*5}}=\left( \begin{matrix}  {{\Gamma }_{11}} & \ldots  & {{\Gamma }_{15}}  \\   \vdots  & \ddots  & \vdots   \\   {{\Gamma }_{51}} & \cdots  & {{\Gamma }_{55}}  \\ \end{matrix} \right) $$     (Eq 6.3)
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$$  \displaystyle \Gamma _{ij}==\int_{\Omega }b_{i}(x)b_{j}(x)dx $$     (Eq 6.4)
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$$  \displaystyle \Gamma _{11}=\int_{0}^{T}1dx=\frac{2\Pi }{w} $$     (Eq 6.5)
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$$  \displaystyle \Gamma _{21}=\Gamma _{12}=\int_{0}^{T}coswxdx=0 $$     (Eq 6.6)
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$$  \displaystyle \Gamma _{31}=\Gamma _{13}=\int_{0}^{T}cos2wxdx=0 $$     (Eq 6.7)
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$$  \displaystyle \Gamma _{41}=\Gamma _{14}=\int_{0}^{T}sinwxdx=0 $$     (Eq 6.8)
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$$  \displaystyle \Gamma _{51}=\Gamma _{15}=\int_{0}^{T}sin2wxdx=0 $$     (Eq 6.9)
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$$  \displaystyle \Gamma _{22}=\int_{0}^{T}coswx.coswxdx=\frac{\Pi }{w} $$     (Eq 6.10)
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$$  \displaystyle \Gamma _{23}=\Gamma _{32}=\int_{0}^{T}coswx.cos2wxdx=0 $$     (Eq 6.11)
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$$  \displaystyle \Gamma _{24}=\Gamma _{42}=\int_{0}^{T}coswx.sinwxdx=0 $$     (Eq 6.12)
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$$  \displaystyle \Gamma _{25}=\Gamma _{52}=\int_{0}^{T}coswx.sin2wxdx=0 $$     (Eq 6.13)
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$$  \displaystyle \Gamma _{33}=\int_{0}^{T}cos2wx.cos2wxdx=\frac{\Pi }{w} $$     (Eq 6.14)
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$$  \displaystyle \Gamma _{34}=\Gamma _{43}=\int_{0}^{T}cos2wx.sinwxdx=0 $$     (Eq 6.15)
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$$  \displaystyle \Gamma _{35}=\Gamma _{53}=\int_{0}^{T}cos2wx.sin2wxdx=0 $$     (Eq 6.16)
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$$  \displaystyle \Gamma _{44}=\int_{0}^{T}sinwx.sinwxdx=\frac{\Pi }{w} $$     (Eq 6.17)
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$$  \displaystyle \Gamma _{45}=\Gamma _{45}=\int_{0}^{T}sinwx.sin2wxdx=0 $$     (Eq 6.18)
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$$  \displaystyle \Gamma _{55}=\int_{0}^{T}sin2wx.sin2wxdx=\frac{\Pi }{w} $$     (Eq 6.19)
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$$  \displaystyle \underline{\Gamma }(F)=\left[ \begin{matrix} \begin{matrix} \begin{matrix} \begin{matrix} \frac{2\Pi }{w} & 0 \\ \end{matrix} & 0 & 0 & 0 \\ \end{matrix} \\ \begin{matrix} \begin{matrix} 0 & \frac{\Pi }{w} \\ \end{matrix} & 0 & 0 & 0 \\ \end{matrix} \\ \end{matrix} \\ \begin{matrix} \begin{matrix} 0 & 0 \\ \end{matrix} & \frac{\Pi }{w} & 0 & 0  \\ \end{matrix} \\ \begin{matrix} \begin{matrix} 0 & 0 \\ \end{matrix} & 0 & \frac{\Pi }{w} & 0  \\ \end{matrix} \\ \begin{matrix} \begin{matrix} 0 & 0 \\ \end{matrix} & 0 & 0 & \frac{\Pi }{w}  \\ \end{matrix} \\ \end{matrix} \right]
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$$     (Eq 6.20)
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It is shown that gram matrix is diagonal matrix.

2)Determinant of gram matrix has been calculated by wolfram alpha.


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$$  \displaystyle \det \underline{\Gamma }(F)=\frac{2{{\Pi }^{5}}} $$     (Eq 6.21) According to theorem on lecture note meeting 10-2 $$\displaystyle \det \underline{\Gamma }(F)\neq 0$$,so the family of functions are linearly independent. Also they are basis functions.
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3)From the lecture note Eq(2) meeting 7-1


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$$  \displaystyle {{\delta }_{ij}}=\left\{ \begin{matrix} \begin{matrix} 1 & for & i=j \\ \end{matrix} \\ \begin{matrix} 0 & for & i\ne j \\ \end{matrix} \\ \end{matrix} \right.
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$$     (Eq 6.22)
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$$  \displaystyle {{\Gamma }_{ij}}=<{{b}_{i}},{{b}_{j}}>={{\delta }_{ij}} $$     (Eq 6.23)
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As it is seen on above,both elements of gram matrix are satisfying the orthogonal condition.Therefore F is a family of orthogonal basis functions.