User:Eml5526.s11.team4.yang/hw2-6

=Homework 2.6=

Problem
Consider F= { 1, Cos iwx, Sin iwx } on interval [ 0,T ], i.e. i=1,2

1). Construct $$\displaystyle \underline{\Gamma }(F) $$; observe property of $$\displaystyle \underline \Gamma  (F)$$.

2). Find det $$\underline{\Gamma }(F)$$.

3). Conclude that F is orthogonal basis, i.e., $$\displaystyle {{\Gamma }_{ij}}=\left\langle {{b}_{i}},{{b}_{j}} \right\rangle ={{\delta }_{ij}}$$

Solutions
1). Calculate the value of Gram matrix, $$\displaystyle w=\frac{2\pi }{T}$$:
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$$ \displaystyle
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{{\Gamma }_{ij}}=\left\langle {{b}_{i}},{{b}_{j}} \right\rangle =\int\limits_{\Omega }(x){{b}_{j}}(x)dx

$$
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$$ \displaystyle
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{{\Gamma }_{11}}=\int\limits_{\Omega }(x){{b}_{1}}(x)dx=\int_{0}^{T}{1dx}=T

$$
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$$ \displaystyle
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{{\Gamma }_{12}}={{\Gamma }_{21}}=\int\limits_{\Omega }(x){{b}_{2}}(x)dx=\int_{0}^{T}{\cos wxdx}=\frac{\sin wT}{w}=0

$$
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$$ \displaystyle
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\begin{align} & {{\Gamma }_{22}}=\int\limits_{\Omega }(x){{b}_{2}}(x)dx=\int_{0}^{T}{{{\left( \cos wx \right)}^{2}}dx} \\ & =\frac{\sin 2wT+2T}{4}-\frac{\sin 0+2\centerdot 0}{4}=\frac{T}{2} \\ \end{align}

$$
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$$ \displaystyle
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{{\Gamma }_{13}}={{\Gamma }_{31}}=\int\limits_{\Omega }(x){{b}_{3}}(x)dx=\int_{0}^{T}{\cos 2wxdx}=\frac{\sin 2wT}{2w}=0

$$
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$$ \displaystyle
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{{\Gamma }_{23}}={{\Gamma }_{32}}=\int\limits_{\Omega }(x){{b}_{3}}(x)dx=\int_{0}^{T}{\cos wx\centerdot \cos 2wxdx}=\frac{3\sin (wx)+\sin (3wx)}{6w}|_{0}^{T}=0

$$
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$$ \displaystyle
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{{\Gamma }_{33}}={{\Gamma }_{33}}=\int\limits_{\Omega }(x){{b}_{3}}(x)dx=\int_{0}^{T}{{{(\cos 2wx)}^{2}}dx}=\frac{\sin 4wT+4wx}{8w}|_{0}^{T}=\frac{T}{2}

$$
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$$ \displaystyle
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{{\Gamma }_{14}}={{\Gamma }_{41}}=\int\limits_{\Omega }(x){{b}_{4}}(x)dx=\int_{0}^{T}{\sin wxdx}=\frac{-\cos wx}{w}|_{0}^{T}=0

$$
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$$ \displaystyle
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{{\Gamma }_{24}}={{\Gamma }_{42}}=\int\limits_{\Omega }(x){{b}_{4}}(x)dx=\int_{0}^{T}{\cos wx\centerdot \sin wxdx}=\frac{-\cos (2wx)}{4w}|_{0}^{T}=0

$$
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$$ \displaystyle
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{{\Gamma }_{43}}={{\Gamma }_{34}}=\int\limits_{\Omega }(x){{b}_{3}}(x)dx=\int_{0}^{T}{(\cos 2wx)\centerdot \sin wxdx}=-\frac{\cos 3wx-3\cos wx}{6w}|_{0}^{T}=0

$$
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$$ \displaystyle
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{{\Gamma }_{44}}=\int\limits_{\Omega }(x){{b}_{4}}(x)dx=\int_{0}^{T}{{{(\sin wx)}^{2}}dx}=\frac{2wx-\cos 2wx}{4w}|_{0}^{T}=\frac{T}{2}

$$
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$$ \displaystyle
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{{\Gamma }_{15}}={{\Gamma }_{51}}=\int\limits_{\Omega }(x){{b}_{5}}(x)dx=\int_{0}^{T}{\sin 2wxdx}=\frac{-\cos 2wx}{2w}|_{0}^{T}=0

$$
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$$ \displaystyle
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{{\Gamma }_{25}}={{\Gamma }_{52}}=\int\limits_{\Omega }(x){{b}_{5}}(x)dx=\int_{0}^{T}{\cos wx\centerdot \sin 2wxdx}=\frac{-2{{\cos }^{3}}(wx)}{3w}|_{0}^{T}=0

$$
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$$ \displaystyle
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{{\Gamma }_{53}}={{\Gamma }_{35}}=\int\limits_{\Omega }(x){{b}_{3}}(x)dx=\int_{0}^{T}{(\cos 2wx)\centerdot \sin 2wxdx}=\frac{-\cos 4wx}{8w}|_{0}^{T}=0

$$
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$$ \displaystyle
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{{\Gamma }_{54}}={{\Gamma }_{45}}=\int\limits_{\Omega }(x){{b}_{4}}(x)dx=\int_{0}^{T}{(\sin wx)\sin 2wxdx}=\frac{2{{\sin }^{3}}wx}{3w}|_{0}^{T}=0

$$
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$$ \displaystyle
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{{\Gamma }_{55}}=\int\limits_{\Omega }(x){{b}_{5}}(x)dx=\int_{0}^{T}{{{(\sin 2wx)}^{2}}dx}=\frac{4wx-\cos 4wx}{8w}|_{0}^{T}=\frac{T}{2}

$$ So, the Gram matrix is:
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$$ \displaystyle
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\Gamma =\left[ \begin{matrix} T & 0 & 0 & 0 & 0 \\ 0 & T/2 & 0 & 0 & 0 \\ 0 & 0 & T/2 & 0 & 0 \\ 0 & 0 & 0 & T/2 & 0 \\ 0 & 0 & 0 & 0 & T/2 \\ \end{matrix} \right]

$$ It is observed that this matrix is diagonal matrix, i.e. only the diagonal component is not zero.
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2). Calculate the determinate of the Gram matrix,
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$$ \displaystyle
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\det (\Gamma )=\det \left[ \begin{matrix} T & 0 & 0 & 0 & 0 \\ 0 & T/2 & 0 & 0 & 0 \\ 0 & 0 & T/2 & 0 & 0 \\ 0 & 0 & 0 & T/2 & 0 \\ 0 & 0 & 0 & 0 & T/2 \\ \end{matrix} \right]=\frac{16}$

$$
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3). from the matrix, it is found that only the diagonal components, i.e. i=j, have non-zero value.
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$$ \displaystyle
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{{\Gamma }_{ij}}=0\begin{matrix} {} & i\ne j \\ \end{matrix}

$$     (Eq )
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$$ \displaystyle
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{{\Gamma }_{ij}}\ne 0\begin{matrix} {} & i=j \\ \end{matrix}

$$ This agrees with the definition of Kronecker delta function. So, the F is orthogonal basis.
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