User:Eml5526.s11.team5.srv/hw4

Given

 * that is ,


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$$F_{p} = \left \{x^{j}; \quad j= 0,1,2,... \right \} $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *  $$ \displaystyle (Eq. 4.5.2)
 * }


 * {| style="width:100%" border="0"

$$F_{c} = \left \{cos (jx);\quad j= 0,1,2,... \right \} $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *  $$ \displaystyle (Eq. 4.5.3)
 * }


 * {| style="width:100%" border="0"

$$F_{s} = \left \{1, sin (jx); \quad j= 1,2,... \right \} $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *  $$ \displaystyle (Eq. 4.5.4)
 * }


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$$F_{f} = \left \{F_{p} = \left \{cos(jx); \quad j= 0,1,2,...                \qquad sin (kx); \quad j = 1,2,... \right \}\right \} $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *  $$ \displaystyle (Eq. 4.5.5)
 * }


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$$F_{p} = \left \{e^{jx}; \quad j = 0,1,2,... \right \}$$ $$
 * style="width:95%" |
 * style="width:95%" |
 *  $$ \displaystyle (Eq. 4.5.6)
 * }

Solution
1.

Consider

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Consider

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Consider

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Consider

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Consider

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2.

The plot showing the phase shift in the basis function according to the constraint breaking solution is given below.

(i) Polynomial basis function



(ii) Cosine basis function



(iii) Sine basis function



(iv) Fourier basis function



(v) Exponential basis function



The Matlab code for plotting the graphs shown above is given below.

3.

The Gram Matrix is defined as:

For this case,


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$$ b_1(x)=1 \quad $$ $$
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 * style="width:95%" |
 *  $$ \displaystyle                                                       (Eq. 4.5.15)
 * }


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$$ b_2(x)= \ e^{x} \quad $$ $$
 * style="width:90%" |
 * style="width:90%" |
 *  $$ \displaystyle                                                       (Eq. 4.5.16)
 * }


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$$  b_3(x)= \ e^{2x} \quad $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *  $$ \displaystyle                                                       (Eq. 4.5.17)
 * }


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Now we need to calculate each element in the $$ \Gamma $$ Matrix: $$
 * style="width:95%" |
 * style="width:95%" |
 *  $$ \displaystyle\!
 * }


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Since $$ \int_\Omega b_i(x)b_j(x)\,dx = \int_\Omega b_j(x)b_i(x)\,dx$$,  $$\Gamma_{ij}$$ will be equivalent to $$\Gamma_{ji}$$ $$
 * style="width:95%" |
 * style="width:95%" |
 *  $$ \displaystyle                                                       (Eq. 4.5.18)
 * }


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$$ \Gamma_{1,1} = \int_{-2}^{4} 1*1*dx = \left [ x\right ]_{0}^{1} = 1 $$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \displaystyle\!
 * }

<span id="(1)">
 * {| style="width:100%" border="0"

$$ \Gamma_{1,2} = \Gamma_{2,1} = \int_{-2}^{4} 1*e^{x}*dx = \left [ e^{x}\right ]_{-2}^{4} = 54.463 $$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \displaystyle\!
 * }

<span id="(1)">
 * {| style="width:100%" border="0"

$$ \Gamma_{1,3} = \Gamma_{3,1} = \int_{-2}^{4} 1*e^{2x}*dx = \left [ e^{2x}/2 \right ]_{-2}^{4} = 1490.4699 $$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \displaystyle\!
 * }

<span id="(1)">
 * {| style="width:100%" border="0"

$$ \Gamma_{2,2} = \int_{-2}^{4} e^{x}*e^{x}*dx = \left [ e^{2x}/2\right ]_{-2}^{4} = 1490.4699 $$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \displaystyle\!
 * }

<span id="(1)">
 * {| style="width:100%" border="0"

$$ \Gamma_{2,3} = \Gamma_{3,2} = \int_{-2}^{4} e^{x}*e^{2x}*dx = \left [ e^{3x}/3\right ]_{-2}^{4} = 54251.6 $$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \displaystyle\!
 * }

<span id="(1)">
 * {| style="width:100%" border="0"

$$ \Gamma_{3,3} = \int_{-2}^{4} e^{2x}*e^{2x}*dx = \left [ e^{4x}/4 \right ]_{-2}^{4} = 2221527.63 $$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \displaystyle\!
 * }

<span id="(1)">

For the value of the determinant, refer http://www.wolframalpha.com

The determinant of the Gram Matrix is not equal to zero.

Therefore,