User:EGM6341.s10.Team1.toddmock/HW6

= Problem 3: Determine 3rd and 4th rows of the Matrix =

Given
The matrix is given to be:
 * {| style="width:70%" border="0" align="center"




 * $$\underbrace{\begin{bmatrix}

1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 1 & 1 & 1 & 1 \\ 0 & 1 & 2 & 3 \end{bmatrix}.}_{A} \underbrace{\begin{bmatrix}

C_0 \\ C_1 \\ C_2 \\ C_3 \end{bmatrix}}_{C_i} = \begin{bmatrix}

z_i \\ z'_i \\ z_{i+1} \\ z'_{i+1} \end{bmatrix} $$
 * }.
 * }.

Find
Need to determine the 3rd and 4th row of the matrix using


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

$$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle \bar{d_3}=z_{i+1}=z(s=1)
 * }.
 * }.

and


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

$$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle \bar{d_4}=z'_{i+1}=z'(s=1)
 * }.
 * }.

Solution
We know that


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$$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle z(s)=\sum_{i=0}^{3} C_i s^i= C_0+ C_1s+ C_2s^2+ C_3s^3
 * }.
 * }.

and


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$$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle z'(s)= \sum_{i=1}^{3} iC_i s^{i-1}= C_1+ 2C_2s+ 3C_3s^2
 * }.
 * }.

thus using the above relations where $$s=1$$ in both cases we then have


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$$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle z_{i+1}= C_0+ C_1+ C_2+ C_3
 * }.
 * }.

and


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$$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle z'_{i+1}= C_1+ 2C_2+ 3C_3
 * }.
 * }.

Therefore from here it is easy to surmise that the 3rd and 4th rows are $$[1 1 1 1]$$ and $$[0 1 2 3]$$ respectively.

Author --Egm6341.s10.team1.toddmock 19:13, 7 April 2010 (UTC)

= Problem 4: verify the inverse of the Matrix =

Given
From the matrix


 * {| style="width:70%" border="0" align="center"




 * $$A=\begin{bmatrix}

1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 1 & 1 & 1 & 1 \\ 0 & 1 & 2 & 3 \end{bmatrix} $$
 * }.
 * }.

Find
Use Matlab to show that the inverse is


 * {| style="width:70%" border="0" align="center"




 * $$A^{-1}=\begin{bmatrix}

1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ -3 & -2 & 3 & -1 \\ 2 & 1 & -2 & 1 \end{bmatrix} $$
 * }.
 * }.

Solution
This yields the result

Which is exactly what we were hoping for.

Author --Egm6341.s10.team1.toddmock 19:13, 7 April 2010 (UTC)

= Problem 9: Proof of $$z_{i+1}$$ =

Given
We know that


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$$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle \dot{z}_{i+1/2}=\dfrac {-3}{2h}(z_i-z_{i+1})-\dfrac {1}{4}(f_i+f_{i+1})
 * }.
 * }.

and we want to use collocation at $$t_{i+1/2}$$ which gives the relation


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

$$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle \triangle = 0 = \dot{z}_{i+1/2}-f_{i+1/2}
 * }.
 * }.

Find
The mission here is to find a relation for $$z_{i+1}$$

Solution
Therefore using the collocation we have
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$$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle f_{i+1/2}= -\dfrac {3}{2h}(z_i-z_{i+1})- \dfrac {1}{4}(f_i+ f_{i+1})
 * }.
 * }.

which can be rearranged to give


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$$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle \dfrac{1}{4}f_i+\dfrac{1}{4}f_{i+1}+f_{i+1/2}=\dfrac{3}{2h}(z_{i+1}-z_i)
 * }.
 * }.

or


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$$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle \dfrac{1}{4}(f_i+4f_{i+1/2}+f_{i+1})=\dfrac{3}{2h}(z_{i+1}-z_i)
 * }.
 * }.

rearranging this gives


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$$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle z_{i+1}-z_i=\dfrac{h}{6}(f_i+4f_{i+1/2}+f_{i+1})
 * }.
 * }.

here it should be noted that the right hand side is Simpson's rule, proceeding further we obtain the expression


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

$$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle z_{i+1}=z_i+\dfrac{h/2}{3}(f_i+4f_{i+1/2}+f_{i+1})
 * }.
 * }.

Author --Egm6341.s10.team1.toddmock 19:14, 7 April 2010 (UTC)