User:Egm6341.s2010.Team1/HW6

HW 6 Problem Set  =Problem 1: Compute the arc length of an ellipse on $$[0^{\circ},60^{\circ}]$$=

Statement
An arc is denoted $$\widehat{PQ}$$ and lies on an ellipse defined by,
 * $$r(\theta)=\frac{a(1-e^2)}{1-e \cos{(\theta)}}$$

where $$a=1$$ and $$e=\sin{(\frac{\pi}{12})}$$. The equation for the arc length is,
 * $$PQ=\int_a^bdl=\int_{\theta_1}^{\theta_2}d\theta \sqrt{r^2+\left (\frac{dr}{d\theta} \right )^2}$$

1.) Estimate the number of nodes required for the composite trapezoidal estimate of the integral to have an error $$O(10^{-10})$$ 2.)  Using successive numerical integrations, find the actual number of nodes required for the trapezoidal technique by considering the stopping criterion $$\left |I_{2n}-I_n \right |<10^{-10}$$ 3.) How many nodes are required for the Romberg table (verify vs trapezoidal), clencurt.m (verify with a similar stopping criterion), and sum (chebfun - compare to trapezoidal).

Solution
The integrating function is,
 * $$F=\sqrt{r^2+\left (\frac{dr}{d\theta} \right )^2}=\frac{1-e^2}{(1-e\cos{\theta})^2}\sqrt{1-2e\cos{\theta}+e^2}$$

1.) An estimate for the error in the composite trapezoidal technique is,
 * $$\left | E_n \right |\leq \frac{(b-a)^3}{12n^2}M_2$$

Rearranging for the number of nodes, $$n$$, gives,
 * $$n=\sqrt{\frac{(b-a)^3}{12(10^{-10})}M_2}$$

where $$M_2=max\left | F^{(2)}(\zeta) \right |, \zeta\in [a,b]$$ and $$[a,b]=[0,60\frac{\pi}{180}]$$. The following MATLAB code solves for $$M_2$$ and determines $$n<16,755$$

2.) Numerically integrating $$F$$ using the trapezoidal rule detemines that at $$n=16,384$$ we get $$\left | I_{16,384}-I_{8,192} \right | < 10^{-10}$$. The following MATLAB code is used.  Notice that in step sizes of 2n, the previous summation can be utilized as shown in the code.

3.) Numerically integrate using the Romberg, clencurt, and sum (chebfun).

Romberg gives $$O(10^{-10})$$ compared to itself at $$n=2^7=128$$, see code.

For the sum (chebfun) the code is shown next. Comparing this to the composite trapezoidal result shows a difference $$O(10^{-11})$$ so is considered accurate enough.

And finally for clencurt.m is shown last. The result is compared to itself and at n=10 has an error less than $$O(10^{-10})$$

Problem solved by EGM6341.s10.Team1.Kumanchik 22:52, 7 April 2010 (UTC)

=Problem 2-Consider v+dv,show lin.mom.compo. =

Find
At $$ t+dt\,$$,consider $$ v+dv\,$$, show $$ dp_{\bar{y}}=mvd\gamma\,$$ by neglecting hot.

Given
Refer Lecture slide (1) [[media:Egm6341.s10.mtg33.djvu|p.33-2]] for problem statement



[Figure 1]

$$ dp_{\bar{y}}=mvd\gamma\,$$

Solution


[Figure 2]

From Fig.2,it can be seen that

$$ dp_{\bar{y}}=m(v+dv)d\gamma\,$$

$$ =mvd\gamma+ \cancelto{}{mdvd\gamma}\,$$

By neglecting higher order term$$ dvd\gamma\,$$,just retention the lowest term

It can be derived that

$$ dp_{\bar{y}}=mvd\gamma\,$$

Egm6341.s10.team1.lei15:00, 7 April 2010 (UTC)

= 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


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

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

and


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

$$ $$
 * $$\displaystyle
 * $$\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


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

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

and


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

$$ $$
 * $$\displaystyle
 * $$\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.

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.

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

=Problem 5: Identify the basis functions $${N_i(s)}\,$$, $$i=1,2,3,4\,$$. Plot $$\bar{N_i}\,$$'s=

Solution
$$ \sum_{i=0}^{3} C_is^i = \sum_{i=1}^{4} N_i(s)d_i\,$$ $$ C_0+C_1s+C_2s^2+C_3s^3=N_1(s)d_1+N_2(s)d_2+N_3(s)d_3+N_4(s)d_4\,$$ $$ C_0=z_i \,$$ $$ C_1=z_i' \,$$ $$ C_2=-3z_i-2z_i'+3z_{i+1}-z_{i+1}' \,$$ $$ C_3=2z_i+z_i'-2z_{i+1}+z_{i+1}' \,$$ $$ d_1=z_i\,$$ $$ d_2=\dot{z_i}=z_i'\frac{1}{h}\,$$ $$ d_3=z_{i+1}\,$$ $$ d_4=\dot{z_{i+1}}=z_{i+1}'\frac{1}{h} \,$$ $$ \Rightarrow z_i+z_i's+(-3z_i-2z_i'+3z_{i+1}-z_{i+1}')s^2+(2z_i+z_i'-2z_{i+1}+z_{i+1}')s^3=N_1(s)z_i+N_2(s)z_i'\frac{1}{h}+N_3(s)z_{i+1}+N_4(s)z_{i+1}'\frac{1}{h} \,$$ $$ \Rightarrow z_i(1-3s^2+2s^3)+z_i'(s-2s^2+s^3)+z_{i+1}(3s^2-2s^3)+z_{i+1}'(-s^2+s^3)=z_i(N_1(s))+z_i'(N_2(s)\frac{1}{h})+z_{i+1}(N_3(s))+z_{i+1}'(N_4(s)\frac{1}{h}) \,$$

$$ \bar{N_1}(s) = 1-3s^2+2s^3 \,$$ $$ \bar{N_2}(s) = s-2s^2+s^3 \,$$ $$ \bar{N_3}(s) = 3s^2-2s^3 \,$$ $$ \bar{N_4}(s) = -s^2+s^3 \,$$

Note for verification that: $$ \bar{N_1}(0)=1, \dot{\bar{N_1}}(0)=0, \bar{N_1}(1)=0, \dot{\bar{N_1}}(1)=0 \,$$ $$ \bar{N_2}(0)=0, \dot{\bar{N_2}}(0)=1, \bar{N_2}(1)=0, \dot{\bar{N_2}}(1)=0 \,$$ $$ \bar{N_3}(0)=0, \dot{\bar{N_3}}(0)=0, \bar{N_3}(1)=1, \dot{\bar{N_3}}(1)=0 \,$$ $$ \bar{N_4}(0)=0, \dot{\bar{N_4}}(0)=0, \bar{N_4}(1)=0, \dot{\bar{N_4}}(1)=1 \,$$



Egm6341.s10.team1.andrewdugan 21:27, 7 April 2010 (UTC)

=Problem 6- Find s=s(t) =

Find
Find $$ s=s(t)\,$$ such that $$ Z(t)=\sum_{i=1}^{4}N_{i}(t)d_{i}\,$$

Given
(1) [[media:Egm6341.s10.mtg35.djvu|p.35-1]]

Local coord. for $$ [t_{i},t_{i+1}]:\,$$

$$ t(s)=(1-s)t_{i}+st_{i+1}\, $$

Solution
Convert

$$ t(s)=(1-s)t_{i}+st_{i+1}\, $$

We can get

$$ s=\frac{t-t_{i}}{t_{i+1}-t_{i}}\,$$

thus

$$ s=s(t)=\frac{t-t_{i}}{t_{i+1}-t_{i}}\,$$

plug$$ s=s(t)\,$$ into

$$ Z(s)=\sum_{i=1}^{4}N_{i}(s)d_{i}\,$$

$$ \Rightarrow Z(T)=\sum_{i=1}^{4}N_{i}(T)d_{i}\,$$

Egm6341.s10.team1.lei15:00, 7 April 2010 (UTC)

=Problem 7: Find the equation of motion $$\mathbf{Z}$$ when local coordinate vector $$ s = \frac{1}{2}$$=

Problem Statement
Prove $$\begin{align} \mathbf{Z}_{i+\frac{1}{2}}& = \ \mathbf{Z}(s = \frac{1}{2}) & = \ \frac{1}{2}(\mathbf{Z}_{i}+\mathbf{Z}_{i+1}+\frac{h}{8}(f_{i}-f_{i+1}))\end{align}$$ Given :
 * {| style="width:100%" border="0" align="left"

C_{i}s^{i}$$
 * (a): (2) [[media:Egm6341.s10.mtg35.djvu|p.35-2]] $$\mathbf{Z}(s) = \ \sum_{i=0}^{3}
 * (a): (2) [[media:Egm6341.s10.mtg35.djvu|p.35-2]] $$\mathbf{Z}(s) = \ \sum_{i=0}^{3}
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

C_{0} \\ C_{1} \\ C_{2} \\ C_{3} \\ \end{bmatrix} = \begin{bmatrix} 1&0&0&0 \\ 0&1&0&0 \\ -3&-2&3&-1 \\ 2&1&-2&1 \\ \end{bmatrix} \cdot \!\, \begin{bmatrix} \mathbf{Z}_{i} \\ \mathbf{Z}_{i}^{'} \\ \mathbf{Z}_{i+1} \\ \mathbf{Z}_{i+1}^{'} \\ \end{bmatrix} $$
 * (b): (1) [[media:Egm6341.s10.mtg35.djvu|p.35-4]] $$\begin{bmatrix}
 * (b): (1) [[media:Egm6341.s10.mtg35.djvu|p.35-4]] $$\begin{bmatrix}
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"


 * (c): (1) [[media:Egm6341.s10.mtg36.djvu|p.36-1]] $$ \mathbf{Z}^{'} = h \dot{\mathbf{Z}}$$
 * }
 * }
 * }

Ref: Lecture Notes [[media:Egm6341.s10.mtg35.djvu|p.35-2&4]]  [[media:Egm6341.s10.mtg36.djvu|p.36-1]]   

Solution
$$\begin{align} \mathbf{Z}(s = \frac{1}{2})& = \ \sum_{i=0}^{3}C_{i}s^{i} \\ & =\ C_{0}(\frac{1}{2})^{0}+C_{1}(\frac{1}{2})^{1}+C_{2}(\frac{1}{2})^{2}+C_{3}(\frac{1}{2})^{3} \\ & =\ \mathbf{Z}_{i}+\mathbf{Z}_{i}^{'}(\frac{1}{2})+(3\mathbf{Z}_{i+1}-3\mathbf{Z}_{i}-2\mathbf{Z}_{i}^{'}-\mathbf{Z}_{i+1}^{'})(\frac{1}{4})+(\mathbf{Z}_{i+1}^{'}-2\mathbf{Z}_{i+1}+2\mathbf{Z}_{i}+\mathbf{Z}_{i}^{'})(\frac{1}{8}) \\ & = \ (1-\frac{3}{4}+\frac{1}{4})\mathbf{Z}_{i}+(\frac{1}{2}-\frac{1}{2}+\frac{1}{8})\mathbf{Z}_{i}^{'}+(\frac{3}{4}-\frac{1}{4})\mathbf{Z}_{i+1}+(-\frac{1}{4}+\frac{1}{8})\mathbf{Z}_{i+1}^{'} \\ & = \ \frac{1}{2} \mathbf{Z}_{i}+\frac{1}{8}\mathbf{Z}_{i}^{'}+\frac{1}{2}\mathbf{Z}_{i+1}-\frac{1}{8}\mathbf{Z}_{i+1}^{'} \\ & = \ \frac{1}{2}(\mathbf{Z}_{i}+\mathbf{Z}_{i+1})+\frac{1}{8}(\mathbf{Z}_{i}^{'}-\mathbf{Z}_{i+1}^{'}) \\ & = \ \frac{1}{2}(\mathbf{Z}_{i}+\mathbf{Z}_{i+1})+\frac{h}{8}(\frac{1}{h}\mathbf{Z}_{i}^{'}-\frac{1}{h}\mathbf{Z}_{i+1}^{'}) \\ & = \ \frac{1}{2}(\mathbf{Z}_{i}+\mathbf{Z}_{i+1})+\frac{h}{8}(\dot{\mathbf{Z}_{i}}-\dot{\mathbf{Z}_{i+1}}) \\ & =\ \frac{1}{2}(\mathbf{Z}_{i}+\mathbf{Z}_{i+1})+\frac{h}{8}(f_{i}-f_{i+1}) \end{align}$$

EGM6341.s10.Team1.Ya-Chiao Chang 02:57, 7 April 2010 (UTC)

=Problem 8: Verify the expression for $$ z_{i+\frac{1}{2}}' \,$$=

Find
Verify that $$ z_{i+\frac{1}{2}}'=-\frac{3}{2}(z_i-z_{i+1})-\frac{1}{4}(z_i'+z_{i+1}')\,$$

Solution
$$ z_{i+\frac{1}{2}}'=z'(s=\frac{1}{2}) \,$$ Recall that $$ z'(s)=\sum_{i=1}^{3}iC_is^{i-1} \,$$ Thus $$ z'(s=\frac{1}{2})= (1)(C_1)(\frac{1}{2})^{1-1}+(2)(C_2)(\frac{1}{2})^{2-1}+(3)(C_3)(\frac{1}{2})^{3-1} \,$$ $$ z'(s=\frac{1}{2}) = (1)(C_1)(1)+(2)(C_2)(\frac{1}{2})+(3)(C_3)(\frac{1}{4}) = C_1+C_2+\frac{3}{4}C_3 \,$$ Recall that $$ C_1=z_i', C_2=-3z_i-2z_i'+3z_{i+1}-z_{i+1}', C_3=2z_i+z_i'-2z_{i+1}+z_{i+1}' \,$$ Then, $$ z'(s=\frac{1}{2}) = z_i'+(-3z_i-2z_i'+3z_{i+1}-z_{i+1}')+\frac{3}{4}(2z_i+z_i'-2z_{i+1}+z_{i+1}') \,$$ $$ z'(s=\frac{1}{2}) = z_i(-3+\frac{3}{2})+z_i'(1-2+\frac{3}{4})+z_{i+1}(3-\frac{3}{2})+z_{i+1}'(-1+\frac{3}{4}) \,$$ $$ z'(s=\frac{1}{2}) = -\frac{3}{2}z_i-\frac{1}{4}z_i'+\frac{3}{2}z_{i+1}-\frac{1}{4}z_{i+1}' \,$$

Egm6341.s10.team1.andrewdugan 21:33, 7 April 2010 (UTC)

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

Given
We know that


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

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

$$ $$
 * $$\displaystyle
 * $$\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


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

$$ $$
 * $$\displaystyle
 * $$\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


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

$$ $$
 * $$\displaystyle
 * $$\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


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

$$ $$
 * $$\displaystyle
 * $$\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})
 * }.
 * }.

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

=Problem 10: Use Kessler's code to generate his table=

Statement
Use Kessler's code to generate his table and then follow the code line-by-line to get $$(P_2,P_3)(P_4,P_5)(P_6,P_6)$$. Comment on the code.

Solution
Kessler's table was generated with the code below for n=8:

The code computes the Trapezoidal error polynomials evaluated at t=1. The polynomial equation is,
 * $$p_{2k}(t)=c_1\frac{t^{2k}}{(2k)!}+c_3\frac{t^{2k-2}}{(2k-2)!}+c_5\frac{t^{2k-4}}{(2k-4)!}+...+c_{2k-3}\frac{t^4}{4!}+c_{2k-1}\frac{t^2}{2!}+c_{2k+1}\frac{t^0}{0!}$$

When evaluated at t=1, the equation is,
 * $$p_{2k}(1)=\frac{c_1}{(2k)!}+\frac{c_3}{(2k-2)!}+\frac{c_5}{(2k-4)!}+...+\frac{c_{2k-3}}{4!}+\frac{c_{2k-1}}{2!}+\frac{c_{2k+1}}{0!}$$

The coefficients, $$c$$, are computed by solving the below equation using $$c_1=-1$$ as a starting point,
 * $$\frac{c_1}{(2k-1)!}+\frac{c_3}{(2k-3)!}+\frac{c_5}{(2k-5)!}+...+\frac{c_{2k-5}}{5!}+\frac{c_{2k-3}}{3!}+\frac{c_{2k-1}}{1!}=0$$

For illustration, the first few coefficients and polynomial equations are shown below.

Notice that the factorials in the denominators of the polynomials differ by 1 from those in the coefficients. Also, to solve for the next coefficient, the preceding coefficients only need to be summed and then moved to the other side of the equal sign, i.e $$c_7=-(\frac{c_1}{7!}+\frac{c_3}{5!}+\frac{c_5}{3!})$$.

The following code will be commented line-by-line to explain the algorithm of Kessler:

Finally, the code will be followed to generate $$(P_2,P_3)(P_4,P_5)(P_6,P_6)$$: The polynomials are written as,
 * $$p_2(t)=c_1\frac{t^2}{2!}+c_3$$
 * $$p_3(t)=c_1\frac{t^3}{3!}+c_3t$$
 * $$p_4(t)=c_1\frac{t^4}{4!}+c_3\frac{t^2}{2!}+c_5$$
 * $$p_5(t)=c_1\frac{t^5}{5!}+c_3\frac{t^3}{3!}+c_5t$$
 * $$p_6(t)=c_1\frac{t^6}{6!}+c_3\frac{t^4}{4!}+c_5\frac{t^2}{2!}+c_7$$
 * $$p_7(t)=c_1\frac{t^7}{7!}+c_3\frac{t^5}{5!}+c_5\frac{t^3}{3!}+c_7t$$

Following the code to get $$c_{1-7}$$, n=3 because 3 loops will generate all 7 coefficients f=1, g=2, cn=-1, cd=1 entering loop which runs 3 times f=1*2*3 (3! = 6) enter fracsum with (1,6) and will sum the fractions w/o leaving a decimal div = gcd(1,6) = 1 n = 1 d = 6 dsum = 1 enter loop (runs once) dsum = 6 leave loop nsum = 6*(1/6)= 1 div = gcd (1,6) = 1 nsum = 1 dsum = 6 leave fracsum with (1,6) cn = [-1; 1] cd = [1; 6] f = [6;1] g = [4;2] enter fracsum with ([-3;1],[6;6]) div = [3;1] n=[-1;1] d=[2;6] enter loop dsum = 2 then 6 leave loop nsum=6*-1/2+6*1/6=-3+1=-2; div=2 nsum=-1; dsum=3; leave fracsum with (-1,3) pn(1,1)=-1; pd(1,1)=3; move on to next loop k=2

At this point it is apparent that fracsum will sum up fractions without reducing them to decimals Also, the second fracsum call does not affect cn so we will skip calculating it And, cn=[-1;1] cd=[1;6] so we have c1 and c3

now k=2 f=[6;1].*[4;2].*[5;3] = [24;2].*[5;3] = [120;6] enter fracsum with (-[-1;1],[1;6].*[120;6])=([1;-1],[120;36]) this is like 1/120-1/36 = -7/360 leave fracsum cn = [-1;1;-7]; cd = [1;6;360]; f = [120;6;1] g = [2+4;[4;2]] = [6;4;2] skip to k=3 f=[120;6;1].*[6;4;2].*[7;5;3] = [720;24;2].*[7;5;3] = [5040;120;6] enter fracsum with ([1;-1;7],[1;6;360].*[5040;120;6]) = ([1;-1;7],[5040;720;2160]) this is 1/5040-1/720+6/2160 = 31/15120 leave fracsum cn = [-1;1;-7;31]; cd = [1;6;360;15120]

The result is,
 * $$c_1=-1 \!$$
 * $$c_3=-\frac{1}{6}$$
 * $$c_5=-\frac{-7}{360}$$
 * $$c_7=-\frac{31}{15120}$$

Plugging into the expressions for $$p_1(t) -> p_7(t)$$ generates the correct result.

Problem solved by EGM6341.s10.Team1.Kumanchik 22:53, 7 April 2010 (UTC)

=Problem 11: Comparison of the elliptic integral codes of Team 3 with those of ours=

Problem Statement
verify the results circumference formula with the complete elliptic integral codes

Solution
1Composite Trapezoidal Rule (a)team 3's codes : subfunction is (b)Ours

2Use Clenshaw-Curtis quadrature : (a)Team 3 (b)Team1(ours) 3Use Romberg Table: (a)Team 3 (b)Team 1 (ours) EGM6341.s10.Team1.Ya-Chiao Chang 02:53, 7 April 2010 (UTC)

=Contributing Members= Egm6341.s10.team1.lei15:03, 7 April 2010 (UTC) Author of problem 2 & 6 EGM6341.s10.Team1.Ya-Chiao Chang 15:23, 7 April 2010 (UTC) Author of problem 7 & 11 Egm6341.s10.team1.toddmock 19:24, 7 April 2010 (UTC) Author of problem 3, 4, & 9 Egm6341.s10.team1.andrewdugan 21:35, 7 April 2010 (UTC) Author of problems 5 & 8