User:Egm6341.s10.team3.sa/HW6

= Computation of arc length an ellipse $$\displaystyle \overset{\frown}{PQ} $$ using successive numerical integration = Ref: Lecture notes [[media:Egm6341.s10.mtg32.djvu|p.32-1]]

Problem Statement
Compute arc length of ellipse $$\displaystyle \overset{\frown}{PQ} $$ to $$\displaystyle O(10^{-10}) $$ from $$\displaystyle \theta_{P} = 0 $$ to $$\displaystyle \theta_{Q} = \pi/3 $$. Successive numerical integration results should be used as stopping criterion, i.e. $$\displaystyle |I_{2n} - I_{n}| < O(10^{-10}) $$.

i) Use error estimate for composite trapezoidal rule
 * {| style="width:100%" border="0" align="left"

$$ $$ to find $$\displaystyle n $$.
 * $$\displaystyle
 * E_{n}| \leq \frac{(b-a)h^2}{12}M_{2}
 * E_{n}| \leq \frac{(b-a)h^2}{12}M_{2}
 * $$\displaystyle (Eq. 1)
 * }
 * }

ii)

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

\overset{\frown}{PQ} = \int_{\theta_{P}}^{\theta_{Q}} dl $$ $$ where $$\displaystyle dl = d\theta\left[r^2 + \left(\frac{dr}{d\theta}\right)^2\right]^{\frac{1}{2}} $$ and $$\displaystyle r(\theta) = \frac{(1-e^2)}{1-e\cos\theta} $$ to obtain the arc length from composite trapezoidal rule, Romberg table, clencurt in Matlab and sum in Chebfun.
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 2)
 * }
 * }

iii)

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

\overset{\frown}{PQ} = \int_{\theta_{P}}^{\theta_{Q}} \left[1-e^2\sin^2\theta\right]^{\frac{1}{2}}d\theta. $$ $$ to obtain the arc length from composite trapezoidal rule, Romberg table, clencurt in Matlab and sum in Chebfun.
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 3)
 * }
 * }

Solution
i) Find $$\displaystyle n $$ In (Eq. 1), recall that $$\displaystyle h = \frac{b-a}{n} $$ and $$\displaystyle M_{2} = \max_{\zeta \in [a,b]}|f^{(2)}(\zeta)| $$. Here, $$\displaystyle a = 0 $$ and $$\displaystyle b = \pi/3 $$.

Also, rearrange (Eq. 1) to get the following equation.
 * {| style="width:100%" border="0" align="left"

n^2 \leq \frac{(b-a)^3}{12}\frac{\max_{\zeta \in [a,b]}|f^{(2)}(\zeta)|}{|E_{n}|}. $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 4)
 * }
 * }

Now, put the below values into (Eq. 4). $$\displaystyle a = 0 $$

$$\displaystyle b = \pi/3 $$

$$\displaystyle |E_{n}| = 10^{-11} (\because |I_{2n} - I_{n}| < O(10^{-10}) $$

$$\displaystyle M_{2} = \max_{\zeta \in [a,b]}|f^{(2)}(\zeta)| = \max_{\theta \in [0,\frac{\pi}{3}]}|dl^{(2)}(\theta)| = 0.2934 $$ for (Eq. 2) and

$$\displaystyle M_{2} = \max_{\zeta \in [a,b]}|f^{(2)}(\zeta)| = \max_{\theta \in [0,\frac{\pi}{3}]}|\left(\sqrt{1-e^2\sin^2(\theta)}\right)^{(2)}| = 0.0670 $$ for (Eq. 3)

Then, $$\displaystyle n^2 \leq 2.8077e+09 \quad \Rightarrow \quad n \leq 52988.59 $$ for (Eq. 2) and

$$\displaystyle n^2 \leq 6.4105e+08 \quad \Rightarrow \quad n \leq 25319.00 $$ for (Eq. 3).

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

$$\displaystyle n \leq 53988 $$ for (Eq. 2) and
 * style="width:10%; padding:10px; border:2px solid #8888aa" |
 * style="width:10%; padding:10px; border:2px solid #8888aa" |
 * style = |
 * }
 * }
 * {| style="width:100%" border="0" align="left"

$$\displaystyle n \leq 25319 $$ for (Eq. 3).
 * style="width:10%; padding:10px; border:2px solid #8888aa" |
 * style="width:10%; padding:10px; border:2px solid #8888aa" |
 * style = |
 * }
 * }

ii) Based on (Eq. 2) In what follows, we are using the follow matlab function as a given function $$\displaystyle dl(\theta) $$.

 (a) Composite trapezoidal rule 

Recall the formula of composite trapezoidal rule.


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

T_{0}(n) = h\left[\frac{1}{2}f(x_{0}) + f(x_{1}) + f(x_{2}) + \cdot\cdot\cdot + f(x_{n-1}) + \frac{1}{2}f(x_{n}) \right] $$ where $$\displaystyle h = \frac{b-a}{n} $$ and $$\displaystyle x_{0} = a,\quad x_{n} = b $$.
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }

Also, realize the fact that if we want to double the number of sub-intervals, we don't have to calculate the functions at even point, i.e. $$\displaystyle f(x_{0}), f(x_{2}), \cdot\cdot\cdot, f(x_{n-2}), f(x_{n}) $$ because we already calculated the functions at these even point at the previous step. Therefore, all we have to do is evaluating the functions only at the odd points and combine them with the calculated values at the previous step. As a result, we can use the following inductive formula.
 * {| style="width:100%" border="0" align="left"

T_{0}(2n) = T_{0}(n) + h\sum_{i=1}^{n}f(x_{2i-1}). $$
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }

The obtained number of nodes to achieve $$\displaystyle O(10^{-10}) $$ is $$\displaystyle n = 32768 $$ which satisfies the result of part i) $$\displaystyle n \leq 53988 $$. The integration result, final error and computational time are summarized in Table at the last part of this problem.

 (b) Romberg Table 

Since the first column of Romberg table corresponds to Trap. rule, we can use the below formulas in order to obtain the first column of Romberg talbe.


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

T_{0}(n) = h\left[\frac{1}{2}f(x_{0}) + f(x_{1}) + f(x_{2}) + \cdot\cdot\cdot + f(x_{n-1}) + \frac{1}{2}f(x_{n}) \right] $$
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }


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

T_{0}(2n) = T_{0}(n) + h\sum_{i=1}^{n}f(x_{2i-1}). $$ where $$\displaystyle h = \frac{b-a}{n} $$ and $$\displaystyle x_{0} = a,\quad x_{n} = b $$.
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }

Also, we gradually set up the Romberg table in parallel, as soon as the result for $$\displaystyle T_0(2^j) $$ becomes available using the below formula.


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

T_{k}(2n) = \frac{2^{2k}T_{k-1}(2n) - T_{k-1}(n)}{2^{2k}-1}. $$
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }

Result

The integration result, final error and computational time are summarized in Table at the last part of this problem.

 (c) clencurt 

To achieve the desired tolerance of error, we use the following Matlab code.

The integration result, final error, final number of nodes and computational time are summarized in Table at the last part of this problem.

 (d) chebfun sum command 

The 'sum' command in chebfun performs integration of a given function up to the machine accuracy. Therefore, we can achieve the tolerance of error of numerical integration for the given function $$\displaystyle dl $$ up to the machine accuracy by using the following Matlab code.

The integration result, final error, final number of nodes and computational time are summarized in Table at the last part of this problem.

 (e) Summary 

In the previous HW, recall that composite trapezoidal rule was a very efficient method to perform integration due to the periodicity of the given function. In this HW, however, note that composite trapezoidal rule requires a very large number of nodes and long computation time. The reason is that the given function does not show a periodicity within a given range $$\displaystyle \theta \in [0, \pi/3] $$  even though it shows a periodicity for the range of $$\displaystyle \theta \in [0, 2\pi] $$.

As shown above, both Romberg table and clencurt require only 32 nodes to achieve the desired tolerance of error and the computation times of the both methods are very short. Therefore, both Romberg Table and clencurt are efficient methods to integrate the given function.

iii) Based on (Eq. 3) In what follows, we are using the follow matlab function as a given function $$\displaystyle dl(\theta) $$.

 (a) Composite trapezoidal rule 

We use the same Matlab code with that of part ii).

The obtained number of nodes to achieve $$\displaystyle O(10^{-10}) $$ is $$\displaystyle n = 16384 $$ which satisfies the result of part i) $$\displaystyle n \leq 25319 $$. The integration result, final error and computational time are summarized in Table at the last part of this problem.

 (b) Romberg Table 

We use the same Matlab code with that of part ii).

The integration result, final error and computational time are summarized in Table at the last part of this problem.

 (c) clencurt 

We use the same Matlab code with that of part ii).

The integration result, final error, final number of nodes and computational time are summarized in Table at the last part of this problem.

 (d) chebfun sum command 

The 'sum' command in chebfun performs integration of a given function up to the machine accuracy. Therefore, we can achieve the tolerance of error of numerical integration for the given function $$\displaystyle \sqrt{(1 - e^2\sin^2\theta)} $$ up to the machine accuracy by using the following Matlab code.

The integration result, final error, final number of nodes and computational time are summarized in Table at the last part of this problem.

 (e) Summary 

As we already explained in summary of part i), composite trapezoidal rule is not a efficient method to integrate the given function due to the same reason (i.e. The given function does not show a periodicity within a given range $$\displaystyle \theta \in [0, \pi/3] $$). Romberg table and clencurt require only 16 nodes to achieve the desired tolerance of error and the computation times of the both methods are very short. Therefore, both Romberg Table and clencurt are efficient methods to integrate the given function.

In addition, note that the integration result of part iii) is different from the integration result of part ii) although the given ranges of the integrations are same with each other (i.e. $$\displaystyle \theta \in [0, \pi/3] $$). As the figures show in problem statement, the definition of $$\displaystyle \theta $$ of part ii) is different from that of part iii). Therefore, it is very clear that the integration results of the both parts are not same with each other even though the range of $$\displaystyle \theta $$ is same.

Yong Nam Ahn 02:11, 7 April 2010 (UTC) Abhishekksingh 09:13, 7 April 2010 (UTC)

=Show that $$ dp_y=mVdr$$ = Ref: Lecture Notes [[media:Egm6341.s10.mtg33.djvu|p.33-2]]

Problem Statement
At t+dt, consider V+dV Show that $$ dp_{\overline{y}}=mVdr$$

Solution


At t+dt, V+dV,see the above figure
 * {| style="width:100%" border="0" align="left"

$$ By neglecting higher order terms, mdVdr, We have: At t+dt, V+dV,see the above figure
 * $$\displaystyle dp_{\overline{y}}=m(V+dV)dr=mVdr+mdVdr
 * $$\displaystyle dp_{\overline{y}}=m(V+dV)dr=mVdr+mdVdr
 * }
 * }
 * {| style="width:100%" border="0" align="left"

$$ End the proof
 * $$\displaystyle dp_{\overline{y}}=mVdr
 * $$\displaystyle dp_{\overline{y}}=mVdr
 * }
 * }

Min Zhong 12:00, 7 Arip 2010 (UTC)

= Evaluate the $$z_{i+1}$$ and $$z'_{i+1}$$ in terms of $$c_0,c_1,c_2,c_3$$ and represent it in the coefficient matrix = Ref Lecture Notes [[media:Egm6341.s10.mtg32.djvu|p.35-3]]

Problem Statement
Evaluate the $$z_{i+1}$$ and z'_{i+1} in terms of $$c_0,c_1,c_2,c_3$$ and represent it in the coefficient matrix i.e, Evaluate the remaining coefficient of Matrix by using degrees of Freedom

Solution
We have

$$ Z(s)=\sum_{i=0}^{3}c_is^i=\sum_{i=0}^{3}N_i(s)d_i $$

such that

$$ d_1=Z_i$$

$$ d_2=\dot{Z_i}$$

$$ d_3=Z_{i+1}$$

$$ d_4={\dot{Z}_{i+1}}$$

where

$$ \dot{Z}=\frac{dZ}{dt}={\frac{dZ}{ds}} \frac{ds}{dt} $$

such that $$ \frac{ds}{dt} = \frac {1}{h} $$

We know the coefficient of matrix for first two rows from lecture notes   [[media:Egm6341.s10.mtg32.djvu|p.35-3]]

$$\begin{bmatrix} 1 &0 &0  &0 \\  0&1  &0  &0 \end{bmatrix}$$

Using the equations above we have

$$  d_3=Z_{i+1}=Z(s=1)=c_0+c_1+c_2+c_3 $$

$$  d_4=\dot{Z}_{i+1}=\frac{1}{h}Z'(s=1)=c_1+2c_2+3c_3 $$

Putting the results in matrix form we obtain

$$\begin{bmatrix} 1 &0 &0  &0 \\ 0 &1  &0  &0 \\ 1 &1  &1  &1 \\ 0 &1  &2  &3 \end{bmatrix}\begin{Bmatrix} c_0\\ c_1\\ c_2\\

c_3\end{Bmatrix}= \begin{Bmatrix} Z_i\\ Z'_i\\ Z_{i+1}\\ Z'_{i+1} \end{Bmatrix}$$

Abhishekksingh 09:13, 7 April 2010 (UTC)

= Verify the inverse of matrix using Matlab = Ref Lecture Notes [[media:Egm6341.s10.mtg32.djvu|p.35-4]]

Problem Statement
Find the inverse of given Matrix

A =

1    0     0     0     0     1     0     0     1     1     1     1     0     1     2     3

Solution
which is same as the one given on [[media:Egm6341.s10.mtg32.djvu|p.35-4]]

Hence Verified

Abhishekksingh 09:13, 7 April 2010 (UTC)

= Identify basis functions and plot them=

Ref Lecture Notes [[media:Egm6341.s10.mtg32.djvu|p.35-4]]

Problem Statement
Identify the basis functions

$$ N_i (s) $$

where $$ i = 1,2,3,4$$

Solution
We have

$$ z(s) = \sum_{i=0}^3 c^i s^i = \sum_{i=1}^4 N_i(s) d_i $$

Expanding above we obtain

$$ N_1 d_1 + N_2 d_2 + N_3 d_3 + N_4 d_4 = C_0 s^0 + C_1 s^1 + C_2 s^2 + C_3 s^3$$

$$                                      = C_0  + C_1 s^1 + C_2 s^2 + C_3 s^3$$

$$d_1 = z_i = z(s=0) = C_0$$

$$d_2 = \dot z_i = \dot z_i (s=0) = C_1$$

$$d_3 = z_{i+1}= z_{i+1} (s=1) = C_0 + C_1 + C_2  + C_3 $$

$$ d_4 = \dot z_{i+1} =C_1 + 2 C_2 + 3 C_3 $$

Inserting above values in first eq we obtain

$$(N_1 + N_3) C_0 + (N_2+N_3+N_4) C_1 + (N_3+2 N_4)C_2 + (N_3 + 3 N_4) C_3 = C_0 + C_1 s + C_2 s^2 + C_3 s^3$$

Comparing both LHS and RHS we obtain

$$N_1 + N_3 = 1$$

$$N_2+N_3+N_4 = s$$

$$N_3+2 N_4=s^2$$

$$N_3 + 3 N_4 = s^3$$

Solving above we obtain basis functions

$$N_1 = 1- 3 s^2 +2 s^3$$

$$N_2 = s^3 - 2 s^2 + s $$

$$N_3 = 3 s^2 - 2 s^3$$

Below is the plot of above basis functions



$$ N_4 = s^3 - s^2$$

Abhishekksingh 09:13, 7 April 2010 (UTC)

= Express $$ t(s) $$ in terms of $$ s(t) $$=

Ref Lecture Notes [[media:Egm6341.s10.mtg32.djvu|p.36-1]]

Problem Statement
We have to show that s is the function of t (s = s(t) )

Solution
We have (from [[media:Egm6341.s10.mtg32.djvu|p.35-1]] eq (1))

$$ t(s)= (1-s) t_i + s t_{i+1} $$

$$    = t_i + s (t_{i+1} - t_i ) $$

$$ but \frac {ds}{dt} = \frac{1}{h} $$

$$ or \frac {dt}{ds} = h $$

so $$ t_{i+1} - t_i = \frac{t_{i+1} - t_i}{1-0}= h $$

$$ t(s) = t_i +h s $$ $$ h s = t - t_i $$ $$ s = \frac {(t - t_i)} {h} $$ $$ s = s(t) $$

Hence Proved

Abhishekksingh 09:13, 7 April 2010 (UTC)

= Show that $$ Z(s= 1/2)= (1/2)(Z_i-Z_{i+1})+ (h/8)(f_i-f_{i+1}) $$  = Ref: Lecture Notes [[media:Egm6341.s10.mtg36.djvu|p.36-2]]

Problem Statement
Proof that $$ Z(\varsigma=2)= \frac{1}{2}(Z_i-Z_{i+1})+ \frac{h}{8}(f_i-f_{i+1}) $$

In Lecture Notes [[media:Egm6341.s10.mtg35.djvu|p.35-2]](2), We have:
 * {| style="width:100%" border="0" align="left"

$$ $$
 * $$\displaystyle Z(\varsigma)= \sum_{n=0}^{3}C_i{\varsigma}^i
 * $$\displaystyle Z(\varsigma)= \sum_{n=0}^{3}C_i{\varsigma}^i
 * $$\displaystyle (Eq. 7-1)
 * }
 * }

In lecture notes [[media:Egm6341.s10.mtg35.djvu|p.35-4]](1), we have:


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

\begin{Bmatrix} 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} \begin{Bmatrix} Z_i \\ Z_i^\prime \\ Z_{i+1} \\ Z_{i+1}^\prime \\ \end{Bmatrix}
 * $$\displaystyle
 * $$\displaystyle

$$ $$ In lecture notes [[media:Egm6341.s10.mtg35.djvu|p.36-1]](1), we have:
 * $$\displaystyle (Eq. 7-2)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

$$ $$
 * $$\displaystyle Z^\prime= h\dot{Z}=hf
 * $$\displaystyle Z^\prime= h\dot{Z}=hf
 * $$\displaystyle (Eq. 7-3)
 * }
 * }

Solution
We start the proof from Equation 1.
 * {| style="width:100%" border="0" align="left"

\begin{align} Z(\varsigma= \frac{1}{2})& = \sum_{n=0}^{3}C_i{ \frac{1}{2}}^i\\ &=C_0+ \frac{1}{2}C_1+ \frac{1}{4}C_2+ \frac{1}{8}C_3\\ \end{align} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 7-4)
 * }
 * }

Next, we will find the expression of Ci by Zi and fi From the matrix Eq. 7-2, we have:
 * {| style="width:100%" border="0" align="left"

$$ $$
 * $$\displaystyle C_0=Z_i
 * $$\displaystyle C_0=Z_i
 * $$\displaystyle (Eq. 7-5)
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

$$ $$
 * $$\displaystyle C_1=Z_i^\prime
 * $$\displaystyle C_1=Z_i^\prime
 * $$\displaystyle (Eq. 7-6)
 * }
 * }


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

$$ $$
 * $$\displaystyle C_2=-3Z_i-2Z_i^\prime+3Z_{i+1}-Z_{i+1}^\prime
 * $$\displaystyle C_2=-3Z_i-2Z_i^\prime+3Z_{i+1}-Z_{i+1}^\prime
 * $$\displaystyle (Eq. 7-7)
 * }
 * }


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

$$ $$
 * $$\displaystyle C_3=2Z_i+Z_i^\prime-2Z_{i+1}+Z_{i+1}^\prime
 * $$\displaystyle C_3=2Z_i+Z_i^\prime-2Z_{i+1}+Z_{i+1}^\prime
 * $$\displaystyle (Eq. 7-8)
 * }
 * }

Plug Eq.7-5,Eq.7-6,Eq.7-7,Eq.7-8, into Eq.7-4,We can get:
 * {| style="width:100%" border="0" align="left"

\begin{align} Z(\varsigma= \frac{1}{2})& =Z_i+ \frac{1}{2}Z_i^\prime + \frac{1}{4}(-3Z_i-2Z_i^\prime+3Z_{i+1}-Z_{i+1}^\prime)+ \frac{1}{8}(2Z_i+Z_i^\prime-2Z_{i+1}+Z_{i+1}^\prime)\\ &=\frac{1}{2}Z_i+\frac{1}{2}Z_{i+1}+\frac{1}{8}Z_i^\prime-\frac{1}{8}Z_{i+1}^\prime\\ &=\frac{1}{2}(Z_i+Z_{i+1})+\frac{1}{8}(Z_i^\prime-Z_{i+1}^\prime)\\ \end{align} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 7-9)
 * }
 * }

Plug Eq 3 into above equation, we have


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

$$\displaystyle Z(\varsigma=\frac{1}{2})=\frac{1}{2}(Z_i+Z_{i+1})+\frac{h}{8}(f_i-f_{i+1}) $$
 * style="width:10%; padding:10px; border:2px solid #8888aa" |
 * style="width:10%; padding:10px; border:2px solid #8888aa" |
 * style = |
 * }
 * }

Min Zhong 12:00, 7 Apri 2010 (UTC)

= Show that $$ Z(s=1/2)'=-(3/2)(Z_i-Z_{i+1})-(1/4)(Z'_i-Z'_{i+1})$$ = Ref: Lecture Notes [[media:Egm6341.s10.mtg36.djvu|p.36-2]]

Problem Statement
Proof that $$\displaystyle Z(\varsigma=\frac{1}{2})^\prime=-\frac{3}{2}(Z_i-Z_{i+1})-\frac{1}{4}(Z_i^\prime-Z_{i+1}^\prime) $$ From Lecture Notes [[media:Egm6341.s10.mtg35.djvu|p.35-1]](2)


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

$$ $$
 * $$\displaystyle Z(\varsigma)=C_0+ C_1\varsigma+ C_2{\varsigma}^2 + C_3{\varsigma}^3
 * $$\displaystyle Z(\varsigma)=C_0+ C_1\varsigma+ C_2{\varsigma}^2 + C_3{\varsigma}^3
 * $$\displaystyle (Eq. 8-1)
 * }
 * }

Solution
We start the proof from Eq.8-1,


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

$$ $$
 * $$\displaystyle Z(\varsigma)^\prime=C_1+ 2C_2\varsigma+ 3C_3{\varsigma}^2
 * $$\displaystyle Z(\varsigma)^\prime=C_1+ 2C_2\varsigma+ 3C_3{\varsigma}^2
 * $$\displaystyle (Eq. 8-2)
 * }
 * }


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

\begin{align} Z(\varsigma={\frac{1}{2}}^\prime)&=C_1+ 2C_2\frac{1}{2}+ 3C_3{\frac{1}{2}}^2\\ &=C_1+ C_2 +\frac{3}{4}C_3\\ \end{align} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 8-3)
 * }
 * }

Plug Eq.7-5,Eq.7-6,Eq.7-7,Eq.7-8, into Eq.8-3, we get:


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

\begin{align} Z(\varsigma=\frac{1}{2})^\prime & = Z_i^\prime-3Z_i-2Z_i^\prime+3Z_{i+1}-Z_{i+1}^\prime+\frac{1}{2}(2Z_i+Z_i^\prime-2Z_{i+1}+Z_{i+1}^\prime) \\ &=-\frac{3}{2}Z_i+\frac{3}{2}Z_{i+1}-\frac{1}{4}Z_i^\prime-\frac{1}{4}Z_{i+1}^\prime\\ &=-\frac{3}{2}(Z_i-Z_{i+1})-\frac{1}{4}(Z_i^\prime+Z_{i+1}^\prime)\\ \end{align}
 * $$\displaystyle
 * $$\displaystyle

$$ Min Zhong 12:00, 7 Apri 2010 (UTC)
 * }
 * }

=Show that $$ Z_{i+1}=Z_i+(h/2)/(3)[f_i+4f_{i+1/2}+f_{i+1}]$$ = Ref: Lecture Notes [[media:Egm6341.s10.mtg36.djvu|p.36-3]]

Problem Statement
Proof that $$ Z_{i+1}=Z_i+\frac{h/2 }{3}[f_i+4f_{i+\frac{1}{2}}+f_{i+1}]$$

From Lecture Notes [[media:Egm6341.s10.mtg36.djvu|p.36-3]](1)


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

$$ $$
 * $$\displaystyle \triangle=\dot Z_{i+ \frac{1}{2}}-f_{i+ \frac{1}{2}}
 * $$\displaystyle \triangle=\dot Z_{i+ \frac{1}{2}}-f_{i+ \frac{1}{2}}
 * <p style="text-align:right;">$$\displaystyle (Eq. 9-1)
 * }
 * }

From Lecture Notes [[media:Egm6341.s10.mtg36.djvu|p.36-2]]


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

\dot Z_{i+ \frac{1}{2}}=- \frac{3}{2h}(Z_i-Z_{i+1})- \frac{1}{4}(f_i+f_{i+1}) $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 9-2)
 * }
 * }

Solution
When $$\triangle=0$$, from Eq.9-1, we have :
 * {| style="width:100%" border="0" align="left"

$$ $$
 * $$\displaystyle \dot Z_{i+ \frac{1}{2}}=f_{i+ \frac{1}{2}}
 * $$\displaystyle \dot Z_{i+ \frac{1}{2}}=f_{i+ \frac{1}{2}}
 * <p style="text-align:right;">$$\displaystyle (Eq. 9-3)
 * }
 * }

Plug Eq. 9-3 into Eq, 9-2, we can get:
 * {| style="width:100%" border="0" align="left"

f_{i+ \frac{1}{2}}=- \frac{3}{2h}(Z_i-Z_{i+1})- \frac{1}{4}(f_i+f_{i+1}) $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 9-4)
 * }
 * }

Rearrange Eq. 9-4 we have:


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

\begin{align} Z_{i+1}&=Z_i+ \frac{h}{6}(f_i+4f_{i+\frac{1}{2}}+f_{i+1})\\ &=Z_i+ \frac{h/2}{3}(f_i+4f_{i+\frac{1}{2}}+f_{i+1})\\ \end{align} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 9-2)
 * }
 * }

= Kessler's code =

Problem Statement
1. Run Kessler's code to reproduce the table of constants and the values of polynomials at t=1 obtained by Kessler. 2. Explain Kessler's code by giving comments for every line. 3. Obtain $$ \displaystyle (p_2,p_3) \; (p_4,p_5) \; (p_6,p_7)$$ by understanding Kessler's code line by line.

Ref: Lecture Notes [[media:Egm6341.s10.mtg30.djvu|p.30-2]] and [[media:Egm6341.s10.mtg37.djvu|p.37-1]]

Ref: Trapezoidal rule error

Solution
Part I : Generation of the Table To generate the table given in Kessler's paper, the following modifications are to be carried out.

1. The first half of the code is the main code. Give that a name called kessler.m

2. The second half of the code is a subroutine with the name fracsum. This should not be part of the file kessler.m   We need to create a new file called fracsum.m in the same folder that contains the file kessler.m

3. Set n=8. This is to ensure that we get the constants $$ c_1 \; through \; c_{17} $$ and polynomials $$ p_2(1) \; through\; p_{16}(1) $$.

4. The most important part is that the subroutine fracsum should end with the line, dsum=dsum/div;

5. The following lines have been appended to kessler.m to generate the table.

Result

Part II: Line by Line Explanation of Kessler's MATLAB Code - Adding Comments

Part III : Generating $$\displaystyle (p_2,p_3),(p_4,p_5),(p_6,p_7)$$
 * {| style="width:100%" border="0" align="left"


 * Recall the following:
 * }
 * }


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

$$
 * $$\displaystyle p_1(t) = c_1 t $$
 * <p style="text-align:right;">$$\displaystyle (Eq. 1)
 * <p style="text-align:right;">$$\displaystyle (Eq. 1)
 * }


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


 * $$\displaystyle p_2(t) = c_1 \frac{t^2}{2!} + c_3 $$
 * <p style="text-align:right;">$$\displaystyle (Eq. 2) $$
 * }
 * }


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


 * $$\displaystyle p_3(t) = c_1 \frac{t^3}{3!} + c_3 t $$
 * <p style="text-align:right;">$$\displaystyle (Eq. 3) $$
 * }
 * }


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


 * $$\displaystyle p_4(t) = c_1 \frac{t^4}{4!} + c_3 \frac{t^2}{2!} + c_5 $$
 * <p style="text-align:right;">$$\displaystyle (Eq. 4) $$
 * }
 * }


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


 * $$\displaystyle p_5(t) = c_1 \frac{t^5}{5!} + c_3 \frac{t^3}{3!} + c_5 t $$
 * <p style="text-align:right;">$$\displaystyle (Eq. 5) $$
 * }
 * }


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


 * $$\displaystyle p_6(t) = c_1 \frac{t^6}{6!} + c_3 \frac{t^4}{4!} + c_5 \frac{t^2}{2!} + c_7 $$
 * <p style="text-align:right;">$$\displaystyle (Eq. 6) $$
 * }
 * }


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


 * $$\displaystyle p_7(t) = c_1 \frac{t^7}{7!} + c_3 \frac{t^5}{5!} + c_5 \frac{t^3}{3!} + c_7 t $$
 * <p style="text-align:right;">$$\displaystyle (Eq. 7) $$
 * }
 * }


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

$$\displaystyle c_1, c_3, c_5 \;and\; c_7 $$
 * Hence we need to find out the constants,
 * Hence we need to find out the constants,
 * }


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

And then we need to have n=3 ,because,
 * So we start off with setting $$ c_1 = -1 $$
 * So we start off with setting $$ c_1 = -1 $$
 * }


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


 * $$\displaystyle n=1 \;will\; yield\; c_3 $$
 * }
 * }


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


 * $$\displaystyle n=2 \;will\; yield\; c_5 $$
 * }
 * }


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


 * $$\displaystyle n=3 \;will\; yield\; c_7 $$
 * }
 * }


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

$$\displaystyle f=1 \,\,\, g=2$$ $$\displaystyle n=3; $$
 * $$\displaystyle c_1 = -1 $$
 * $$\displaystyle c_1 = -1 $$
 * }


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


 * Consider the first iteration: i.e, k=1
 * }
 * }


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

f= 1*2*3 =6
 * $$ \displaystyle f=f.*g.*(g+1) $$ yields
 * $$ \displaystyle f=f.*g.*(g+1) $$ yields
 * }


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

$$\displaystyle f=6 $$
 * Hence,
 * style="width:10%; padding:10px; border:2px solid #8888aa" |
 * style="width:10%; padding:10px; border:2px solid #8888aa" |
 * style = |
 * }
 * }


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


 * $$\displaystyle [newcn,newcd]=fracsum(-1*cn,cd.*f);$$
 * }
 * }


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

fracsum(-1*1, 1*6) i.e, fracsum(-1,6)
 * $$ \displaystyle fracsum(-1*cn,cd.*f) $$ yields
 * $$ \displaystyle fracsum(-1*cn,cd.*f) $$ yields
 * }


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

$$\displaystyle fracsum(-1,6) $$
 * Hence,we have
 * style="width:10%; padding:10px; border:2px solid #8888aa" |
 * style="width:10%; padding:10px; border:2px solid #8888aa" |
 * style = |
 * }
 * }


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

$$\displaystyle div=gcd(round(n),round(d))$$
 * Now looking into the fracsum(n,d) function,
 * Now looking into the fracsum(n,d) function,
 * }


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

because, round(-1) = -1 round(6) = 6 and hence, gcd(-1,6) = 1.
 * will yield div = 1.
 * will yield div = 1.
 * }


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

$$n=round(n./div);$$ will give n=1 becasue round(1/1) =1
 * Now,
 * Now,
 * }

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

will give d=6 becasue round(6/1) =6
 * $$\displaystyle d=round(d./div);$$
 * $$\displaystyle d=round(d./div);$$
 * }


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

$$\displaystyle n=1\; and\; d=6 $$
 * Thus,
 * style="width:15%; padding:10px; border:2px solid #8888aa" |
 * style="width:15%; padding:10px; border:2px solid #8888aa" |
 * style = |
 * }
 * }


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

$$\displaystyle for k=1:length(d) $$ $$\displaystyle dsum=lcm(dsum,d(k)); $$ $$\displaystyle end $$
 * $$\displaystyle dsum=1; $$
 * $$\displaystyle dsum=1; $$
 * }
 * }


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

and lcm (1,6) = 6, we have, dsum =6.
 * Since length(d) = 1
 * Since length(d) = 1
 * }
 * }


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

$$\displaystyle dsum = 6 $$
 * Thus,
 * style="width:15%; padding:10px; border:2px solid #8888aa" |
 * style="width:15%; padding:10px; border:2px solid #8888aa" |
 * style = |
 * }
 * }


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

$$ \displaystyle div=gcd(round(nsum),round(dsum)); $$
 * $$ \displaystyle nsum=dsum*sum(n./d);$$
 * $$ \displaystyle nsum=dsum*sum(n./d);$$
 * }
 * }

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

and
 * nsum = 6* sum(1/6) = 1
 * nsum = 6* sum(1/6) = 1
 * }

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

since, round(nsum) = round(1) = 1 round(dsum) = round(6) = 6 gcd(1,6) =1 Now,
 * div = 1
 * div = 1
 * }
 * {| style="width:100%" border="0" align="left"

$$ \displaystyle dsum=dsum/div; $$
 * $$ \displaystyle nsum=nsum/div; $$
 * $$ \displaystyle nsum=nsum/div; $$
 * }

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

dsum = 6
 * nsum = 1
 * nsum = 1
 * }


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

$$\displaystyle [newcn \; newcd] = [1 \; 6] $$
 * Now going back,
 * Now going back,
 * }


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

$$\displaystyle \therefore c_3 = \frac{1}{6} $$
 * Thus,
 * style="width:10%; padding:10px; border:2px solid #8888aa" |
 * style="width:10%; padding:10px; border:2px solid #8888aa" |
 * style = |
 * }
 * }


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

$$\displaystyle cn=[cn;newcn]; cd=[cd;newcd] $$ $$\displaystyle f=[f;1]; $$
 * Next, we have,
 * Next, we have,
 * }


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

cn = [-1;1]; cd=[1;6]
 * Hence,
 * Hence,
 * }


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


 * and,
 * and,

So, f=[6;1]
 * }

This is followed by,


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

which means, g = [2+2, 2] = [4;2]
 * $$\displaystyle g=[2+g(1);g] $$ ;
 * $$\displaystyle g=[2+g(1);g] $$ ;
 * }

Now consider the second iteration - i.e, k=2


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


 * Consider the second iteration: i.e, k=2
 * }
 * }


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

f= (6*4*5; 1*2*3)
 * $$ \displaystyle f=f.*g.*(g+1) $$ yields
 * $$ \displaystyle f=f.*g.*(g+1) $$ yields
 * }


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

$$\displaystyle f=(120;6) $$
 * Hence,
 * style="width:10%; padding:10px; border:2px solid #8888aa" |
 * style="width:10%; padding:10px; border:2px solid #8888aa" |
 * style = |
 * }
 * }


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


 * $$\displaystyle [newcn,newcd]=fracsum(-1*cn,cd.*f);$$
 * }
 * }


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

fracsum(-1*-1, 1*120) i.e, fracsum(1,120)
 * $$ \displaystyle fracsum(-1*cn,cd.*f) $$ yields
 * $$ \displaystyle fracsum(-1*cn,cd.*f) $$ yields
 * }

and,


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

fracsum(-1*1, 6*6) i.e, fracsum(-1,36)
 * $$ \displaystyle fracsum(-1*cn,cd.*f) $$ yields
 * $$ \displaystyle fracsum(-1*cn,cd.*f) $$ yields
 * }


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

$$\displaystyle fracsum(1,120)\; and\; fracsum(-1,36) $$
 * Hence,we have
 * style="width:10%; padding:10px; border:2px solid #8888aa" |
 * style="width:10%; padding:10px; border:2px solid #8888aa" |
 * style = |
 * }
 * }


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

$$\displaystyle div=gcd(round(n),round(d))$$
 * Now looking into the fracsum(1,120) function,
 * Now looking into the fracsum(1,120) function,
 * }


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

because, round(1) = 1 round(120) = 120 and hence, gcd(1,120) = 1.
 * will yield div = 1.
 * will yield div = 1.
 * }


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

$$n=round(n./div);$$ will give n=1 becasue round(1/1) =1
 * Now,
 * Now,
 * }

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

will give d=120 becasue round(36/1) =120
 * $$\displaystyle d=round(d./div);$$
 * $$\displaystyle d=round(d./div);$$
 * }and,


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

$$\displaystyle div=gcd(round(n),round(d))$$
 * Now looking into the fracsum(-1,36) function,
 * Now looking into the fracsum(-1,36) function,
 * }


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

because, round(-1) = -1 round(36) = 36 and hence, gcd(-1,36) = 1.
 * will yield div = 1.
 * will yield div = 1.
 * }


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

$$n=round(n./div);$$ will give n=-1 becasue round(-1/1) =-1
 * Now,
 * Now,
 * }

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

will give d=6 becasue round(36/1) =36
 * $$\displaystyle d=round(d./div);$$
 * $$\displaystyle d=round(d./div);$$
 * }


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

$$\displaystyle n=[1;-1]\; and\; d=[120;36] $$
 * Thus,
 * style="width:35%; padding:10px; border:2px solid #8888aa" |
 * style="width:35%; padding:10px; border:2px solid #8888aa" |
 * style = |
 * }
 * }


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

$$\displaystyle for k=1:length(d) $$ $$\displaystyle dsum=lcm(dsum,d(k)); $$ $$\displaystyle end $$
 * $$\displaystyle dsum=1; $$
 * $$\displaystyle dsum=1; $$
 * }
 * }


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

and For d(1), dsum = lcm (1,120) = 120, and once again for d(2), dsum = lcm (120,36) = 360 Finally we have, dsum = 360.
 * Since length(d) = 2
 * Since length(d) = 2
 * }
 * }


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

$$\displaystyle dsum = 360$$
 * Thus,
 * style="width:15%; padding:10px; border:2px solid #8888aa" |
 * style="width:15%; padding:10px; border:2px solid #8888aa" |
 * style = |
 * }
 * }


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

$$ \displaystyle div=gcd(round(nsum),round(dsum)); $$
 * $$ \displaystyle nsum=dsum*sum(n./d);$$
 * $$ \displaystyle nsum=dsum*sum(n./d);$$
 * }
 * }

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

and
 * nsum = 360* sum(1/120 - 1/36) = -7
 * nsum = 360* sum(1/120 - 1/36) = -7
 * }

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

since, round(nsum) = round(-7) = -7 round(dsum) = round(360) = 360 gcd(-7,360) =1 Now,
 * div = 1
 * div = 1
 * }
 * {| style="width:100%" border="0" align="left"

$$ \displaystyle dsum=dsum/div; $$
 * $$ \displaystyle nsum=nsum/div; $$
 * $$ \displaystyle nsum=nsum/div; $$
 * }

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

dsum = 360
 * nsum = -7
 * nsum = -7
 * }


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

$$\displaystyle [newcn\; newcd] = [-7 \;360] $$
 * Now going back,
 * Now going back,
 * }


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

$$\displaystyle \therefore c_5 = -\frac{7}{360} $$
 * Thus,
 * style="width:10%; padding:10px; border:2px solid #8888aa" |
 * style="width:10%; padding:10px; border:2px solid #8888aa" |
 * style = |
 * }
 * }


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

$$\displaystyle cn=[cn;newcn]; cd=[cd;newcd] $$ $$\displaystyle f=[f;1]; $$
 * Next, we have,
 * Next, we have,
 * }


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

cn = [-1;1;7]; cd=[1;6;360]
 * Hence,
 * Hence,
 * }


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


 * and,
 * and,

So, f=[120;6;1]
 * }

This is followed by,


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

which means, g = [2+4;4; 2] = [6;4;2]
 * $$\displaystyle g=[2+g(1);g] $$ ;
 * $$\displaystyle g=[2+g(1);g] $$ ;
 * }


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

we will obtain,
 * Again following the same procedure for Iteration 3, i.e for k=3
 * Again following the same procedure for Iteration 3, i.e for k=3
 * }


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

$$\displaystyle \therefore c_7 = \frac{31}{15120} $$
 * style="width:10%; padding:10px; border:2px solid #8888aa" |
 * style="width:10%; padding:10px; border:2px solid #8888aa" |
 * style = |
 * }
 * }


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


 * On substituting, the values of
 * }
 * }


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

in (Eq.1) through (Eq.7) we obtain,
 * $$\displaystyle c_1\;c_3\;c_5\;and\;c_7$$
 * $$\displaystyle c_1\;c_3\;c_5\;and\;c_7$$
 * }


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


 * $$\displaystyle p_1(t) = - t $$
 * }
 * }


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


 * $$\displaystyle p_2(t) = - \frac{t^2}{2!} + \frac{1}{6} $$
 * }
 * }


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


 * $$\displaystyle p_3(t) = - \frac{t^3}{3!} +\frac{t}{6}$$
 * }
 * }


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


 * $$\displaystyle p_4(t) = - \frac{t^4}{4!} + \frac{t^2}{12} -\frac{7}{360} $$
 * }
 * }


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


 * $$\displaystyle p_5(t) = - \frac{t^5}{5!} + \frac{t^3}{36} -\frac{7t}{360} $$
 * }
 * }


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


 * $$\displaystyle p_6(t) = - \frac{t^6}{6!} + \frac{t^4}{144} - \frac{7 t^2}{720} + \frac{31}{15120}$$
 * }
 * }


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


 * $$\displaystyle p_7(t) = - \frac{t^7}{7!} + \frac{t^5}{720} - \frac{7 t^3}{2160} + \frac{31t}{15120} $$
 * }
 * }

Subramanian Annamalai 15:31, 7 April 2010 (UTC)

=  Verification of results for the question 10 of HW 5 =

Problem Statement
$$ \displaystyle Compare\, results\, of\, Team 3\, to\, Team4\, $$

$$ \displaystyle i)\, Integrate\, the\, ellipse\, arc\, length\, by\, Quad.,\, Comp.\, Trap.,\, Romberg,\, Clencurt\, and\, the\, complete\, elliptic\, integral\, of\, the\, second\, kind\, method\, $$

$$ \displaystyle ii)\, Evaluate\, the\, given\, circumference\, formular\, by\, Quad.,\, Comp. Trap.,\, Romberg\, and\, Clencurt\, $$

$$ \int_{\theta=0}^{\theta=2\pi} dl = C $$

$$ \displaystyle Where\,\,\,\, \begin{align} dl &=d\theta [r^2+(\frac{dr}{d\theta})^2]^(\frac{1}{2})\\ r(\theta) &= \frac{1-e^2}{1-ecos(\theta)} \\ e &= sin(\frac{\pi}{12}) \\\end{align}\ $$

Ref: Lecture Notes [[media:Egm6341.s10.mtg25.djvu|p.25-2]] and [[media:Egm6341.s10.mtg30.djvu|p.30-3]]

$$ \displaystyle iii)\, Evaluate\, the\, complete\, elliptic\, integral\, of\, the\, second\, kind\, method\, $$

$$ \displaystyle C = 4a E (e) $$

$$ \displaystyle E(e) = \int_{0}^{\pi /2} [1-e^2sin^2(\theta)]^{(\frac{1}{2})} d\theta $$

Ref: Team 4 in HW 4 [|problem_11] and Wikipedia[|Elliptic_integral]

$$ \displaystyle iiii)\, Compare\, and\, analyze\, the\, results\, $$

Solution
$$ \displaystyle i)\, Summary\, of\, Results\, given\, by\, Team4\, and\, Team3\, $$

$$ \displaystyle \dagger\, Note:\,\, In\, the\, result\, of\, Clencurt,\, X\, means\, ocillating\, $$

$$ \displaystyle ii)\, Comparision\, $$

$$ \displaystyle a)\, The\, big\, difference\, was\, caused\, by\, the\, calculation\, of\, error.\, $$

$$ \displaystyle :\, Team 4\, computed\, I_n\, to\, the\, error\, 10^{-10}\, b/w\, the\, exact\, I\, and\, I_n.\, $$

$$ \displaystyle \,\, However,\, Team 3\, computed\, I_n\, to\, the\, error\, 10^{-10}\, b/w\, I_{n-1}\, and\, I_n.\, $$

$$ \displaystyle \,\, In\, order\, to\, clearly\, know\, the\, exact\, I,\,\, Team 3\, directly\, computed\, I\, with\, Quad\, again\, $$

$$ \displaystyle b)\, Team 4\, exactly\, found\, I_n\, with\, Clencurt\, but\, the\, result\, of\, Team 3\, is\, oscillating\, on\, the\, place\, for\, 10^{-15}\, $$

$$ \displaystyle c)\, Correction\, for\, Clencurt\, and\, 2nd\, method\, $$

$$ \displaystyle :\, Since\, the\, Clencurt's\, result\, of\, Team 3\, is\, going\, in\, busy\, mode\, and\, the\, input\, of\, 2nd\, method\, was\, not\, correctly\, uploaded,\, it\, is\, not\, easy\, to\, clearly\, see\, the\, results.\, $$

$$ \displaystyle Thus,\, the\, inputs\, for\, above\, two\, are\, attached\, below\, again.\, $$

$$ \displaystyle c-1)\, Clencurt\, $$

$$ \displaystyle c-2)\, 2nd\, method\, $$

$$ \displaystyle iii)\, Analyzing\, the\, Results\, of\, Clencurt\, $$

$$ \displaystyle The\, oscillating\, on\, the\, place\, for\, 10^{-15}\, is\, caused\, by\, the\, difference\, b/w\, I_{exact}\, and\, I_{cl}\, in\, the\, result\, of\, team 3.\, $$

$$ \displaystyle Thus,\, we\, verified\, the\, Clencurt's\, result\, with\, a\, different\, order\, of\, errors\, $$

$$ \displaystyle a)\, Correctional\, Clencurt\, $$

$$ \displaystyle b)\, The\, difference\, when\, I_{exact}\, -\, I_{cl}\, <\, 10^{-8}\, is\, 7.591901329817574*10^{-9}.\, $$

$$ \displaystyle This\, value\, is\, pretty\, small\, and\, can\, be\, thought\, that\, the\, Clencurt\, integration\, close\, up\, to\, the\, I_{exact},\, the\, Quad\, integration.\, $$

--Heejun Chung 13:07, 7 April 2010 (UTC)

=Contributing Team Members=

1.Abhishekksingh 09:16, 7 April 2010 (UTC) Authored 3,4,5 and 6 Proof read 2,7,8 and 9 2.Yong Nam Ahn 14:12, 7 April 2010 (UTC) Authored 1, Proof read 11 3.Min 14:20, 7 April 2010 (UTC) Authored 2,7,8,9, Proof read 3,4,5 and 6 4.Heejun Chung 14:53, 7 April 2010 (UTC) Authored 11 Proof read 10 5.Subramanian Annamalai 15:10, 7 April 2010 (UTC) Authored 10 Proof read 1

=References= 1. Introduction to Numerical Methods,2nd Edition by Kendall E Atkinson Second Edition 2. Numerical Methods for Engineers,5th Edition by Steven C Chapra and Raymond P Canale