University of Florida/Egm6341/s10.Team2/HW5

Statement
P. 26-3

To show that the value of the constants $$\; c_{6}=0$$ and $$c_{5}=\frac{-7}{360}$$ in the polynomials $$P_{4}\left ( t \right )$$ and $$P_{5}\left ( t \right )$$

Solution
From | page 26-2, we know that:

$$P_{5}\left ( t \right )=\frac{-t^{5}}{120}+\frac{t^{3}}{36}+ C_{5}t+C_{6}$$

We select $$\; P_{5}\left ( t \right )$$ such that it is a odd function and $$\; P_{5}\left ( 1 \right )=P_{5}\left ( -1 \right )=P_{5}\left ( 0 \right )=0$$

Therefore, using $$P_{5}\left ( 0 \right )=0$$, we get:

$$P_{5}\left ( 0 \right )=\frac{-0^{5}}{120}+\frac{0^{3}}{36}+ C_{5}*0+C_{6}=0$$


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$$  \displaystyle \boldsymbol \therefore \; C_{6}=0 $$     (1)
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Similarly, using $$\; P_{5}\left ( 1 \right )=0$$, we get:

$$P_{5}\left ( 1 \right )=\frac{-1^{5}}{120}+\frac{1^{3}}{36}+ C_{5}*1+C_{6}=\frac{-1}{120}+\frac{1}{36}+C_{5}=0 $$


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$$  \displaystyle \boldsymbol \therefore \; C_{5}=\frac{1}{120}-\frac{1}{36}=-\frac{7}{360} $$     (2)
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Author
Egm6341.s10.team2.lee 08:58, 24 March 2010 (UTC) proofread --Egm6341.s10.team2.niki 14:53, 24 March 2010 (UTC)

Statement
P. 27-1

 Continue the proof of Trapezoidal Rule Error to steps 4a and 4b and determine $$\boldsymbol{ P_6(t)} $$ and $$\boldsymbol{ P_7(t) }$$ 

Solution
From steps 3a and 3b( P.26-3) we get the expression


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$$  \displaystyle \boldsymbol{E=[P_2(t)g^{(1)}(t)+P_4(t)g^{(3)}(t)]_{-1}^{+1}-\underbrace{\int_{-1}^{+1}P_5(t)g^{(5)}(t)dt}_D} $$     (P.26-3)
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$$  \displaystyle P_4(t)=\frac{-t^4}{24}+\frac{t^2}{12}-\frac{7}{360}=c_1(\frac{t^4}{4!})+c_3(\frac{t^2}{2!})+c_5$$
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$$

P_5(t)=\frac{-t^5}{120}+\frac{t^3}{36}-\frac{7t}{360}=c_1(\frac{t^5}{5!})+c_3(\frac{t^3}{3!})+c_5(t) $$     (P.26-3)
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Step 4a

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$$  \displaystyle \int_{-1}^{+1}\underbrace{P_5(t)}_{v'}\underbrace{g^{(5)}(t)}_udt=[g^{(5)}(t)P_6(t)]_{-1}^{+1}-\underbrace{\int_{-1}^{+1}[{P_6(t)g^{(6)}(t)}}_E]dt $$ where, $$P_6(t)=\int P_5(t)=c_1(\frac{t^6}{6!})+c_3(\frac{t^4}{4!})+c_5(\frac{t^2}{2!})+c_7$$
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$$ P_6(t)=\frac{-t^6}{720}+\frac{t^4}{144}-\frac{7t^2}{720}+\alpha $$

Step 4b

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$$  \displaystyle \int_{-1}^{+1}[{\underbrace{P_6(t)}_{v'}\underbrace{g^{(6)}(t)}_u}]dt=E=[g^{(6)}(t)P_7(t)]_{-1}^{+1}-\int_{-1}^{+1}P_7(t)g^{(7)}(t)dt $$ where, $$P_7(t)=\int P_6(t)==c_1(\frac{t^7}{7!})+c_3(\frac{t^5}{5!})+c_5(\frac{t^3}{3!})+c_7(t)+c_8$$
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$$P_7(t)=\frac{-t^7}{5040}+\frac{t^5}{720}-\frac{7t^3}{2160}+\alpha t+\beta$$

Selecting $$\displaystyle{P_7(t)}$$ such that

$$\displaystyle{P_7(+1)=P_7(-1)=P_7(0)=0}$$ ,we get

$$\displaystyle{P_7(t=0)=0\Rightarrow \beta=c_8=0}$$

$$P(-1)=0=-\frac{(-1)^7}{7!}+\frac{1}{6}(\frac{(-1)^5}{5!})-\frac{7}{360}(\frac{(-1)^3}{3!})-\alpha\Rightarrow \alpha=c_7=\frac{31}{15120}$$

Summary

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$$  \displaystyle P_6(t)=\frac{-t^6}{720}+\frac{t^4}{144}-\frac{7t^2}{720}+\frac{31}{15120} $$
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$$

P_7(t)=\int P_6(t)=\frac{-t^7}{5040}+\frac{t^5}{720}-\frac{7t^3}{2160}+\frac{31}{15120} t $$
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Author
--Egm6341.s10.team2.niki 22:04, 23 March 2010 (UTC)

--Proofread by Egm6341.s10.team2.lee 10:14, 26 March 2010 (UTC)

Statement
find $$\; t_k(x)$$ on P. 27-1

Solution
From Page 27-1, we know:

$$x\left ( t_{k} \right )=\frac{1}{2}\left ( x_{k} + x_{k+1} \right ) + t_{k}\frac{h}{2}$$

Now rearranging the terms on both sides to write $$\; t_{k}$$ in terms of $$\; x$$, we get:

$$t_{k}\left ( x \right )=\frac{2x-\left (x_{k}+x_{k+1} \right )}{h}$$ $$ \; \;$$ where the size of the interval, $$\; h=x_{k+1}-x_{k}$$

It can also be easily verified by substituting $$\; x=x_{k}$$ which yields $$t_{k}\left ( x=x_{k} \right )= \frac{x_{k}-x_{k+1}}{x_{k+1}-x_{k}}=-1$$

Also when $$\; x=x_{k+1}$$ yields $$t_{k}\left ( x=x_{k+1} \right )= \frac{x_{k+1}-x_{k}}{x_{k+1}-x_{k}}=1$$

Author
Srikanth Madala (SM)

Statement
P. 27-2

Find the values of $$d_{1}=\bar{d_{2}}, d_{2}=\bar{d_{4}}$$ and $$d_{3}=\bar{d_{6}} $$

Clue : Use the formula: $$d_{i}=\bar{d_{2i}}= \frac{P_{2i}\left ( 1 \right )}{2^{2i}}$$

Solution
We know from the formula:

$$d_{i}=\bar{d_{2i}}= \frac{P_{2i}\left ( 1 \right )}{2^{2i}}$$

From | Mtg 26,slide 26-1, we know the formula:

$$P_{2}\left ( t \right )= \frac{-t^{2}}{2!}+\frac{1}{6}$$

$$P_{2}\left ( 1 \right )= \frac{-1^{2}}{2!}+\frac{1}{6}$$


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$$d_{1}=\bar{d_{2}}= \frac{P_{2}\left ( 1 \right )}{2^{2}}=\frac{\frac{-1}{3}}{4}=\frac{-1}{12} $$

(1)
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From | Mtg 26,slide 26-3, we know the formula:

$$P_{4}\left ( t \right )= \frac{-t^{4}}{24}+\frac{t^{2}}{12}-\frac{7}{360}$$

$$P_{4}\left ( 1 \right )= \frac{-1^{4}}{24}+\frac{1^{2}}{12}-\frac{7}{360}= \frac{1}{45}$$


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$$d_{2}=\bar{d_{4}}= \frac{P_{4}\left ( 1 \right )}{2^{4}}=\frac{\frac{1}{45}}{16}=\frac{1}{720}$$

(2)
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From problem 2 on this HW, where we deduced that:

$$P_6(t)=\frac{-t^6}{720}+\frac{t^4}{144}-\frac{7t^2}{720}+\frac{31}{15120} $$

$$P_6(1)=\frac{-1^6}{720}+\frac{1^4}{144}-\frac{7*1^2}{720}+\frac{31}{15120}=\frac{-2}{945}$$


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$$d_{3}=\bar{d_{6}}= \frac{P_{6}\left ( 1 \right )}{2^{6}}=\frac{\frac{-2}{945}}{64}=\frac{-1}{30240}$$

(3)
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Author
--Srikanth Madala 01:44, 24 March 2010 (UTC) Srikanth Madala (SM)

Statement
P. 28-2 derive

Derive the Following Equation for the error of the trapezoidal rule:

$$ E_{n}^{1} = I - T_{0}(n) = \sum_{r=1}^{l} h^{2r}\tilde{d}_{2r} \left[ f^{2r-1}(b) - f^{2r-1}(a) \right] - \frac{h^{2l}}{2^{2l}} \sum_{k=0}^{n-1} \int_{x_{k}}^{x_{k+1}} P_{2}l(t_{k}(x))f^{2l}(x)dx $$

Solution
The Following Equation is to be derived:

$$ E_{n}^{1} = I - T_{0}(n) = \sum_{r=1}^{l} h^{2r}\tilde{d}_{2r} \left[ f^{2r-1}(b) - f^{2r-1}(a) \right]  -  \frac{h^{2l}}{2^{2l}} \sum_{k=0}^{n-1} \int_{x_{k}}^{x_{k+1}} P_{2}l(t_{k}(x))f^{2l}(x)dx  $$

$$ E= \int_{-1}^{+1} P_{1}(t)g^{1}(t) dt = \underbrace{ \frac{h}{2} \left[ P_{2}g^{1} + P_{4}g^{3} + ... + P_{2l}g^{2l-1} \right]_{-1}^{+1} }_{A} - \underbrace{ \int_{-1}^{+1} P_{2l}(t)g^{2l}(t)dt }_{B} $$

Since $$ P_{2} $$ is an even function then:

$$ \left[ P_{2}(t)g^{1}(t) \right]_{-1}^{+1} = P_{2}(+1) \left[ g^{1}(+1) - g^{1}(-1) \right] $$

With this property then A can be rewritten as follows:

$$ \sum_{i}^{l}P_{2r}(+1)\left[g^{2r-1}(+1) - g^{2r-1}(-1) \right] $$

A transformation of variable is needed from 't' to 'x' and is as follows:

$$ g_{k}^{i}(t) = (\frac{h}{2})^{i}f^{i}(x(t)) \quad \quad x \epsilon [x_{k}, x_{k+1} ] $$

With this transformation A can be rewritten as follows:

$$ \frac{h}{2} \sum_{i=1}^{l} P_{2r}(+1)(\frac{h}{2})^{2r-1} \left[f^{2r-1}(b) - f^{2r-1}(a) \right] $$

The Following is then defined:

$$ \tilde{d}_{2r} = \frac{P_{2r}(+1)}{2^{2r}} $$ as discussed in [28-2] of the lectures.

A is then rewritten as follows:

$$ \frac{h}{2} \sum_{i=1}^{l} \tilde{d}_{2r} 2^{2r} \left(\frac{h}{2} \right) ^{2r-1} \left[f^{2r-1}(b) - f^{2r-1}(a) \right] $$

$$ \frac{h}{2} \sum_{i=1}^{l} \tilde{d}_{2r} 2 h^{2r-1} \left[f^{2r-1}(b) - f^{2r-1}(a) \right] $$

$$ \sum_{i=1}^{l} \tilde{d}_{2r} h^{2r} \left[f^{2r-1}(b) - f^{2r-1}(a) \right] $$

The Next Step is to simplify the term 'B'

$$ B= \int{-1}^{+1} P_{2l}(t)g^{2l}(t) dt $$

Using the previously mentioned transformation of variable the following can be stated:

$$ \int_{-1}^{+1}P_{2l}(t) \left(\frac{h}{2}\right)^{2l}f^{2l}(x(t))dt $$

With the previously performed simplifications the following is written:

$$ E_{n}^{1} = A - B = \sum_{i=1}^{l} \tilde{d}_{2r} h^{2r} \left[f^{2r-1}(b) - f^{2r-1}(a) \right] - \frac{h^{2l}}{2^{2l}}\int_{-1}^{+1}P_{2l}(t) \left(\frac{h}{2}\right)^{2l}f^{2l}(x(t))dt $$

This is recognized as the Error for the trapezoidal rule.

Author
--Egm6341.s10.Team2.GV 22:09, 25 March 2010 (UTC)

--Proofread by Egm6341.s10.team2.lee 10:13, 26 March 2010 (UTC)

Statement
P. 28-2 bernoulli nos

Solution




Author
Egm6341.s10.team2.lee 10:12, 26 March 2010 (UTC)

Statement
P. 28-2

 Redo steps in the proof of the Trapezoidal Rule error by trying to cancel terms with odd order derivatives of "g" 

Solution
We begin with equation (5) on P. 21-1 which is the result of transformation of variables on equation (1) P. 21-1

(| Prob 4 HW4) (P. 21-1)


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$$  \displaystyle E_n^1=\frac{h}{2}\sum _{k=0}^{n-1}[\int_{-1}^{1}g_k(t)dt-[g_k(-1)+g_k({+1})]] $$   (5) on P.21-1 From | Prob 5 HW4, we can express the above equation as:
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$$  \displaystyle E_n^1=\frac{h}{2}\sum _{k=0}^{n-1}[\int_{-1}^{+1}(-t)g^{(1)}(t)dt] $$   (1)
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Integrating the term withing the square brackets by "Integration by parts" | Prob 7 HW4 we can rewrite (1) as follows
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$$  \displaystyle E_n^1=\frac{h}{2}\sum _{k=0}^{n-1}[[P_2(t)g^{(1)}(t)]_{-1}^{+1}-\int_{-1}^{+1}P_2(t)g^{(2)}(t)dt] $$   (2) In order to eliminate terms with even powers of $$\displaystyle{h}$$ we need to remove terms with odd derivatives of $$\displaystyle{g(t)}$$.Therefore, the boundary term in eqn (2) above must be set to zero by selection of $$\displaystyle{P_2(t)}$$.
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We have $$\displaystyle{P_2(t)}$$ from eqns (1 and 2)P. 21-3
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$$  \displaystyle P_2(t)=c_1(\frac{t^2}{2!})+c_3 $$   (1)p21-3
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$$  \displaystyle c_2=0 $$   (2)p21-3 Setting $$\displaystyle{P_2(-1)=0}$$ gives $$\displaystyle{c_3=1/2}$$ and hence we get
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$$  \displaystyle P_2(t)=c_1(\frac{t^2}{2!})+c_3 $$   (3)
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So following this method, the next term to be eliminated will have P4(t)
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$$  \displaystyle P_4(t)=c_1(\frac{t^4}{4!})+c_3(\frac{t^2}{2!})+c_4(t)+c_5 $$   (4)
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Setting $$\displaystyle{P_4(-1)=0 and P_4(1)=0 }$$ and solving we get $$\displaystyle{c_4=0 and c_4=-5/24 }$$.Continuing on these lines we get the eqn (2) in the form


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$$  \displaystyle E_n^1=\frac{h}{2}\sum_{k=0}^{n-1}{[P_2g^{(1)}]_{-1}^{+1}-[P_3g^{(2)}]_{-1}^{+1}}+[P_4g^{(3)}]_{-1}^{+1}-[P_5g^{(4)}]_{-1}^{+1}+[P_6g^{(5)}]_{-1}^{+1}-[P_7g^{(6)}]_{-1}^{+1}-\int_{-1}^{+1}P_7g^{(7)}dt $$   (5) Dropping all the terms with odd order derivatives of $$g(t)$$ we get
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$$  \displaystyle E_n^1=\frac{h}{2}\sum_{k=0}^{n-1}{-[P_3g^{(2)}]_{-1}^{+1}}+-[P_5g^{(4)}]_{-1}^{+1}+-[P_7g^{(6)}]_{-1}^{+1}-\int_{-1}^{+1}P_7g^{(7)}dt$$ (6) In general on integrating "m" times we get:
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$$  \displaystyle E_n^1=\frac{h}{2}\sum_{k=0}^{n-1}(\sum_{i=1}^{m}-[P_{(2i+1)}g^{(2i)}]_{-1}^{+1})-\int_{-1}^{+1}P_{(2m+1)}g^{(2m+1)}dt $$   (7) manipulating the terms yields,
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$$  \displaystyle E_n^1=\frac{h}{2}\sum_{k=0}^{n-1}[\sum_{i=1}^{m}-[(P_{(2i+1)}(+1)g^{(2i)}(+1))-(P_{(2i+1)}(-1)g^{(2i)}(-1))]-\int_{-1}^{+1}P_{(2m+1)}g^{(2m+1)}dt $$   (8)
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Transforming g(t) back to f(x) we get [see [| prob 6 HW4]


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$$  \displaystyle E_n^1=\sum_{i=1}^{m}(\frac{h}{2})^{2i+1}\sum_{k=0}^{n-1}[(P_{(2i+1)}(-1)( f^{(2i)}(x_k))-(P_{(2i+1)}(+1) f^{(2i)}(x_{k+1}))]-(\frac{h}{2})^{2m+1}\sum_{k=0}^{n-1}(\int_{x_k}^{x_{k+1}}P_{(2m+1)}(x) f^{(2m+1)}(x))dx) $$   (9)
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Since the polynomial is odd, it can be extracted out to give the eqn in the form
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$$  \displaystyle E_n^1=\sum_{i=1}^{m}(\frac{h}{2})^{2i+1}P_{(2i+1)}(-1)\left [\sum_{k=0}^{n-1}[( f^{(2i)}(x_k)+f^{(2i)}(x_{k+1})]\right ]-(\frac{h}{2})^{2m+2}\sum_{k=0}^{n-1}(\int_{x_k}^{x_{k+1}}P_{(2m+1)}(x) f^{(2m+1)}(x))dx) $$   (9)
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The problem with this formula is evident at first sight, from this expression.The HO derivatives add up in the first term of the expression and since they will not cancel out (9) cannot be reduced to a simpler form. To see the difference in the two approaches more clearly we compare the equations from the two methods. From (1) P. 27-1 we have
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$$  \displaystyle E_n^1=\sum_{r=1}^{l}h^{2r}d_{2r}[f^{(2r-1)}(b)-f^{(2r-1)}(a)]-\frac{h^{(2l)}}{2^{(2l)}}\sum_{k=0}^{n-1}\int_{x_k}^{x_{k+1}}P_{2l}(t_k(x))f^{(2l)}(x)dx $$   (10)
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$$  \displaystyle E_n^1=\sum_{i=1}^{m}(\frac{h}{2})^{2i+1}P_{(2i+1)}(-1)\left [\sum_{k=0}^{n-1}[( f^{(2i)}(x_k)+f^{(2i)}(x_{k+1})]\right ]-(\frac{h}{2})^{2m+2}\sum_{k=0}^{n-1}(\int_{x_k}^{x_{k+1}}P_{(2m+1)}(x) f^{(2m+1)}(x))dx) $$   (9) Comparing the eqns 9 and 10 we see that the first term on 9 has a summation of HO derivatives at different points in the square brackets as opposed to 10 which has the difference of the higher order derivative at only the end points. In this lies the difference of the two methods. 10 is general and easier to apply as we need to evaluate the differential of F(x) at only the end points while 9 is computationally cumbersome since it needs the derivative evaluated at all points.
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Author
--Egm6341.s10.team2.niki 14:52, 24 March 2010 (UTC)

===Problem 8: Calculating the coefficients $$C_{3},C_{5} \; and \; C_{7}$$ and the pairs of polynomials $$\left (P_{2},P_{3} \right ),\; \left (P_{4},P_{5}  \right ),\; \left (P_{6},P_{7}  \right )$$ using the Recurrence formula===

Statement
P. 29-3

Use the recurrence formula on P.29-2 to obtain the coefficients and expressions of the pairs of polynomials $$\left (P_{2},P_{3} \right ),\; \left (P_{4},P_{5}  \right ),\; \left (P_{6},P_{7}  \right )$$

Solution
We know the recurrence formulae:

$$P_{2i}(t)=\sum_{j=0}^{i} C_{2j+1} \; \frac{t^{2(i-j)}}{[2(i-j)]!}$$

$$P_{2i+1}(t)=\int P_{2i}(t)$$

$$P_{2i+1}(t)=\sum_{j=0}^{i} C_{2j+1} \; \frac{t^{2(i-j)+1}}{[2(i-j)+1]!}+C_{2i+2}$$

Since all the $$\; P_{2i+1}(t)$$ polynomials are odd functions and are set to zero at +1 and -1, we have:

$$P_{2i+1}(1)=\sum_{j=0}^{i} C_{2j+1} \; \frac{1}{[2(i-j)+1]!}=0$$

$$\sum_{j=0}^{1} C_{2j+1} \; \frac{1}{[2(1-j)+1]!}=0\Rightarrow \frac{C_{1}}{3!}+\frac{C_{3}}{1!}=0 \;\Rightarrow \frac{C_{3}}{1!} = -\frac{C_{1}}{3!}= -\frac{(-1)}{3!}=\frac{1}{6}$$

similarly,

$$\sum_{j=0}^{2} C_{2j+1} \; \frac{1}{[2(2-j)+1]!}=0\Rightarrow \frac{C_{1}}{5!}+\frac{C_{3}}{3!}+\frac{C_{5}}{1!}=0 \;\Rightarrow \frac{C_{5}}{1!} = -\frac{C_{1}}{5!}-\frac{C_{3}}{3!}= -\frac{(-1)}{5!}-\frac{\frac{1}{6}}{3!}=\frac{-7}{360}$$

$$\sum_{j=0}^{3} C_{2j+1} \; \frac{1}{[2(3-j)+1]!}=0\Rightarrow \frac{C_{1}}{7!}+\frac{C_{3}}{5!}+\frac{C_{5}}{3!}+\frac{C_{7}}{1!}=0 \;\Rightarrow \frac{C_{7}}{1!} = -\frac{C_{1}}{7!}-\frac{C_{3}}{5!}-\frac{C_{5}}{3!}= -\frac{(-1)}{7!}-\frac{\frac{1}{6}}{5!}-\frac{\frac{-7}{360}}{3!}=\frac{31}{15120}$$

Now evaluating the polynomials using the coefficients:

$$P_{2}(t)=\sum_{j=0}^{1}C_{2j+1} \; \frac{t^{2(1-j)}}{[2(1-j)]!}=C_{1}\frac{t^{2}}{2!}+C_{3}=-\frac{t^{2}}{2}+\frac{1}{6}$$

$$P_{3}(t)=\sum_{j=0}^{1}C_{2j+1} \; \frac{t^{[2(1-j)+1]}}{[2(1-j)+1]!}=C_{1}\frac{t^{3}}{3!}+C_{3}\frac{t^{1}}{1!}=-\frac{t^{3}}{6}+\frac{t}{6}$$

Similarly,

$$P_{4}(t)=\sum_{j=0}^{2}C_{2j+1} \; \frac{t^{2(2-j)}}{[2(2-j)]!}=C_{1}\frac{t^{4}}{4!}+C_{3}\frac{t^{2}}{2!}+C_{5}\frac{t^{0}}{0!}=-\frac{t^{4}}{24}+\frac{t^{2}}{12}-\frac{7}{360}$$

$$P_{5}(t)=\sum_{j=0}^{2}C_{2j+1} \; \frac{t^{[2(2-j)+1]}}{[2(2-j)+1]!}=C_{1}\frac{t^{5}}{5!}+C_{3}\frac{t^{3}}{3!}+C_{5}\frac{t^{1}}{1!}=-\frac{t^{5}}{120}+\frac{t^{3}}{36}-\frac{7t}{360}$$

$$P_{6}(t)=\sum_{j=0}^{3}C_{2j+1} \; \frac{t^{2(3-j)}}{[2(3-j)]!}=C_{1}\frac{t^{6}}{6!}+C_{3}\frac{t^{4}}{4!}+C_{5}\frac{t^{2}}{2!}+C_{7}=-\frac{t^{6}}{720}+\frac{t^{4}}{144}-\frac{7t^{2}}{720}+\frac{31}{15120}$$

$$P_{7}(t)=\sum_{j=0}^{3}C_{2j+1} \; \frac{t^{[2(3-j)+1]}}{[2(3-j)+1]!}=C_{1}\frac{t^{7}}{7!}+C_{3}\frac{t^{5}}{5!}+C_{5}\frac{t^{3}}{3!}+C_{7}\frac{t^{1}}{1!}=-\frac{t^{7}}{5040}+\frac{t^{5}}{720}-\frac{7t^{3}}{2160}+\frac{31t}{15120}$$

Author
Srikanth Madala (SM)

Statement
To write a explanatory commentary on kessler's Matlab code for the evaluation of polynomials and their coefficients in the Trapezoidal error P. 30-2

Solution
Matlab code

Author
Srikanth Madala (SM)

Statement
P. 30-4

An Ellipse as given:

$$ r (\theta) = \frac{a(1-e^{2})}{1-e \cos( \theta)} $$

where:

$$ eccentricity=e= \sin(\frac{\pi}{12}) $$

$$ a=1 $$

Calculate the Circumference of the elipse as follows:

Method 1:

$$ C= \int_{0}^{ 2 \pi } \left[r^{2} + (\frac{dr}{d\theta})^{2} \right] ^{\frac{1}{2}} $$

Method 2:

$$ C = 4aE(e^{2}) $$

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

Method 1
The Circumference will be calculated by using the following:

$$ C= \int_{0}^{ 2 \pi } \left[r^{2} + (\frac{dr}{d\theta})^{2} \right] ^{\frac{1}{2}} $$

And then integrated numerically using 3 methods:

- Composite Trapezoidal - Clencurt - Romberg Table

The approximation for the exact integral was calculated using the "quad" Matlab command. It was approximated to be 6.176517851674653

Composite Trapezoidal
Matlab code for composite Trapezoidal calculation

Matlab code for calculation of the integrand

Clencurt
Matlab code

Romberg Table
Romberg Table for n=16

6.176517900218236  6.176517900385847   6.176517900341150   6.176517900343991   6.176517900343945   6.176517900343941   6.176517900343945   6.176517900343943   6.176517900343947   6.176517900343945   6.176517900343941                   0   6.176517900343944   6.176517900343946   6.176517900343945   6.176517900343941                   0                   0   6.176517900343946   6.176517900343946   6.176517900343941                   0                   0                   0   6.176517900343946   6.176517900343942                   0                   0                   0                   0   6.176517900343943                   0                   0                   0                   0                   0

Error of Romberg Table


 * 1.0e-007 *

-0.485435833752490 -0.487111941893659  -0.486664966103945   -0.486692917078813  -0.486692899315244  -0.486692934842381   -0.486692908197028  -0.486692925960597  -0.486692917078813   -0.486692925960597  -0.486692925960597  -0.486692881551676   -0.486692925960597  -0.486692890433460                   0   -0.486692899315244                   0                   0

Columns 4 through 6

-0.486693378931591 -0.486692917078813  -0.486692881551676   -0.486692917078813  -0.486692881551676                   0   -0.486692881551676                   0                   0                    0                   0                   0                    0                   0                   0                    0                   0                   0

Matlab code

Method 2
$$ \Complex = 4aE(e^{2}) $$

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

The integral is calculated using 3 methods:

- Composite Trapezoidal - "Clencurt" Matlab Command - Romberg Table

It will then by used to find the circumference as previously stated.

Composite Trapezoidal
$$ \Complex = 4 \times I = 6.147423421522488 $$

Romberg Table
1.536876035753892  1.536849128589532   1.536849129666574   1.536849129666558   1.536849129666558   1.536849129666558   1.536855855380622   1.536849129599258   1.536849129666558   1.536849129666558   1.536849129666558                   0   1.536850811044599   1.536849129662352   1.536849129666558   1.536849129666558                   0                   0   1.536849550007914   1.536849129666295   1.536849129666558                   0                   0                   0   1.536849234751700   1.536849129666542                   0                   0                   0                   0   1.536849155937831                   0                   0                   0                   0                   0

$$ I = 1.536849129666558 $$

$$ \Complex = 4 \times I = 4 \times 1.536849129666558 =  6.147396518666232 $$

Both Methods yield similar but not equal results. Also it is noted that the error associated with method 1 is less than that of method 2

Author
--Egm6341.s10.Team2.GV 22:07, 25 March 2010 (UTC)

Statement
Derive the infinitesimal arc length using the Law of cosine

Author
Egm6341.s10.team2.lee 08:57, 24 March 2010 (UTC)

Signatures
Solved problems 4 and 9. Proofread 5,10--Srikanth Madala (SM)

Solved problem 2 and 7. Proofread 1,3,8,10--Egm6341.s10.team2.niki| Niki Nachappa (NN) 17:17, 24 March 2010 (UTC)

Proofread problems 3, 8, 11--Egm6341.s10.team2.patodon 20:57, 24 March 2010 (UTC)

Solved problems 5 and 10. Proofread 1, 4, 11 --Egm6341.s10.Team2.GV 22:14, 25 March 2010 (UTC)

Solved problems 1,6, 11. Proofread 2, 7, 9 --Egm6341.s10.team2.lee 16:28, 26 March 2010 (UTC)