User:Egm6341.s10.team3.sa/HW5

Problem Set 5 =Trapezoidal Error - Derivation: Evaluation of the constants$$c_5 \,and\, c_6$$ in the polynomial $$p_{5}(t) $$= Ref: Lecture Notes [[media:Egm6341.s10.mtg26.djvu|p.26-3]]

Problem Statement
Given that:
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p_{5}(t) = -\frac{t^5}{120} + \frac{t^3}{36} + c_{5} t + c_{6} $$ $$
 * $$ \displaystyle
 * $$ \displaystyle
 * $$\displaystyle (Eq. 1)
 * }
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 * with,
 * }
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p_{5}(0) =0 $$ $$
 * $$ \displaystyle
 * $$\displaystyle (Eq. 2)
 * }
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p_{5}(\pm 1) = 0 $$ $$
 * $$ \displaystyle
 * $$\displaystyle (Eq. 3)
 * }

Solution

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$$ \displaystyle p_{5}(0) = -\frac{0}{120} + \frac{0}{36} + c_{5} (0) + c_{6} = 0 $$
 * Substitute Eqn(2) in Eqn(1),
 * }
 * }


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$$ \displaystyle c_{6} = 0 $$
 * This yields,
 * }


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$$ \displaystyle p_{5}(1) = -\frac{1}{120} + \frac{1}{36} + c_{5}  = 0 $$
 * Substitute Eqn(3) in Eqn(1),
 * }


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$$ \displaystyle c_{5} = -\frac{7}{360} $$
 * This yields,
 * }


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 * Substituting $$\displaystyle p_{5}(-1) = 0$$ also will give the same value of $$ \displaystyle c_5 $$
 * }
 * }
 * }

Hence,
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$$ \displaystyle c_6 = 0 \, and\, c_5 = -\frac{7}{360} $$ Subramanian Annamalai 01:13, 24 March 2010 (UTC)
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= Proof of Trapezoidal error - Evaluating the polynomials $$p_{6}(t)$$ and $$p_{7}(t)$$ =

Problem Statement
$$ \displaystyle i)\, Do\, steps\, 4a\, and\, 4b\, in\, order\, to\, determine\, P_{6}(t)\, and\, P_{7}(t)\, $$

Ref: Lecture Notes [[media:Egm6341.s10.mtg27.djvu|p.27-1]]

$$ \displaystyle ii)\, Steps\, 3a\, and\, 3b\, are\, given\, by\, Lecture\, 26\,$$

$$ E=[P_2(t)g^{(1)}(t)+P_4(t)g^{(3)}(t)]_{-1}^{+1}-\underbrace{\int_{-1}^{+1}P_5(t)g^{(5)}(t)dt}_{Call\, D\,} $$

$$ \begin{align} P_4(t)&=-\frac{t^4}{4!}+\frac{t^2}{12}+\alpha \\ &=c_{1}\frac{t^4}{4!}+c_{3}\frac{t^2}{2!}+c_{5}\, (c_{5}=\alpha) \\ &=-\frac{t^4}{24}+\frac{t^2}{12}-\frac{7}{360} \\\end{align}\ $$

$$ \begin{align} P_5(t)&=-\frac{t^5}{120}+\frac{t^3}{36}+\alpha t + \beta \\ &=c_{1}\frac{t^5}{5!}+c_{3}\frac{t^3}{3!}+c_{5}t+c_{6}\, (c_{5}=\alpha\, and\, c_{6}=\beta) \\ &=-\frac{t^5}{120}+\frac{t^3}{36}-\frac{7t}{360} \\\end{align}\ $$

Ref: Lecture Notes [[media:Egm6341.s10.mtg26.djvu|p.26-2&3]]

Solution
$$ \displaystyle i)\, Step\, 4a\, :\, Recall\, D\, and\, integrate\, by\, parts\, $$

$$ D=:\int_{-1}^{+1}\underbrace{P_5(t)}_{v'}\underbrace{g^{(5)}(t)}_{u}dt = [P_6(t)g^{(5)}(t)]_{-1}^{+1}-\underbrace{\int_{-1}^{+1}{P_6(t)g^{(6)}(t)}\, dt}_{Call\, E} $$

$$P_6(t) = c_1(\frac{t^6}{6!})+c_3(\frac{t^4}{4!})+c_5(\frac{t^2}{2!})+c_7 = -\frac{t^6}{720}+\frac{t^4}{144}-\frac{7t^2}{720}+\alpha $$

$$ \displaystyle ii)\, Step\, 4b\, :\, Recall\, E\, and\, integrate\, by\, parts\, $$

$$ E=:\int_{-1}^{+1}\underbrace{P_6(t)}_{v'}\underbrace{g^{(6)}(t)}_{u}dt = [P_7(t)g^{(6)}(t)]_{-1}^{+1}-\underbrace{\int_{-1}^{+1}{P_7(t)g^{(7)}(t)}\, dt}_{Call\, F} $$

$$P_7(t) = c_1(\frac{t^7}{7!})+c_3(\frac{t^5}{5!})+c_5(\frac{t^3}{3!})+c_7(t)+c_8 = -\frac{t^7}{5040}+\frac{t^5}{720}-\frac{7t^3}{2160}+\alpha t+\beta $$

$$ \displaystyle iii)\, Select\, points\, such\, that\, $$

$$ \displaystyle P_{7}(\pm 1)=0 $$

$$ \displaystyle P_{7}(0)=0 $$

$$\displaystyle Recall\, P_7(t)\, and\, plug\, in\, t=0\, and\, t=-1\,$$ $$P_7(t=0) = c_1(\frac{0}{7!})+c_3(\frac{0}{5!})+c_5(\frac{0}{3!})+c_7(0)+c_8 = -\frac{0}{5040}+\frac{0}{720}-\frac{0}{2160}+\alpha \times 0+\beta $$ $$\displaystyle Hence\, c_8\, =\, \beta\, =\, 0\,$$ $$P_7(t=-1) = c_1(\frac{(-1)}{7!})+c_3(\frac{(-1)}{5!})+c_5(\frac{(-1)}{3!})+c_7(-1) = -\frac{(-1)}{5040}+\frac{(-1)}{720}-\frac{(-1)}{2160}+\alpha \times (-1) $$ $$\displaystyle Hence\, c_7\, =\, \alpha\, =\, \frac{31}{15120}\,$$

$$\displaystyle Therefore\, $$

$$ P_6(t)=-\frac{t^6}{600}+\frac{t^4}{108}-\frac{7t^2}{720}+\frac{31}{15120} $$

$$ P_7(t)=-\frac{t^7}{4200}+\frac{t^5}{540}-\frac{7t^3}{2160}+\frac{31}{15120} t $$

$$ E= [P_2(t)g^{(1)}(t)+P_4(t)g^{(3)}(t)+P_6(t)g^{(5)}(t)]_{-1}^{+1}-F $$

$$ \displaystyle where\, F=:\int_{-1}^{+1}{P_7(t)g^{(7)}(t)}\, dt $$

--Heejun Chung 13:06, 26 March 2010 (UTC)

= Find $$ {t}_{k}(x) $$ = Ref: Lecture Notes [[media:Egm6341.s10.mtg27.djvu|p.27-1]]

Problem Statement
Find $$ {t}_{k}(x) $$ by the reversion of $$ x ({t}_{k}) $$

where
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$$ $$
 * $$\displaystyle x(t_k)=\frac{1}{2}(x_k+x_{k+1})+t_k \frac{h}{2}
 * $$\displaystyle x(t_k)=\frac{1}{2}(x_k+x_{k+1})+t_k \frac{h}{2}
 * $$\displaystyle (Eq. 1)
 * }
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$$
 * $$\displaystyle h=x_{k+1}-x_k
 * $$\displaystyle h=x_{k+1}-x_k
 * }
 * }

Solution
Revise Eq.1 and express $$t(x_k)$$by x:


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$$\displaystyle \Rightarrow t_k(x)=\frac{2}{h}(x- \frac{x_k+x_{k+1}}{2}) $$ $$
 * $$\displaystyle (Eq. 2)
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 * }

When $$ x = x_k $$
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$$ When $$ x = x_{k+1} $$
 * $$\displaystyle t_k(x_k)=\frac{2}{h}(x_k- \frac{x_k+x_{k+1}}{2})=\frac{2}{x_{k+1}-x_k}\frac{x_k-x_{k+1}}{2}=-1
 * $$\displaystyle t_k(x_k)=\frac{2}{h}(x_k- \frac{x_k+x_{k+1}}{2})=\frac{2}{x_{k+1}-x_k}\frac{x_k-x_{k+1}}{2}=-1
 * }
 * }
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$$
 * $$\displaystyle t_k(x_{k+1})=\frac{2}{h}(x_{k+1}- \frac{x_k+x_{k+1}}{2})=\frac{2}{x_{k+1}-x_k}\frac{x_{k+1}-x_k}{2}=1
 * $$\displaystyle t_k(x_{k+1})=\frac{2}{h}(x_{k+1}- \frac{x_k+x_{k+1}}{2})=\frac{2}{x_{k+1}-x_k}\frac{x_{k+1}-x_k}{2}=1

-- By Min Zhong 12:00, 23 March 2010 (UTC)

= Find coefficients of error terms of Trapezoidal rule $$\bar{d_{2}}, \; \bar{d_{4}}, \; \bar{d_{6}}$$ from polynomials $$P_{2}, P_{4}, P_{6}$$ = Ref: Lecture notes [[media:Egm6341.s10.mtg27.djvu|p.27-2]]

Problem Statement
Find coefficients $$\displaystyle \bar{d_{2}}, \; \bar{d_{4}}, \; \bar{d_{6}}$$ using the following relations.


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d_{r} = \bar{d_{2r}} = \frac{P_{2r}(1)}{2^{2r}} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 1)
 * }
 * }

Solution
We already obtained the polynomials $$\displaystyle P_{2}(t), \; P_{4}(t)$$ in class meeting 26. (see lecture notes [[media:Egm6341.s10.mtg26.djvu|p.26-1]] and [[media:Egm6341.s10.mtg26.djvu|p.26-3]].)


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


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P_{4}(t) = -\frac{t^4}{24} + \frac{t^2}{12} - \frac{7}{360} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 3)
 * }
 * }

Also, at the problem 2 in the HW5 (Ref:[[media:Egm6341.s10.mtg27.djvu|p.27-1]]), $$\displaystyle P_{6}(t) $$ have been obtained.


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

From the (Eq.1) and (Eq.2),


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$$\displaystyle d_{1} = \bar{d_{2}} = \frac{P_{2}(1)}{2^2}=\frac{1}{4}\times\frac{-3+1}{6}=-\frac{1}{12} $$
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From the (Eq.1) and (Eq.3),


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$$\displaystyle d_{2} = \bar{d_{4}} = \frac{P_{4}(1)}{2^4}=\frac{1}{16}\times\frac{-15+30-7}{360}=\frac{1}{720} $$
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From the (Eq.1) and (Eq.4),


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$$\displaystyle d_{3} = \bar{d_{6}} = \frac{P_{6}(1)}{2^6}=\frac{1}{64}\times\frac{-21+105-147+31}{15120}=-\frac{1}{30240} $$
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Yong Nam Ahn 14:46, 24 March 2010 (UTC)

= Derive Trapezoidal Error expression = Ref: Lecture Notes [[media:Egm6341.s10.mtg28.djvu|p.28-1]]

Problem Statement
Derive $$ E_n^1 = \sum_{r=1}^{l} \bar{d}_{2r}\big[ f^{2r-1}(b) - f^{2r-1}(a) \big] - \frac{h^{2l}}{2l} \sum_{k=0}^{n-1} \int_{x_k}^{x_{k+1}} p_{2l}(t_k(x))f^{(2l)}(x) \,dx$$ Ref:Lecture Notes [[media:Egm6341.s10.mtg27.djvu|p.27-1]]

Solution
We have

$$ E = \sum_{r=1}^l P_{(2r)}(+1) [g^{(2r-1)}- g^{(2r-1)}(-1)] - \int\limits_{-1}^{+1} P_{(2l)}(t) g^{(2l)}(t) dt]$$

and $$ g_k^i(t) = (\frac{h}{2})^i f^i(x(t)) $$

$$ g^{(2r-1)}(t) = (\frac{h}{2})^{(2r-1)} f^{(2r-1)} (x(t)) $$

$$ g^{(2r-1)}(+1) = (\frac{h}{2})^{(2r-1)} f^{(2r-1)} (x(1)) $$

and $$ x(1) = b $$

$$ g^{(2r-1)}(+1) = (\frac{h}{2})^{(2r-1)} f^{(2r-1)} (b) $$

Similarly

$$ g^{(2r-1)}(-1) = (\frac{h}{2})^{(2r-1)} f^{(2r-1)} (a) $$

and $$ g^{(2l)} = (\frac{h}{2})^{(2l)} f^{(2l)} (x) $$

using above equations into first equation we obtain

$$ E = \sum_{r=1}^l P_{(2r)}(+1) \frac{h}{2} (\frac{h}{2})^{(2r-1)} [f^{(2r-1)}(b)- f^{(2r-1)}(a)] - \sum_{k=1}^{n-1} \int\limits_{-1}^{+1} P_{(2l)}(t_k) (\frac{h}{2})^{2l} f^{(2l)}(x) dx]$$

$$ = \sum_{r=1}^l h^{2r} \frac{P_{(2r)}(+1)}{2^{2r}}[f^{(2r-1)}(b)- f^{(2r-1)}(a)] - (\frac{h}{2})^{2l} \sum_{k=1}^{n-1} \int\limits_{x_k}^{x_{k+1}} P_{(2l)}(t_k(x))  f^{(2l)}(x) dx]$$

but $$ \frac{P_{(2r)}(+1)}{2^{2r}} = \overline{d}_{2r}$$

So finally we obtain

$$ = \sum_{r=1}^l h^{2r} \overline{d}_{2r} [f^{(2r-1)}(b)- f^{(2r-1)}(a)] - (\frac{h}{2})^{2l} \sum_{k=1}^{n-1} \int\limits_{x_k}^{x_{k+1}} P_{(2l)}(t_k(x))  f^{(2l)}(x) dx]$$

Abhishekksingh 06:18, 24 March 2010 (UTC)

= Find coefficients of error terms of Trapezoidal rule $$\bar{d_2}, \; \bar{d_4}, \; \bar{d_6}$$ from Bernoulli numbers = Ref: Lecture notes [[media:Egm6341.s10.mtg28.djvu|p.28-2]]

Problem Statement
Using Bernoulli numbers obtained by the Taylor series expansion of $$\displaystyle x\,coth(x)$$,


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x\,coth(x) = \sum_{r=0}^{\infty}\underbrace{\frac{B_{2r}}{(2r)!}}_{-\bar{d_{2r}}}x^{2r} $$
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }

i) Verify $$\displaystyle \bar{d_{2}}, \; \bar{d_{4}}, \; \bar{d_{6}}$$ by comparing the results from $$\displaystyle \bar{d_{2r}} = \frac{P_{2r}(1)}{2^{2r}}$$. ii) Compute $$\displaystyle \bar{d_{8}}, \; \bar{d_{10}}$$

Solution
i)

Let's explicitly express first several terms of the Taylor series expansion of $$\displaystyle x\,coth(x)$$,


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x\,coth(x) = B_{0} + \frac{B_2}{2!}x^{2} + \frac{B_4}{4!}x^{4} + \frac{B_6}{6!}x^{6} + \frac{B_8}{8!}x^{8} + \frac{B_{10}}{10!}x^{10} + \cdot\cdot\cdot. $$
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }

Note that $$\displaystyle B_{0}=1, \, B_{2}=\frac{1}{6}, \, B_{4}=-\frac{1}{30}, \, B_{6}=\frac{1}{42}$$.

Then, based on the following relation between $$\displaystyle \bar{d_{2r}} $$ and Bernoulli numbers,


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\frac{B_{2r}}{(2r)!} = -\bar{d_{2r}}, $$
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }

we can obtain $$\displaystyle \bar{d_{2}}, \,\bar{d_{4}}, \,\bar{d_{6}}$$ as follows.


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$$\displaystyle \bar{d_{2}}=-\frac{B_{2}}{2!}=-\frac{1}{2}\times\frac{1}{6}=-\frac{1}{12} $$
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$$\displaystyle \bar{d_{4}}=-\frac{B_{4}}{4!}=\frac{1}{24}\times\frac{1}{30}=\frac{1}{720} $$
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$$\displaystyle \bar{d_{6}}=-\frac{B_{6}}{6!}=-\frac{1}{720}\times\frac{1}{42}=-\frac{1}{30240} $$
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 * }
 * }

Above results are exactly same with the results from $$\displaystyle \bar{d_{2r}} = \frac{P_{2r}(1)}{2^{2r}}$$ as we obtained at problem 4 in HW5.

ii)

Since $$\displaystyle B_{8}=-\frac{1}{30}, \, B_{10}=\frac{5}{66}$$,


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$$\displaystyle \bar{d_{8}}=-\frac{B_{8}}{8!}=\frac{1}{40320}\times\frac{1}{30}=\frac{1}{1209600} $$
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$$\displaystyle \bar{d_{10}}=-\frac{B_{10}}{10!}=-\frac{1}{3628800}\times\frac{5}{66}=-\frac{1}{47900160} $$
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Yong Nam Ahn 14:46, 24 March 2010 (UTC)

= Cancel odd order of derivative of g in the proof of Trap error  = Ref: Lecture Notes [[media:Egm6341.s10.mtg28.djvu|p.28-2]]

Problem Statement
Redo steps in proof of Trap. error by trying to cancel terms with odd order of derivative of g. From Lecture Notes [[media:Egm6341.s10.mtg21.djvu|p.21-1,p.21-2,p.21-3]], the composite trap error is given by:
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E_n^1=\frac{h}{2} \sum_{k=0}^{n-1} \left[ \underbrace{\int_{-1}^{1}g_k(t)dt-(g_k(-1)+g_k(+1))}_{E}\right] $$ Integrate E by part we have:
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }
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\begin{align} E &=\int_{-1}^{1}p_{1}(t)g_k^{(1)}(t)dt \\ &= \left[ P_2(t)g^{(1)}(t) \right]_{-1}^{+1} - \int_{-1}^{1}P_2(t)g^{(2)}(t)dt \\ &= \left[ P_2(t)g^{(1)}(t)\right]_{-1}^{+1}-\left[ P_3(t)g^{(2)}(t)\right]_{-1}^{+1}+ \int_{-1}^{+1} P_3(t)g^{(3)}(t)dt \\ \end{align} $$ $$ Where
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 1)
 * }
 * }
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 * $$\displaystyle P_1(t)=-t
 * $$\displaystyle P_1(t)=-t

$$
 * }
 * }


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$$ $$
 * $$\displaystyle P_2(t)=-\frac{t^2}{2}+ \alpha
 * $$\displaystyle P_2(t)=-\frac{t^2}{2}+ \alpha
 * $$\displaystyle (Eq. 2)
 * }
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 * $$\displaystyle P_3(t)=-\frac{t^3}{6}+ \alpha t+ \beta
 * $$\displaystyle P_3(t)=-\frac{t^3}{6}+ \alpha t+ \beta

$$
 * }
 * }

$$ \alpha $$ and $$\beta$$ are the integration constants

Solution
We start the proof from Equation 1. i) Step 1 we try to cancel the term contains first order of derivative of g in Eq. 1 by having


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$$ Then
 * $$\displaystyle \left[ P_2(t)g^{(1)}(t)\right]_{-1}^{+1}=0
 * $$\displaystyle \left[ P_2(t)g^{(1)}(t)\right]_{-1}^{+1}=0
 * }
 * }
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$$ From Equation 2.
 * $$\displaystyle P_2(+1)= P_2(-1)=0
 * $$\displaystyle P_2(+1)= P_2(-1)=0
 * }
 * }
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$$ Summary:
 * $$\displaystyle P_2(+1)=-\frac{1^2}{2}+ \alpha \Rightarrow \alpha=\frac{1}{2}
 * $$\displaystyle P_2(+1)=-\frac{1^2}{2}+ \alpha \Rightarrow \alpha=\frac{1}{2}
 * }
 * }
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$$
 * $$\displaystyle P_2(t)=-\frac{t^2}{2}+ \frac{1}{2}
 * $$\displaystyle P_2(t)=-\frac{t^2}{2}+ \frac{1}{2}
 * }
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$$
 * $$\displaystyle E= -\left[ P_3(t)g^{(2)}(t)\right]_{-1}^{+1}+ \underbrace {\int_{-1}^{+1} P_3(t)g^{(3)}(t)dt}_{A}
 * $$\displaystyle E= -\left[ P_3(t)g^{(2)}(t)\right]_{-1}^{+1}+ \underbrace {\int_{-1}^{+1} P_3(t)g^{(3)}(t)dt}_{A}
 * }
 * }

From Lecture Notes [[media:Egm6341.s10.mtg26.djvu|p.26-2]], The result of integration by part of A is


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$$
 * $$\displaystyle A= \left[ P_4(t)g^{(3)}(t)\right]_{-1}^{+1}-\int_{-1}^{+1} P_4(t)g^{(4)}(t)dt
 * $$\displaystyle A= \left[ P_4(t)g^{(3)}(t)\right]_{-1}^{+1}-\int_{-1}^{+1} P_4(t)g^{(4)}(t)dt


 * }
 * }

Then we have
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$$ $$ Where
 * $$\displaystyle E= \left[ P_3(t)g^{(2)}(t)\right]_{-1}^{+1}+ \left[ P_4(t)g^{(3)}(t)\right]_{-1}^{+1}-\int_{-1}^{+1} P_4(t)g^{(4)}(t)dt
 * $$\displaystyle E= \left[ P_3(t)g^{(2)}(t)\right]_{-1}^{+1}+ \left[ P_4(t)g^{(3)}(t)\right]_{-1}^{+1}-\int_{-1}^{+1} P_4(t)g^{(4)}(t)dt
 * $$\displaystyle (Eq. 3)
 * }
 * }
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$$
 * $$\displaystyle P_3(t)=-\frac{t^3}{6}+ \alpha t+ \beta = -\frac{t^3}{6}+ \frac{1}{2}t+ \beta
 * $$\displaystyle P_3(t)=-\frac{t^3}{6}+ \alpha t+ \beta = -\frac{t^3}{6}+ \frac{1}{2}t+ \beta
 * }
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$$ $$ where $$\omega$$ is the integration constant
 * $$\displaystyle P_4(t)= \int P_3= -\frac{t^4}{24}+ \frac{1}{4}t^2+ \beta t + \omega
 * $$\displaystyle P_4(t)= \int P_3= -\frac{t^4}{24}+ \frac{1}{4}t^2+ \beta t + \omega
 * $$\displaystyle (Eq. 4)
 * }
 * }

ii) Step 2

we try to cancel the term contains third order of derivative of g in Eq. 3

Let
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$$
 * $$\displaystyle \left[ P_4(t)g^{(3)}(t)\right]_{-1}^{+1}=0
 * $$\displaystyle \left[ P_4(t)g^{(3)}(t)\right]_{-1}^{+1}=0
 * }
 * }

Then
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$$ From Equation 4, then
 * $$\displaystyle P_4(+1)= P_4(-1)=0
 * $$\displaystyle P_4(+1)= P_4(-1)=0
 * }
 * }
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$$
 * $$\displaystyle P_4(+1)= -\frac{1}{24}+ \frac{1}{4}+ \beta  + \omega =0
 * $$\displaystyle P_4(+1)= -\frac{1}{24}+ \frac{1}{4}+ \beta  + \omega =0
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

$$ Solve the above two equations, we can find
 * $$\displaystyle P_4(-1)= -\frac{1}{24}+ \frac{1}{4}- \beta  + \omega =0
 * $$\displaystyle P_4(-1)= -\frac{1}{24}+ \frac{1}{4}- \beta  + \omega =0
 * }
 * }
 * {| style="width:100%" border="0" align="left"

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

$$
 * $$\displaystyle  \omega = -\frac{5}{24}
 * $$\displaystyle  \omega = -\frac{5}{24}
 * }
 * }

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

$$
 * $$\displaystyle P_3(t)= -\frac{t^3}{6}+ \frac{1}{2}t
 * $$\displaystyle P_3(t)= -\frac{t^3}{6}+ \frac{1}{2}t
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

$$
 * $$\displaystyle P_4(t)=-\frac{t^4}{24}+ \frac{1}{4}t^2 -\frac{5}{24}
 * $$\displaystyle P_4(t)=-\frac{t^4}{24}+ \frac{1}{4}t^2 -\frac{5}{24}
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

$$ $$ From Lecture Notes [[media:Egm6341.s10.mtg26.djvu|p.26-2]], The result of integration by part of B is
 * $$\displaystyle E= -\left[ P_3(t)g^{(2)}(t)\right]_{-1}^{+1}- \underbrace{\int_{-1}^{+1} P_4(t)g^{(4)}(t)dt}_{B}
 * $$\displaystyle E= -\left[ P_3(t)g^{(2)}(t)\right]_{-1}^{+1}- \underbrace{\int_{-1}^{+1} P_4(t)g^{(4)}(t)dt}_{B}
 * $$\displaystyle (Eq. 5)
 * }
 * }
 * {| style="width:100%" border="0" align="left"


 * $$\displaystyle B= \left[ P_5(t)g^{(4)}(t)\right]_{-1}^{+1}-\int_{-1}^{+1} P_5(t)g^{(5)}(t)dt
 * $$\displaystyle B= \left[ P_5(t)g^{(4)}(t)\right]_{-1}^{+1}-\int_{-1}^{+1} P_5(t)g^{(5)}(t)dt

$$
 * }
 * }

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


 * $$\displaystyle E= -\left[ P_3(t)g^{(2)}(t)\right]_{-1}^{+1}-\left[ P_5(t)g^{(4)}(t)\right]_{-1}^{+1}+\int_{-1}^{+1} P_5(t)g^{(5)}(t)dt
 * $$\displaystyle E= -\left[ P_3(t)g^{(2)}(t)\right]_{-1}^{+1}-\left[ P_5(t)g^{(4)}(t)\right]_{-1}^{+1}+\int_{-1}^{+1} P_5(t)g^{(5)}(t)dt

$$
 * }
 * }

iii) Step 3

We can continually cancel the items with odd order derivative of g and find the general expression of E as


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

$$ $$
 * $$\displaystyle E= -\left[ \underbrace{P_3(t)g^{(2)}(t)}_{C}+P_5(t)g^{(4)}(t)+ P_7(t)g^{(6)}(t)+...+P_{2l+1}(t)g^{(2l)}(t)\right]_{-1}^{+1} - \int_{-1}^{+1} P_{2l+1}(t)g^{(2l+1)}(t)dt
 * $$\displaystyle E= -\left[ \underbrace{P_3(t)g^{(2)}(t)}_{C}+P_5(t)g^{(4)}(t)+ P_7(t)g^{(6)}(t)+...+P_{2l+1}(t)g^{(2l)}(t)\right]_{-1}^{+1} - \int_{-1}^{+1} P_{2l+1}(t)g^{(2l+1)}(t)dt
 * $$\displaystyle (Eq. 6)
 * }
 * }

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

$$ Since $$P_3(t) and P_{3i}(t) $$ are odd function, Then
 * $$\displaystyle C = \left[ P_3(t)g^{(2)}(t)\right]_{-1}^{+1}=P_3(+1)g^{(2)}(+1)-P_3(-1)g^{(2)}(-1)
 * $$\displaystyle C = \left[ P_3(t)g^{(2)}(t)\right]_{-1}^{+1}=P_3(+1)g^{(2)}(+1)-P_3(-1)g^{(2)}(-1)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

$$
 * $$\displaystyle P_3(+1)=-P_3(-1)
 * $$\displaystyle P_3(+1)=-P_3(-1)
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

$$
 * $$\displaystyle C = - P_3(-1)\left[(g^{(2)}(+1)+g^{(2)}(-1))\right]
 * $$\displaystyle C = - P_3(-1)\left[(g^{(2)}(+1)+g^{(2)}(-1))\right]
 * }
 * }

Thus Equation 6 can be simplified as follows:


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

$$\displaystyle \Rightarrow  E= \sum_{r=1}^{l}P_{2r+1}(-1) \left[g^{2r}(+1)+g^{2r}(-1) \right] - \int_{-1}^{+1} P_{2l+1}(t)g^{(2l+1)}(t)dt $$ $$
 * style="width:10%; padding:10px; border:2px solid #8888aa" |
 * style="width:10%; padding:10px; border:2px solid #8888aa" |
 * $$\displaystyle (Eq. 7)
 * style = |
 * }
 * }

iv) Step 4 From the equation (5) in lecture note [[media:Egm6341.s10.mtg21.djvu|p.21-2]], We find the composite trap error $$E_n^1$$ through Equation 7


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

$$\displaystyle E_n^1 = \frac{h}{2} \sum_{r=1}^{l} \sum_{k=0}^{n-1}P_{2r+1}(-1) \left[g_k^{2r}(+1)+g_k^{2r}(-1) \right]-\frac{h}{2} \sum_{k=0}^{n-1} \int_{-1}^{1}P_{2l+1}(t)g_k^{2l+1}(t)dt $$ $$ From the lecture note [[media:Egm6341.s10.mtg21.djvu|p.21-2]], $$\displaystyle g_k(t)$$can be changed to $$\displaystyle f_(x_k)$$ and below was obtained.
 * <p style="text-align:right;">$$\displaystyle (Eq. 8)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

\begin{align} &g_k^{(2r)}(+1)=\left(\frac{h}{2}\right)^{2r}f^{(2r)}(x_{k+1}),\\ &g_k^{(2r)}(-1)=\left(\frac{h}{2}\right)^{2r}f^{(2r)}(x_{k}),\\ &g_k^{(2l+1)}(t)=\left(\frac{h}{2}\right)^{2l+1}f^{(2l+1)}(x_{t}) \end{align} $$
 * $$\displaystyle
 * $$\displaystyle
 * }.
 * }.

Plug the above equations into Equation 8. We have
 * {| style="width:100%" border="0" align="left"

$$\displaystyle E_n^1 = \sum_{r=1}^{l} \sum_{k}^{n-1}\left(\frac{h}{2}\right)^{2r+1}P_{2r+1}(-1) \left[f^{2r}(x_{k+1})+f^{2r}(x_{k})\right] - \left(\frac{h}{2}\right)^{2r+2}\sum_{k=0}^{n-1} \int_{x_k}^{x_{k+1}}P_{2l+1}(t)f_{2l+1}(x_t)dx $$ $$
 * style="width:10%; padding:10px; border:2px solid #8888aa" |
 * style="width:10%; padding:10px; border:2px solid #8888aa" |
 * <p style="text-align:right;">$$\displaystyle (Eq. 9)
 * style = |
 * }
 * }

When we derive the composite trap error, we use the simple trap error which contains additive term $$ g^{2r}(+1)+g^{2r}(-1) $$. Because of the additive sign, no term could be canceled in the summation of $$E_n^1$$. We have to add all points $$\sum_{k}^{n-1}\left[f^{2r}(x_{k+1})+f^{2r}(x_{k})\right]$$ when evaluate the error. It is very complicated compared with that derived by canceling items with even oder derivative of g. In that case, we can get simple trap with item $$ g^{2r-1}(+1)-g^{2r-1}(-1) $$ which can simplify the expression of composite trap error by cancel same term when summation and result in only initial and end point $$[f^{(2r-1)}(x_{n})-f^{(2r-1)}(x_0)$$. Thus Trap error proofed by canceling terms with the even order derivative of g is more effective. -- By Min Zhong 12:00, 23 March 2010 (UTC)

= Obtain expressions for $$(P_2,P_3),(P_4,P_5),(P_6,P_7)$$ Using recurrence formula  =

Ref: Lecture Notes [[media:Egm6341.s10.mtg29.djvu|p.29-3]]

Problem Statement
Obtain expressions for $$ (P_2,P_3),(P_4,P_5),(P_6,P_7) $$ using $$ \frac{c_1}{(2i+1)!} + \frac{c_3}{(2i-1)!} +\frac{c_5}{(2i-3)!} + ............ + \frac{c_{2i-1}}{3!} + c_{2i+1}$$ Ref:Lecture Notes [[media:Egm6341.s10.mtg29.djvu|p.29-2]]

Solution
$$ (P_2,P_3)$$

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

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

$$ P_{2}(t) = C_1 \frac{t^{2)}}{2!} + C_3 $$

Using eq (6) from Lecture Notes [[media:Egm6341.s10.mtg29.djvu|p.29-3]], we obtain

$$ \frac {C_1}{3!}+ C_3 = 0 $$

$$ C_3 = - \frac {C_1}{3!} $$

$$    = - \frac{-1}{3!} = \frac{1}{6} $$

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

$$ P_3 = \sum_{j=0}^i C_{2j+1} \frac{t^{2(i-1)}}{(2(i-j))!} + C_{2i+2} $$

where $$ C_{2i+2} = 0  $$

$$ P_3 = C_1 \frac{t}{3!} + C_3 \frac{t}{1!}$$

where $$ C_1 = -1 and C_3 = \frac{1}{6} $$ from previous result

$$ P_3 = - \frac{t}{3!} + \frac{t}{6} = 0 $$

$$ P_4, P_5 $$

$$ P_4 = C_1 \frac{t^4}{4!} + C_3 \frac{t^2}{2!}+ C_5 $$

Using eq (6) from Lecture Notes [[media:Egm6341.s10.mtg29.djvu|p.29-3]], we obtain

$$ \frac{C_1}{5!} + \frac {C_3}{3!}+\frac {C_5}{1!} = 0 $$

$$ C_5 = \frac {1}{5!} - \frac{1}{36} = \frac {-7}{360}$$

$$ P_4 = - \frac{t^4}{24} + \frac{t^2}{12} - \frac{7}{360} $$

$$ P_5 = C_1 \frac{t^3}{5!} + C_3 \frac{t^3}{3!}+ C_5 \frac{t^3}{1!} $$

$$ P_5 = (\frac {C_1}{5!} + \frac {C_3}{3!}+ \frac {C_5}{1!})t^3 $$

$$ P_5 = 0 $$

$$ P_6, P_7$$

$$ P_6 = C_1 \frac{t^6}{12!} + C_3 \frac{t^4}{10!} + C_5 \frac{t^2}{8!} + C_7 \frac{1}{6!} $$

By using eq 6 from Lecture Notes [[media:Egm6341.s10.mtg29.djvu|p.29-3]], we obtain

$$ \frac{C_1}{7!} + \frac {C_3}{5!}+\frac {C_5}{3!} + C_7 = 0 $$

$$ C_7 = \frac{1}{7!} - \frac{1}{6!} - \frac{7}{2160}$$

$$   = \frac{11} {15120} $$

$$ P_6 = - \frac{t^6}{12!} + \frac{t^4}{6*10!}- \frac{7 t^2}{360 * 8!} + \frac {11}{15120} $$

$$ P_7 = C_1 \frac {t^5}{7!} + C_3 \frac {t^5}{5!} + C_5 \frac{t^5}{3!}+C_7 \frac{t^5}{1!} $$

$$ P_7 = (\frac {C_1}{7!} + \frac {C_3}{5!} + \frac {C_5}{3!}+ \frac {C_7}{1!}) t^5 $$

$$ P_7 = 0 * t^5 $$

$$ P_7 = 0 $$

Abhishekksingh 06:18, 24 March 2010 (UTC)

=Kessler's MATLAB code: Line by line explanation= Ref: Lecture Notes [[media:Egm6341.s10.mtg30.djvu|p.30-2]]

Problem Statement

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

Provide an explanation to the code used by the author to find the error in the trapezoidal rule.
 * Refer to the paper by P.Kessler on Trapezoidal rule error, Math 128a, Berkeley, 2006.
 * Refer to the paper by P.Kessler on Trapezoidal rule error, Math 128a, Berkeley, 2006.
 * }
 * }

Solution

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

$$ c_1 $$ = -1 $$ f=1 \,\,\, g=2$$ $$ f=f.*g.*(g+1) $$ yields f=6. $$ fracsum(-1*cn,cd.*f) $$ yields fracsum(-1, 6) Now looking into the fracsum function,

div=gcd(round(n),round(d))

will yield div = 1. because, round(-1) = -1 round(6) = 6 gcd(-1,6) = 1.

n=round(n./div); will give n=1

d=round(d./div); will give d=6

Next, for k=1:length(d) dsum=lcm(dsum,d(k)); end

will yield dsum = 6.

nsum=dsum*sum(n./d); is, nsum = 6*sum(1/6) = 6*(1/6) =1.

div=gcd(round(nsum),round(dsum)); gives div = 1.

So. nsum = 1 and  dsum =6 Hence $$ c_3 = \frac{1}{6} $$

Now going back, [newcn newcd] = [1 6] cn=[cn;newcn]; cd=[cd;newcd]

cn = [-1;1]; cd=[1;6]

f=[f;1]; So, f=[6;1]

and

g=[2+g(1);g]; which means, g = [2+2, 2] = [4;2]

[newpn,newpd]=fracsum((g-1).*cn,f.*cd); i.e, fracsum(-3,6) and fracsum(-1,1)

Again repeating the exercise for fracsum as specified above will give, would yield, $$-\frac{1}{2} \,\,\, and  \frac{1}{6} $$ respectively. Hence,

$$p_2(1) = -\frac{1}{2} + \frac{1}{6} = -\frac{1}{3} $$

So this explains the way to calculate $$ c_3 \,\,\, and p_2(1) $$.

The other constants and the values of polynomial at t=1 can be similarly calculated. It is to be noted that once one constant is found the succeeding constants can be calculated from the previous ones.

Subramanian Annamalai 01:13, 24 March 2010 (UTC)
 * }

=  Compare the result of the given circumference formular with the complete elliptic integral of the second kind method =

Problem Statement
$$ \displaystyle Compare\, the\, results\, i)\, and\, ii)\, $$

$$ \displaystyle i)\, Evaluate\, the\, given\, circumference\, formular\, with\, Comp. Trap,\, Romberg\, and\, Clencurt\, $$

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

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

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

$$ \displaystyle ii)\, The\, complete\, elliptic\, integral\, of\, the\, second\, kind\, method\, $$

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

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

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

Solution
$$ \displaystyle i)\, Quad\, function\, $$

$$ \displaystyle a)\, Define\, function\,\, dl\, $$

$$ \displaystyle b)\, Integrate\,\, dl\,\, with\, Quad\, $$

$$ \displaystyle Therefore\,\,\, \int_{\theta=0}^{\theta=2\pi} dl = 6.176601938988738 $$

$$ \displaystyle ii)\, Comp.\, Trap.\, $$

$$ \displaystyle a)\, Define\, function\,\, dl\, $$

$$ \displaystyle b)\, Integrate\,\, dl\,\, with\, Comp.\, Trap.\, $$

$$ \displaystyle Therefore\,\,\, dl = 6.176601987658693 $$

$$ \displaystyle iii)\, Romberg\, $$

$$ \displaystyle a)\, Define\, function\,\, dl\, $$

$$ \displaystyle b)\, Integrate\,\, dl\,\, with\, Romberg\, $$

$$ \displaystyle Therefore\,\,\, dl = 6.176601987700596 $$

$$ \displaystyle iiii)\, Clencurt\, $$

$$ \displaystyle iiiii)\, The\, complete\, elliptic\, integral\, of\, the\, second\, kind\, method\, $$

$$ \displaystyle Therefore\,\,\, C = 4aE(e) = 6.176601933080789\,\,\, when\, a=1\, $$

$$ \displaystyle Therefore\, $$

--Heejun Chung 13:11, 26 March 2010 (UTC)

= Derivation of the formula to obtain arc length of ellipse from cosine raw = Ref: Lecture notes [[media:Egm6341.s10.mtg31.djvu|p.31-1]]

Problem Statement
Derive the following formula for arc length of ellipse from cosine raw.


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

Arclength(PQ) = \int_{\theta_{P}}^{\theta_{Q}}d\theta\left[r^2+\left(\frac{dr}{d\theta}\right)^{2}\right]^{\frac{1}{2}} $$
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }

Solution


Recall the cosine raw:


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

\overline{AB}^2=\overline{OA}^2+\overline{OB}^2-2\overline{OA}\,\overline{OB}\cos d\theta $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 1)
 * }
 * }

To simply the derivation, let's use the following notations.

$$\displaystyle \overline{AB} = dl $$

$$\displaystyle \overline{OA} = r(\theta) = r $$

$$\displaystyle \overline{OB} = r(\theta+d\theta) \approx r + dr $$

Also, from the Taylor series expansion of $$\displaystyle \cos d\theta $$,
 * {| style="width:100%" border="0" align="left"

\cos d\theta = \sum_{n=0}^{\infty}\frac{(-1)^n}{(2n)!}d\theta^{2n} = 1 - \frac{d\theta^2}{2!} + \frac{d\theta^4}{4!} - \cdots, $$
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }

we can use the following approximated expression since the angle $$\displaystyle d\theta $$ is extremely small,
 * {| style="width:100%" border="0" align="left"

\cos d\theta \approx 1 - \frac{d\theta^2}{2} $$
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }

As a consequence, (Eq.1) can be expressed as below,


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

\begin{align} dl^2 &= r(\theta)^2+r(\theta+d\theta)^2-2r(\theta)r(\theta+d\theta)\cos d\theta\\ &\approx r^2 + (r+dr)^2 -2r(r+dr)\left(1-\frac{1}{2}d\theta^2\right)\\ &=2r^2+2rdr+dr^2-(2r^2+2rdr)+(r^2+rdr)d\theta^2\\ &=dr^2+r^2d\theta^2+\underbrace{rdrd\theta^2}_{\approx 0}\\ &\approx d\theta^2\left[r^2+\left(\frac{dr}{d\theta}\right)^2\right] \end{align} $$
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }

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

dl = d\theta\left[r^2+\left(\frac{dr}{d\theta}\right)^2\right]^{\frac{1}{2}} $$
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }

Finally, the arc length PQ can be obtained by


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

$$\displaystyle Arclength(PQ)=\int_{\theta_{P}}^{\theta_{Q}}dl = \int_{\theta_{P}}^{\theta_{Q}}d\theta\left[r^2+\left(\frac{dr}{d\theta}\right)^2\right]^{\frac{1}{2}} $$
 * style="width:10%; padding:10px; border:2px solid #8888aa" |
 * style="width:10%; padding:10px; border:2px solid #8888aa" |
 * style = |
 * }
 * }

Above result is exactly same with (2) [[media:Egm6341.s10.mtg30.djvu|p.30-3]].

Yong Nam Ahn 14:46, 24 March 2010 (UTC)

= Clarification for HW4 problem 8 =

The I_exact is obtained from the online integration calculator. There are several online sites and one of them is on line integral

Using such online sites is not a good way to get the integration value because we don't know what kind of algorithm is behind such calculators. However, we needed the integration value which comes from the different method with our investigated methods in HW4 as a criterion in order to compare the integration values from our investigated methods with it. So, this online calculator is used.

There is a better way:

In HW4, we compared the integration value with the criterion which is obtained from online calculator and increased the number of nodes if the difference b/w these two values is larger than O(10^-10).

Instead of comparing the obtained value with the criterion from online, we can compare the obtained value with the value of the previous step. For example, compare the value from 4 nodes with the value from the 2 nodes (which is the values of the previous step). If the difference b/w them is greater than O(10^-10), obtain the value from 8 nodes and compare it with the values from 4 nodes (which is the values of the previous step) and so on... By doing this, we can get integration value with the error less than O(10^-10) without using a separate criterion.

-- Uploaded by Min Zhong 12:00, 23 March 2010 (UTC)

= Contributing Team Members = 1.Subramanian Annamalai 01:09, 24 March 2010 (UTC) Authored: 1 and 9 Proof-read: 4,6 and 11

2. Abhishekksingh 06:18, 24 March 2010 (UTC) Authored: 5 and 8 Proof-read: 3 and 7

3. Yong Nam Ahn 14:46, 24 March 2010 (UTC) Authored: 4, 6, and 11 Proof-read: 2 and 10

4. Min Zhong 16:46, 24 March 2010 (UTC) Authored: 3, 7, Proof-read: 5 and 8

5. Heejun Chung 13:08, 26 March 2010 (UTC) Authored: 2 and 10 Proof-read: 1 and 9

=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