User:Egm6341.s10.team3.heejun/HW5

=  Proof of Trapezoidal error =

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

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

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

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

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

$$P_7(t) = c_1(\frac{t^7}{7!})+c_3(\frac{t^5}{5!})+c_5(\frac{t^3}{3!})+c_7(\frac{t}{1!})+c_8 = -\frac{t^7}{5040}+\frac{t^5}{720}-\frac{7t^3}{2160}+\alpha t+\beta $$

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

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

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

$$\displaystyle Therefore\, $$

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

$$ P_7(t)=-\frac{t^7}{5040}+\frac{t^5}{720}-\frac{7t^3}{2160}+\frac{31}{15120} t $$

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

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

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

Problem Statement
$$ \displaystyle i)\, The\, given\, circumference\, formular\, $$

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

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

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

$$ \displaystyle ii)\, Integrate\,\, dl\, $$

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

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

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

$$ \displaystyle iiii)\, The\, difference\, between\, 2)\, and\, 3)\, is\, 5.907948796846085\times 10^{-9}\, $$

$$ \displaystyle It\, is\, pretty\, much\, small\, so\, can\, be\, negligible.\, $$

$$ \displaystyle Therefore\, above\, two\, methods\, have\, the\, same\, results.\, $$