User:Egm6341.s10.team3.heejun/Meeting 42 transcript

 '''MTG 42: Tue. 13 Apr 10'''


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Note:  Typeset of Transparencies, not lecture transcript  Heejun Chung 20:57, 11 August 2010 (UTC)


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HW: Show that $$\displaystyle \bar{CD}=\bar{AB} $$ + higher order terms of $$\displaystyle d\gamma $$
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Parameterization of ellipse:




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HW: Show $$ \displaystyle \frac{x^2}{a^2}+\frac{y^2}{b^2}=1\, $$
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$$ \displaystyle dl = [dx^{2} + dy^{2}]^{\frac{1}{2}} $$
 * Arc length:
 * Arc length:
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$$ \displaystyle e=[1- \frac{b^{2}}{a^{2}}]^{\frac{1}{2}} $$
 * Eccentricity:
 * Eccentricity:
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 * $$ \displaystyle \Rightarrow C= \int dl = a \int^{2 \pi}_{t=0} [1- e^{2} cos^{2}t]^{\frac{1}{2}} dt = 4a \underbrace{\int^{\frac{\pi}{2}}_{\alpha=0} [1- e^{2} sin^{2} \alpha]^{\frac{1}{2}}d \alpha}_{\color{blue}{elliptic\,\, integral\,\, of\,\, the\,\, 2nd\,\, kind}} $$
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HW: Show $$ \displaystyle dl = [dx^{2} + dy^{2}]^{\frac{1}{2}} $$
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HW: Show that the circumference of an ellipse is given by, $$\displaystyle C = \int dl = a \int_{0}^{2\pi} [1-e^2cos^2t]^{1/2}\,dt$$ where $$ \displaystyle e = \Bigg[1-\frac{b^2}{a^2}\Bigg]^{1/2} $$
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HW: Prove that $$ a \int\limits_{t=0}^{2\pi}[1-e^2 cos^2 t]^{1/2}dt = 4a \int\limits_{\alpha=0}^{\pi/2} [1-e^2 sin^2 \alpha]^{1/2}d\alpha   $$
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Clenshow-Curtis:

* Periodicity (trapezoidal error) * Runge phenomenon


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$$ \displaystyle x = \underbrace{cos \theta}_{\color{blue}{Chebyshev}} $$
 * Change of variable:
 * Change of variable:
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 * $$ \displaystyle I= \int^{+1}_{-1} f(x) dx = \int^{\pi}_{0} f(cos \theta) \underbrace{sin \theta d \theta}_{\color{blue}{- dx}} $$
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$$ \displaystyle f(cos \theta) $$ known
 * Initial simpler if cosine series of
 * Initial simpler if cosine series of
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HW: By Changing the variable of integration, $$ \displaystyle x=cos \theta $$ show that $$\displaystyle I= \int_{-1}^{+1}f(x)dx= \int_{0}^{\pi}f(cos \theta)sin \theta d\theta $$
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 * $$ \displaystyle f(cos \theta) = \underbrace{\frac{a_{0}}{2} + \sum^{\infty}_{k=1} a_{k} cos (k \theta)}_{\color{blue}{cosine series}} $$
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 * $$ \displaystyle a_{k} = \frac{2}{\pi} \int^{\pi}_{0} f(cos \theta) cos(k \theta) d \theta $$
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 * $$ \displaystyle I = a_{0} + \sum^{\infty}_{j=1} \frac{2 a_{2j}}{1-(2j)^{2}} $$ known
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 * $$ \displaystyle a_{k} $$   can be evaluated numerically using Trapezoidal rule which corresponds to the Discrete Cosine Transform(DCT). (which can be computed efficiently using FFT: Fast Fourier Transform)


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HW: The known cosine series of $$ \displaystyle f(x) $$ is $$\displaystyle  f(cos \theta)= \frac{a_0}{2}+ \sum_{k=1}^{\infty}a_kcos \left( k\theta\right) $$ Find out the expression for $$ \displaystyle a_k $$ $$\displaystyle a_k= \frac{2}{\pi} \int_{0}^{\pi}f(cos\theta)cos(k\theta)d\theta $$
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HW: The following is known: $$\displaystyle I= \int_{-1}^{+1}f(x)dx= \int_{0}^{\pi}f(cos\theta)sin\theta d\theta $$ The known cosine series of $$\displaystyle f(x) $$ is as follows: $$\displaystyle  f(cos \theta)= \frac{a_0}{2}+ \sum_{k=1}^{\infty}a_kcos \left( k\theta\right) $$ Find the expression of Integration as: $$\displaystyle I=a_0+ \sum_{j=1}^{\infty} \frac{2a_{2j}}{1- \left( 2j\right)^2} $$
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