User:EGM6341.s11.team1.Chiu/Mtg32

Mtg 32: Fri, 18 Mar 11

[[media: Nm1.s11.mtg32.djvu| page32-1]]

Composition of $$ \color{blue} \bar{d}_{ \color{red} 2r} \ $$ and recurrence relation for $$ \color{blue} p_{ \color{red} 2i} \ $$ and $$ \color{blue} p_{ \color{red} 2i+1} \ $$ 

Recall $$ \bar{d}_{ \color{red} 2r} \ $$ can be composed using c, [[media: Nm1.s11.mtg31.djvu| (2) page31-1]], [[media: Nm1.s11.mtg31.djvu| (1) page31-2]]


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$$  \displaystyle { \color{blue} p_{ \color{red} 2i}(t)=} \sum_{j=0}^{ \color{red} i} c_{2j+1} \frac{t^{2({ \color{red}i}-j)}}{[2({ \color{red}i}-j)]!} $$     (1)
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$$  \displaystyle \underbrace_{ \color{blue} \overset{odd \ function}{\downarrow}}{ \color{blue} =} \int P_{ \color{red} 2i}(t)dt $$     (2)
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$$  \displaystyle \color{blue} P_{ \color{red}2i+1}( {\color{red}0})={ \color{red}0} \to \to \to \to \to \to \to \downarrow $$
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$$  \displaystyle \qquad \underset{ \color{red} \underset{(3)}{\uparrow}}= \sum_{j=0}^{ \color{red}i}c_{2j+1} \underbrace{ \frac{t^{2({ \color{red}i}-j)+1}}{[2({ \color{red}i}-j)+1]!}}_{ \color{blue} odd \ function}+ \underbrace{c_{2i+2}}_{ \color{red}=0 \ (4)} $$
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$$ (P_{ \color{red} 2i}, P_{ \color{red} 2i+1}) \ $$ have same number of terms.
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$$  \displaystyle P_{ \color{red}2i+1}(+1)=0 \ \overset{\Rightarrow} \ \sum_{j=0}^{ \color{red} i} c_{2j+1} \frac{1}{[2({ \color{red}i}-j)+1]!} \ \overset{=} \ 0 $$
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[[media: Nm1.s11.mtg32.djvu| page32-2]]

[[media: Nm1.s11.mtg32.djvu| (5) page32-1]]:
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$$ \sum_{{ \color{blue}j}=0}^{ \color{red} i} \frac{c_{2{ \color{blue} j}+1}}{[2({ \color{red}i}-{ \color{blue} j})+1]!} \ = \ 0 $$     (1)
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$$ \underbrace{\frac{c_{ \color{blue} 1}}{(2{ \color{red}i}{ \color{blue}+1})!}}_{ \color{blue} j=0} + \underbrace{\frac{c_{ \color{blue} 3}}{(2{ \color{red}i}{ \color{blue}-1})!}}_{ \color{blue} j=1} + \frac{c_{ \color{blue} 5}}{(2{ \color{red}i}{ \color{blue}-3})!} + \cdots + \underbrace{ \frac{c_{ \color{blue} 2i-1}}{ \color{blue} 3!}}_{ \color{blue} j=i-1} + \underbrace{ \color{blue} c_{ \color{red} 2i+1}}_{ \color{blue} j=i} = 0 $$     (2)
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$$ c_{1}, \cdots, c_{2i-1} \ $$ known
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$$ \ c_{2i+1} \ $$ unknown
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HW*6.5:


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With $$ p_{1}(t)=-t \ $$, use (2) to find


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$$ (p_{2}, p_{3}), (p_{4},p_{5}), (p_{6},p_{7}) \ $$ s10


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$$ (p_{8}, p_{9}) \ $$ s11


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References:

Kessler 2006, Suli & Meyers 2003 : proof not constructive as above proof.

End Reference

[[media: Nm1.s11.mtg32.djvu| page32-3]]


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 Summary / algo ( al-Kharizmi, history ): 


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 * 1) $$ \ { \color{red} i=0:} \ p_{ \color{red}1}(t)=c_{ \color{red}1}t, \ c_{ \color{red}1}=-1 \ $$


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 * 2) $$ \ { \color{red} i=1,2,3, \cdots :} \ $$


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 * 2a) Compose $$ c_{ \color{red} 2i+1} \ $$ based on $$ \left \{ c_{ \color{blue} 2j+1}, { \color{blue} j=0,1,\cdots,} { \color{red} i} \right \} \ $$ using [[media: Nm1.s11.mtg32.djvu| (2) page32-2]]


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 * 2b) Obtain $$ p_{ \color{red} 2i}(t) \ $$ using [[media: Nm1.s11.mtg32.djvu| (1) page32-1]]


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 * 2c) Compose $$ \bar{d}_{ \color{red} 2i}(t) \ $$ using [[media: Nm1.s11.mtg32.djvu| (2) page31-1]]


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HW*6.6: Kessler's code

Using   $$ (p_{ \color{red} 2i}, p_{ \color{red} 2i+1}), { \color{red} i=1,2,3} \ $$    to understand Kessler's code line by line, starting with the best of s10. Reproduce Kessler's results.


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Application: Composition of arc length (e.g., orbital travel distance of spacecraft at zero thrust)

[[media: Nm1.s11.mtg32.djvu| page32-4]]


 * nm1.s11.Mtg32.pg4.fig1.svg$$ \overline{OA}=r( \theta), \ \overline{OB}=r( \theta + d \theta), \ \overline{AB}=dl, \ $$ infinite arc length

Goal: Find $$ dl \ $$ in terms of $$ (r, \theta), d \theta. \ $$


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$$ \overline{AB}^{2}=dl^{2}= \overline{AA'}^{2}+ \overline{A'B}^{2} \ $$ (named after, but not by, Pythagoras ! NPR, 2011.03.09)


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$$ \overline{AA'}=r( \theta)d \theta, A'B=dr \cong r( \theta + d \theta)-r( \theta)= \frac{dr( \theta)}{d \theta} d \theta + hot \ $$


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


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$$ \overset{\frown}{PQ}= \int^{ \theta_{Q}}_{ \theta_{P}} \left [ r^{2}+( \frac{dr}{d \theta})^{2} \right ]^{1/2}d \theta $$     (1)
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