User:Egm6341.s10.team3.sa/Mtg29

Mtg 29: Tue, 16 Mar 10

 Typeset of transparencies, not lecture transcript. Subramanian Annamalai 13:57, 12 August 2010 (UTC)

[[media: Egm6341.s10.mtg29.djvu | Page 29-1]] 

Summary: $$\displaystyle p_{2i} $$ is an even function. ([[media: Egm6341.s10.mtg28.djvu | Page 28-1]]) $$\displaystyle p_{(2i+1)} $$ is a odd function. ([[media: Egm6341.s10.mtg26.djvu | Page 26-1]])

Recall the following: $$\displaystyle p_1(t) = -t $$

From [[media: Egm6341.s10.mtg26.djvu | Page 26-1]] $$\displaystyle p_2(t) = -\frac{t^2}{2!} + \frac{1}{6} $$  $$\displaystyle p_3(t) = -\frac{t^3}{3!} + \frac{t}{6} $$ 

From [[media: Egm6341.s10.mtg26.djvu | Page 26-3]] $$\displaystyle p_4(t) = -\frac{t^4}{4!} + \frac{t^2}{12} - \frac{7}{360}$$  $$\displaystyle p_5(t) = -\frac{t^5}{5!} + \frac{t^3}{36} - \frac{7t}{360}$$ 

From [[media: Egm6341.s10.mtg26.djvu | Page 26-2]] $$\displaystyle p_{k+1}(t) = \int p_k(t) \,dt $$ 

Comparison of $$\displaystyle \bar{d}_{2r}: $$ From [[media: Egm6341.s10.mtg27.djvu | Page 27-1]], $$\displaystyle (1) \bar{d}_{2r} = \frac{p_{2r}(1)}{2^{2r}} $$

From [[media: Egm6341.s10.mtg28.djvu | Page 28-2]], $$\displaystyle (1) \bar{d}_{2r} = -\frac{B_{2r}(1)}{(2r)!} $$ 

[[media: Egm6341.s10.mtg29.djvu | Page 29-2]]   RECURRENCE FORMULA:

$$\displaystyle p_{2i}(t) = \sum_{j=0}^{i} c_{2j+1} \frac{t^{(2(i-j))}}{[2(i-j)]!}$$ <p style="text-align:right;">$$\displaystyle (Eq. 1) $$

$$\displaystyle p_{2i}(t) = \int p_{2i}(t) \,dt$$ <p style="text-align:right;">$$\displaystyle (Eq. 2) $$

$$\displaystyle p_{2i+1} = \sum_{j=0}^{i} c_{2j+1} \frac{t^{(2(i-j)+1)}}{[2(i-j)+1]!}$$ <p style="text-align:right;">$$\displaystyle (Eq. 3) $$

$$\displaystyle p_{2i+1}(0) = 0 \Rightarrow \;\; p_{2i+1}$$ is an odd function $$\displaystyle c_{2i+2} =0 $$ <p style="text-align:right;">$$\displaystyle (Eq. 4) $$

$$\displaystyle (p_{2i},p_{2i+1})$$ have the same number of terms. Also, $$\displaystyle p_{2i+1}(1) = 0 \Rightarrow = \sum_{j=0}^{i} c_{2j+1} \frac{1}{2(i-j)+1]!} =0 $$ <p style="text-align:right;">$$\displaystyle (Eq. 5) $$

Expanding (Eq.5), we have , $$\displaystyle \frac{c_1}{(2i+1)!} + \frac{c_3}{(2i-1)!} + \frac{c_5}{(2i-3)!} + \cdots + \frac{c_{2i-1}}{3!} + \underbrace{c_{2i+1}}_{unknown} = 0 $$ <p style="text-align:right;">$$\displaystyle (Eq. 6) $$

$$\displaystyle c_1,c_3,\cdots c_{2i-1}$$ are known. <br\><br\>

<p style="text-align:left;">[[media: Egm6341.s10.mtg29.djvu | Page 29-3]] 


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HW:
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Obtain expressions for $$\displaystyle\color{blue} (p_2,p_3),(p_4,p_5),(p_6,p_7) $$ using $$ \displaystyle\color{blue} \frac{c_1}{(2i+1)!} + \frac{c_3}{(2i-1)!} +\frac{c_5}{(2i-3)!} + \cdots + \frac{c_{2i-1}}{3!} + c_{2i+1} = 0 \;\; and\;\;\ p_1(t)=-t \; i.e, \; c_1 =-1.$$ Ref: Kessler(2006) and Suli & Meyers(2003)


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