User:Egm6341.s2010.Team1.lei/HW5

For HW5,I am responsible for 3 problems: (1)27-1; (1)28-2 and (1)29-3.

=Problem 2-Pf of Trap.error =

Find
Do steps 4ab to find $$ p_{6}(t),p_{7}(t)\,$$

Given
[[media:Egm6341.s10.mtg26.djvu|p.26-1]],[[media:Egm6341.s10.mtg26.djvu|p.26-2]],[[media:Egm6341.s10.mtg26.djvu|p.26-3]]

Solution
Step 4a [[media:Egm6341.s10.mtg26.djvu|p.26-3]]

$$ E= \,[\,p_{2}(t)g^{(1)}(t)+p_{4}(t)g^{(3)(t)}\,]_{-1}^{+1}\,-\,\underbrace{\left (\int_{-1}^{+1}p_{5}(t)g^{(5)}(t)dt\right)}_{B}\,$$

$$ B=\,[\,p_{6}(t)g^{(5)}(t)\,]_{-1}^{+1}\,-\,\underbrace{\left (\int_{-1}^{+1}p_{6}(t)g^{(6)}(t)dt\right)}_{C}\,$$

$$ p_{6}(t)=\int p_{5}(t)dt = \int -\frac{t^{5}}{120}+\frac{t^{3}}{36}-\frac{7t}{360}dt \,$$

$$ =-\frac{t^{6}}{720}+\frac{t^{4}}{144}-\frac{7t^{2}}{720}+\alpha \,$$

Step 4b [[media:Egm6341.s10.mtg26.djvu|p.26-2]]

$$ C=\,[\,p_{7}(t)g^{(6)}(t)\,]_{-1}^{+1}\,-\,\int_{-1}^{+1}p_{7}(t)g^{(7)}(t)dt\,$$

$$ p_{7}(t)=\int p_{6}(t)dt = \int -\frac{t^{6}}{720}+\frac{t^{4}}{144}-\frac{7t^{2}}{720}+\alpha dt \,$$

$$ =-\frac{t^{7}}{5040}+\frac{t^{5}}{720}-\frac{7t^{3}}{2160}+\alpha t+\beta \,$$

Select $$ p_{7}(t)\,$$ st $$ p_{7}(+1)=0,p_{7}(-1)=0,p_{7}(0)=0 \, $$

$$ \Rightarrow \beta=0, \alpha=\frac{31}{15120}\, $$

Summary: steps 4ab

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

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

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{31t}{15120}\,$$ $$
 * $$\displaystyle
 * }
 * }

=Problem 5-Deriv.Trap.rule error =

Find
Deriv.(1)[[media:Egm6341.s10.mtg27.djvu|p.29-1]]

Given
(4)[[media:Egm6341.s10.mtg21.djvu|p.21-2]]

$$ E= \int_{-1}^{+1}p_{1}(t)g^{(1)}(t)dt = [P_2(t)g^{(1)}(t)]_{-1}^{+1}-\int_{-1}^{+1}P_2(t)g^{(2)}(t)dt \,$$

Solution
From the (4)[[media:Egm6341.s10.mtg21.djvu|p.21-2]],we can get the generalized form:

$$ E= \int_{-1}^{+1}p_{1}(t)g^{(1)}(t)dt \,$$

$$ =[p_{2}g^{1}+p_{4}g^{3}+...+p_{2l}g^{(2l-1)}]_{-1}^{+1}-\int_{-1}^{+1}p_{2l}(t)g^{(2l)}(t)dt \,$$

Recall that

$$ [p_{2}(t)g^{1}]_{-1}^{+1}=p_{2}(+1)g^{1}(+1)-p_{2}(-1)g^{1}(-1)\,$$

where $$ p_{2}\,$$ is even func.

$$\Rightarrow p_{2}(-1)=p_{2}(+1)\, $$

thus

$$ [p_{2}(t)g^{1}]_{-1}^{+1}=p_{2}(+1)[g^{1}(+1)-g^{1}(-1)]\, $$

and $$ p_{2i}\,$$ is even func.

$$\Rightarrow E=\underbrace{\left (\sum_{r=1}^{l} p_{2r}(+1)[g^{2r-1}(+1)-g^{2r-1}(-1)]\right)}_{A}-\underbrace{\left (\int_{-1}^{+1}p_{2l}(t)g^{2l}(t)dt\right)}_{B}\, $$

next, we should change varible t into x

from (3) [[media:Egm6341.s10.mtg21.djvu|p.21-2]]

$$ g_{k}^{(i)}(t)=(\frac{h}{2})^{i}f^{(i)}(x(t)), [\,x_k,x_{k+1}\,]\,$$

$$ \Rightarrow g^{(2r-1)}(+1)=(\frac{h}{2})^{2r-1}f^{(2r-1)}(x_{k+1})\, $$

$$ \Rightarrow g^{(2r-1)}(-1)=(\frac{h}{2})^{2r-1}f^{(2r-1)}(x_{k})\, $$

Recall (3) [[media:Egm6341.s10.mtg27.djvu|p.27-1]]

$$ \overline{d_{2r}} = \frac{p_{2r}(+1)}{2^{2r}}\, $$

$$ \Rightarrow p_{2r}(+1)=\overline {d_{2r}}*2^{2r}\, $$

thus $$ A \,$$ can be changed into

$$ A=\sum_{r=1}^{l} 2\overline{d_{2r}} h^{2r-1}[f^{2r-1}(x_{k+1})-f^{2r-1}(x_{k})]\,$$

from (3) [[media:Egm6341.s10.mtg21.djvu|p.21-2]]

$$ g_{k}^{(i)}(t)=(\frac{h}{2})^{i}f^{(i)}(x(t)), [\,x_k,x_{k+1}\,]\,$$

$$ g_{k}^{(2l)}(t)=(\frac{h}{2})^{2l}f^{(2l)}(x)\, $$

thus $$ B\, $$ can be changed into

$$ B=\frac{h^{2l}}{2^{2l}}\int_{x_k}^{x_{k+1}}p_{2l}(t_{k}(x))f^{(2l)}(x)(\frac{2}{h})dx \, $$

Recall trap.rule error is in (5)[[media:Egm6341.s10.mtg21.djvu|P21-1]]

$$ E_{n}^{1}=\frac{h}{2}\sum_{k=0}^{n-1}[\int_{-1}^{+1}g_{k}(t)dt-(g_{k}(-1)+g_{k}(+1))]\,$$

Plug A and B into the equation $$\displaystyle E_n^1$$
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\begin{align} E^1_n & = \left(\frac{h}{2}\right)\sum_{r=1}^{\ell} p_{2r}(+1) \left(\frac{h}{2}\right)^{2r-1} \sum_{k=0}^{n-1} \Big[f^{(2r-1)}(x_{k+1})-f^{(2r-1)}(x_k)\Big]- \frac{h}{2}\sum_{k=0}^{n-1} \left[   \int_{x_k}^{x_{k+1}}p_{2l}(t_k(x))\left(\frac{h}{2}\right)^{2\ell}f^{(2\ell)}(x)\frac{2}{h} dx \right]\\ &= \sum_{r=1}^{\ell} h^{2r} \frac{p_{2r}(+1)}{2^{2r}} \Big[f^{(2r-1)}(x_{n})-f^{(2r-1)}(x_0)\Big]-  \left(\frac{h}{2}\right)^{2\ell} \sum_{k=0}^{n-1} \left[ \int_{x_k}^{x_{k+1}}p_{2\ell}(t_k(x))f^{(2\ell)}(x) dx \right] \end{align} $$
 * $$\displaystyle
 * $$\displaystyle
 * }.
 * }.


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E^1_n = \sum_{r=1}^{\ell} h^{2r} \overline{d}_{2r} \Big[f^{(2r-1)}(b)-f^{(2r-1)}(a)\Big]-  \left(\frac{h}{2}\right)^{2\ell} \sum_{k=0}^{n-1} \left[ \int_{x_k}^{x_{k+1}}p_{2\ell}(t_k(x))f^{(2\ell)}(x) dx \right] $$
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }

=Problem 9-Recurrence formula =

Find
Use (6)[[media:Egm6341.s10.mtg29.djvu|p.29-2]] to obtain $$ (p_{2},p_{3}),(p_{4},p_{5}),(p_{6},p_{7})\,$$

Given
Using $$ p_{1}(t)=-t,i.e.,c_{1}=-1 \, $$

Solution
(6)[[media:Egm6341.s10.mtg29.djvu|p.29-2]]

$$ \frac{c_{1}}{(2i+1)!}+\frac{c_{3}}{(2i-1)!}+\frac{c_{5}}{(2i-3)!}+...+\frac{c_{2i-1}}{3!}+c_{2i+1}=0\, $$

when $$ i=1 \,$$

$$ \frac{c_{1}}{3!}+c_{3}=0\, $$

Recall $$ p_{1}(t)=-t,i.e.,c_{1}=-1 \, $$

$$ \Rightarrow c_{3}=\frac{1}{6}\, $$

when $$ i=2 \,$$

$$ \frac{c_{1}}{5!}+\frac{c_{3}}{3!}+c_{5}=0\, $$

Recall $$ c_{3}=\frac{1}{6}\, $$

$$ \Rightarrow c_{5}=-\frac{7}{360}\, $$

when $$ i=3 \,$$

$$ \frac{c_{1}}{7!}+\frac{c_{3}}{5!}+\frac{c_{5}}{3!}+c_{7}=0\, $$

Recall $$ c_{5}=-\frac{7}{360}\, $$

$$ \Rightarrow c_{7}=\frac{31}{15120}\, $$

From (1)and (3)[[media:Egm6341.s10.mtg29.djvu|p.29-2]]

$$ p_{2i}(t)=\sum_{j=0}^{i}c_{2j+1}\frac{t^{2(i-j)}}{[2(i-j)]!}\, $$

$$ p_{2i+1}(t)=\sum_{j=0}^{i}c_{2j+1}\frac{t^{2(i-j)+1}}{[2(i-j)+1]!}+c_{2i+2}\, $$

where $$ c_{2i+2}=0\, $$

thus when $$ i=1 \,$$

$$ p_{2}=c_{1}\frac{t^{2}}{2!}+c_{3}=-\frac{t^{2}}{2}+\frac{1}{6}\,$$

$$ p_{3}=c_{1}\frac{t^{3}}{3!}+c_{3}t=-\frac{t^{3}}{6}+\frac{1t}{6}\,$$

when $$ i=2 \,$$

$$ p_{4}=c_{1}\frac{t^{4}}{4!}+c_{3}\frac{t^{2}}{2!}+c_{5}=-\frac{t^{4}}{24}+\frac{t^{2}}{12}-\frac{7}{360}\,$$

$$ p_{5}=c_{1}\frac{t^{5}}{5!}+c_{3}\frac{t^{3}}{3!}+c_{5}t=-\frac{t^{5}}{120}+\frac{t^{3}}{36}-\frac{7t}{360}\,$$

when $$ i=3 \,$$

$$ p_{6}=c_{1}\frac{t^{6}}{6!}+c_{3}\frac{t^{4}}{4!}+c_{5}\frac{t^{2}}{2!}+c_{7}=-\frac{t^{6}}{720}+\frac{t^{4}}{144}-\frac{7t^{2}}{720}+\frac{31}{15120}\,$$

$$ p_{7}=c_{1}\frac{t^{7}}{7!}+c_{3}\frac{t^{5}}{5!}+c_{5}\frac{t^{3}}{3!}+c_{7}t=-\frac{t^{7}}{5040}+\frac{t^{5}}{720}-\frac{7t^{3}}{2160}+\frac{31t}{15120}\,$$


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 * style="width:50%; padding:10px; border:2px solid #8888aa" |
 * $$ \displaystyle
 * $$ \displaystyle

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


 * style="width:50%; padding:10px; border:2px solid #8888aa" |
 * $$ \displaystyle
 * $$ \displaystyle

p_{4}=-\frac{t^{4}}{24}+\frac{t^{2}}{12}-\frac{7}{360}, p_{5}=-\frac{t^{5}}{120}+\frac{t^{3}}{36}-\frac{7t}{360}\,$$ $$
 * $$\displaystyle
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"


 * style="width:50%; padding:10px; border:2px solid #8888aa" |
 * $$ \displaystyle
 * $$ \displaystyle

p_{6}=-\frac{t^{6}}{720}+\frac{t^{4}}{144}-\frac{7t^{2}}{720}+\frac{31}{15120}, p_{7}=-\frac{t^{7}}{5040}+\frac{t^{5}}{720}-\frac{7t^{3}}{2160}+\frac{31t}{15120}\,$$ $$
 * $$\displaystyle
 * }
 * }