User:Egm6341.s10.Team4.nimaa&m/HW5

= Problem 2: Steps 4a and 4b in the proof of higher order derivation of Trap. error =

Given
Envisage the acquired expression for E at the end of step 3 on [[media:Egm6341.s10.mtg26.djvu|26-3]] as following:
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$$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle E=[p_2(t).g^{(1)}(t)+p_4(t).g^{(3)}(t)]_{-1}^{+1}-\int\limits_{-1}^{+1} p_5(t).g^{(5)}(t)dt


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Find
What are the expressions for $$\displaystyle p_6(t) $$, and $$\displaystyle p_7(t) $$ ?

Solution
At the end of step 3 of this procedure, we have obtained:


 * $$\displaystyle

p_4(t)=-\dfrac{t^4}{24}+\dfrac{t^2}{12}-\dfrac{7}{360} $$


 * $$\displaystyle

p_5(t)=-\dfrac{t^5}{120}+\dfrac{t^3}{36}-\dfrac{7}{360}t $$

Step 4a:



E=[p_2(t).g^{(1)}(t)+p_4(t).g^{(3)}(t)]_{-1}^{+1}-\int\limits_{-1}^{+1} p_5(t).g^{(5)}(t)dt $$

If we call the second term (integration term) as:



D:=\int\limits_{-1}^{+1} p_5(t).g^{(5)}(t)dt $$

By integrating by parts we have:



dv=p_5(t)dt\Rightarrow v=p_6(t)=\int p_5(t)dt $$



u=g^{(5)}(t)\Rightarrow du=g^{(6)}(t)dt $$



\Rightarrow D=[uv]_{-1}^{+1}-\int\limits_{-1}^{+1} vdu=[p_6(t).g^{(5)}(t)]_{-1}^{+1}-\int\limits_{-1}^{+1} p_6(t).g^{(6)}(t)dt $$



p_6(t)=\int p_5(t)dt=\int (-\dfrac{t^5}{120}+\dfrac{t^3}{36}-\dfrac{7}{360}t)dt=-\dfrac{t^6}{6!}+\dfrac{t^4}{144}-\dfrac{7}{720}t^2+\alpha $$

Step 4b:

Then by defining F as:



F:=\int\limits_{-1}^{+1} p_6(t).g^{(6)}(t)dt $$

We repeat integration by parts method;



dv=p_6(t)dt\Rightarrow v=p_7(t)=\int p_6(t)dt $$



u=g^{(6)}(t)\Rightarrow du=g^{(7)}(t)dt $$



\Rightarrow F=[p_7(t).g^{(6)}]_{-1}^{+1}-\int\limits_{-1}^{+1} p_7(t).g^{(7)}(t)dt $$



p_7(t)=-\dfrac{t^7}{7!}+\dfrac{t^5}{720}-\dfrac{7t^3}{2160}+\alpha t+\beta $$

We want to make $$\displaystyle [p_7(t).g^{(6)}(t)]_{-1}^{+1} $$ to be equal to zero. So, for an odd function like $$\displaystyle p_7(t) $$, we have:



p_7(0)=0\Rightarrow \beta=0 $$



p_7(\pm 1)=0 $$



\Rightarrow p_7(1)=-\dfrac{1}{7!}+\dfrac{1}{720}-\dfrac{7}{2160}+\alpha=0 $$



\Rightarrow \alpha=\dfrac{3!}{7!\times 3}=\dfrac{31}{15120} $$


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$$\displaystyle p_6(t)=-\dfrac{t^6}{6!}+\dfrac{t^4}{144}-\dfrac{7t^2}{720}+\dfrac{31}{15120} $$
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$$\displaystyle p_7(t)=-\dfrac{t^7}{7!}+\dfrac{t^5}{720}-\dfrac{7t^3}{2160}+\dfrac{31}{15120}t $$
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Author
Solved and typed by - Egm6341.s10.Team4.nimaa&amp;m 23:25, 18 March 2010 (UTC) .

= Problem 8: Using Recurrence formula to find $$\displaystyle c_i $$ =

Given
Consider the Recurrence formula on slide [[media:Egm6341.s10.mtg29.djvu|29-2]] as follows:
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$$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle c_{2i+1}=-\dfrac{c_1}{(2i+1)!}-\dfrac{c_3}{(2i-1)!}-\dfrac{c_5}{(2i-3)!}-...-\dfrac{c_{2i-1}}{3!}


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Where;
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$$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle p_1(t)=-t\Rightarrow c_1=-1


 * }
 * }

Find
Obtain $$\displaystyle (p_2,p_3) $$, $$\displaystyle (p_4,p_5) $$ and $$\displaystyle (p_6,p_7) $$:

Solution

 * $$\displaystyle

p_1(t)=-t\Rightarrow p_2(t)=-\frac{t^2}{2}+c_3\Rightarrow p_3(t)=-\frac{t^3}{3!} $$

For $$\displaystyle (p_2,p_3) $$ we can get $$\displaystyle i=1 $$. Thus,


 * $$\displaystyle

c_{(2\times 1+1)}=c_3=-\frac{c_1}{(2\times 1+1)!} $$


 * $$\displaystyle

\Rightarrow c_3=\frac{1}{3!} $$


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$$\displaystyle p_2(t)=-\frac{t^2}{2!}+\dfrac{1}{6} $$
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$$\displaystyle p_3(t)=-\frac{t^3}{3!}+\dfrac{1}{6}t $$
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Likewise, for $$\displaystyle (p_4,p_5) $$, we can write:



p_4(t)=\int p_3(t)dt=\int (-\frac{t^3}{3!}+c_3t)dt=\int (-\frac{t^3}{3!}+\frac {t}{6})dt=-\frac{t^4}{4!}+\frac{t^2}{12}+c_5 $$



p_5(t)=\int p_4(t)dt=\int (-\frac{t^4}{4!}+\frac{t^2}{12}+c_5)dt=-\frac{t^5}{5!}+\frac{t^3}{36}+c_5t $$

By setting $$\displaystyle i=2 $$ in Recurrence equation:



c_{2\times 2+1}=c_5=-\frac{-1}{(2\times 2+1)!}-\frac{\frac{1}{6}}{(2\times 2-1)!}=\frac{1}{5!}-\frac{1}{36}=-\frac{7}{360} $$


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$$\displaystyle p_4(t)=-\frac{t^4}{4!}+\frac{t^2}{12}-\frac{7}{360} $$
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$$\displaystyle p_5(t)=-\frac{t^5}{5!}+\frac{t^3}{36}-\frac{7}{360}t $$
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Ultimately, for $$\displaystyle (p_6,p_7) $$, we have:



p_6(t)=\int p_5(t)dt=\int (-\frac{t^5}{5!}+\frac{t^3}{36}-\frac{7}{360}t)dt=-\frac{t^6}{6!}+\frac{t^4}{144}-\frac{7t^2}{720}+c_7 $$



p_7(t)=\int p_6(t)dt=\int (-\frac{t^6}{6!}+\frac{t^4}{144}-\frac{7t^2}{720}+c_7)dt=-\frac{t^7}{7!}+\frac{t^5}{720}-\frac{7t^3}{2160}+c_7t $$

By setting $$\displaystyle i=3 $$ in Recurrence equation:



c_{2\times 3+1}=c_7=-\frac{-1}{(2\times 3+1)!}-\frac{\frac{1}{6}}{(2\times 3-1)!}-\frac{\frac{-7}{360}}{(2\times 3-3)!}=\frac{1}{7!}-\frac{1}{720}+\frac{7}{2160}=\frac{31}{15120} $$


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$$\displaystyle p_6(t)=-\dfrac{t^6}{6!}+\dfrac{t^4}{144}-\dfrac{7t^2}{720}+\dfrac{31}{15120} $$
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$$\displaystyle p_7(t)=-\dfrac{t^7}{7!}+\dfrac{t^5}{720}-\dfrac{7t^3}{2160}+\dfrac{31}{15120}t $$
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Author
Solved and typed by - Egm6341.s10.Team4.nimaa&amp;m 00:44, 19 March 2010 (UTC) .