University of Florida/Egm6341/s10.team2.niki/HW5

Statement
P. 27-1

 Continue the proof of Trapezoidal Rule Error to steps 4a and 4b and determine $$\boldsymbol{ P_6(t)} $$ and $$\boldsymbol{ P_7(t) }$$ 

Solution
From steps 3a and 3b we get the expression


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$$  \displaystyle \boldsymbol{E=[P_2(t)g^{(1)}(t)+P_4(t)g^{(3)}(t)]_{-1}^{+1}-\underbrace{\int_{-1}^{+1}P_5(t)g^{(5)}(t)dt}_D} $$     (Summary p 26-3)
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$$  \displaystyle P_4(t)=\frac{-t^4}{24}+\frac{t^2}{12}-\frac{7}{360}=c_1(\frac{t^4}{4!})+c_3(\frac{t^2}{2!})+c_5$$
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$$

P_5(t)=\frac{-t^5}{120}+\frac{t^3}{36}-\frac{7t}{360}=c_1(\frac{t^5}{5!})+c_3(\frac{t^3}{3!})+c_5(t) $$     (Summary p 26-3)
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Step 4a

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$$  \displaystyle \int_{-1}^{+1}\underbrace{P_5(t)}_{v'}\underbrace{g^{(5)}(t)}_udt=[g^{(5)}(t)P_6(t)]_{-1}^{+1}-\underbrace{\int_{-1}^{+1}[{P_6(t)g^{(6)}(t)}}_E]dt $$ where, $$P_6(t)=\int P_5(t)=c_1(\frac{t^6}{6!})+c_3(\frac{t^4}{4!})+c_5(\frac{t^2}{2!})+c_7$$
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$$ P_6(t)=\frac{-t^6}{720}+\frac{t^4}{144}-\frac{7t^2}{720}+\alpha $$

Step 4b

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$$  \displaystyle \int_{-1}^{+1}[{\underbrace{P_6(t)}_{v'}\underbrace{g^{(6)}(t)}_u}]dt=D=[g^{(6)}(t)P_7(t)]_{-1}^{+1}-\int_{-1}^{+1}P_7(t)g^{(7)}(t)dt $$ where, $$P_7(t)=\int P_6(t)==c_1(\frac{t^7}{7!})+c_3(\frac{t^5}{5!})+c_5(\frac{t^3}{3!})+c_7(t)+c_8$$
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$$P_7(t)=\frac{-t^7}{5040}+\frac{t^5}{720}-\frac{7t^3}{2160}+\alpha t+\beta$$

Selecting $$\displaystyle{P_7(t)}$$ such that

$$\displaystyle{P_7(+1)=P_7(-1)=P_7(0)=0}$$ ,we get

$$\displaystyle{P_7(t=0)=0\Rightarrow \beta=c_8=0}$$

$$P(-1)=0=-\frac{(-1)^7}{7!}+\frac{1}{6}(\frac{(-1)^5}{5!})-\frac{7}{360}(\frac{(-1)^3}{3!})-\alpha\Rightarrow \alpha=c_7=\frac{31}{15120}$$

Summary

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$$  \displaystyle P_6(t)=\frac{-t^6}{600}+\frac{t^4}{108}-\frac{7t^2}{720}+\frac{31}{15120} $$
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$$

P_7(t)=\int P_6(t)=\frac{-t^7}{4200}+\frac{t^5}{540}-\frac{7t^3}{2160}+\frac{31}{15120} t $$ (Summary p 26-3)
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Author
--Egm6341.s10.team2.niki 21:38, 23 March 2010 (UTC)

Statement
P. 28-2

 Redo steps in the proof of the Trapezoidal Rule error by trying to cancel terms with odd order derivatives of "g" 

Solution
We begin with equation (5) on P. 21-1 which is the result of transformation of variables on equation (1) P. 21-1

(| Prob 4 HW4) (P. 21-1)


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$$  \displaystyle E_n^1=\frac{h}{2}\sum _{k=0}^{n-1}[\int_{-1}^{1}g_k(t)dt-[g_k(-1)+g_k({+1})]] $$   (5) on P.21-1 From | Prob 5 HW4, we can express the above equation as:
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$$  \displaystyle E_n^1=\frac{h}{2}\sum _{k=0}^{n-1}[\int_{-1}^{+1}(-t)g^{(1)}(t)dt] $$   (1)
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Step2a
Integrating the term withing the square brackets by "Integration by parts" | Prob 7 HW4 we can rewrite (1) as follows
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$$  \displaystyle E_n^1=\frac{h}{2}\sum _{k=0}^{n-1}[[P_2(t)g^{(1)}(t)]_{-1}^{+1}-\int_{-1}^{+1}P_2(t)g^{(2)}(t)dt] $$   (2) In order to eliminate terms with even powers of $$\displaystyle{h}$$ we need to remove terms with odd derivatives of $$\displaystyle{g(t)}$$.Therefore, the boundary term in eqn (2) above must be set to zero by selection of $$\displaystyle{P_2(t)}$$.
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We have $$\displaystyle{P_2(t)}$$ from eqns (1 and 2)P. 21-3
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$$  \displaystyle P_2(t)=c_1(\frac{t^2}{2!})+c_3 $$   (1)p21-3
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$$  \displaystyle c_2=0 $$   (2)p21-3 Setting $$\displaystyle{P_2(-1)=0}$$ gives $$\displaystyle{c_3=1/2}$$ and hence we get
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$$  \displaystyle P_2(t)=c_1(\frac{t^2}{2!})+c_3 $$   (3)
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So following this method, the next term to be eliminated will have P4(t)
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$$  \displaystyle P_4(t)=c_1(\frac{t^4}{4!})+c_3(\frac{t^2}{2!})+c_4(t)+c_5 $$   (4)
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Setting $$\displaystyle{P_4(-1)=0 and P_4(1)=0 }$$ and solving we get $$\displaystyle{c_4=0 and c_4=-5/24 }$$.Continuing on these lines to we get the eqn (2) in the form


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$$  \displaystyle E_n^1=\frac{h}{2}\sum_{k=0}^{n-1}{[P_2g^{(1)}]_{-1}^{+1}-[P_3g^{(2)}]_{-1}^{+1}}+[P_4g^{(3)}]_{-1}^{+1}-[P_5g^{(4)}]_{-1}^{+1}+[P_6g^{(5)}]_{-1}^{+1}-[P_7g^{(6)}]_{-1}^{+1}-\int_{-1}^{+1}P_7g^{(7)}dt $$   (5)
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$$  \displaystyle E_n^1=\frac{h}{2}\sum_{k=0}^{n-1}{-[P_3g^{(2)}]_{-1}^{+1}}+-[P_5g^{(4)}]_{-1}^{+1}+-[P_7g^{(6)}]_{-1}^{+1}-\int_{-1}^{+1}P_7g^{(7)}dt$$ (6)
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$$  \displaystyle E_n^1=\frac{h}{2}\sum_{k=0}^{n-1}(\sum_{i=1}^{m}-[P_{(2i+1)}g^{(2i)}]_{-1}^{+1})-\int_{-1}^{+1}P_{(2m+1)}g^{(2m+1)}dt $$   (7) manipulating the terms yields,
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$$  \displaystyle E_n^1=\frac{h}{2}\sum_{k=0}^{n-1}[\sum_{i=1}^{m}-[(P_{(2i+1)}(+1)g^{(2i)}(+1))-(P_{(2i+1)}(-1)g^{(2i)}(-1))]-\int_{-1}^{+1}P_{(2m+1)}g^{(2m+1)}dt $$   (8)
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$$  \displaystyle E_n^1=\frac{h}{2}\sum_{k=0}^{n-1}[\sum_{i=1}^{m}[(P_{(2i+1)}(-1)g^{(2i)}(-1))-(P_{(2i+1)}(+1)g^{(2i)}(+1))]-\int_{-1}^{+1}P_{(2m+1)}g^{(2m+1)}dt $$   (9) Transforming g(t) back to f(x) we get [see [| prob 6 HW4]
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$$  \displaystyle E_n^1=\frac{h}{2}\sum_{k=0}^{n-1}[\sum_{i=1}^{m}[(P_{(2i+1)}(-1)(\frac{h^{2i}}{2})f^{(2i)}(x_k))-(P_{(2i+1)}(+1)(\frac{h^{2i}}{2})f^{(2i)}(x_{k+1}))]-\int_{x_k}^{x_{k+1}}P_{(2m+1)}(x)(\frac{h^{2m+1}}{2})f^{(2m+1)}(x))dt $$   (10)
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To see the difference in the two approaches we must compare the equations from the two methods. From (1) P. 27-1 we have
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$$  \displaystyle E_n^1=\sum_{r=1}^{l}h^{2r}d_{2r}[f^{(2r-1)}(b)-f^{(2r-1)}(a)]-\frac{h^{(2l)}}{2^{(2l)}}\sum_{k=0}^{n-1}\int_{x_k}^{x_{k+1}}P_{2l}(t_k(x))f^{(2l)}(x)dx $$   (11)
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It is seen that the first term in eqn 10 is a summation as against the term of eqn (11) which is dependent only on the end points.