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

Problem 2:Continuation of proof of trapezoidal error

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Statement

P. 27-1

Continue the proof of Trapezoidal Rule Error to steps 4a and 4b and determine ${𝑷}_6\mathit{(}𝒕\mathit{)}$ and ${𝑷}_7\mathit{(}𝒕\mathit{)}$

Solution

From steps 3a and 3b we get the expression

$𝑬\mathit{=}\mathit{[}{𝑷}_2\mathit{(}𝒕\mathit{)}{𝒈}^{\mathit{(}1\mathit{)}}\mathit{(}𝒕\mathit{)}\mathit{+}{𝑷}_4\mathit{(}𝒕\mathit{)}{𝒈}^{\mathit{(}3\mathit{)}}\mathit{(}𝒕\mathit{)}{]}_{\mathit{-}1}^{\mathit{+}1}\mathit{-}\underset{𝑫}{\underbrace{\underset{\mathit{-}1}{\overset{\mathit{+}1}{\int }}{𝑷}_5\mathit{(}𝒕\mathit{)}{𝒈}^{\mathit{(}5\mathit{)}}\mathit{(}𝒕\mathit{)}𝒅𝒕}}$

(Summary p 26-3)

$P_4(t)=\frac{-t^4}{24}+\frac{t^2}{12}-\frac{7}{360}=c_1(\frac{t^4}{4!})+c_3(\frac{t^2}{2!})+c_5$ $P_5(t)=\frac{-t^5}{120}+\frac{t^3}{36}-\frac{7t}{360}=c_1(\frac{t^5}{5!})+c_3(\frac{t^3}{3!})+c_5(t)$

(Summary p 26-3)

Step 4a

$\underset{-1}{\overset{+1}{\int }}\underset{v^{\prime }}{\underbrace{P_5(t)}}\underset{u}{\underbrace{g^{(5)}(t)}}dt=[g^{(5)}(t)P_6(t){]}_{-1}^{+1}-\underset{E}{\underbrace{\underset{-1}{\overset{+1}{\int }}[P_6(t)g^{(6)}(t)}}]dt$

where, $P_6(t)=\int P_5(t)=c_1(\frac{t^6}{6!})+c_3(\frac{t^4}{4!})+c_5(\frac{t^2}{2!})+c_7$

          $P_6(t)=\frac{-t^6}{720}+\frac{t^4}{144}-\frac{7t^2}{720}+\alpha$

Step 4b

$\underset{-1}{\overset{+1}{\int }}[\underset{v^{\prime }}{\underbrace{P_6(t)}}\underset{u}{\underbrace{g^{(6)}(t)}}]dt=D=[g^{(6)}(t)P_7(t){]}_{-1}^{+1}-\underset{-1}{\overset{+1}{\int }}{P}_7(t)g^{(7)}(t)dt$

where, $P_7(t)=\int P_6(t)==c_1(\frac{t^7}{7!})+c_3(\frac{t^5}{5!})+c_5(\frac{t^3}{3!})+c_7(t)+c_8$

          $P_7(t)=\frac{-t^7}{5040}+\frac{t^5}{720}-\frac{7t^3}{2160}+\alpha t+\beta$

Selecting $P_7(t)$ such that

$P_7(+1)=P_7(-1)=P_7(0)=0$ ,we get

$P_7(t=0)=0\Rightarrow \beta =c_8=0$

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

Summary

$P_6(t)=\frac{-t^6}{600}+\frac{t^4}{108}-\frac{7t^2}{720}+\frac{31}{15120}$ $P_7(t)=\int P_6(t)=\frac{-t^7}{4200}+\frac{t^5}{540}-\frac{7t^3}{2160}+\frac{31}{15120}t$

(Summary p 26-3)

Author

--Egm6341.s10.team2.niki 21:38, 23 March 2010 (UTC)

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Problem 7: Understanding the derivation of the proof of Trapezoidal Error

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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)

$E_n^1=\frac{h}{2}\sum _{k=0}^{n-1}[\underset{-1}{\overset{1}{\int }}{g}_k(t)dt-[g_k(-1)+g_k(+1)]]$

(5) on P.21-1

From Prob 5 HW4, we can express the above equation as:

$E_n^1=\frac{h}{2}\sum _{k=0}^{n-1}[\underset{-1}{\overset{+1}{\int }}(-t)g^{(1)}(t)dt]$

(1)

Step2a

Integrating the term withing the square brackets by "Integration by parts" Prob 7 HW4 we can rewrite (1) as follows

$E_n^1=\frac{h}{2}\sum _{k=0}^{n-1}[[P_2(t)g^{(1)}(t){]}_{-1}^{+1}-\underset{-1}{\overset{+1}{\int }}{P}_2(t)g^{(2)}(t)dt]$

(2)

In order to eliminate terms with even powers of $h$ we need to remove terms with odd derivatives of $g(t)$.Therefore, the boundary term in eqn (2) above must be set to zero by selection of $P_2(t)$.

We have $P_2(t)$ from eqns (1 and 2)P. 21-3

$P_2(t)=c_1(\frac{t^2}{2!})+c_3$

(1)p21-3

$c_2=0$

(2)p21-3

Setting $P_2(-1)=0$ gives $c_3=1/2$ and hence we get

$P_2(t)=c_1(\frac{t^2}{2!})+c_3$

(3)

So following this method, the next term to be eliminated will have P4(t)

$P_4(t)=c_1(\frac{t^4}{4!})+c_3(\frac{t^2}{2!})+c_4(t)+c_5$

(4)

Setting $P_4(-1)=0and{P}_4(1)=0$ and solving we get $c_4=0and{c}_4=-5/24$.Continuing on these lines to we get the eqn (2) in the form

$E_n^1=\frac{h}{2}\sum _{k=0}^{n-1}[P_2{g}^{(1)}{]}_{-1}^{+1}-[P_3{g}^{(2)}{]}_{-1}^{+1}+[P_4{g}^{(3)}{]}_{-1}^{+1}-[P_5{g}^{(4)}{]}_{-1}^{+1}+[P_6{g}^{(5)}{]}_{-1}^{+1}-[P_7{g}^{(6)}{]}_{-1}^{+1}-\underset{-1}{\overset{+1}{\int }}{P}_7{g}^{(7)}dt$

(5)

$E_n^1=\frac{h}{2}\sum _{k=0}^{n-1}-[P_3{g}^{(2)}{]}_{-1}^{+1}+-[P_5{g}^{(4)}{]}_{-1}^{+1}+-[P_7{g}^{(6)}{]}_{-1}^{+1}-\underset{-1}{\overset{+1}{\int }}{P}_7{g}^{(7)}dt$

(6)

$E_n^1=\frac{h}{2}\sum _{k=0}^{n-1}(\sum _{i=1}^m-[P_{(2i+1)}{g}^{(2i)}{]}_{-1}^{+1})-\underset{-1}{\overset{+1}{\int }}{P}_{(2m+1)}{g}^{(2m+1)}dt$

(7)

manipulating the terms yields,

$E_n^1=\frac{h}{2}\sum _{k=0}^{n-1}[\sum _{i=1}^m-[(P_{(2i+1)}(+1)g^{(2i)}(+1))-(P_{(2i+1)}(-1)g^{(2i)}(-1))]-\underset{-1}{\overset{+1}{\int }}{P}_{(2m+1)}{g}^{(2m+1)}dt$

(8)

$E_n^1=\frac{h}{2}\sum _{k=0}^{n-1}[\sum _{i=1}^m[(P_{(2i+1)}(-1)g^{(2i)}(-1))-(P_{(2i+1)}(+1)g^{(2i)}(+1))]-\underset{-1}{\overset{+1}{\int }}{P}_{(2m+1)}{g}^{(2m+1)}dt$

(9)

Transforming g(t) back to f(x) we get [see [prob 6 HW4]

$E_n^1=\frac{h}{2}\sum _{k=0}^{n-1}[\sum _{i=1}^m[(P_{(2i+1)}(-1)(\frac{h^{2i}}{2})f^{(2i)}(x_k))-(P_{(2i+1)}(+1)(\frac{h^{2i}}{2})f^{(2i)}(x_{k+1}))]-\underset{x_k}{\overset{x_{k+1}}{\int }}{P}_{(2m+1)}(x)(\frac{h^{2m+1}}{2})f^{(2m+1)}(x))dt$

(10)

To see the difference in the two approaches we must compare the equations from the two methods. From (1) P. 27-1 we have

$E_n^1=\sum _{r=1}^l{h}^{2r}{d}_{2r}[f^{(2r-1)}(b)-f^{(2r-1)}(a)]-\frac{h^{(2l)}}{2^{(2l)}}\sum _{k=0}^{n-1}\underset{x_k}{\overset{x_{k+1}}{\int }}{P}_{2l}(t_k(x))f^{(2l)}(x)dx$

(11)

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.