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 ${\boldsymbol {P_{6}(t)}}$ and ${\boldsymbol {P_{7}(t)}}$

Solution

From steps 3a and 3b we get the expression

$\displaystyle {\boldsymbol {E=[P_{2}(t)g^{(1)}(t)+P_{4}(t)g^{(3)}(t)]_{-1}^{+1}-\underbrace {\int _{-1}^{+1}P_{5}(t)g^{(5)}(t)dt} _{D}}}$

(Summary p 26-3)

$\displaystyle P_{4}(t)={\frac {-t^{4}}{24}}+{\frac {t^{2}}{12}}-{\frac {7}{360}}=c_{1}({\frac {t^{4}}{4!}})+c_{3}({\frac {t^{2}}{2!}})+c_{5}$ $P_{5}(t)={\frac {-t^{5}}{120}}+{\frac {t^{3}}{36}}-{\frac {7t}{360}}=c_{1}({\frac {t^{5}}{5!}})+c_{3}({\frac {t^{3}}{3!}})+c_{5}(t)$

(Summary p 26-3)

Step 4a

$\displaystyle \int _{-1}^{+1}\underbrace {P_{5}(t)} _{v'}\underbrace {g^{(5)}(t)} _{u}dt=[g^{(5)}(t)P_{6}(t)]_{-1}^{+1}-\underbrace {\int _{-1}^{+1}[{P_{6}(t)g^{(6)}(t)}} _{E}]dt$

where, $P_{6}(t)=\int P_{5}(t)=c_{1}({\frac {t^{6}}{6!}})+c_{3}({\frac {t^{4}}{4!}})+c_{5}({\frac {t^{2}}{2!}})+c_{7}$

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

Step 4b

$\displaystyle \int _{-1}^{+1}[{\underbrace {P_{6}(t)} _{v'}\underbrace {g^{(6)}(t)} _{u}}]dt=D=[g^{(6)}(t)P_{7}(t)]_{-1}^{+1}-\int _{-1}^{+1}P_{7}(t)g^{(7)}(t)dt$

where, $P_{7}(t)=\int P_{6}(t)==c_{1}({\frac {t^{7}}{7!}})+c_{3}({\frac {t^{5}}{5!}})+c_{5}({\frac {t^{3}}{3!}})+c_{7}(t)+c_{8}$

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

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

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

(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)

$\displaystyle E_{n}^{1}={\frac {h}{2}}\sum _{k=0}^{n-1}[\int _{-1}^{1}g_{k}(t)dt-[g_{k}(-1)+g_{k}({+1})]]$

(5) on P.21-1

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

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

(1)

Step2a

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

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

(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 $\displaystyle {P_{2}(t)}$.

We have $\displaystyle {P_{2}(t)}$ from eqns (1 and 2)P. 21-3

$\displaystyle P_{2}(t)=c_{1}({\frac {t^{2}}{2!}})+c_{3}$

(1)p21-3

$\displaystyle c_{2}=0$

(2)p21-3

Setting $\displaystyle {P_{2}(-1)=0}$ gives $\displaystyle {c_{3}=1/2}$ and hence we get

$\displaystyle P_{2}(t)=c_{1}({\frac {t^{2}}{2!}})+c_{3}$

(3)

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

$\displaystyle P_{4}(t)=c_{1}({\frac {t^{4}}{4!}})+c_{3}({\frac {t^{2}}{2!}})+c_{4}(t)+c_{5}$

(4)

Setting $\displaystyle {P_{4}(-1)=0andP_{4}(1)=0}$ and solving we get $\displaystyle {c_{4}=0andc_{4}=-5/24}$.Continuing on these lines to we get the eqn (2) in the form

$\displaystyle E_{n}^{1}={\frac {h}{2}}\sum _{k=0}^{n-1}{[P_{2}g^{(1)}]_{-1}^{+1}-[P_{3}g^{(2)}]_{-1}^{+1}}+[P_{4}g^{(3)}]_{-1}^{+1}-[P_{5}g^{(4)}]_{-1}^{+1}+[P_{6}g^{(5)}]_{-1}^{+1}-[P_{7}g^{(6)}]_{-1}^{+1}-\int _{-1}^{+1}P_{7}g^{(7)}dt$

(5)

$\displaystyle E_{n}^{1}={\frac {h}{2}}\sum _{k=0}^{n-1}{-[P_{3}g^{(2)}]_{-1}^{+1}}+-[P_{5}g^{(4)}]_{-1}^{+1}+-[P_{7}g^{(6)}]_{-1}^{+1}-\int _{-1}^{+1}P_{7}g^{(7)}dt$

(6)

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

(7)

manipulating the terms yields,

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

(8)

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

(9)

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

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

(10)

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

$\displaystyle E_{n}^{1}=\sum _{r=1}^{l}h^{2r}d_{2r}[f^{(2r-1)}(b)-f^{(2r-1)}(a)]-{\frac {h^{(2l)}}{2^{(2l)}}}\sum _{k=0}^{n-1}\int _{x_{k}}^{x_{k+1}}P_{2l}(t_{k}(x))f^{(2l)}(x)dx$

(11)

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.