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

Problem 2:Continuation of proof of trapezoidal error

Statement

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)

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

Statement

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)

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