University of Florida/Egm6341/s10.team3.aks/HW5

(5) Derive Trapezoidal Error expression

Ref: Lecture Notes p.28-1

Problem Statement

Derive equation 1 from Lecture Notes p.27-1

Solution

We have

$E=\sum _{r=1}^l{P}_{(2r)}(+1)[g^{(2r-1)}-g^{(2r-1)}(-1)]-\underset{-1}{\overset{+1}{\int }}{P}_{(2l)}(t)g^{(2l)}(t)dt]$

and $g_k^i(t)=(\frac{h}{2}{)}^i{f}^i(x(t))$

$g^{(2r-1)}(t)=(\frac{h}{2}{)}^{(2r-1)}{f}^{(2r-1)}(x(t))$

$g^{(2r-1)}(+1)=(\frac{h}{2}{)}^{(2r-1)}{f}^{(2r-1)}(x(1))$

and $x(1)=b$

$g^{(2r-1)}(+1)=(\frac{h}{2}{)}^{(2r-1)}{f}^{(2r-1)}(b)$

Similarly

$g^{(2r-1)}(-1)=(\frac{h}{2}{)}^{(2r-1)}{f}^{(2r-1)}(a)$

and $g^{(2l)}=(\frac{h}{2}{)}^{(2l)}{f}^{(2l)}(x)$

using above equations into first equation we obtain

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

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

but $\frac{P_{(2r)}(+1)}{2^{2r}}={\bar{d}}_{2r}$

So finally we obtain

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

(8) Obtain expressions for (P₂,P₃),(P₄,P₅),(P₆,P₇) Using recurrence formula

Ref: Lecture Notes p.29-3

Problem Statement

Obtain expressions for $(P_2,P_3),(P_4,P_5),(P_6,P_7)$ using eq (6) Lecture Notes p.29-2

Solution

$(P_2,P_3)$

$P_{2i}(t)=\sum _{j=0}^i{C}_{2j+1}\frac{t^{2(i-j)}}{(2(i-j))!}$

$P_2(t)=\sum _{j=0}^i{C}_{2j+1}\frac{t^{2(1-j)}}{(2(1-j))!}$

$P_2(t)=C_1\frac{t^{2)}}{2!}+C_3$

Using eq (6) from Lecture Notes p.29-3 , we obtain

$\frac{C_1}{3!}+C_3=0$

$C_3=-\frac{C_1}{3!}$

$=-\frac{-1}{3!}=\frac{1}{6}$

so $P_2(t)=-\frac{t^2}{2}+\frac{1}{6}$

$P_3=\sum _{j=0}^i{C}_{2j+1}\frac{t^{2(i-1)}}{(2(i-j))!}+C_{2i+2}$

where $C_{2i+2}=0$

$P_3=C_1\frac{t}{3!}+C_3\frac{t}{1!}$

where $C_1=-1and{C}_3=\frac{1}{6}$ from previous result

$P_3=-\frac{t}{3!}+\frac{t}{6}=0$

$P_4,P_5$

$P_4=C_1\frac{t^4}{4!}+C_3\frac{t^2}{2!}+C_5$

Using eq (6) from Lecture Notes p.29-3 , we obtain

$\frac{C_1}{5!}+\frac{C_3}{3!}+\frac{C_5}{1!}=0$

$C_5=\frac{1}{5!}-\frac{1}{36}=\frac{-7}{360}$

$P_4=-\frac{t^4}{24}+\frac{t^2}{12}-\frac{7}{360}$

$P_5=C_1\frac{t^3}{5!}+C_3\frac{t^3}{3!}+C_5\frac{t^3}{1!}$

$P_5=(\frac{C_1}{5!}+\frac{C_3}{3!}+\frac{C_5}{1!})t^3$

$P_5=0$

$P_6,P_7$

$P_6=C_1\frac{t^6}{12!}+C_3\frac{t^4}{10!}+C_5\frac{t^2}{8!}+C_7\frac{1}{6!}$

By using eq 6 from Lecture Notes p.29-3 , we obtain

$\frac{C_1}{7!}+\frac{C_3}{5!}+\frac{C_5}{3!}+C_7=0$

$C_7=\frac{1}{7!}-\frac{1}{6!}-\frac{7}{2160}$

$=\frac{11}{15120}$

$P_6=-\frac{t^6}{12!}+\frac{t^4}{6*10!}-\frac{7t^2}{360*8!}+\frac{11}{15120}$

$P_7=C_1\frac{t^5}{7!}+C_3\frac{t^5}{5!}+C_5\frac{t^5}{3!}+C_7\frac{t^5}{1!}$

$P_7=(\frac{C_1}{7!}+\frac{C_3}{5!}+\frac{C_5}{3!}+\frac{C_7}{1!})t^5$

$P_7=0*t^5$

$P_7=0$