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

(5) Derive Trapezoidal Error expression
Ref: Lecture Notes [[media:Egm6341.s10.mtg28.djvu|p.28-1]]

Problem Statement
Derive equation 1 from Lecture Notes [[media:Egm6341.s10.mtg27.djvu|p.27-1]]

Solution
We have

$$ E = \sum_{r=1}^l P_{(2r)}(+1) [g^{(2r-1)}- g^{(2r-1)}(-1)] - \int\limits_{-1}^{+1} P_{(2l)}(t) g^{(2l)}(t) dt]$$

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

$$ E = \sum_{r=1}^l P_{(2r)}(+1) \frac{h}{2} \frac{h}{2})^{(2r-1)} [f^{(2r-1)}(b)- f^{(2r-1)}(a)] - \sum_{k=1}^{n-1} \int\limits_{-1}^{+1} P_{(2l)}(t_k) (\frac{h}{2})^{2l} f^{(2l)}(x) dx]$$

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

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

So finally we obtain

$$ = \sum_{r=1}^l h^{2r} \overline{d}_{2r} [f^{(2r-1)}(b)- f^{(2r-1)}(a)] - (\frac{h}{2})^{2l} \sum_{k=1}^{n-1} \int\limits_{x_k}^{x_{k+1}} P_{(2l)}(t_k(x))  f^{(2l)}(x) dx]$$

(8) Obtain expressions for (P2,P3),(P4,P5),(P6,P7) Using recurrence formula
Ref: Lecture Notes [[media:Egm6341.s10.mtg29.djvu|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 [[media:Egm6341.s10.mtg29.djvu|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 [[media:Egm6341.s10.mtg29.djvu|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 $$ C_1 = -1 and C_3 = \frac{1}{6} $$ 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 [[media:Egm6341.s10.mtg29.djvu|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 [[media:Egm6341.s10.mtg29.djvu|p.29-3]], we obtain

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

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

$$   = \frac{11} {15120} $$

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

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

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

$$ P_7 = 0 * t^5 $$

$$ P_7 = 0 $$