# 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

$E=\sum _{r=1}^{l}P_{(2r)}(+1)[g^{(2r-1)}-g^{(2r-1)}(-1)]-\int \limits _{-1}^{+1}P_{(2l)}(t)g^{(2l)}(t)dt]$ 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 $(P_{2},P_{3}),(P_{4},P_{5}),(P_{6},P_{7})$ 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 $C_{1}=-1andC_{3}={\frac {1}{6}}$ 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

${\frac {C_{1}}{7!}}+{\frac {C_{3}}{5!}}+{\frac {C_{5}}{3!}}+C_{7}=0$ $C_{7}={\frac {1}{7!}}-{\frac {1}{6!}}-{\frac {7}{2160}}$ $={\frac {11}{15120}}$ $P_{6}=-{\frac {t^{6}}{12!}}+{\frac {t^{4}}{6*10!}}-{\frac {7t^{2}}{360*8!}}+{\frac {11}{15120}}$ $P_{7}=C_{1}{\frac {t^{5}}{7!}}+C_{3}{\frac {t^{5}}{5!}}+C_{5}{\frac {t^{5}}{3!}}+C_{7}{\frac {t^{5}}{1!}}$ $P_{7}=({\frac {C_{1}}{7!}}+{\frac {C_{3}}{5!}}+{\frac {C_{5}}{3!}}+{\frac {C_{7}}{1!}})t^{5}$ $P_{7}=0*t^{5}$ $P_{7}=0$ 