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 (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 $C_{1}=-1andC_{3}={\frac {1}{6}}$ 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

${\frac {C_{1}}{7!}}+{\frac {C_{3}}{5!}}+{\frac {C_{5}}{3!}}+C_{7}=0$

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

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

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

$P_{7}=C_{1}{\frac {t^{5}}{7!}}+C_{3}{\frac {t^{5}}{5!}}+C_{5}{\frac {t^{5}}{3!}}+C_{7}{\frac {t^{5}}{1!}}$

$P_{7}=({\frac {C_{1}}{7!}}+{\frac {C_{3}}{5!}}+{\frac {C_{5}}{3!}}+{\frac {C_{7}}{1!}})t^{5}$

$P_{7}=0*t^{5}$

$P_{7}=0$