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

Problem Statement

Derive equation 1 from Lecture Notes p.27-1

Solution

We have

${\displaystyle E=\sum _{r=1}^{l}P_{(2r)}(+1)[g^{(2r-1)}-g^{(2r-1)}(-1)]-\int \limits _{-1}^{+1}P_{(2l)}(t)g^{(2l)}(t)dt]}$

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

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

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

and ${\displaystyle x(1)=b}$

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

Similarly

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

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

using above equations into first equation we obtain

${\displaystyle E=\sum _{r=1}^{l}P_{(2r)}(+1){\frac {h}{2}}{\frac {h}{2}})^{(2r-1)}[f^{(2r-1)}(b)-f^{(2r-1)}(a)]-\sum _{k=1}^{n-1}\int \limits _{-1}^{+1}P_{(2l)}(t_{k})({\frac {h}{2}})^{2l}f^{(2l)}(x)dx]}$

${\displaystyle =\sum _{r=1}^{l}h^{2r}{\frac {P_{(2r)}(+1)}{2^{2r}}}[f^{(2r-1)}(b)-f^{(2r-1)}(a)]-({\frac {h}{2}})^{2l}\sum _{k=1}^{n-1}\int \limits _{x_{k}}^{x_{k+1}}P_{(2l)}(t_{k}(x))f^{(2l)}(x)dx]}$

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

So finally we obtain

${\displaystyle =\sum _{r=1}^{l}h^{2r}{\overline {d}}_{2r}[f^{(2r-1)}(b)-f^{(2r-1)}(a)]-({\frac {h}{2}})^{2l}\sum _{k=1}^{n-1}\int \limits _{x_{k}}^{x_{k+1}}P_{(2l)}(t_{k}(x))f^{(2l)}(x)dx]}$

Problem Statement

Obtain expressions for ${\displaystyle (P_{2},P_{3}),(P_{4},P_{5}),(P_{6},P_{7})}$ using eq (6) Lecture Notes p.29-2

Solution

${\displaystyle (P_{2},P_{3})}$

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

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

${\displaystyle P_{2}(t)=C_{1}{\frac {t^{2)}}{2!}}+C_{3}}$

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

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

${\displaystyle C_{3}=-{\frac {C_{1}}{3!}}}$

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

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

${\displaystyle P_{3}=\sum _{j=0}^{i}C_{2j+1}{\frac {t^{2(i-1)}}{(2(i-j))!}}+C_{2i+2}}$

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

${\displaystyle P_{3}=C_{1}{\frac {t}{3!}}+C_{3}{\frac {t}{1!}}}$

where ${\displaystyle C_{1}=-1andC_{3}={\frac {1}{6}}}$ from previous result

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

${\displaystyle P_{4},P_{5}}$

${\displaystyle P_{4}=C_{1}{\frac {t^{4}}{4!}}+C_{3}{\frac {t^{2}}{2!}}+C_{5}}$

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

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

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

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

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

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

${\displaystyle P_{5}=0}$

${\displaystyle P_{6},P_{7}}$

${\displaystyle P_{6}=C_{1}{\frac {t^{6}}{12!}}+C_{3}{\frac {t^{4}}{10!}}+C_{5}{\frac {t^{2}}{8!}}+C_{7}{\frac {1}{6!}}}$

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

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

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

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

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

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

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

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

${\displaystyle P_{7}=0}$