User:EGM6341.s11.TEAM1.Yoon.HW6

2
We know,

At $$\displaystyle t = \pm 1$$,

We also know,

And,

At $$\displaystyle t = \pm 1$$,

Functions $$\displaystyle p_{3}(t), p_{5}(t), p_{7}(t), \cdots, $$ are odd functions.

Since functions $$\displaystyle p_{2}(t), p_{4}(t), p_{6}(t), \cdots, $$ are made zero at $$\displaystyle t = \pm 1$$, Equation (3) becomes,

Since $$\displaystyle p_{(2r-1)}(t) $$ are odd functions, we know $$\displaystyle p_{(2r-1)}(1) + p_{(2r-1)}(-1) = 0 $$.

And so $$\displaystyle E $$ becomes,

Using the below relationship

In Equation (4) $$\displaystyle g_k(t)$$can be transferred to $$\displaystyle f(x_k)$$ as shown below:

And so, Equation (4) becomes,

Substituting Equation (5) in Equation (1),

In the above error equation $$\displaystyle \beta \neq f^{(2r-2)}(x_n) - f^{(2r-2)}(x_0)$$, but it contains terms that involve all the points from $$\displaystyle x_0 \ to\ x_n $$.

But instead of canceling out terms that involve odd order derivatives of $$\displaystyle g(t)$$, if we had canceled only terms that involve even order derivatives of $$\displaystyle g(t)$$ then the term $$\displaystyle \beta $$ would involve only first and last points, i.e., $$\displaystyle x_0 = a $$ and $$\displaystyle x_n = b $$, as we have got in Problem 5 above.

1
Part.a

Above result is exactly same with $$\displaystyle \bar{d_{2r}} = \frac{P_{2r}(1)}{2^{2r}}$$

Result table

Part.b Since $$\displaystyle \bar{d_{2r}} = \frac{P_{2r}(1)}{2^{2r}}$$,

Result table

Part.c

Then, based on the relation between $$\displaystyle \bar{d_{2r}} $$ and Bernoulli numbers, we can obtain $$\displaystyle \bar{d_{2}}, \,\bar{d_{4}}, \,\bar{d_{6}}$$ as follows.

Since $$\displaystyle B_{2}=\frac{1}{6}, \, B_{4}=-\frac{1}{30}, \, B_{6}=\frac{1}{42}, B_{8}=-\frac{1}{30}, \, B_{10}=\frac{5}{66}$$,

Result table

--EGM6341.s11.TEAM1.Yoon 17:01, 3 April 2011 (UTC)

Given
The results calculated from the $$ xcoth \,$$ method in Problem 6 are: $$\overline{d_{2}}=\frac{-1}{12}, \overline{d_{4}}=\frac{1}{720}, \overline{d_{6}}=\frac{-1}{30240}, \overline{d_{8}}=\frac{1}{1209600}, \overline{d_{10}}=\frac{-1}{47900160} \,$$ Using the $$ \frac{P_{2r}(1)}{2^{2r}}\,$$ method, remember that: $$\overline{d_{2}}=\frac{P_2^{(1)}}{2^2}\,$$ $$\overline{d_{4}}=\frac{P_4^{(1)}}{2^4}\,$$ $$\overline{d_{6}}=\frac{P_6^{(1)}}{2^6}\,$$ $$\overline{d_{8}}=\frac{P_8^{(1)}}{2^8}\,$$ $$\overline{d_{10}}=\frac{P_{10}^{(1)}}{2^{10}}\,$$

Solution
$$ P_2=C_1\frac{t^2}{2!}+C_3\,$$ $$ P_4=C_1\frac{t^4}{4!}+C_3\frac{t^2}{2!}+C_5\,$$ $$ P_6=C_1\frac{t^6}{6!}+C_3\frac{t^4}{4!}+C_5\frac{t^2}{2!}+C_7\,$$ $$ P_8=C_1\frac{t^8}{8!}+C_3\frac{t^6}{6!}+C_5\frac{t^4}{4!}+C_7\frac{t^2}{2!}+C_9\,$$ $$ P_{10}=C_1\frac{t^{10}}{10!}+C_3\frac{t^8}{8!}+C_5\frac{t^6}{6!}+C_7\frac{t^4}{4!}+C_9\frac{t^2}{2!}+C_{11}\,$$ where $$ C_1=-1, C_3=\frac{1}{6}, C_5=\frac{-7}{360}, C_7=\frac{31}{15120}, C_9=\frac{1}{4725}, C_{11}=\frac{73}{3421440} \,$$
 * Note: Coefficients $$ C_7, C_9, C_{11} \,$$ can be found using the same method as in Problem 1.

$$ P_2(1)=(-1)(\frac{1}{2})+\frac{1}{6}=\frac{-1}{3}\Rightarrow\overline{d_{2}}=(\frac{-1}{3})(\frac{1}{4})=\frac{-1}{12}\,$$ $$ P_4(1)=(-1)(\frac{1}{24})+(\frac{1}{6})(\frac{1}{2})-\frac{7}{360}=\frac{1}{45}\Rightarrow\overline{d_{4}}=(\frac{1}{45})(\frac{1}{16})=\frac{1}{720}\,$$ $$ P_6(1)=(-1)(\frac{1}{720})+(\frac{1}{6})(\frac{1}{24})+(\frac{-7}{360})(\frac{1}{2})+\frac{31}{15120}=\frac{-2}{945}\Rightarrow\overline{d_{6}}=(\frac{-2}{945})(\frac{1}{64})=\frac{-1}{30240}\,$$ $$ P_8(1)=(-1)(\frac{1}{40320})+(\frac{1}{6})(\frac{1}{720})+(\frac{-7}{360})(\frac{1}{24})+(\frac{31}{15120})(\frac{1}{2})-\frac{127}{604800}=\frac{1}{4725}\Rightarrow\overline{d_{8}}=(\frac{1}{4725})(\frac{1}{256})=\frac{1}{1209600}\,$$ $$ P_{10}(1)=(-1)(\frac{1}{3628800})+(\frac{1}{6})(\frac{1}{40320})+(\frac{-7}{360})(\frac{1}{720})+(\frac{31}{15120})(\frac{1}{24})-(\frac{127}{604800})(\frac{1}{2})+\frac{73}{3421440}=\frac{-2}{93555}\Rightarrow\overline{d_{8}}=(\frac{-2}{93555})(\frac{1}{1024})=\frac{-1}{47900160}\,$$

Solved by: -Egm6341.s10.team1.andrewdugan 08:17, 24 March 2010 (UTC)