User:Egm6341.s2010.Team1.lei/HW1

For HW 1, I am responsible for 3 problems: 5-1(2),6-1(1)and 8-3(1).

The details of solutions of each problem is following:

=Problem 5-1b=

Find
Use the Intermediate Mean Value Theorem to show Eq. (5) [[media:Egm6341.s10.mtg2.pdf|Lec. 2-2]] and Eq. (1) [[media:Egm6341.s10.mtg2.pdf|Lec. 2-3]]

Given
Eq. (5)
 * $$R_{n+1}\left ( x \right )=\frac{1}{n!}\int_{x_{0}}^{x}\left ( x-t \right )^{n}f^{(n+1)}\left ( t \right )dt$$

Eq. (1)
 * $$R_{n+1}\left ( x \right )=\frac{(x-x_{0})^{n+1}}{(n+1)!}f^{(n+1)}\left ( \xi \right )$$

IMVT
 * $$\int_{a}^{b}w\left ( x \right )f\left ( x \right )dx=f\left ( \xi \right )\int_{a}^{b}w\left ( x \right )dx$$

Solution [[media:HW1_P5_1_2.jpg|P5-1b]]
Let $$w\left ( x \right )$$ be nonnegative in the integral on $$[a,b] \,$$ and let $$f\left ( x \right )$$ be continuous on $$[a,b] \,$$. Then,


 * $$\int_{a}^{b}w\left ( x \right )f\left ( x \right )dx=f\left ( \xi \right )\int_{a}^{b}w\left ( x \right )dx$$

for
 * $$\xi \in [a,b]$$

Using a Taylor series expansion,
 * $$f\left ( x \right )=P_{n}\left ( x \right )+R_{n+1}\left ( x \right )$$

where
 * $$R_{n+1}\left ( x \right )=\frac{1}{n!}\int_{x_{0}}^{x}\left ( x-1 \right )^{n}f^{(n+1)}\left ( t \right )dt$$
 * $$\Rightarrow \frac{1}{n!}f^{(n+1)}\left ( \xi \right )\int_{x_{0}}^{x}\left ( x-t \right )^{n}dt$$
 * $$\Rightarrow \frac{1}{n!}f^{(n+1)}\left ( \xi \right )\left ( \frac{-1}{n+1} \right )\left ( x-t \right )^{n+1}|_{x_{0}}^{x}$$
 * {| style="width:100%" border="0" align="left"


 * style="width:50%; padding:10px; border:2px solid #8888aa" |
 * $$ \displaystyle
 * $$ \displaystyle

\Rightarrow \frac{(x-x_{0})^{n+1}}{(n+1)!}f^{(n+1)}\left ( \xi \right ) $$ $$
 * $$\displaystyle
 * }
 * }

=Problem 6-1=

Find
Construct the Taylor series of $$f(x) \,$$ around $$x_0=\frac{\pi}{4}$$ for $$n=0,1,...,10 \,$$.

Plot these series for each $$n \,$$. Then estimate the maximum of $$R(x) \,$$ at $$x=\frac{\pi}{2}$$

Given

 * $$f\left ( x \right )=\sin x$$ for $$x\in [0,\pi ]$$

Solution [[media:HW1_P6_1_1.jpg|P6-1]]

 * $$f\left ( x \right )=\sin x=P_n(x)+R_{n+1}(x)$$
 * $$P_n(x)=f(x_0)+\frac{(x-x_0)}{1!}f^{(1)}(x_0)+\frac{(x-x_0)^2}{2!}f^{(2)}(x_0)+...+\frac{(x-x_0)^n}{n!}f^{(n)}(x_0)$$
 * $$R_{n+1}(x)=\frac{(x-x_0)^{n+1}}{(n+1)!}f^{(n_1)}(\xi)$$

For
 * $$f(x)=\sin x, f^{(1)}(x)=\cos x, f^{(2)}(x)=-\sin x, f^{(3)}(x)=-\cos x, f^{(4)}(x)=\sin x \,$$

When $$n=0 \,$$
 * $$\Rightarrow \sin x=P_0(x)+R_1(x)=\frac{\sqrt{2}}{2}+(x-\frac{\pi}{4})\cos \xi$$

When $$n=1 \,$$
 * $$\Rightarrow \sin x=P_1(x)+R_2(x)=\frac{\sqrt{2}}{2}\left [ 1+\left ( x-\frac{\pi}{4} \right ) \right ]-\frac{(x-\frac{\pi}{4})^2}{2!}\sin \xi$$

When $$n=2 \,$$
 * $$\Rightarrow \sin x=P_2(x)+R_3(x)=\frac{\sqrt{2}}{2}\left [ 1+\left ( x-\frac{\pi}{4} \right )-\frac{(x-\frac{\pi}{4})^2}{2!} \right ]-\frac{(x-\frac{\pi}{4})^3}{3!}\cos \xi$$

When $$n=3 \,$$
 * $$\Rightarrow \sin x=P_3(x)+R_4(x)=\frac{\sqrt{2}}{2}\left [ 1+\left ( x-\frac{\pi}{4} \right )-\frac{(x-\frac{\pi}{4})^2}{2!}-\frac{(x-\frac{\pi}{4})^3}{3!} \right ]+\frac{(x-\frac{\pi}{4})^4}{4!}\sin \xi$$

...

When $$n=10 \,$$
 * $$\Rightarrow \sin x=P_{10}(x)+R_{11}(x)=\frac{\sqrt{2}}{2}\left [ 1+\left ( x-\frac{\pi}{4} \right )-\frac{(x-\frac{\pi}{4})^2}{2!}-\frac{(x-\frac{\pi}{4})^3}{3!}...- \frac{(x-\frac{\pi}{4})^{10}}{10!}\right ]-\frac{(x-\frac{\pi}{4})^{11}}{11!}\cos \xi$$

Finally, the maximum can be computed with,
 * $$R_{n+1}(x)=\frac{(x-x_0)^{n+1}}{(n+1)!}f^{(n+1)}(\xi)$$ for $$\xi \in [0,\pi] \,$$
 * $$\left | R_{n+1}(x) \right |\leq \frac{(x-x_0)^{n+1}}{(n+1)!}\max \left \{ f^{(n+1)}(\xi) \right \} \leq \frac{(x-x_0)^{n+1}}{(n+1)!}=\frac{(\frac{\pi}{4})^{n+1}}{(n+1)!}\leq \frac{\pi}{4}$$

=Problem 8-3=

Find
Prove $$P_2(x_j)=f(x_j)\,$$

Given
$$P_2(x_j)=\sum _{i=0}^{2}l_i(x_j)f(x_i)$$

Solution [[media:HW1_P8_3_1.jpg|P8-3]]

 * $$P_2(x_j)=\sum _{i=0}^{2}l_i(x_j)f(x_i)=l_0(x_j)f(x_0)+l_1(x_j)f(x_1)+l_2(x_j)f(x_2)$$

Also,
 * $$l_1(x_j)=\delta _{ij}=\left\{\begin{matrix}

1(i=j)\\ 0(i\neq j) \end{matrix}\right.$$ And
 * $$l_0(x_0)=1\,$$
 * $$l_0(x_1)=l_0(x_2)=0\,$$


 * $$l_1(x_1)=1\,$$
 * $$l_1(x_0)=l_1(x_2)=0\,$$


 * $$l_2(x_2)=1\,$$
 * $$l_2(x_0)=l_2(x_1)=0\,$$

Therefore,
 * $$P_2(x_0)=l_0(x_0)f(x_0)+l_1(x_0)f(x_1)+l_2(x_0)f(x_2)=f(x_0)\,$$
 * $$P_2(x_1)=l_0(x_1)f(x_0)+l_1(x_1)f(x_1)+l_2(x_1)f(x_2)=f(x_1)\,$$
 * $$P_2(x_2)=l_0(x_2)f(x_0)+l_1(x_2)f(x_1)+l_2(x_2)f(x_2)=f(x_2)\,$$

...


 * {| style="width:100%" border="0" align="left"


 * style="width:50%; padding:10px; border:2px solid #8888aa" |
 * $$ \displaystyle
 * $$ \displaystyle

\Rightarrow P_2(x_j)=\sum _{i=0}^{2}l_i(x_j)f(x_i)=f(x_j) $$ $$
 * $$\displaystyle
 * }
 * }