User:Egm6341.s2010.Team1.lei/HW2

For HW 2, I am responsible for 3 problems: 9-1(a),12-3(a)and 13-2(a).

=Problem 9-1(a)=

Find
Use Eq. (3) [[media:Egm6341.s10.mtg8.pdf|Lec. 8-3]] and Eq. (1) [[media:Egm6341.s10.mtg8.pdf|Lec. 8-3]] to find expression for $$C_0,C_1,C_2 \,$$ in terms of $$(x_i,f(x_i)), i=0,1,2 \,$$

Given
Eq. (3) [[media:Egm6341.s10.mtg8.pdf|Lec. 8-3]]
 * $$P_2(x)=\sum _{i=0}^{2}l_i(x)f(x_i)$$

Eq. (1) [[media:Egm6341.s10.mtg8.pdf|Lec. 8-3]]
 * $$f_{2}\left ( x \right )=p_{2}\left ( x \right )=C_2 x^2+C_1 x^1+C_0, C_0,C_1,C_2= 3 unknowns \,$$

Solution
From Eq. (2) [[media:Egm6341.s10.mtg7.pdf|Lec. 7-3]]
 * $$l_{i,n}\left ( x \right )=l_{i}\left ( x \right )=\prod_{j=0,j\neq i}^n\frac{x-x_j}{x_i-x_j}$$

Then
 * $$l_{0}\left ( x \right )=\frac{(x-x_1)(x-x_2)}{(x_0-x_1)(x_0-x_2)}$$
 * $$l_{1}\left ( x \right )=\frac{(x-x_0)(x-x_2)}{(x_1-x_0)(x_1-x_2)}$$
 * $$l_{2}\left ( x \right )=\frac{(x-x_0)(x-x_1)}{(x_2-x_0)(x_2-x_1)}$$

Also From Eq. (3) [[media:Egm6341.s10.mtg8.pdf|Lec. 8-3]]
 * $$P_2(x)=\sum _{i=0}^{2}l_i(x)f(x_i)$$


 * $$\Rightarrow P_2(x)=\sum _{i=0}^{2}l_i(x)f(x_i)=l_{0}\left ( x \right )f(x_0)+l_{1}\left ( x \right )f(x_1)+l_{2}\left ( x \right )f(x_2)=C_2 x^2+C_1 x^1+C_0 $$


 * $$\Rightarrow C_2=\frac{f(x_0)}{(x_0-x_1)(x_0-x_2)}+\frac{f(x_1)}{(x_1-x_0)(x_1-x_2)}+\frac{f(x_2)}{(x_2-x_0)(x_2-x_1)}$$
 * $$\Rightarrow C_1=\frac{-f(x_0)(x_1+x_2)}{(x_0-x_1)(x_0-x_2)}+\frac{-f(x_1)(x_0+x_2)}{(x_1-x_0)(x_1-x_2)}+\frac{-f(x_2)(x_0+x_1)}{(x_2-x_0)(x_2-x_1)}$$
 * $$\Rightarrow C_0=\frac{f(x_0)(x_1x_2)}{(x_0-x_1)(x_0-x_2)}+\frac{f(x_1)(x_0x_2)}{(x_1-x_0)(x_1-x_2)}+\frac{f(x_2)(x_0x_1)}{(x_2-x_0)(x_2-x_1)}$$


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


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

C_2=\frac{f(x_0)}{(x_0-x_1)(x_0-x_2)}+\frac{f(x_1)}{(x_1-x_0)(x_1-x_2)}+\frac{f(x_2)}{(x_2-x_0)(x_2-x_1)}$$
 * $$ \displaystyle

C_1=\frac{-f(x_0)(x_1+x_2)}{(x_0-x_1)(x_0-x_2)}+\frac{-f(x_1)(x_0+x_2)}{(x_1-x_0)(x_1-x_2)}+\frac{-f(x_2)(x_0+x_1)}{(x_2-x_0)(x_2-x_1)}$$
 * $$ \displaystyle

C_0=\frac{f(x_0)(x_1x_2)}{(x_0-x_1)(x_0-x_2)}+\frac{f(x_1)(x_0x_2)}{(x_1-x_0)(x_1-x_2)}+\frac{f(x_2)(x_0x_1)}{(x_2-x_0)(x_2-x_1)}$$

$$
 * $$\displaystyle
 * }
 * }

=Problem 12-3(a)=

Find
Prove $$P_{n+1}^{(n+1)}\left ( x \right )={(n+1)}!$$

Given

 * $$q_{n+1}\left ( x \right ):=(x-x_{0})(x-x_{1})(x-x_{2})...(x-x_{n})$$

Solution

 * $$n=0 \,$$
 * $$q_{1}\left ( x \right ):=(x-x_{0})$$


 * $$q_{1}^{(1)}\left ( x \right )=1$$


 * $$n=1 \,$$
 * $$q_{2}\left ( x \right ):=(x-x_{0})(x-x_{1})$$


 * $$q_{2}^{(2)}\left ( x \right )=2\times 1=2!$$


 * $$n=2 \,$$
 * $$q_{3}\left ( x \right ):=(x-x_{0})(x-x_{1})(x-x_{2})$$


 * $$q_{3}^{(3)}\left ( x \right )=3\times 2\times 1=3!$$


 * $$n=3 \,$$
 * $$q_{4}\left ( x \right ):=(x-x_{0})(x-x_{1})(x-x_{2})(x-x_{3})$$


 * $$q_{4}^{(4)}\left ( x \right )=4\times 3\times 2\times 1=4!$$

...


 * $$n=n \,$$
 * $$q_{n+1}\left ( x \right ):=(x-x_{0})(x-x_{1})(x-x_{2})(x-x_{3}...(x-x_{n})$$


 * $$q_{n+1}^{(n+1)}\left ( x \right )=n\times (n-1)\times (n-2)...\times 1=(n+1)!$$


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


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

\Rightarrow P_{n+1}^{(n+1)}\left ( x \right )={(n+1)}!$$ $$
 * $$\displaystyle
 * }
 * }

=Problem 13-2(a)=

Find

 * $$\left | E_{1} \right |\leq \frac{(b-a)^{3}}{12}M_2=\frac{h^{3}}{12}M_2)$$

Given

 * $$n=1, \, q_{2}\left ( x \right )=(x-a)(x-b)$$
 * $$h:=b-a\,$$

Solution
Let:$$M_2:= max f^{(2)} (\xi), \xi \in [a,b]$$
 * $$\left | E_{1} \right |\leq \frac{M_2}{2!}\int_{a}^{b}\left |q_{2}\left ( x \right )\right |dx $$
 * $$=\frac{M_2}{2!}\int_{a}^{b}\underbrace{\left ( x-a \right )}_{\geq 0}\underbrace{\left ( b-x \right )}_{\geq 0}dx $$
 * $$=\frac{M_2}{2!}\int_{a}^{b}[(b+a)x-ab-x^2] dx$$
 * $$=\frac{M_2}{2!}[\frac{1}{2}(b+a)x^2-ab x-\frac{1}{3}x^3]|_{a}^{b}$$
 * $$=\frac{M_2}{2!}\times \frac{1}{6}(b^3-3ab^2+3a^2b-a^3) $$
 * $$=\frac{M_2}{2!}\times \frac{1}{6}(b^2-2ba+a^2)(b-a) $$
 * $$=\frac{M_2}{2!}\times \frac{1}{6}(b-a)^2(b-a) $$
 * $$=\frac{M_2}{2!}\times \frac{1}{6}(b-a)^3 $$
 * $$=\frac{M_2}{12}(b-a)^3 $$

and
 * $$h:=b-a\,$$
 * $$=\frac{M_2}{12}h^3 $$
 * {| style="width:100%" border="0" align="left"


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

\Rightarrow \left | E_{1} \right |\leq \frac{(b-a)^{3}}{12}M_2=\frac{h^{3}}{12}M_2$$ $$
 * $$\displaystyle
 * }
 * }