User:EGM6341.s10.Team1.Kumanchik/HW3

=Homework 3=

Find
Where does the proof of bounding the error for the simple Simpson's rule break down?

Given
1) $$G(t):= e(t)-t^4e(1) \,$$

2) $$G(t):= e(t)-t^6e(1) \,$$

3) $$G(t):= e(t)-t^5e(1) \,$$, where for this case continue differentiating to get $$G^{(3)}(0) \,$$

Solution
Start by performing derivatives of $$e(t) \,$$,


 * $$e^{(0)}(t)=\int_{-t}^{t}F(t)dt - \frac{t}{3}[F(-t)+4F(0)+F(t)]$$
 * $$e^{(1)}(t)=F(-t)+F(t)-\frac{1}{3}[F(-t)+4F(0)+F(t)]-\frac{t}{3}[-F^{(1)}(-t)+F^{(1)}(t)]$$
 * $$e^{(2)}(t)=\frac{1}{3}[-F^{(1)}(-t)+F^{(1)}(t)]-\frac{t}{3}[F^{(2)}(-t)+F^{(2)}(t)]$$
 * $$e^{(3)}(t)=-\frac{t}{3}[-F^{(3)}(-t)+F^{(3)}(t)]$$
 * $$e^{(4)}(t)=-\frac{1}{3}[-F^{(3)}(-t)+F^{(3)}(t)]-\frac{t}{3}[F^{(4)}(-t)+F^{(4)}(t)]$$

Note that $$e^{(n)}(0)=0 \,$$ for $$n=0,1,2,3,4 \,$$.

Case 1:


 * $$G^{(2)}(t)=e^{(2)}(t)-12t^2e(1) \,$$

and,
 * $$G^{(2)}(0)=0 \,$$

Plugging in $$\zeta_2 \,$$,
 * $$G^{(2)}(\zeta_2)=\frac{1}{3}\underbrace{[-F^{(1)}(-\zeta_2)+F^{(1)}(\zeta_2)]}_{2\zeta_2F^{(2)}(\zeta_3)}-\frac{\zeta_2}{2}[F^{(2)}(-\zeta_2)+F^{(2)}(\zeta_2)]-12\zeta_2^2e(1)=0$$

Solving for $$e(1) \,$$,


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


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

e(1)=\frac{1}{18\zeta_2}F^{(2)}(\zeta_3)-\frac{1}{36}F^{(2)}\underbrace{(-\zeta_2)}_{problem}-\frac{1}{36}F^{(2)}(\zeta_2) $$ This variable is not on the interval between 0 and $$\zeta_2 \,$$ $$
 * $$\displaystyle
 * }
 * }

Case 2:


 * $$G^{(4)}(t)=e^{(4)}(t)-360t^2e(1) \,$$

and,
 * $$G^{(4)}(0)=0 \,$$

Plugging in $$\zeta_4 \,$$
 * $$G^{(4)}(\zeta_4)=-\frac{1}{3}[-F^{(3)}(-\zeta_4)+F^{(3)}(\zeta_4)]-\frac{\zeta_4}{3}[F^{(4)}(-\zeta_4)+F^{(4)}(\zeta_4)]-360(\zeta_4^2)e(1)=0 \,$$

Solving for $$e(1) \,$$,


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


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

e(1)=\frac{-1}{540\zeta_4}F^{(4)}(\zeta_5)-\frac{1}{1080\zeta_4}[F^{(4)}(\underbrace{-\zeta_4)}_{problem}+F^{(4)}(\zeta_4)] $$ This variable is not on the interval between 0 and $$\zeta_4 \,$$ $$
 * $$\displaystyle
 * }
 * }

Case 3:


 * $$G^{(3)}(t)=e^{(3)}(t)-60t^2e(1) \,$$

and,
 * $$G^{(3)}(0)=0 \,$$

We continue taking derivatives of $$G \,$$
 * $$G^{(4)}(t)=e^{(4)}(t)-120te(1) \,$$

to get,
 * $$G^{(4)}(\zeta_4)=-\frac{1}{3}[-F^{(3)}(-\zeta_4)+F^{(3)}(\zeta_4)]-\frac{\zeta_4}{3}[F^{(4)}(-\zeta_4)+F^{(4)}(\zeta_4)]-120(\zeta_4^2)e(1)=0 \,$$

Solving for $$e(1) \,$$


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


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

e(1)=\frac{-3}{360}F^{(4)}(\zeta_4)-\frac{1}{360}F^{(4)}\underbrace{(-\zeta_4)}_{problem} $$ This variable is not on the interval between 0 and $$\zeta_4 \,$$ $$
 * $$\displaystyle
 * }
 * }