User:Egm6341.s11.team5.oh/HW4

= Problem 4.1 - Simpson's rule integrates the 3rd order polynomial exactly (I and I_2) = From (meeting 18 page 2)

Given

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

f(x) = P_3(x) = 1 + 3x -9x^2 +12x^3 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (1-1)
 * }
 * }

Find
 Find, I (exact) and I_2 (Simple simpson) 

Solution
 We solved on our own 


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

\begin{align} I &= \int_{-2}^{1} f(x) \ dx \\ &= \left[ x + \frac{3}{2} x^2 -3x^3 +3x^4\right]_{-2}^{1} \\ &= -73.5 \end{align} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (1-2)
 * }
 * }


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


 * }
 * }
 * }


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

\begin{align} I_2 &= \frac{h}{3} \left[ f(x_0) +4f(x_1) + f(x_2) \right] \\ &= \frac{1}{2} \left[ f(-2) +4f(- \frac{1}{2}) + f(1) \right] \\ &= \frac{1}{2} \left[ -137 -17 +7 \right] \\ &= -73.5 \end{align} $$
 * $$\displaystyle
 * $$\displaystyle

Where, $$\displaystyle \ \left( h = \frac{1-(-2)}{2} = \frac{3}{2}, \qquad x_0 = -2, \ x_1 = - \frac{1}{2}, \ x_2 = 1 \right)$$

$$
 * $$\displaystyle (1-3)
 * }
 * }

Author and proof-reader
[Author] Oh

[Proof-reader]

=Problem 4-2: Evaluation of $$ \alpha^{(1)}(t)$$=

Given
$$ \displaystyle\ \alpha(t)=\int_{-t}^{t} f(x(t'))dt'$$ where $$ \displaystyle\ f(x(t))=F(t)$$.

Find
Show that $$ \displaystyle\ \alpha^{(1)}(t)=F(-t)+F(t) $$.

Solution
We are asked to show that for $$ \displaystyle\ \alpha(t)=\int_{-t}^{t} f(x(t'))dt'=\int_{-t}^{t} F(t')dt',\alpha^{(1)}(t)=F(-t)+F(t)$$, where $$ \displaystyle\ f(x(t))=F(t)$$.

Using The Fundamental Theorem of Calculus (Second Part), we know that for some differentiable function $$ \displaystyle\ G(t)$$ with with derivative $$ \displaystyle\ g(t)=G'(t)$$, $$ \displaystyle\ \int_a^b g(t)dt=G(b)-G(a)$$.

Allowing $$ \displaystyle\ f(x(t))=F(t)$$ and $$ \displaystyle\ G'(t)=F(t)$$, then $$ \displaystyle\ \alpha(t)=\int_{-t}^{t} F(t')dt'=G(t)-G(-t)$$.

If the derivative of $$ \displaystyle \alpha $$ is now taken, the $$ \displaystyle\ \alpha^{(1)}(t)=\frac{d}{dt}(G(t)-G(-t))=G'(t)+G'(-t)$$.

Substituting in our definitions $$ \displaystyle\ F(t)=G'(t)$$ we see that

$$ \displaystyle\ \alpha^{(1)}(t)=F(-t)+F(t) $$.

Author
Solved and Typed by - Egm6341.s10.Team4.riherd 14:55, 18 February 2010 (UTC) .

= Problem 4-3: Evaluation of $$ G^{2}(0) $$ =

Given
Refer Lecture slide 15-2 for problem statement

From Lecture slide 15-1,


 * {| style="width:90%" border="0" align="center"



G^{1}(t) = e^{1}(t) - 5t^4e(1) $$ $$
 * $$\displaystyle
 * $$\displaystyle \longrightarrow (1)


 * }
 * }

where,


 * {| style="width:90%" border="0" align="center"



e(t) := \alpha(t) - \alpha_{2}(t) $$ $$
 * $$\displaystyle
 * $$\displaystyle \longrightarrow (2)


 * }
 * }

where $$\displaystyle \alpha(t), \alpha_{2}(t) $$ are given by,


 * {| style="width:90%" border="0" align="center"



\alpha(t) := \int_{-t}^{t}F(t)dt $$ $$
 * $$\displaystyle
 * $$\displaystyle \longrightarrow (3)


 * }
 * }


 * {| style="width:90%" border="0" align="center"



\alpha^{2}(t) := \frac{t}{3} \left[F(-t) + 4F(0)+ F(t)\right] $$ $$
 * $$\displaystyle
 * $$\displaystyle \longrightarrow (4)


 * }
 * }

and,


 * {| style="width:90%" border="0" align="center"



F(t) := f(x(t)) $$ $$
 * $$\displaystyle
 * $$\displaystyle \longrightarrow (5)


 * }.
 * }.

Find
Show that,


 * {| style="width:90%" border="0" align="center"



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


 * }.
 * }.

Solution
On differentiating Equation (3) [Please refer differentiating an integral and Problem 14 Solution below],


 * {| style="width:90%" border="0" align="center"



\alpha^{1}(t) = F(-t) + F(t) $$ $$
 * $$\displaystyle
 * $$\displaystyle \longrightarrow (6)

.
 * }
 * }

Again differentiating Equation (6),


 * {| style="width:90%" border="0" align="center"



\alpha^{2}(t) = -F(-t) + F(t) $$ $$
 * $$\displaystyle
 * $$\displaystyle \longrightarrow (7)

.
 * }
 * }

On differentiating Equation (4),
 * {| style="width:90%" border="0" align="center"



\begin{align} \Rightarrow \alpha_{2}^{1}(t) &= \frac{1}{3} \left[ F(-t) + 4F(0) + F(t) \right] + \frac{t}{3} \left[-F^{1}(-t) + \cancelto{0}{4F^{1}(0)} + F^{1}(t)\right] \\ &= \frac{1}{3} \left[ F(-t) + 4F(0) + F(t) \right] + \frac{t}{3} \left[-F^{1}(-t) + F^{1}(t)\right] \end{align} $$
 * $$\displaystyle


 * }.
 * }.

Again on differentiating the above equation,


 * {| style="width:90%" border="0" align="center"



\begin{align} \Rightarrow \alpha_{2}^{2}(t) &= \frac{1}{3} \left[ -F^{1}(-t) + \cancelto{0}{4F^{1}(0)} + F^{1}(t) \right] + \frac{1}{3} \left[-F^{1}(-t) + F^{1}(t)\right] + \frac{t}{3} \left[F^{2}(-t) + F^{2}(t)\right]\\ &= \frac{2}{3} \left[ -F^{1}(-t) + F^{1}(t) \right] + \frac{t}{3} \left[F^{2}(-t) + F^{2}(t)\right] \end{align} $$
 * $$\displaystyle


 * }.
 * }.

So,


 * {| style="width:90%" border="0" align="center"




 * $$\displaystyle

\Rightarrow \alpha_{2}^{2}(t) = -\frac{2}{3} F^{1}(-t) + \frac{2}{3} F^{1}(t) + \frac{t}{3} \left[F^{2}(-t) + F^{2}(t)\right]

$$ $$
 * $$\displaystyle \longrightarrow (8)


 * }.
 * }.

On differentiating Equation (1) we get,


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G^{2}(t) = e^{2}(t) - 20t^3e(1) $$ $$
 * $$\displaystyle
 * $$\displaystyle \longrightarrow (9)


 * }.
 * }.

At point $$\displaystyle t = 0 $$,


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G^{2}(0) = e^{2}(0) - \cancelto{0}{20(t)^3e(1)} \Rightarrow G^{2}(0) = e^{2}(0) $$ $$
 * $$\displaystyle
 * $$\displaystyle \longrightarrow (10)


 * }.
 * }.

$$\displaystyle e^{2}(t) $$ is given by differentiating Equation (2) twice,


 * {| style="width:90%" border="0" align="center"



e^{2}(t) = \alpha^{2}(t) - \alpha_{2}^{2}(t) $$ $$
 * $$\displaystyle
 * $$\displaystyle \longrightarrow (11)


 * }.
 * }.

Substituting Equations (7) and (8) in (11),


 * {| style="width:90%" border="0" align="center"



\Rightarrow e^{2}(t) = -F(-t) + F(t) +\frac{2}{3} F^{1}(-t) - \frac{2}{3} F^{1}(t) - \frac{t}{3} \left[F^{2}(-t) + F^{2}(t)\right] $$
 * $$\displaystyle


 * }.
 * }.

At point $$\displaystyle t = 0 $$,


 * {| style="width:90%" border="0" align="center"



\begin{align} \Rightarrow e^{2}(0) &= -F(-0) + F(0) +\frac{2}{3} F^{1}(-0) - \frac{2}{3} F^{1}(0) - \cancelto{0}{\frac{t}{3} \left[F^{2}(-t) + F^{2}(t)\right]} \\ &= -\frac{1}{3} F^{1}(-0) + \frac{1}{3} F^{1}(0) \\ &= \frac{1}{3} \left[ -F^{1}(0) + F^{1}(0) \right] \end{align} $$
 * $$\displaystyle


 * }.
 * }.

So,


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 * $$\displaystyle

\Rightarrow e^{2}(0) = 0

$$


 * }.
 * }.

which means from Equation (10),


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$$\displaystyle
 * style="width:20%; padding:10px; border:2px solid #8888aa" |
 * style="width:20%; padding:10px; border:2px solid #8888aa" |

\Rightarrow G^{2}(0) = 0

$$


 * style= |


 * }.
 * }.

Author
Solved and Typed by - Egm6341.s10.Team4.andy 21:56, 17 February 2010 (UTC) .

= Problem 4-4: Third derivative of e(t) in Lagrange Intp. Error =

Given
If we define function e(t) as the following,


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$$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle e(t)=\int\limits_{-t}^{t} F(t)\, dt-\dfrac{t}{3}[F(-t)+4F(0)+F(t)]


 * }
 * }

Find
Show that:


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$$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle e^{(3)}(t)=\dfrac{-t}{3}[F^{(3)}(t)-F^{(3)}(-t)]


 * }
 * }

Solution
Let e(t) be,


 * $$\displaystyle

e(t)=\alpha(t)-\alpha_2(t) $$

Basically, we have the following in the calculus for differentiating an integral:



\left({\int\limits_{g(x)}^{h(x)} f(x)\, dx}\right)'=\dfrac{d}{dx}\int\limits_{g(x)}^{h(x)} f(x)\, dx=f(h(x)).h'(x)-f(g(x)).g'(x) $$ Assuming k a point in the [-t,t] interval:



\alpha (t)=\int\limits_{-t}^{k} F(t)\, dt+\int\limits_{k}^{t} F(t)\, dt $$



\Rightarrow \alpha^{(1)}(t)=[F(k)\times0-F(-t)\times(-1)]+[F(t)\times 1-F(k)\times 0]=F(-t)+F(t) $$



\Rightarrow \alpha^{(2)}(t)=[F^{(1)}(-t)\times(-1)+F^{(1)}(t)\times 1]=F^{(1)}(t)-F^{(1)}(-t) $$



\Rightarrow \alpha^{(3)}(t)=F^{(2)}(t)+F^{(2)}(-t) $$



\alpha_2^{(1)}(t)=\dfrac{1}{3}[F(-t)+4F(0)+F(t)]-\dfrac{t}{3}F^{(1)}(-t)+0+\dfrac{t}{3}F^{(1)}(t) $$



\Rightarrow \alpha_2^{(2)}(t)=-\dfrac{1}{3}F^{(1)}(-t)+\dfrac{1}{3}F^{(1)}(t)-\dfrac{1}{3}F^{(1)}(-t)+\dfrac{t}{3}F^{(2)}(-t)+\dfrac{1}{3}F^{(1)}(t)+\dfrac{t}{3}F^{(2)}(t) $$



=-\dfrac{2}{3}F^{(1)}(-t)+\dfrac{2}{3}F^{(1)}(t)+\dfrac{t}{3}F^{(2)}(-t)+\dfrac{t}{3}F^{(2)}(t) $$



\Rightarrow \alpha_2^{(3)}(t)=\dfrac{2}{3}F^{(2)}(-t)+\dfrac{2}{3}F^{(2)}(t)+\dfrac{1}{3}F^{(2)}(-t)-\dfrac{t}{3}F^{(3)}(-t)+\dfrac{1}{3}F^{(2)}(t)+\dfrac{t}{3}F^{(3)}(t) $$



=F^{(2)}(-t)+F^{(2)}(t)+\dfrac{t}{3}[F^{(3)}(t)-F^{(3)}(-t)] $$



\Rightarrow e^{(3)}(t)=\alpha^{(3)}(t)-\alpha_2^{(3)}(t)=F^{(2)}(t)+F^{(2)}(-t)-F^{(2)}(-t)-F^{(2)}(t)+\dfrac{t}{3}[F^{(3)}(-t)-F^{(3)}(t)]= $$



=\dfrac{t}{3}[F^{(3)}(-t)-F^{(3)}(t)]=\dfrac{-t}{3}[F^{(3)}(t)-F^{(3)}(-t)] $$


 * {| style="width:60%" border="0"

$$ e^{(3)}(t)=\dfrac{-t}{3}[F^{(3)}(t)-F^{(3)}(-t)] $$
 * style="width:30%; padding:10px; border:1px solid #8888aa" |
 * style="width:30%; padding:10px; border:1px solid #8888aa" |
 * style= |
 * }

Author
Solved and Typed by - Egm6341.s10.Team4.nimaa&amp;m 02:42, 18 February 2010 (UTC) .

= Problem 4-5: Relationship between $$ \xi$$ and $$ \zeta_4$$=

Given

 * {| style="width:70%" border="0" align="center"




 * $$ \displaystyle x=\alpha + ht$$.


 * }.
 * }.

where,


 * {| style="width:70%" border="0" align="center"




 * $$ \displaystyle h= \frac{(b-a)}{2}$$.


 * }.
 * }.

Find
Prove that
 * {| style="width:70%" border="0" align="center"




 * $$\displaystyle \frac{-F^4(\zeta_4)}{90}= \frac{-(b-a)^4}{1440}f^4(\xi) $$


 * }.
 * }.

and hence deduce a relationship between $$\displaystyle \xi$$ and $$\displaystyle \zeta_4$$

Solution
We have F(t)= f(x(t))= f(x). Given


 * {| style="width:70%" border="0" align="center"




 * $$ \displaystyle x=\alpha + ht$$.


 * }.
 * }.

where,


 * {| style="width:70%" border="0" align="center"




 * $$ \displaystyle h= \frac{(b-a)}{2}$$.


 * }.
 * }.


 * {| style="width:70%" border="0" align="center"




 * $$\displaystyle F^'(t)= \frac{dF(t)}{dt}= \frac{df(x(t))}{dt}= \frac{df(x)}{dt}= \frac{df(x)}{dx}\cdot\frac{dx}{dt}= f^'(x)\cdot h$$


 * }.
 * }.


 * {| style="width:70%" border="0" align="center"




 * $$\displaystyle F^{}(t)= \frac{d}{dt}\cdot \frac{dF(t)}{dt}= \frac{d}{dt}\cdot(f^'(x)h) = h\cdot \frac{df^'(x)}{dt}= h\cdot \frac{df^'(x)}{dx}\cdot \frac{dx}{dt}= h^2\cdot f^{}(x)$$


 * }.
 * }.


 * {| style="width:70%" border="0" align="center"




 * $$\displaystyle F^{}(t)= \frac{d}{dt}\cdot (F^{}(t))= \frac{d}{dt}\cdot(h^2\cdot f^{}(x)) = h^2\cdot \frac{df^{}(x)}{dt}= h^2\cdot \frac{df^{}(x)}{dx}\cdot \frac{dx}{dt}= h^3\cdot f^{}(x)$$


 * }.
 * }.


 * {| style="width:70%" border="0" align="center"




 * $$\displaystyle F^{'}(t)= \frac{d}{dt}\cdot (F^{}(t))= \frac{d}{dt}\cdot(h^3\cdot f^{}(x)) = h^3\cdot \frac{df^{}(x)}{dt}= h^3\cdot \frac{df^{}(x)}{dx}\cdot \frac{dx}{dt}= h^4\cdot f^{'}(x)$$


 * }.
 * }.


 * {| style="width:70%" border="0" align="center"




 * $$\Rightarrow \displaystyle F^4(t)=h^4\cdot f^4(x)$$
 * }.
 * }.


 * {| style="width:70%" border="0" align="center"




 * $$\Rightarrow \displaystyle F^4(t)=\frac{(b-a)}{16} ^4\cdot f^4(x)$$
 * }.
 * }.

Substituting $$\displaystyle \xi$$ for $$\displaystyle x$$ and $$\displaystyle \zeta_4$$ for $$\displaystyle t$$


 * {| style="width:70%" border="0" align="center"




 * $$\Rightarrow \displaystyle F^4(\zeta_4)=\frac{(b-a)}{16} ^4\cdot f^4(\xi)$$
 * }.
 * }.


 * {| style="width:70%" border="0" align="center"




 * $$\Rightarrow \displaystyle \frac{-F^4(\zeta_4)}{90}= \frac{-(b-a)^4}{1440}f^4(\xi) $$
 * }.
 * }.

Since $$ \displaystyle x=\alpha + ht$$, We have $$ \displaystyle \xi=\alpha + h\zeta_4$$

This is the required relation between $$\displaystyle \xi$$ and $$\displaystyle \zeta_4$$.

Hence proved.

Author
Solved and Typed by - Egm6341.s10.team4.anandankala 13:43, 19 February 2010 (UTC) .

= Problem 4-6: Runge phenomenon =

Given
Refer Lecture slide 16-1 for problem statement
 * {| style="width:100%" border="0" align="left"

I = \int^{5}_{-5} \frac{1}{1+x^2} dx $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle \longrightarrow (1)


 * }.
 * }.

Find
'''(1) Find $$\displaystyle I_n$$ using Newton-Cotes for n=1, 2,. . ., 15'''

(2) Plot $$\displaystyle f, f_n $$ $$\mbox{,  }$$ for n=1, 2, 3, 8, 12

(3) Plot $$\displaystyle I_n$$ vs. $$\displaystyle n$$ and observe that $$\displaystyle I_n$$ does not converge as $$\displaystyle n \rightarrow \infty $$

(4) Observe weight $$\displaystyle W_{i,n}:=\int^{b}_{a} l_{i,n}(x),\ dx$$ are not all positive for n $$ \ge $$ 8. '''Plot $$\displaystyle l_{i,n}(x)$$ for i=1, 2, ... ,8 & n=8.'''


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


 * $$\displaystyle
 * $$\displaystyle

$$


 * }.
 * }.

Solution
'''(1) Find $$\displaystyle I_n$$ using Newton-Cotes for n=1, 2,. . ., 15'''

Newton-cotes formula $$\displaystyle P_n$$,


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

P_n = \sum_{i=0}^n l_{i,n}(x)f(x_i)$$
 * $$\displaystyle
 * $$\displaystyle


 * }.
 * }.

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

l_{i,n}(x) = \prod_{j=0,j\ne i}^n \frac{x-x_j}{x_i-x_j}$$
 * $$\displaystyle
 * $$\displaystyle


 * }.
 * }.


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

I_n = \int^{5}_{-5}P_n(x) dx = \sum_{i=0}^n \underbrace{\int^{5}_{-5}l_{i,n}(x) dx}_{W_i} f(x_i) = \sum_{i=0}^n w_i f(x_i) $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle \longrightarrow (2)
 * }.
 * }.

From the lecture slide 13-1,

Although this integration can be done numerically, there ia an another approach(analyitical solution) that derive the w(i) weight values directly.

the Error of integration of Newton-Cotes formula


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

\left | E_n(f(x)) \right | = \left | I-I_n \right | \le \int^{b}_{a} \left | f(x)-P_n(x)\right |  dx \le \frac {\int^{b}_{a}\left | q_{n+1}(x) \right| dx}{(n+1)!}\Big( max f^{(n+1)}(\xi), \xi \in [a,b] \Big)$$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle \longrightarrow (3)


 * }.
 * }.

From the equation (2),

We know the error is zero when the $$\displaystyle f(x) $$ is the polynomial of degree less than $$n$$


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

E_n(f(x)) = 0 $$
 * $$\displaystyle
 * $$\displaystyle


 * }.
 * }.

because, for the polynomial of degree less than $$\displaystyle n (\mathcal P_n)$$


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

if\quad f(x) \in \mathcal P_n, \qquad f^{(n+1)}(x)=0 $$
 * $$\displaystyle
 * $$\displaystyle


 * }.
 * }.

In order to find the $$\displaystyle w_i$$ in equation (2), we can use n equations and solve using matrix method like below,

for the $$\displaystyle f(x)=1,\; x, \; x^2, \; ...... \; x^n$$ (simplist and representative polynomial less than nth order), plug f(x) into equation(3)


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

E_n(f(x))= E_n(1)=E_n(x)=E_n(x^2)=......=E_n(x^n)=0 $$
 * $$\displaystyle
 * $$\displaystyle


 * }.
 * }.


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



\begin{align} &E_n(1)= \int^{5}_{-5}1 dx -[w_1(1)+w_2(1)+w_3(1)+......+w_n(1)] =0\\ &E_n(x)= \int^{5}_{-5}x dx -[w_1(x_0)+w_2(x_2)+w_3(x_3)+......+w_n(x_n)] =0\\ &E_n(x^2)= \int^{5}_{-5}x^2 dx -[w_1(x_0^2)+w_2(x_1^2)+w_3(x_3^2)+......+w_n(x_n^2)] =0\\ &E_n(x^3)= \int^{5}_{-5}x^3 dx -[w_1(x_0^3)+w_2(x_1^3)+w_3(x_3^3)+......+w_n(x_n^3)] =0\\ &.......................\\ &..............................\\ &......................................\\ &E_n(x^n)= \int^{5}_{-5}x^n dx -[w_1(x_0^n)+w_2(x_1^n)+w_3(x_3^n)+......+w_n(x_n^n)] =0 \end{align} $$
 * $$\displaystyle
 * }.
 * }.

Hence, the $$\displaystyle w_i$$ may be determined from


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



\underbrace{ \begin{bmatrix} 1 & 1 & 1 & 1 & ...... & 1\\ x_0 & x_1 & x_2 & x_3 & ...... & x_n\\ x_0^2 & x_1^2 & x_2^2 & x_3^2 & ...... & x_n^2\\ x_0^3 & x_1^3 & x_2^3 & x_3^3 & ...... & x_n^3\\ \vdots & \vdots& \vdots& \vdots& \vdots & \vdots \\ x_0^n & x_1^n & x_2^n & x_3^n & ...... & x_n^n\\ \end{bmatrix} }_{A} \underbrace{ \begin{bmatrix} w_0\\ w_1\\ w_2\\ w_3\\ \vdots\\ w_n\\ \end{bmatrix} }_{W} = \underbrace{ \begin{bmatrix} \int^{5}_{-5}1 dx\\ \int^{5}_{-5}x dx\\ \int^{5}_{-5}x^2dx\\ \int^{5}_{-5}x^3dx\\ \vdots \\ \int^{5}_{-5}x^ndx\\ \end{bmatrix} = \begin{bmatrix} (b-a)\\ \frac{(b^2-a^2)}{2}\\ \frac{(b^3-a^3)}{3}\\ \frac{(b^4-a^4)}{4}\\ \vdots \\ \frac{(b^{n+1}-a^{n+1})}{n+1} \end{bmatrix} }_{B} $$
 * $$\displaystyle
 * }.
 * }.

$$\displaystyle W_i$$ can be obtained like this,
 * {| style="width:100%" border="0" align="left"

\begin{align} &AW=B\\ &W_i=A^{-1}B \end{align} $$
 * $$\displaystyle
 * $$\displaystyle
 * }.
 * }.

plug obtained $$\displaystyle W_i$$ and \displaystyle f(x)=\frac{1}{1+x^2} into equation(2), $$\displaystyle I_n$$ can be calculated,
 * {| style="width:100%" border="0" align="left"

I_n = \sum_{i=0}^n w_i f(x_i) $$
 * $$\displaystyle
 * $$\displaystyle
 * }.
 * }.

$$ \begin{array}{|c||c|c|} n & I_n \\ \hline 1&0.384615\\ 2&6.794872\\	3&2.081448\\ 4&2.374005\\ 5&2.307692\\ 6&3.870449\\	7&2.898994\\	8&1.500489\\ 9&2.398618\\ 10&4.673301\\ 11&3.244773\\ 12&-0.31294\\ 13&1.919797\\ 14&7.899545\\ 15&4.155559\\ \hline \end{array} $$

 Matlab Code: 

(2) Plot $$\displaystyle f, f_n $$ $$\mbox{,  }$$ for n=1, 2, 3, 8, 12

$$\displaystyle f, f_n $$ were plotted by below matlab code.

 

 Matlab Code: 

(3) Plot $$\displaystyle I_n$$ vs. $$\displaystyle n$$ and observe that $$\displaystyle I_n$$ does not converge as $$\displaystyle n \rightarrow \infty $$

 



(4) Observe weight $$\displaystyle W_{i,n}:=\int^{b}_{a} l_{i,n}(x),\ dx$$ are not all positive for n $$ \ge $$ 8. '''Plot $$\displaystyle l_{i,n}(x)$$ for i=1, 2, ... ,8 & n=8.'''

This is the table that show the $$\displaystyle W_{i,n}:=\int^{b}_{a} l_{i,n}(x),\ dx$$ for n=8


 * Plot $$\displaystyle l_{i,n}(x)$$ for i=1, 2, ... ,8 & n=8.
 * Plot $$\displaystyle l_{i,n}(x)$$ for i=1, 2, ... ,8 & n=8.


 *  Matlab Code: 

.

Author
Solved - --Egm6341.s10.Team4.roni 05:15, 18 February 2010 (UTC)

Typed and reviewed by - Egm6341.s10.Team4.yunseok 02:06, 18 February 2010 (UTC) .

=Problem 4.7: Determining the Upper Bound of the Error in the Composite Simpson's Rule =

''' This problem was solved without referring to S10 homework. '''

Given: The Error as the Difference between Actual and Numeric
The error $$ \displaystyle E_n $$ in the Composite Simpson's rule


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

$$ \displaystyle \begin{align} {E_n} &= I-I_n \\ &= \int\limits_{a}^{b} f(x)dx - \dfrac{h}{3}[f(x_0)+4f(x_1)+2f(x_2)+...+2f(x_{n-2})+4f(x_{n-1})+f(x_n)]\\ \end{align}$$ (7.0)
 * style="width:98%" |
 * style="width:98%" |
 * <p style="text-align:right">
 * }

Show: The Error for the the Composite Simpson's Rule is Bounded Above
That the error for the Composite Simpson's Rule is given by,
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$$ \displaystyle \left| {E}_{n}^{2} \right| \leq \frac{(b-a)^{5}}{2880 n^{4}} M_{4} = \frac{(b-a)h^{4}}{2880} M_{4} \quad where \quad M_{4} := max \left| f^{4}(\xi) \right| \quad s.t. \quad \xi \in [a,b] $$     (7.1)
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Solution: Relating the Error in the Simple Simpson's Rule to the Partitioning of Intervals in the Composite Simpson's Rule
Consider the error $$ \displaystyle E_n $$ which is defined as the difference between the actual value of the integral, $$ \displaystyle I $$ and the value obtained from numerically integrating using the Composite Simpson's Rule, $$ \displaystyle I_n $$ where $$ \displaystyle n \geq 2 $$ is an even integer,


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$$ \displaystyle \begin{align} {E_n} &= I-I_n \\ &= \int\limits_{a}^{b} f(x)dx - \dfrac{h}{3}[f(x_0)+4f(x_1)+2f(x_2)+...+2f(x_{n-2})+4f(x_{n-1})+f(x_n)]\\ \end{align}$$ (7.2) Observe that the following pattern arises from the terms $$ \displaystyle I_n $$ in (7.2)
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$$ \begin{align}\displaystyle & f(x_0)+4f(x_1)+{\color{red}2f(x_2)}+...+{\color{green}2f(x_{n-2})}+4f(x_{n-1})+f(x_n) = \\ & \left[ f(x_0)+4f(x_1) + {\color{red}f(x_2)} \right] + \left[ {\color{red} f(x_2)}+4f(x_3) + f(x_4) \right] + \cdots \\ & \left[ f(x_{n-4})+4f(x_{n-3}) + {\color{green}f(x_{n-2})} \right] + \left[ {\color{green}f(x_{n-2})}+4f(x_{n-1}) + f(x_{n}) \right] \end{align} $$ The above relation allows us to write the terms inside of the summation as what is shown in (7.3). We must also take precautions to ensure that the partitioning of not only the summation, but also the partitioning of the integral contain the same subintervals namely, $$ \displaystyle [x_{2i-2},x_{2i}] $$. Both are accounted for defining $$ \displaystyle h:=\frac{b-a}{n} $$ and letting $$ \displaystyle x_i = a + ih $$ where $$ \displaystyle i = 0,1,2,...,n $$.
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$$ \displaystyle \begin{align} {E_n} &= \sum_{i=1}^{n/2}\left[ \int\limits_{x_{2i-2}}^{x_{2i}} f(x)dx \right] - \dfrac{h}{3}\sum_{i=1}^{n/2} \left[ f(x_{2i-2})+4f(x_{2i-1})+f(x_{2i}) \right]\\ &= \sum_{i=1}^{n/2} \left[ \int\limits_{x_{2i-2}}^{x_{2i}} f(x)dx-\dfrac{h}{3}[f(x_{2i-2})+4f(x_{2i-1})+f(x_{2i})]\right]\\ where \quad h&=\frac{b-a}{n} = \frac{x_i - x_{i-1}}{n} \end{align}$$ (7.3)
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We know that for the Simple Simson's Rule the error associated with the Lagrange Interpolation in, $$ \displaystyle E_2 $$ is,
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$$ \displaystyle {E_2} = \frac{x_{i+1}-x_{i-1}}{2}f^{(4)}(\xi) \quad s.t. \quad \xi \in [x_{i-1},x_{i+1}] $$     (7.4)
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Which implies for the Composite Simpson's Rule
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$$ \displaystyle \begin{align} \left| {E_n} \right| &\leq  \sum_{i=1}^{n/2} max \left| \frac{(x_{i+1}-x_{i-1})^5}{90\cdot 2^5}f^{(4)}(\xi) \right| \quad s.t. \quad \xi \in [x_{i-1},x_{i+1}]\\ &= \frac{(x_{i+1}-x_{i-1})^5}{90} \cdot \sum_{i=1}^{n/2} {\color{red}\underbrace{max \left| f^{(4)}(\xi)\right|}_{M_4}} \quad s.t. \quad \xi \in [x_{i-1},x_{i+1}] \end{align} $$ (7.5)
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Where $$ M_{4} $$ is defined on $$ \displaystyle [a,b] $$ as
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$$ \displaystyle M_{4} := max \left| f^{(4)}(\xi)\right| \quad s.t. \quad \xi \in [a,b] $$     (7.6) Next we will consider $$ \displaystyle \bar{M}_{4} \leq n \cdot M_{4} $$ allowing us to write $$ \displaystyle E_n $$ as,
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$$ \begin{align} \displaystyle \left| {E_n} \right| & \leq \left| \frac{(b-a)^{5}}{2880n^{5}}nM_{4} \right| \\ &= \left| \left(\frac{(b-a)}{2880}\right) {\color{red} \left( \frac{(b-a)^4} {n^4}\right)} M_{4}\right| \\ &= \left| \frac{(b-a){\color{red}h^{4}}}{2880}M_{4} \right| \end{align} $$ (7.7)
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Therefore showing that the error associated with the Composite Simpson's Rule is

=Problem 4.9 proof of SSET in different cases= Refer to lecture slide [[media:nm1.s11.mtg22.djvu|mtg-22]] for the problem statement.

Objectives
A). Redo the proof for two cases:

1).
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$$ \displaystyle
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G(t):=e(t)-{{t}^{4}}e(1)

$$ 2). :{| style="width:100%" border="0" $$ \displaystyle
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G(t):=e(t)-{{t}^{6}}e(1)

$$ Point out where proof breaks down.
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B). for G(t) as in (1) p.19-1(w/t5), find G(3)(0) and follow same steps in proof to see what happen.

Solutions
A). Redo the proof for two cases: 1). $$\displaystyle G(t):=e(t)-{{t}^{4}}e(1)$$
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$$ \displaystyle
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e(t)=\int\limits_{-t}^{+t}{F(t)dt}-\frac{t}{3}[F(-t)+4F(0)+F(t)]

$$ Where $$\displaystyle F(t)=f(x(t))$$
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$$ \displaystyle
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\text{G}\left( 0 \right)=e(0)-{{0}^{4}}e(1)=\int\limits_{-0}^{+0}{F(t)dt}-\frac{0}{3}[F(-0)+4F(0)+F(0)]=0

$$
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$$ \displaystyle
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\text{G}\left( 1 \right)=e(1)-{{1}^{4}}e(1)=0

$$ Apply Rolle ‘ s theorem, it yields:
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$$ \displaystyle
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\exists {{\xi }_{1}}\in \left( 0,1 \right)\begin{matrix} {} & st. \\ \end{matrix}{{G}^{\left( 1 \right)}}({{\xi }_{1}})=0

$$
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$$ \displaystyle
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\begin{align} & {{\text{G}}^{(1)}}\left( t \right)={{e}^{(1)}}(t)-4{{t}^{3}}e(1)={{A}^{(1)}}(t)-A{{_{2}^{L}}^{(1)}}(t)-4{{t}^{3}}e(1) \\ & \begin{matrix} \begin{matrix} {} & {} \\ \end{matrix} & {}  \\ \end{matrix}=F(-t)+F(t)-\{\frac{1}{3}[F(-t)+4F(0)+F(t)]+\frac{t}{3}[-{{F}^{(1)}}(-t)+{{F}^{(1)}}(t)]\}-4{{t}^{3}}e(1) \\ \end{align}

$$ So,
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$$ \displaystyle
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\begin{align} & {{\text{G}}^{(1)}}\left( 0 \right)=F(-0)+F(0)-\{\frac{1}{3}[F(-0)+4F(0)+F(0)]+\frac{0}{3}[-{{F}^{(1)}}(-t)+{{F}^{(1)}}(t)]\}-4\centerdot {{0}^{3}}e(1) \\ & \begin{matrix} {} & {} & {} \\ \end{matrix}=0 \\ \end{align}

$$ Apply Rolle ‘ s theorem, it yields:
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$$ \displaystyle
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\exists {{\xi }_{2}}\in \left( 0,{{\xi }_{1}} \right)\begin{matrix} {} & st. \\ \end{matrix}{{G}^{\left( 2 \right)}}({{\xi }_{2}})=0

$$
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$$ \displaystyle
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\begin{align} & {{\text{G}}^{(2)}}\left( t \right)={{e}^{(2)}}(t)-12{{t}^{2}}e(1)={{A}^{(2)}}(t)-A{{_{2}^{L}}^{(2)}}(t)-12{{t}^{2}}e(1) \\ & \begin{matrix} \begin{matrix} {} & {} \\ \end{matrix} & {}  \\ \end{matrix}=-{{F}^{(1)}}(-t)+{{F}^{(1)}}(t)-\{\frac{1}{3}[-{{F}^{(1)}}(-t)+{{F}^{(1)}}(t)] \\ & \begin{matrix} {} & {} & {} & {} \\ \end{matrix}+\frac{1}{3}[-{{F}^{(1)}}(-t)+{{F}^{(1)}}(t)]+\frac{t}{3}[{{F}^{(2)}}(-t)+{{F}^{(2)}}(t)]\}-12{{t}^{2}}e(1) \\ & \begin{matrix} \begin{matrix} {} & {} \\ \end{matrix} & {}  \\ \end{matrix}=\frac{1}{3}[-{{F}^{(1)}}(-t)+{{F}^{(1)}}(t)]-\frac{t}{3}[{{F}^{(2)}}(-t)+{{F}^{(2)}}(t)]\}-12{{t}^{2}}e(1) \\ \end{align}

$$ So,
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$$ \displaystyle
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\begin{align} & {{\text{G}}^{(2)}}\left( 0 \right)=\frac{1}{3}[-{{F}^{(1)}}(-0)+{{F}^{(1)}}(0)]-\frac{0}{3}[{{F}^{(2)}}(-0)+{{F}^{(2)}}(0)]\}-12\centerdot {{0}^{2}}e(1) \\ & \begin{matrix} {} & {} & = \\ \end{matrix}0 \\ \end{align}

$$ Apply Rolle ‘ s theorem, it yields:
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$$ \displaystyle
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\exists {{\xi }_{3}}\in \left( 0,{{\xi }_{2}} \right)\begin{matrix} {} & st. \\ \end{matrix}{{G}^{\left( 3 \right)}}({{\xi }_{3}})=0

$$
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$$ \displaystyle
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\begin{align} & {{\text{G}}^{(3)}}\left( t \right)=\frac{1}{3}[-{{F}^{(1)}}(-t)+{{F}^{(1)}}(t)]-\frac{t}{3}[{{F}^{(2)}}(-t)+{{F}^{(2)}}(t)]-24te(1) \\ & \begin{matrix} {} & {} & = \\ \end{matrix}\frac{1}{3}[{{F}^{(2)}}(-t)+{{F}^{(2)}}(t)]-\frac{1}{3}[{{F}^{(2)}}(-t)+{{F}^{(2)}}(t)]-\frac{t}{3}[-{{F}^{(3)}}(-t)+{{F}^{(3)}}(t)]-24te(1) \\ & \begin{matrix} {} & {} & = \\ \end{matrix}-\frac{t}{3}[-{{F}^{(3)}}(-t)+{{F}^{(3)}}(t)]-24te(1) \\ \end{align}

$$ So, $$\displaystyle {{\text{G}}^{(3)}}\left( {{\xi }_{3}} \right)=\frac{3}[{{F}^{(3)}}(-{{\xi }_{3}})-{{F}^{(3)}}({{\xi }_{3}})]-24\left( {{\xi }_{3}} \right)e(1)=0$$, then using DMVT:
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$$ \displaystyle
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e(1)=\frac{\frac{3}[{{F}^{(3)}}(-{{\xi }_{3}})-{{F}^{(3)}}({{\xi }_{3}})]}{24\left( {{\xi }_{3}} \right)}=\frac{\frac{3}[-2{{\xi }_{3}}{{F}^{(4)}}({{\xi }_{4}})]}{24\left( {{\xi }_{3}} \right)}=\frac{-{{\xi }_{3}}{{F}^{(4)}}({{\xi }_{4}})}{36}

$$ Where, $$\displaystyle {{\xi }_{4}}\in \left( -{{\xi }_{3}},{{\xi }_{3}} \right)$$
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$$ \displaystyle
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{{\text{E}}_{2}}=h\centerdot e(1)=h\frac{-{{\xi }_{3}}{{F}^{(4)}}({{\xi }_{4}})}{36}

$$     (Eq ) The proof will break down here because the magnitude of error is related to $$\displaystyle {{\xi }_{3}}$$in the above equation.
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2)

$$\displaystyle G(t):=e(t)-{{t}^{6}}e(1)$$
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$$ \displaystyle
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e(t)=\int\limits_{-t}^{+t}{F(t)dt}-\frac{t}{3}[F(-t)+4F(0)+F(t)]

$$ Where $$\displaystyle F(t)=f(x(t))$$
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$$ \displaystyle
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\text{G}\left( 0 \right)=e(0)-{{0}^{6}}e(1)=\int\limits_{-0}^{+0}{F(t)dt}-\frac{0}{3}[F(-0)+4F(0)+F(0)]=0

$$
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$$ \displaystyle
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\text{G}\left( 1 \right)=e(1)-{{1}^{6}}e(1)=0

$$ Apply Rolle ‘ s theorem, it yields:
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$$ \displaystyle
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\exists {{\xi }_{1}}\in \left( 0,1 \right)\begin{matrix} {} & st. \\ \end{matrix}{{G}^{\left( 1 \right)}}({{\xi }_{1}})=0

$$
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$$ \displaystyle
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\begin{align} & {{\text{G}}^{(1)}}\left( t \right)={{e}^{(1)}}(t)-6{{t}^{5}}e(1)={{A}^{(1)}}(t)-A{{_{2}^{L}}^{(1)}}(t)-6{{t}^{5}}e(1) \\ & \begin{matrix} \begin{matrix} {} & {} \\ \end{matrix} & {}  \\ \end{matrix}=F(-t)+F(t)-\{\frac{1}{3}[F(-t)+4F(0)+F(t)]+\frac{t}{3}[-{{F}^{(1)}}(-t)+{{F}^{(1)}}(t)]\}-6{{t}^{5}}e(1) \\ \end{align}

$$ So,
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$$ \displaystyle
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\begin{align} & {{\text{G}}^{(1)}}\left( 0 \right)=F(-0)+F(0)-\{\frac{1}{3}[F(-0)+4F(0)+F(0)]+\frac{0}{3}[-{{F}^{(1)}}(-t)+{{F}^{(1)}}(t)]\}-6\centerdot {{0}^{5}}e(1) \\ & \begin{matrix} {} & {} & {} \\ \end{matrix}=0 \\ \end{align}

$$ Apply Rolle ‘ s theorem, it yields:
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$$ \displaystyle
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\exists {{\xi }_{2}}\in \left( 0,{{\xi }_{1}} \right)\begin{matrix} {} & st. \\ \end{matrix}{{G}^{\left( 2 \right)}}({{\xi }_{2}})=0

$$
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$$ \displaystyle
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\begin{align} & {{\text{G}}^{(2)}}\left( t \right)={{e}^{(2)}}(t)-30{{t}^{4}}e(1)={{A}^{(2)}}(t)-A{{_{2}^{L}}^{(2)}}(t)-30{{t}^{4}}e(1) \\ & \begin{matrix} \begin{matrix} {} & {} \\ \end{matrix} & {}  \\ \end{matrix}=-{{F}^{(1)}}(-t)+{{F}^{(1)}}(t)-\{\frac{1}{3}[-{{F}^{(1)}}(-t)+{{F}^{(1)}}(t)] \\ & \begin{matrix} {} & {} & {} & {} \\ \end{matrix}+\frac{1}{3}[-{{F}^{(1)}}(-t)+{{F}^{(1)}}(t)]+\frac{t}{3}[{{F}^{(2)}}(-t)+{{F}^{(2)}}(t)]\}-30{{t}^{4}}e(1) \\ & \begin{matrix} \begin{matrix} {} & {} \\ \end{matrix} & {}  \\ \end{matrix}=\frac{1}{3}[-{{F}^{(1)}}(-t)+{{F}^{(1)}}(t)]-\frac{t}{3}[{{F}^{(2)}}(-t)+{{F}^{(2)}}(t)]\}-30{{t}^{4}}e(1) \\ \end{align}

$$ So,
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$$ \displaystyle
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\begin{align} & {{\text{G}}^{(2)}}\left( 0 \right)=\frac{1}{3}[-{{F}^{(1)}}(-0)+{{F}^{(1)}}(0)]-\frac{0}{3}[{{F}^{(2)}}(-0)+{{F}^{(2)}}(0)]\}-30\centerdot {{0}^{4}}e(1) \\ & \begin{matrix} {} & {} & = \\ \end{matrix}0 \\ \end{align}

$$ Apply Rolle ‘ s theorem, it yields:
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$$ \displaystyle
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\exists {{\xi }_{3}}\in \left( 0,{{\xi }_{2}} \right)\begin{matrix} {} & st. \\ \end{matrix}{{G}^{\left( 3 \right)}}({{\xi }_{3}})=0

$$
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$$ \displaystyle
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\begin{align} & {{\text{G}}^{(3)}}\left( t \right)=\frac{1}{3}[-{{F}^{(1)}}(-t)+{{F}^{(1)}}(t)]-\frac{t}{3}[{{F}^{(2)}}(-t)+{{F}^{(2)}}(t)]-120{{t}^{3}}e(1) \\ & \begin{matrix} {} & {} & = \\ \end{matrix}\frac{1}{3}[{{F}^{(2)}}(-t)+{{F}^{(2)}}(t)]-\frac{1}{3}[{{F}^{(2)}}(-t)+{{F}^{(2)}}(t)]-\frac{t}{3}[-{{F}^{(3)}}(-t)+{{F}^{(3)}}(t)]-120{{t}^{3}}e(1) \\ & \begin{matrix} {} & {} & = \\ \end{matrix}-\frac{t}{3}[-{{F}^{(3)}}(-t)+{{F}^{(3)}}(t)]-120{{t}^{3}}e(1) \\ \end{align}

$$ So, $$\displaystyle {{\text{G}}^{(3)}}\left( {{\xi }_{3}} \right)=\frac{3}[{{F}^{(3)}}(-{{\xi }_{3}})-{{F}^{(3)}}({{\xi }_{3}})]-120{{\left( {{\xi }_{3}} \right)}^{3}}e(1)=0$$, then using DMVT:
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$$ \displaystyle
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e(1)=\frac{\frac{3}[{{F}^{(3)}}(-{{\xi }_{3}})-{{F}^{(3)}}({{\xi }_{3}})]}{120{{\left( {{\xi }_{3}} \right)}^{3}}}=\frac{\frac{3}[-2{{\xi }_{3}}{{F}^{(4)}}({{\xi }_{4}})]}{120{{\left( {{\xi }_{3}} \right)}^{3}}}=\frac{-{{F}^{(4)}}({{\xi }_{4}})}{180{{\xi }_{3}}}

$$ Where, $$\displaystyle {{\xi }_{4}}\in \left( -{{\xi }_{3}},{{\xi }_{3}} \right)$$
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$$ \displaystyle
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{{\text{E}}_{2}}=h\centerdot e(1)=h\frac{-{{F}^{(4)}}({{\xi }_{4}})}{180{{\xi }_{3}}}

$$
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 * }The proof will break down here because the magnitude of error is related to $$\displaystyle {{\xi }_{3}}$$in the above equation.

B)
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$$ \displaystyle
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G(t):=e(t)-{{t}^{5}}e(1)

$$
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$$ \displaystyle
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e(t)=\int\limits_{-t}^{+t}{F(t)dt}-\frac{t}{3}[F(-t)+4F(0)+F(t)]

$$ Where $$\displaystyle F(t)=f(x(t))$$
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$$ \displaystyle
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\text{G}\left( 0 \right)=e(0)-{{0}^{5}}e(1)=\int\limits_{-0}^{+0}{F(t)dt}-\frac{0}{3}[F(-0)+4F(0)+F(0)]=0

$$
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$$ \displaystyle
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\text{G}\left( 1 \right)=e(1)-{{1}^{5}}e(1)=0

$$ Apply Rolle ‘ s theorem, it yields:
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$$ \displaystyle
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\exists {{\xi }_{1}}\in \left( 0,1 \right)\begin{matrix} {} & st. \\ \end{matrix}{{G}^{\left( 1 \right)}}({{\xi }_{1}})=0

$$
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$$ \displaystyle
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\begin{align} & {{\text{G}}^{(1)}}\left( t \right)={{e}^{(1)}}(t)-5{{t}^{4}}e(1)={{A}^{(1)}}(t)-A{{_{2}^{L}}^{(1)}}(t)-5{{t}^{4}}e(1) \\ & \begin{matrix} \begin{matrix} {} & {} \\ \end{matrix} & {}  \\ \end{matrix}=F(-t)+F(t)-\{\frac{1}{3}[F(-t)+4F(0)+F(t)]+\frac{t}{3}[-{{F}^{(1)}}(-t)+{{F}^{(1)}}(t)]\}-5{{t}^{4}}e(1) \\ \end{align}

$$ So,
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$$ \displaystyle
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\begin{align} & {{\text{G}}^{(1)}}\left( 0 \right)=F(-0)+F(0)-\{\frac{1}{3}[F(-0)+4F(0)+F(0)]+\frac{0}{3}[-{{F}^{(1)}}(-t)+{{F}^{(1)}}(t)]\}-5\centerdot {{0}^{4}}e(1) \\ & \begin{matrix} {} & {} & {} \\ \end{matrix}=0 \\ \end{align}

$$ Apply Rolle ‘ s theorem, it yields:
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$$ \displaystyle
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\exists {{\xi }_{2}}\in \left( 0,{{\xi }_{1}} \right)\begin{matrix} {} & st. \\ \end{matrix}{{G}^{\left( 2 \right)}}({{\xi }_{2}})=0

$$
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$$ \displaystyle
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\begin{align} & {{\text{G}}^{(2)}}\left( t \right)={{e}^{(2)}}(t)-20{{t}^{3}}e(1)={{A}^{(2)}}(t)-A{{_{2}^{L}}^{(2)}}(t)-20{{t}^{3}}e(1) \\ & \begin{matrix} \begin{matrix} {} & {} \\ \end{matrix} & {}  \\ \end{matrix}=-{{F}^{(1)}}(-t)+{{F}^{(1)}}(t)-\{\frac{1}{3}[-{{F}^{(1)}}(-t)+{{F}^{(1)}}(t)] \\ & \begin{matrix} {} & {} & {} & {} \\ \end{matrix}+\frac{1}{3}[-{{F}^{(1)}}(-t)+{{F}^{(1)}}(t)]+\frac{t}{3}[{{F}^{(2)}}(-t)+{{F}^{(2)}}(t)]\}-20{{t}^{3}}e(1) \\ & \begin{matrix} \begin{matrix} {} & {} \\ \end{matrix} & {}  \\ \end{matrix}=\frac{1}{3}[-{{F}^{(1)}}(-t)+{{F}^{(1)}}(t)]-\frac{t}{3}[{{F}^{(2)}}(-t)+{{F}^{(2)}}(t)]\}-20{{t}^{3}}e(1) \\ \end{align}

$$ So,
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$$ \displaystyle
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\begin{align} & {{\text{G}}^{(2)}}\left( 0 \right)=\frac{1}{3}[-{{F}^{(1)}}(-0)+{{F}^{(1)}}(0)]-\frac{0}{3}[{{F}^{(2)}}(-0)+{{F}^{(2)}}(0)]\}-20\centerdot {{0}^{3}}e(1) \\ & \begin{matrix} {} & {} & = \\ \end{matrix}0 \\ \end{align}

$$ Apply Rolle ‘ s theorem, it yields:
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$$ \displaystyle
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\exists {{\xi }_{3}}\in \left( 0,{{\xi }_{2}} \right)\begin{matrix} {} & st. \\ \end{matrix}{{G}^{\left( 3 \right)}}({{\xi }_{3}})=0

$$
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$$ \displaystyle
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\begin{align} & {{\text{G}}^{(3)}}\left( t \right)=\frac{1}{3}[-{{F}^{(1)}}(-t)+{{F}^{(1)}}(t)]-\frac{t}{3}[{{F}^{(2)}}(-t)+{{F}^{(2)}}(t)]-60{{t}^{2}}e(1) \\ & \begin{matrix} {} & {} & = \\ \end{matrix}\frac{1}{3}[{{F}^{(2)}}(-t)+{{F}^{(2)}}(t)]-\frac{1}{3}[{{F}^{(2)}}(-t)+{{F}^{(2)}}(t)]-\frac{t}{3}[-{{F}^{(3)}}(-t)+{{F}^{(3)}}(t)]-60{{t}^{2}}e(1) \\ & \begin{matrix} {} & {} & = \\ \end{matrix}-\frac{t}{3}[-{{F}^{(3)}}(-t)+{{F}^{(3)}}(t)]-60{{t}^{2}}e(1) \\ \end{align}

$$ So,
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$$ \displaystyle
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{{\text{G}}^{(3)}}\left( 0 \right)=-\frac{0}{3}[-{{F}^{(3)}}(-t)+{{F}^{(3)}}(t)]-60\centerdot {{0}^{2}}e(1)=0

$$ Apply Rolle ‘ s theorem, it yields:
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$$ \displaystyle
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\exists {{\xi }_{4}}\in \left( 0,{{\xi }_{3}} \right)\begin{matrix} {} & st. \\ \end{matrix}{{G}^{\left( 4 \right)}}({{\xi }_{4}})=0

$$
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$$ \displaystyle
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{{\text{G}}^{(4)}}\left( t \right)=\frac{1}{3}[{{F}^{(3)}}(-t)-{{F}^{(3)}}(t)]-\frac{t}{3}[{{F}^{(4)}}(-t)+{{F}^{(4)}}(t)]-120te(1)

$$ So,
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$$ \displaystyle
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{{\text{G}}^{(4)}}\left( {{\xi }_{4}} \right)=\frac{1}{3}[{{F}^{(3)}}(-{{\xi }_{4}})-{{F}^{(3)}}({{\xi }_{4}})]-\frac{3}[{{F}^{(4)}}(-{{\xi }_{4}})+{{F}^{(4)}}({{\xi }_{4}})]-120{{\xi }_{4}}e(1)=0

$$ Using DMVT:
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$$ \displaystyle
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e(1)=\frac{\frac{1}{3}[-2{{\xi }_{4}}{{F}^{(4)}}({{\xi }_{5}})]-\frac{3}[{{F}^{(4)}}(-{{\xi }_{4}})+{{F}^{(4)}}({{\xi }_{4}})]}{120{{\xi }_{4}}}=\frac{-[2{{F}^{(4)}}({{\xi }_{5}})+{{F}^{(4)}}(-{{\xi }_{4}})+{{F}^{(4)}}({{\xi }_{4}})]}{360}

$$ Where, $$\displaystyle {{\xi }_{5}}\in \left( -{{\xi }_{4}},{{\xi }_{4}} \right)$$
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$$ \displaystyle
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{{\text{E}}_{2}}=h\centerdot e(1)=h\frac{-[2{{F}^{(4)}}({{\xi }_{5}})+{{F}^{(4)}}(-{{\xi }_{4}})+{{F}^{(4)}}({{\xi }_{4}})]}{360}

$$ Here, it is observed that the magnitude of error is related to the value of function at these two points$$\displaystyle {{\xi }_{4}},{{\xi }_{5}}$$in the above equation.
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