User:Egm6341.s11.team4.yang/hw4

=HW 4.2 problem=

Refer to lecture slide [[media:nm1.s11.mtg20.djvu|mtg-20]] for the problem statement.

Given
Given that :
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$$ \displaystyle
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A(t)=\int\limits_{-t}^{+t}{F(x)dx=}\int\limits_{-t}^{k}{F(x)dx+}\int\limits_{k}^{+t}{F(x)dx}

$$     (Eq ) Prove:
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$$ \displaystyle
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{{A}^{(1)}}(t)=F(-t)+F(t)

$$     (Eq )
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=Answer:= Take a derivative on this equation,
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$$ \displaystyle
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A(t)=\int\limits_{-t}^{+t}{F(x)dx=}\int\limits_{-t}^{k}{F(x)dx+}\int\limits_{k}^{+t}{F(x)dx}

$$     (Eq ) We assume that $$\displaystyle \begin{align} & G(t)=\int{F(t)dt},\frac{dG(t)}{dt}=F(t) \\ & \\ \end{align}$$
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$$ \displaystyle
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\int\limits_{-t}^{k}{F(x)dx}=G(k)-G(-t)

$$     (Eq )
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$$ \displaystyle
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\int\limits_{k}^{+t}{F(x)dx}=G(t)-G(k)

$$     (Eq ) Where k is constant value,
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$$ \displaystyle
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\begin{align} & \frac{dA(t)}{dt}=\frac{d\left( G(k)-G(-t) \right)}{dt}+\frac{d\left( G(t)-G(k) \right)}{dt} \\ & =-\frac{d\left( G(-t) \right)}{dt}\frac{d\left( -t \right)}{dt}+\frac{d\left( G(t) \right)}{dt}=F(-t)+F(t) \\ \end{align}

$$     (Eq ) So,
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$$ \displaystyle
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{{A}^{(1)}}(t)=F(-t)+F(t)

$$     (Eq )
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=Problem 4.9= 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)

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