User:Egm6341.s10.team4.anandankala/HW2-15

Prove that

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 * $$\displaystyle e(1)= \frac{-F^4(\zeta_4)}{90}= \frac{-(b-a)^4}{1440}f^4(\xi) $$

Deduce a relation between $$\displaystyle \xi and \zeta_4.$$
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Solution
We have F(t)= f(x(t))= f(x). Let us consider


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 * $$ \displaystyle x=\alpha + ht, where h= \frac{(b-a)}{2}$$.


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


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


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


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


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


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


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


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


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


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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.