User:Egm6341.s10.team3.sa/Mtg15

Mtg 15: Thu, 28 Jan 10

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Proof of Lagrange interpolation error We know that, $$\displaystyle G(t=0) = 0 = G(t=1) $$

Now by applying Rolle's theorem to the function G, we know that, there exists, $$\displaystyle \xi_1 \in (0,1) \,such\, that\, G^{(1)}(\xi_1) = 0.$$.

Note: Here the interval (0,1) is open as the function can take up all values between 0 and 1 EXCLUDING 0 and 1.

$$\displaystyle G^{(1)}(0) = 0 \;- \,Why?$$ We shall understand why from the following derivation.

Recall from [[media: Egm6341.s10.mtg14.djvu | Page 14-2]] that, $$\displaystyle G(t):= e(t) - t^5 e(1)$$ Hence, $$\displaystyle G'(t):= e'(t) - 5t^4\underbrace{e(1)}_{constant}$$

Also, we know from [[media: Egm6341.s10.mtg14.djvu | Page 14-2]] that, $$\displaystyle e(t) = \alpha(t) - \alpha_2(t) $$ where, $$\displaystyle \alpha(t) = \int_{-t}^{t}f(x(t)) \,dt = \int_{-t}^{t}F(t) \,dt$$ and, $$\displaystyle \alpha_2(t) = \frac{t}{3}\big[F(-t) +4F(0) +F(t)\big] $$


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Show that:
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$$
 * $$\displaystyle\color{blue} \alpha^{(1)}(t)= F(-t) + F(t)
 * $$\displaystyle\color{blue} \alpha^{(1)}(t)= F(-t) + F(t)
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Taking the first derivative w.r.t t, we have $$\displaystyle \alpha'(t) = F(-t)+F(t)$$ and, $$\displaystyle \alpha_2'(t) = \frac{1}{3}\big[F(-t) +4F(0) +F(t)\big] + \frac{t}{3}\big[F'(-t)+F'(t) \big]$$

[[media: Egm6341.s10.mtg15.djvu | Page 15-2]]  Thus, $$\displaystyle e'(t) = \alpha'(t) - \alpha_2'(t)$$, and, $$\displaystyle e'(0) = 2F(0) -[2F(0)+0] = 0 $$

Since, $$\displaystyle G'(0) =e'(0)$$, we have $$\displaystyle G'(0)=0$$

Earlier we proved that, $$\displaystyle G'(\xi_1) = 0 $$.

Applying Rolle's theorem once again, we know there exists, $$\displaystyle \xi_2 \in (0,\xi_1) \,such\, that\, G^{(2)}(\xi_2) = 0.$$.

Applying the same logic as above, we have: $$\displaystyle G^{(2)}(0)=0$$ and


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For the Simpson's rule, prove that $$\displaystyle\color{blue} G^{(2)}(0)=0 $$.


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Applying Rolle's theorem once again, we know there exists, $$\displaystyle \xi_3 \in (0,\xi_2) \,such\, that\, G^{(3)}(\xi_3) = 0.$$.

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


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Prove that $$\displaystyle\color{blue} e^{(3)}(t) = - \frac{t}{3}[F^{(3)}(t)- F^{(3)}(-t)]$$ where $$\displaystyle\color{blue} e(t)= \int_{-t}^t f(x(t))\, dt - \frac {t}{3}[F(-t)+4 F(0)+F(t)] $$


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

Now we have, $$\displaystyle G^{(3)}(\xi_3) = -\frac{\xi_3}{3}\big[F^{(3)}(\xi_3) - F^{(3)}(-\xi_3) \big] -60(\xi_3)^2 e(1)=0$$

[[media: Egm6341.s10.mtg15.djvu | Page 15-3]]  Applying derivative mean value theorem, $$\displaystyle G^{(3)}(\xi_3) = -\frac{\xi_3}{3}\big[2\xi_3 F^{(4)}(\xi_4) \big] -60(\xi_3)^2 e(1)=0$$

since $$\displaystyle \xi_3 \neq 0 $$ Note: This is because $$\displaystyle \xi_3 \in (0,\xi_2) $$ which excludes 0.

Solving for e(1) we have, $$\displaystyle \begin{align} e(1) &= -\frac{1}{90} F^{(4)}(\xi_4)\\ &= -\frac{(b-a)^4}{1440} f^4(\xi) \end{align}$$


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Given that$$\displaystyle\color{blue} -\frac{1}{90}F^{(4)}(\zeta _{4})=-\frac{(b-a)^{4}}{1440}f^{(4)}(\xi) $$

Find the relationship between $$\displaystyle\color{blue} \zeta _{4}$$ and $$\displaystyle\color{blue} \xi $$


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