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I'm working on the following exercise

a) Find weights $w_0,\bar{w_0},w_1,\bar{w_1}$ such that $$\int_a^bf(x)dx=w_0f(a)+\bar{w_0}f'(a)+w_1f(b)+\bar{w_1}f'(b)$$ for any $f\in\mathcal{P}_3$.

<p>b) Show that for <span class="math-container">$f\in \mathcal{C}^4$</span>, the error in this quadrature formula, i.e., true integral minus its approximation, is of the form <span class="math-container">$C(b−a)^5f^{(4)}(\xi)$</span> for some <span class="math-container">$\xi\in[a,b]$</span>, and give the constant <span class="math-container">$C$</span>.</p>

For part (a), I used the Hermite interpolation polynomial $p_3$ with interpolation points $x_0=a$ and $x_1=b$, which is the unique polynomial of degree $3$ such that $p_3(x_i)=f(x_i)$ and $p_3'(x_i)=f'(x_i)$, for $i=0,1$. From uniqueness it follows that $f\in\mathcal{P}_3\implies f\equiv p_3$, thus $\int_a^bf(x)dx= \int_a^bp_3(x)dx$. From there I calculated $w_0=w_1=(b-a)/2$ and $\bar{w_0}=(b-a)^2/12=-\bar{w_1}$.

I'm looking for some help for part (b). I tried to follow a proof for a similar result on Simpson's quadrature rule, but my calculations didn't work. I defined the error as $E(f)=\int_a^bf(x)dx-\left(\frac{b-a}{2}f(a)+\frac{(b-a)^2}{12}f'(a)+\frac{b-a}{2}f(b)-\frac{(b-a)^2}{12}f'(b)\right)$, which yields, after the substitution $x(t)=\frac{a+b}{2}+\frac{b-a}{2}t$ for $t\in[-1,1]$: $$E(f)=\frac{b-a}{2}\left(\int_{-1}^1F(\tau)d\tau-\left(F(-1)+F(1)+\frac13(F'(-1)-F'(1))\right)\right).$$ Then I define $$G(t)=\int_{-t}^tF(\tau)d\tau-t\left(F(-t)+F(t)+\frac13(F'(-1)-F'(1))\right),$$ so $E(f)=\frac{b-a}{2}G(1)$. Then I define $H(t)=G(t)-t^5G(1)$. Then by a few applications of Rolle's theorem [EDIT: I think I also applied Rolle's theorem here while it was not even applicaple], there exists an $\xi_3\in(0,1)$ such that $H'''(\xi_3)=0$, but since $G'''(t)=-2(F''(t)+F''(-t))-t(F'''(t)-F'''(-t))$ still involves second derivates of $F$, this does not really help (in the proof I'm trying to imitate we apply the MVT on $F'''(\xi_3)-F'''(-\xi_3)$).

Perhaps I made a mistake, and perhaps another choice for the function $G(t)$ would work, but I did not manage to go further. Any help is much appreciated!

Václav Mordvinov
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1 Answers1

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You seem to be trying to follow a proof I don't like of the error formula for Simpson's rule. Your problems start with your formula for $G(t)$ which should be $$G(t)=\int_{-t}^tF(\tau)d\tau-t\left(F(-t)+F(t)+\frac{t}3(F^{\prime}(-t)-F^{\prime}(t))\right)$$ Then you can differentiate to get $$\begin{align}G^{\prime}(t)&=\frac t3\left(F^{\prime}(-t)-F^{\prime}(t)\right)+\frac{t^2}3\left(F^{\prime\prime}(-t)+F^{\prime\prime}(t)\right)\\ G^{\prime\prime}(t)&=\frac13\left(F^{\prime}(-t)-F^{\prime}(t)\right)+\frac{t}3\left(F^{\prime\prime}(-t)+F^{\prime\prime}(t)\right)+\frac{t^2}3\left(F^{\prime\prime\prime}(t)-F^{\prime\prime\prime}(-t)\right)\\ G^{\prime\prime\prime}(t)&=t\left(F^{\prime\prime\prime}(t)-F^{\prime\prime\prime}(-t)\right)+\frac{t^2}3\left(F^{(4)}(-t)+F^{(4)}(t)\right)\end{align}$$ Then $$G(0)=G^{\prime}(0)=G^{\prime\prime}(0)=G^{\prime\prime\prime}(0)=0$$ So, as in the proof for Cauchy's mean value theorem we let $$H(t)=G(t)-G(1)\cdot t^5$$ Then $H(0)=G(0)=H(1)$ so by Rolle's theorem there is some $0<x_1<1$ such that $H^{\prime}(x_1)=0$. Then $2$ more applications of Rolle's theorem gets us to $H^{\prime\prime}(x_2)=H^{\prime\prime\prime}(x_3)=G^{\prime\prime\prime}(x_3)-60G(1)\cdot x_3^2=0$ for some $0<x_3<x_2<x_1<1$. So $$G(1)=\frac1{60x_3^2}\left(x_3\left(F^{\prime\prime\prime}(x_3)-F^{\prime\prime\prime}(-x_3)\right)+\frac{x_3^2}3\left(F^{(4)}(-x_3)+F^{(4)}(x_3)\right)\right)$$ If we let $$J(t)=F^{\prime\prime\prime}(t)-\frac{F^{\prime\prime\prime}(x_3)-F^{\prime\prime\prime}(-x_3)}{2x_3}t$$ Then $$J(-x_3)=J(x_3)=\frac{F^{\prime\prime\prime}(-x_3)+F^{\prime\prime\prime}(x_3)}2$$ So $$J^{\prime}(x_4)=F^{(4)}(x_4)-\frac{F^{\prime\prime\prime}(x_3)-F^{\prime\prime\prime}(-x_3)}{2x_3}=0$$ for some $-x_3<x_4<x_3$ by Rolle's theorem. Thus $$G(1)=\frac1{30}\left(F^{(4)}(x_4)+\frac16F^{(4)}(-x_3)+\frac16F^{(4)}(x_3)\right)=\frac2{45}F^{(4)}(x_5)$$ For some $-1<-x_3<x_5<x_3<1$ by the intermediate value theorem. Multiply by $\frac{b-a}2$ and restate in terms of $f$ to get $$\int_a^bf(x)dx=\frac{b-a}2\left(f(a)+f(b)\right)+\frac{(b-a)^2}{12}\left(f^{\prime}(a)-f^{\prime}(b)\right)+\frac{(b-a)^5}{720}f^{(4)}(\xi)$$ I think this is a really hard way to prove it and if you had instead followed my proof where I use the error estimate for Hermite interpolation, life would have been much simpler.

user5713492
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  • Thank you very much for you detailed answer, this is really helpful! One question: you mean $J'(x_4)$, by the MVT, right? (instead of $J(X_4)$). Would you also have any suggestions on how to arrive at the correct function $G$ in this kind of derivations? – Václav Mordvinov Feb 19 '19 at 23:33
  • Right, $J^{\prime}(x_4)=0$, but by Rolle's theorem, not mean value theorem. I plugged in my first guess and it worked. As I said, I'm not a fan of this method and my suggestion would be to instead use the error estimate for Hermite interpolation, which leads to a much simpler and more straightforward proof. – user5713492 Feb 19 '19 at 23:46
  • Thank you! I'll also try to write down a proof for myself using Hermite interpolation errors today – Václav Mordvinov Feb 20 '19 at 10:24
  • You're right, it's so much simpler using the Hermite interpolation errors. Using Mathematica for $\int_a^b(x-a)^2(x-b)^2dx$, the proof takes only two lines – Václav Mordvinov Feb 20 '19 at 10:53
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    Why use Mathematica? Tabular integration zaps this one or even let $x=a+(b-a)t$ so $$\begin{align}\int_a^b(x-a)^2(x-b)^2dx&=(b-a)^5\int_0^1t^2(1-t)^2dt=(b-a)^5\text{B}(3,3)\&=(b-a)^5\frac{\Gamma(3)\Gamma(3)}{\Gamma(6)}=(b-a)^5\frac{2!2!}{5!}=\frac{(b-a)^5}{30}\end{align}$$ – user5713492 Feb 20 '19 at 16:06
  • Thanks again! I especially like the one with the Beta function and forgot about tabular integration. I'll this next time – Václav Mordvinov Feb 20 '19 at 19:09