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Let $L$ be the quadratic Lagrange interpolating polynomial for a function $f$ such that $f(-h)=L(-h)$, $f(0)=L(0)$, and $f(h)=L(h)$. Then it is well-known that $$f(x)=L(x)+\frac{f'''(\xi(x))}{6}x(x-h)(x+h),$$ where $\xi(x)\in(-h,h)$.

Integrating, we get $$\int_{-h}^h f(x)\,dx = \frac{h}{3}(f(-h)+4f(0)+f(h))+\int_{-h}^h\frac{f'''(\xi(x))}{6}x(x-h)(x+h)\,dx.$$

Of course, since $x(x-h)(x+h)$ changes signs on $[-h,h]$, we can't use the mean value theorem.

I have seen the following lemma used to fix this problem: $$\int_{-h}^h\frac{f'''(\xi(x))}{6}x(x-h)(x+h)\,dx=\frac{-f^{(4)}(\mu)}{24}\int_{-h}^hx^2(x-h)(x+h)\,dx$$ for some $\mu\in(-h,h)$.

Then the right hand side evaluates to $-\frac{f^{(4)}(\mu)}{90}h^5$ as desired.

What is the proof of this lemma? I imagine it's integration by parts followed by mean value theorem, but I can't seem to figure it out.

Doug Torrance
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1 Answers1

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I don't believe there can be a lemma that magically transforms the integral with an arbitrary $\xi(x)$ into the right hand side. What probably happens is a better error bound: $$f(x)=L(x)+\frac{f^{4}(\mu(x))}{24}x^2(x-h)(x+h) \tag{1}$$ valid for even functions $f$. (Any $f$ can be decomposed into even and odd parts, with odd part disregarded since it contributes $0$ to the integral and to Simpson's rule.)

The proof of (1) follows the proof of the Lagrange interpolation error: introduce $\nu$ such that $$g(t) : = f(t) - L(t) - \nu t^2(t-h)(t+h)$$ vanishes at $t=x$. Observe that $g$ vanishes at $\pm h$, $x$, and also at $0$ together with $g'$. Hence, $g' $ has at least $4$ zeros on $(-h,h)$, $g''$ has at least $3$, etc, $g^{(4)}$ has at least one zero. From $g^{(4)}(\mu)=0$ we get $\nu = f^{(4)}(\mu)/24$.

Then you integrate (1), and that's it. The happy incident is that the Lagrange intepolant $L$ is also a Hermite interpolant, matching the (zero) derivative of $f$ at $x=0$.

  • Can you please clarify how $g(t)$ vanishes at $x$? Is $x$ a dummy variable, as well as $t$ in $g(t)$? – sequence Jan 30 '19 at 18:44