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This is part of a document on numerical analysis. I have included its link below. In the derivation for the error in the linear interpolation approximation on page 84, the author "brings" the ξ(x) term of the integral outside using the mean value theorem generalized for integrals. I cannot see how this theorem is used; I am specifically referring to the highlighted conditions from a Wikipedia article in the other image.

Conditions for mean value theorem for integrals.

I haven't found anything on this thus far on the internet. I appreciate any feedback.

Thanks.

MarianD
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Panteleimon Tassopoulos
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  • A different but much more flexible approach is demonstrated in this answer to a the similar question for Simpson's rule. You will find that is possible to adapt use of the generalized mean value theorem of differentiation to the trapezoidal rule as well. – Carl Christian Mar 22 '19 at 20:34

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This is the weighted or extended version of the mean value theorem, $\int_a^bw(x)g(x)dx=g(c)\int_a^bw(x)dx$ if $w$ has a uniform sign. But you are right, one needs that $f$ is continuous for that.

Here we can show that the function $g$ in $$ f(x)-P_1(x)=(x-a)(x-b)g(x) $$ is continuous on $[a,b]$ if $f$ is continuously differentiable there and we know from the interpolation error formula that its values are equal to $g(x)=\frac12f''(\xi_x)$ for some $\xi_x\in(a,b)$. There is no need for the relation $x\mapsto\xi_x$ to be continuous as well.

Proof of the continuity of $g$: Expanding the linear interpolation one finds an alternative way to write $g(x)$ as \begin{align} g(x)(x-a)(x-b)&=f(x)-f(a)-(x-a)\frac{f(b)-f(a)}{b-a} \\ &=\frac{(f(x)-f(a))(b-x)-(x-a)(f(b)-f(x))}{b-a} \\[1em] \implies g(x)&=\frac{\frac{f(b)-f(x)}{b-x}-\frac{f(x)-f(a)}{x-a}}{b-a}=[a,x,b]f \end{align} so that for the limits of $g$ to exist in $a$ and $b$ we need that the difference quotients of $f$ converge there, that is, that $f$ is differentiable in these points.

Now inserted into the mean value theorem you get $$ \int_a^b(x-a)(x-b)g(x)dx=g(c)\int_a^b(x-a)(x-b)dx=\frac12f''(\xi_c)\cdot (-\frac16)(b-a)^3. $$

Lutz Lehmann
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  • I understand that $g$ is continuous on $(a,b)$. If $f''$ is continuous at $a$, then I understand why $g$ can be extended continuously to the point $a$ and similarly for $b$. Would you comment on the need for differentiability at the end points? In particular, does this approach work if $f''$ is not continuous at $a$ and $b$? In the past, you have used the generalized mean value theorem of differentiation to analyze this and other quadrature rules. That approach did not need differentiability at the endpoints. – Carl Christian Mar 22 '19 at 20:28
  • The for me standard proof of the error formula uses Rolle's theorem repeatedly. It constructs a function that has roots at $a,x,b$ and concludes that the second derivative has at least on root inside the interval $(a,b)$. This root is $\xi_x$. So yes, the second derivative is not needed at the end points. However the first derivative is needed to continuous there to prove continuity of $g$. – Lutz Lehmann Mar 22 '19 at 21:08
  • I am sorry for being unclear. I understand how to derive the error formula $f(x) - p_1(x) = f''(\xi)(x-a)(x-b)$ using Rolle's theorem repeatedly. What I do not understand is why or rather when $g(x) = \frac{f(x) - p_1(x)}{(x-a)(x-b)}$ is continuous at $a$ and $b$. If $f''$ is continuous, then the continuity of $g$ at $a$ or $b$ is immediate. I am sure how to adjust the proof when $f''$ is not assumed to be continuous. – Carl Christian Mar 22 '19 at 21:38
  • I meant to write: "I am not sure how to adjust the proof when $f''$ ..." – Carl Christian Mar 22 '19 at 21:45
  • Dear Carl Christian, how is the continuity of g(x) immediate at a and b, if f''(x) is assumed continuous on [a,b]? – Panteleimon Tassopoulos Aug 04 '19 at 15:08
  • @PanteleimonTassopoulos : You only need that $f'(a)$ and $f'(b)$ exist. I re-ordered the answer so that that may be more obvious. You need the second derivative inside the interval where the interpolation error formula takes its values from. – Lutz Lehmann Aug 04 '19 at 15:45
  • Okay now I see, thanks a lot. – Panteleimon Tassopoulos Aug 04 '19 at 17:08