I'm wondering How to show the following:

$$ f(x)=-\frac{(x-x_1)(x-2x_0+x_1)}{(x_1-x_0)^2}f(x_0)+\frac{(x-x_0)(x-x_1)}{x_0-x_1}f'(x_0)+\frac{(x-x_0)^2}{(x_1-x_0)^2}f(x_1)+R(x) $$

where $R(x)=\frac{1}{6}(x-x_0)^2(x-x_1)f'''(\xi)$

I want to use Taylor to show this, but how can I kill f''(x_1),f''(x_0), it's impossible to do this

so could anyone give me some help? thanks


2 Answers 2


This problem is estimate the trunction error of Hermite interplotion , Taylor series is the particular instance of Hermite interplotation . I solve this problem with the method by defining an Auxiliary function and Rolle's theorem.

Let $$ P_2(x) =-\frac{(x-x_1)(x-2x_0+x_1)}{(x_1-x_0)^2}f(x_0)+\frac{(x-x_0)(x-x_1)}{x_0-x_1}f'(x_0)+\frac{(x-x_0)^2}{(x_1-x_0)^2}f(x_1) $$ we can think $P_2(x)$ is an polynomail of 2th-degree approximation or interplotation of function $f(x)$ with totally three conditions $f(x_0)=P_2(x_0), f(x_1)=P_2(x_1), f'(x_0)=P_2(x_0)$ , two at point $x_0$ (Guarantee better smoothness) and one at $x_1$

Construct the auxiliary function as follows $$ \phi(t) = f(t)-P_2(t) - \frac{(t-x_0)^2(t-x_1)}{(x-x_0)^2(x-x_1)}\left(f(x)-P_2(x)\right) $$

we obsevered that $\phi(x) = \phi(x_0) = \phi(x_1)=\phi'(x_0) = 0$ . Then repeat the Rolle's Theorem , we can conclude that $\phi'''(t) $ must has one root $\xi \in (\min(x,x_0,x_1),\max(x,x_0,x_1))$ such that $\phi'''(\xi) = 0 $ and note that $P_2'''(t) \equiv 0 $ , then we get following $$ 0=\phi'''(\xi) = f'''(\xi) - \frac{6}{(x-x_0)^2(x-x_1)}\left(f(x)-P_2(x)\right) $$ thus we conclude the truction error $R(x) = f(x) - P_2(x) = \frac 16 f'''(\xi)(x-x_0)^2(x-x_1) $

  • $\begingroup$ I hope to inspire you and broaden your thinking $\endgroup$
    – Hugo
    Commented Mar 25, 2021 at 6:02
  • $\begingroup$ Yeah, exactly it’s a problems given by my numerical analysis teacher, thanks for your answers! $\endgroup$
    – user867836
    Commented Mar 25, 2021 at 6:03
  • $\begingroup$ That’s great. I’m also learning numerical analysis. I would be very grateful if I could get your vote for my answer. $\endgroup$
    – Hugo
    Commented Mar 25, 2021 at 6:07
  • $\begingroup$ From your profile I think maybe we are college classmate! What a coincidence. And may I know your name? I’m Bowen_Liu, a student in Information and Computing Science :D $\endgroup$
    – user867836
    Commented Mar 25, 2021 at 7:07
  • $\begingroup$ Tang Zhiguo , we are classmates in the same classroom, hahahahaha , we do the same homework $\endgroup$
    – Hugo
    Commented Mar 25, 2021 at 7:10

These sorts of questions are usually solved using the mean value theorem several times. Let $x^*$ be any fixed $x\in(x_0,x_1)$. Then, to start, let

$$G(x) = f(x) -\left(-\frac{(x-x_1)(x-2x_0+x1)}{(x_1-x_0)^2}f(x_0)+\frac{(x-x_0)(x-x_1)}{x_0-x_1}+\frac{(x-x_0)^2}{x_1-x_0}\right)-\lambda(x-x_0)^2(x-x_1)$$

Since $(x^*-x_0)^2(x^*-x_1)\neq 0$, we can find $\lambda$ such that $G(x^*)=0$.

Now, we know that $G(x)$ has zeros at $x_0$, $x^*$, and $x_1$. Thus by the mean value theorem there exists some point in each of $(x_0,x^*)$ and $(x^*,x_1)$ such that $G'(x)=0$.

Now, we know that $G'(x)=0$ at $x_0$ two other points, just call them $\bar{x}_1$ and $\bar{x}_2$. Again applying the meal value theorem, there exists some $\bar{\bar{x}}_1 \in (x_0,\bar{x}_1)$ and some $\bar{\bar{x}}_2 \in (\bar{x}_1,\bar{x}_2)$ such that $G''(\bar{\bar{x}}_1)=G''(\bar{\bar{x}}_2)=0$.

Finally, we can now apply the mean value theorem one more time to say there exists $\xi \in (\bar{\bar{x}}_1,\bar{\bar{x}}_2)$ such that $G'''(\xi)=0$. Turning our attention back to the definition of $G(x)$, we can see that $0=f'''(\xi)-6\lambda$. (To see this, take the derivative of both sides three times. Note: This is easier than it looks initially because the part inside the large parenthesises is quadradic and will vanish when differentiated three times.)

And thus $\lambda = \frac{1}{6}f'''(\xi)$. Since this process can be repeated for any $x$, we are done.

  • $\begingroup$ yeah i got it, thanks to your detailed response, I think I may focus on Taylor too much! $\endgroup$
    – user867836
    Commented Mar 25, 2021 at 5:07
  • $\begingroup$ I'm glad it was helpful. I would have to think about it to be sure, but I'm pretty sure Taylor's theorem can't be usefully applied in any direct way since $\xi$ depends on our choice of $x$. In practice, we would just say the error in our approximation is bounded by $$ \frac{1}{6}\left[\max_{\xi\in (x_0,x_1)}{f'''(\xi)}\right](x-x_0)^2(x-x_1)$$ By the way, if you like my answer please accept it as the answer to your question. $\endgroup$ Commented Mar 25, 2021 at 5:18

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