# Use two points and a derivative to approximate a function

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

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 $$R(x) = f(x) - P_2(x) = \frac 16 f'''(\xi)(x-x_0)^2(x-x_1)$$

– Hugo
Commented Mar 25, 2021 at 6:02
• Yeah, exactly it’s a problems given by my numerical analysis teacher, thanks for your answers!
– user867836
Commented Mar 25, 2021 at 6:03
• That’s great. I’m also learning numerical analysis. I would be very grateful if I could get your vote for my answer.
– Hugo
Commented Mar 25, 2021 at 6:07
• 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
– user867836
Commented Mar 25, 2021 at 7:07
• Tang Zhiguo , we are classmates in the same classroom, hahahahaha , we do the same homework
– 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.

• yeah i got it, thanks to your detailed response, I think I may focus on Taylor too much!
– user867836
Commented Mar 25, 2021 at 5:07
• 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. Commented Mar 25, 2021 at 5:18