Polynomial interpolant as $\varepsilon \rightarrow 0$

Question

I need a check on the following exercise:

Let $f \colon \mathbb{R} \rightarrow \mathbb{R} $ a smooth function and consider the data $(0,f(0)), (\epsilon,f(\epsilon))$ and $(1,f(1))$. Let $p(x)$ the limit as $\epsilon \rightarrow 0$ of the polynomial interpolant of the data.

i) What is the degree of $p(x)$?

ii) What data does $p(x)$ interpolate?

iii) What data does $p'(x)$ interpolate?

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Here's my attempt:

i,ii) I consider Newton's interpolation for the sets of three points. Using divided difference table, I obtain the Newton polynomial $$P(x)=f(0) + \Bigl(\frac{f(\epsilon)-f(0)}{\epsilon} \Bigr) x + \Bigl( \frac{f(1)-f(\epsilon)}{1-\epsilon} - \frac{f(\epsilon)- f(0)}{\epsilon} \Bigr) x(x-\epsilon)$$

As $\epsilon \rightarrow 0$, I note that there are several differential quotients, so $$p(x)=\lim_{\epsilon \rightarrow 0} P(x) = f(0) + f'(0)x + (f'(1) - f'(0))x^2$$

Therefore:

• The degree of $p(x)$ is two.
• $p(x)$ interpolates the points $(0,f(0))$ and $(1,f(0)+f'(1))$

iii) $$p'(x)=f'(0) + 2 (f'(1)-f'(0))$$ so $p'(x)$ interpolates $(0,f'(0))$ and $(1, 2f'(1)-f'(0))$

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Is everything okay, or am I missing something?

Answer 1

You made a slight error, $$ \lim_{ϵ→0}\frac{f(1)-f(ϵ)}{1-ϵ}=\frac{f(1)-f(0)}{1-0}=f(1)-f(0). $$ You will not get $f'(1)$ at this place, and thus anywhere else in the ensuing formulas. You could apply the mean value theorem, but that would introduce an unknown midpoint.

Answer 2

Intuitively, in the limit $p(x)$ must interpolate $f(0)$ and $f(1)$ (as $f(\epsilon)$ converges to $f(0)$), and $p'(x)$ must interpolate $f'(0)$ as $p$ interpolates two infinitesimally close points.

Fixing your formula,

$$p_\epsilon(x)=f(0) + \frac{f(\epsilon)-f(0)}{\epsilon} x + \Bigl( \frac{f(1)-f(\epsilon)}{1-\epsilon} - \frac{f(\epsilon)- f(0)}{\epsilon} \Bigr) x(x-\epsilon)$$

tends to

$$p(x)=f(0) +f'(0) x + \Bigl( f(1)-f(0) - f'(0) \Bigr) x^2,$$

which fulfills the above findings.

In blue, the function $\sin2x$. In magenta, interpolation with $\epsilon=0.1$ and in green, $\epsilon=0$.