0

I have a function $f:[a,b] \to \mathbb{R}$, where $f \in C^{2}([a,b])$. Admit that $p_{1}$ resulting from Lagrange interpolation of f in $x_{0},x_{1} \in [a,b]$. Then I must show that for any $x \in [a,b]$ we have $\lvert f'(x) - p'_{1}(x) \rvert \leq \frac{(x-x_{0})^{2}+(x-x_{1})^{2}}{2(x_{1}-x_{0})} max_{x \in [a,b]} {\lvert f''(x) \rvert} $. I have tried to use a simillar technique to what is used when doing an estimate for the error of the function and the polinomial, without the derivatives, which is based on the number of zeros of the function $f - p_{1}$ and Rolle's Theorem, but it doesn't seem to be working. Thanks a lot.

EDIT: I think this may be related with the following formula I have from the teacher's notes: $f'(x)-p'_{1}(x) = f_[x_{0},...,x_{n},x,x]W_{n+1}(x) + f_[x_{0},...,x_{n},x]W'_{n+1}(x)$ which can be rewritten as $f'(x)-p'_{1}(x) = \frac{f^{(n+2)}(\xi_{2})}{(n+2)!}W_{n+1}(x) + \frac{f^{(n+1)}(\xi_{1})}{(n+1)!}W'_{n+1}(x)$ but I'm having trouble getting one formula from the other, because in our case n=1 so we should get a third derivative on f... Maybe it's a mistake in the exercise?

Pedro Silva
  • 421
  • Note that $f(x)-p_1(x)=fx_0,x_1,x(x-x_1)$ and thus $f'(x)-p_1'(x)=fx_1,x_1,x,x(x-x_1)+fx_0,x_1,x$, but this would lead to a third derivative in the error term. – Lutz Lehmann Aug 16 '21 at 11:39
  • I'm sorry, I forgot the derivatives you're right @LutzLehmann, methods discussed were exactly what you stated in your last comment for example. We had some theorems about bounding error of the interpolating polynomial, the one that seems to relate the most to this exercise is a generalization of what you just mentioned. I don't have sources because this is from an exercise sheet from my teacher and it's in portuguese. – Pedro Silva Aug 16 '21 at 17:30

2 Answers2

1

Just to prove the formula, take Taylor expansions at $x$, $$ k=0,1:~~f(x_k)=f(x)+f'(x)(x_k-x)+\int_x^{x_k}(x_k-s)f''(s)\,ds $$ and then the difference to eliminate $f(x)$ $$ f(x_1)-f(x_0)=f'(x)(x_1-x_0)+\int_{x_0}^{x_1}w(s)f''(s)\,ds,~~~ w(s)=\begin{cases} x_0-s,&s\in[x_0,x)\\ x_1-s,&s\in[x,x_1] \end{cases} $$ Then $$ \left|\int_{x_0}^{x_1}w(s)f''(s)\,ds\right|\le\|f''\|_{[x_0,x_1],\max}\int_{x_0}^{x_1}|w(s)|\,ds $$ which after integration leads to the given bound.

There might be a more systematic approach that generalizes to higher degree interpolation.

Lutz Lehmann
  • 126,666
  • I did the calculations and it gives indeed the bound we wanted, but does this also work for $x \in [a,x_0]$ and $x \in [x_1,b]$? Because it seems that the way you stated it it's for $x \in [x_0,x_1]$ – Pedro Silva Oct 05 '21 at 12:13
0

Just an obvious suggestion: Comparing this expression with the standard expression for the interpolation error $$ |f(x)-p_1(x)| = \dfrac{f''(\xi)}{2}|(x-x_0)(x-x_1)|, $$

it suffices to show that $$ |(x-x_0)(x-x_1)| \leq \dfrac{(x-x_0)^2+(x-x_1)^2}{(x_1-x_0)}. $$

PierreCarre
  • 20,974
  • 1
  • 18
  • 34