I'm sorry for my bad english, I translated it from a French exercice.

Let $f$ be a function of class $C^3$ defined on $[a,b]$, and $c=(a+b)/2$. We want to approximate the quantity $d(f) = f''(c)$ by an expression of the following form :

$\delta(f) = \lambda_0 f(a) + \lambda_1 f(c) + \lambda_2 f(b)$

which is such as $d(Q) = \delta(Q)$ for any polynomial $Q$ of degree less than or equal to 2.

Let $P$ be the Lagrange interpolation polynomial of $f$ on the points $(a,c,b)$ and $r$ the interpolation error $r(x) = f(x) - P(x)$. Show that $d(f) - \delta(f) = r''(c)$.

Deduce that $|d(f) - \delta(f)| \leq \dfrac{b-a}{2} \sup_{x \in [a,b]} |f'''(x)|$.

(We found previously that the divided difference of the function f on the points $(a,c,b)$ verify $f[a,c,b] = \delta(f)/2$)

I'm blocked on this question. Could someone help me ? Thank you in advance.

  • $\begingroup$ There are no more detail on $c$ in the wording. I think it's $c = (a+b)/2$ throughout the exercise ! $\endgroup$ – Mélanie De la Cheminée Apr 12 '17 at 12:06
  • $\begingroup$ my bad. I misunderstood the question. $\endgroup$ – user251257 Apr 12 '17 at 15:08
  • $\begingroup$ Surely because of my translation ! I just would like to know if it's a formula that I don't know, and what is it, because I study as an autodidact. $\endgroup$ – Mélanie De la Cheminée Apr 12 '17 at 18:37
  • $\begingroup$ I know that for a function $f \in C^{n+1}([a,b])$, $\forall x \in [a,b], |f(x)-P(x)| \leq \dfrac{|(x-x_0)(x-x_1)...(x-x_n)|}{(n+1)!} \sup_{x \in [a,b]} |f^{(n+1)}(x)|$ with $a \leq x_0 < x_1 < ... < x_n \leq b$ and $P$ the Lagrange polynomial at the points $x_0,...,x_n$. But how can I apply it for $|d(f)-\delta(f)| = |r''(c)|$ ? $\endgroup$ – Mélanie De la Cheminée Apr 14 '17 at 13:10

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