My Math Forum Polynomial interpolation, interpolation error

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 April 14th, 2017, 09:34 AM #1 Newbie   Joined: Jun 2016 From: France Posts: 3 Thanks: 0 Polynomial interpolation, interpolation error Hello! To begin, I'm sorry for my bad English, I translated it from a French exercise. 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 that $d(Q) = \delta(Q)$ for any polynomial $Q$ of degree less than or equal to 2. I'm blocked on two questions: a) 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 there exists $y \in [a,b]$ such that $r''(y) = 0$. b) 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 $f[a,c,b] = \delta(f)/2$) Here are my answers: a) $f$ is $3$ times continuously differentiable on $[a,b]$ => $r$ has $3$ roots on $[a,b]$ => $r'$ has $2$ roots on $[a,b]$ => $r''$ has $1$ root on $[a,b]$ (Rolle's theorem). $r(a) = r(c) = r(b)$ $\exists y_1 \in ]a,c[, r'(y_1) = 0$ $\exists y_2 \in ]c,b|, r'(y_2) = 0$ $r'(y_1) = r'(y_2)$ => $\exists y \in ]y_1,y_2[, r''(y)=0$ Is it right? b) Seriously, I don't know! I know that for a $f \in C^{n+1}([a,b]), |f(x) - P(x)| \leq \dfrac{|(x-x_0)(x-x_1)...(x-x_n)|}{(n+1)!} \sup_{x\in [a,b]} |f^{(n+1)}(x)|$ with $P$ Lagrange polynomial at the points $x_0,...,x_n$. But how can I apply it for $|d(f) - \delta(f)| = |r''(c)|$? Could someone help me? Thank you in advance. Last edited by skipjack; April 14th, 2017 at 01:51 PM.
April 16th, 2017, 04:09 AM   #2
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Quote:
 Originally Posted by loveelya a) $f$ is $3$ times continuously differentiable on $[a,b]$ => $r$ has $3$ roots on $[a,b]$ => $r'$ has $2$ roots on $[a,b]$ => $r''$ has $1$ root on $[a,b]$ (Rolle's theorem). $r(a) = r(c) = r(b)$ $\exists y_1 \in ]a,c[, r'(y_1) = 0$ $\exists y_2 \in ]c,b|, r'(y_2) = 0$ $r'(y_1) = r'(y_2)$ => $\exists y \in ]y_1,y_2[, r''(y)=0$ Is it right?
I think so yes. Successive application of Rolles theorem will show $r''(c) = 0$.
Quote:
 Originally Posted by loveelya b) Seriously, I don't know! I know that for a $f \in C^{n+1}([a,b]), |f(x) - P(x)| \leq \dfrac{|(x-x_0)(x-x_1)...(x-x_n)|}{(n+1)!} \sup_{x\in [a,b]} |f^{(n+1)}(x)|$ with $P$ Lagrange polynomial at the points $x_0,...,x_n$. But how can I apply it for $|d(f) - \delta(f)| = |r''(c)|$?
I can't really finish this, but this is all i can come up with...

Define a new function $w(t)$ such that $w(a) = w(b) = w(c) = 0$.

i.e., $w(t) = f(t) - \delta(t) - E_2(x) \cdot \dfrac{(t - a)(t - c)(t - b)}{(x - a)(x - c)(x - b)}$

Differentiating twice with respect to t gives,

$w''(t) = f''(t) - \delta''(t) - \dfrac{3 E_2(x)}{(x - a)(x - c)(x - b)}$

From a), we have $w''(c) = 0$ which gives,

$f''(c) - \delta''(c) = \dfrac{3 E_2(x)}{(x - a)(x - c)(x - b)}$

 May 5th, 2017, 08:45 AM #3 Newbie   Joined: Jun 2016 From: France Posts: 3 Thanks: 0 Thank you !!

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