# Thread: Bound of error(Lagrange Interpolation)

1. ## Bound of error(Lagrange Interpolation)

Hello everybody

Given the Lagrange polynomial with the table:

$\displaystyle \begin{tabular} {|c|c|c|c|} \hline x & 8.3 & 8.6 & 8.7 \\ \hline f(x) & 17.56492 & 18.50515 & 18.32091 \\ \hline \end{tabular}$

Find $\displaystyle P_2(x)$ for $\displaystyle x = 8.4$, the actual error and the bound of error for the function $\displaystyle f(x) = x \ln x$

Ok now I've found $\displaystyle P_2(x)$ and I'm approximating $\displaystyle P_2(8.4)$ which from my polyinomial is 17.877155 (close enough I think).
But I don't know how to find the bound of error using the remainder term! Anyone please show me the steps at least. Thanks in advance.

2. Maybe this thread can help you : http://www.mathhelpforum.com/math-he...-integral.html.
Post #10

3. Ok thanks...I want to know if this works.
Applying the formula:

$\displaystyle |f(x)-P_2(x)|\leq \frac{sup_{[a;b]}|f^{(n+1)}(x)|}{(n+1)!}\prod_{i=0}^{n}{(x-x_i)}$

now setting $\displaystyle a=8.3, b=8.7,$ I've got $\displaystyle f^{(3)}(x)=-\frac{1}{x^2}$

Is it okay to choose $\displaystyle sup_{[a;b]}|f^{(3)}(x)| = \frac{1}{8.3^2}$ since $\displaystyle |f^{(3)}(x)|$ is decreasing on [a, b] ?

4. Originally Posted by javax

Is it okay to choose $\displaystyle sup_{[a;b]}|f^{(3)}(x)| = \frac{1}{8.3^2}$ since $\displaystyle |f^{(3)}(x)|$ is decreasing on [a, b] ?
I think so.