# Thread: Bound of error(Lagrange Interpolation)

1. ## Bound of error(Lagrange Interpolation)

Hello everybody

Given the Lagrange polynomial with the table:

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

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

Ok now I've found $P_2(x)$ and I'm approximating $P_2(8.4)$ 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.

Post #10

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

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

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

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

4. Originally Posted by javax

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