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Let the function $f(x) = \ln(x)$ be approximated by an interpoation polynomial of degree of 9 with 10 nodes uniformly distributed in the interval $[1,2]$. What bound can be placed on the error?

I've been trying to think over the problem for the past day, but i'm still not sure sure how I am supposed to do this problem.

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  • $\begingroup$ How are you interpolating? $\endgroup$
    – user142299
    Apr 14, 2014 at 15:45

1 Answer 1

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On http://en.wikipedia.org/wiki/Polynomial_interpolation#Interpolation_error I find the error expression of $$\frac{f^{(n+1)}(\xi)}{(n+1)!}\prod_{i=0}^n{(x-x_i)}$$ In this case $f(x)=\ln x$ so $$f^{(n+1)}(x)=n! (-1)^{n}x^{-(n+1)}$$ We have $$|f^{(n+1)}(x)|<n!$$ and $$\prod_{i=0}^n{(x-x_i)}<1$$ So in this case an error bound is just $$\frac{n!}{(n+1)!}={1\over 10}$$

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  • $\begingroup$ ah, makes sense! $\endgroup$
    – Ozera
    Apr 14, 2014 at 16:08
  • $\begingroup$ @Ozera See this article: math.okstate.edu/~binegar/4513-F98/4513-l16.pdf It gives a good explanation of that formula. $\endgroup$
    – user142299
    Apr 14, 2014 at 16:16
  • $\begingroup$ Sorry for digging an old post back up, but your second centred line of math should read $f^{n+1}(x) = (n)!*(-1)^n*x^{-(n+1)}$, no? This factor of $9!$ greatly changes the error bound. $\endgroup$
    – NNN
    Mar 15, 2017 at 2:17
  • $\begingroup$ @Alex Yes thanks. $\endgroup$
    – user142299
    Mar 15, 2017 at 5:50

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