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I was given the following nodes:
$x_1=0$
$x_2=1$
$x_3=8$
$x_4=27$
and was asked to interpolate the Lagrange polynomial of the function $f(x)=\sqrt[3]{x}$ (meaning, I have the values: $(0,0)$,$(1,1)$,$(8,2)$,$(27,3)$ and need to interpolate the Lagrange polynomial), and to evaluate the error of interpolation at the given interval $[0,27]$.
So, I constructed the Lagrange polynomials;
$L_0=-\frac{1}{216}(x-1)(x-8)(x-27)$
$L_1=\frac{1}{182}x(x-8)(x-27)$
$L_2=-\frac{1}{1064}x(x-1)(x-27)$
$L_3=\frac{1}{4446}x(x-1)(x-8)$
and found the interpolating polynomial, $P(x)$ using the formula:
$$P(x)=\sum_{i=0}^{n=3}f_i\cdot L_i(x)$$ So far everything seems good (I hope).
The problem arises when I try to evaluate the error...
Using the following formula:
$$E(x)=f(x)-P(x)=\frac{MN}{(n+1)!}$$ Where:
$$M=\max_{x\in[0,27]}|f^{(4)}(x)|$$ $$N=\max_{x\in[0,27]}|x(x-1)(x-8)(x-27)|$$ When I try to evaluate $M$, I get: $f^{(4)}(x)=\frac{10}{27}x^{-\frac{8}{3}}=\frac{10}{27\sqrt[3]{x^8}}$, but this function has no maximum value at $[0,27]$!
How could that be?

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1 Answer 1

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First, your "error formula" is wrong: $MN/(n+1)!$ is an upper-bound of the absolute value of $E(x)$, not its value.

The objective of the exercise you're doing is to show you that this upper-bound can be undefined, while you get an actual error $\max_x E(x)$ which is finite. Thus you should calculate $\max_x E(x)$.

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  • $\begingroup$ But us says here: math.niu.edu/~dattab/MATH435.2013/INTERPOLATION.pdf at page 13 (Theorem 6.7) that "for every $x$ in the interval, there exist a number $\xi$ such that... ", in my case, $0$ is part of the interval $[0,27]$, but $E(0)$ is undefined. $\endgroup$ Jun 3, 2014 at 12:36
  • $\begingroup$ They even gave an example (page 16 example 6.8) that shows how to find it... $\endgroup$ Jun 3, 2014 at 12:39
  • $\begingroup$ @so.very.tired But your $f$ does not fulfill the assumptions of the theorem. Your $f$ is not four times differentiable on $[0,27]$. $\endgroup$
    – martini
    Jun 3, 2014 at 13:17
  • $\begingroup$ @so.very.tired: $E(0)$ is not undefined, if is $f(0)-P(0)$ and thus is $0$. $\endgroup$
    – Joce
    Jun 3, 2014 at 13:20
  • $\begingroup$ @so.very.tired: in your document, there's actually an error: Step 4 p. 226 (=p. 16...) the maximum error $MN/(n+1)!$ is not equal to some $E(\bar{x})$, it is an upper bound for $|E(x)|$ (as they state below in the example). $\endgroup$
    – Joce
    Jun 3, 2014 at 13:21

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