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Given $f(x)=\sqrt[3] x$, find an approximation of $\sqrt[3]{20}$ using Lagrange interpolation method.
$x_0=0$, $x_1=1$, $x_2=8$, $x_3=27$ and $x_4=64$
$f(x_0)=0$, $f(x_1)=1$, $f(x_2)=2$, $f(x_3)=3$ and $f(x_4)=4$

What I've done is, calculating $p(20)=-1.3139$ which is obviously wrong. Why is that?

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It's basically a cautionary tale for interpolating polynomials. They don't necessarily have to behave qualitatively like the functions they're supposed to be approximating. – J. M. Feb 14 '12 at 12:42
up vote 7 down vote accepted

interpolating polynomial and cube root

That's a rather common problem with interpolating polynomials. For starters, the function you are approximating has a vertical tangent. Since polynomials can't have vertical tangents, any attempt to approximate the cube root with an interpolating polynomial will be troublesome. The other thing is that even polynomials of modest degree tend to be wiggly in between widely spaced interpolation points (except in some special cases), so the inaccuracy you observed is par for the course.

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As per J.M.'s answer, you can see why interpolation method does not always result in accurate values. May be if you use more data points you could get a better answer (see note below). In any way, it is important to make sure that you got the function correct. Based on you input ([0, 0], [1, 1], [8, 2], [27, 3],[64,4]), I obtained this function: y=

enter image description here

This value was obtained with the help of this site, in case you want to check this and other methods: Interpolation Calculator

Note: I have edited this answer to include the correct points and to add reference to note below.

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"May be if you use more data points" - but not always. Runge's example comes to mind. – J. M. Feb 14 '12 at 13:21
@J.M. Thanks for the clarification. – NoChance Feb 14 '12 at 15:26

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