Talk:Newton polynomial

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Can we get an algorithm here? —Preceding unsigned comment added by 24.199.203.189 (talk) 20:57, 17 February 2008 (UTC)

I did a complete rewrite of the article to make it clearer and to use the same style, notation as in polynomial interpolation and Lagrange polynomial. MathMartin 13:16, 17 Aug 2004 (UTC)

[edit] Newton series

Hi Guys,

We seem to be missing the textbook definition of a newton series

\sum_n a_n \frac{(x)_n}{n!} = \sum_n a_n {n+x \choose n-1}

Sigh. I'll try to add and harmonize at some point ... linas 16:50, 27 January 2006 (UTC)

Never mind. See Newton series. linas 21:48, 27 January 2006 (UTC)

[edit] Numerical Precision in Example

It's probably worth pointing out that the example provided is only worked out to very low precision. If someone takes that example and works it out on a computer with double precision (or probably even single precision), they will quickly see that the constant and quadratic terms in the interpolating polynomial are 0. Intuitively, this should make sense because the tangent function is odd and the data points are symmetric about x = 0. —Preceding unsigned comment added by 151.200.114.163 (talk) 21:15, 24 October 2007 (UTC)

Right. I added a comment that the computation uses only six digits. -- Jitse Niesen (talk) 12:08, 1 April 2008 (UTC)

Looking at a plot of the function, it appears that all the coefficient are wrong. It is obvious that the coefficient for x^1 should be positive and somewhere close to 1. —Preceding unsigned comment added by 24.58.11.37 (talk) 01:37, 1 April 2008 (UTC)

If the interpolation points are close together, then the coefficient for x^1 is indeed close to 1. However, here the space between the points is fairly wide so there is a large error. I computed the interpolating polynomial myself to check and the end result is correct (to the limited precision of the computation, of course). -- Jitse Niesen (talk) 12:08, 1 April 2008 (UTC)