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I wanted to calculate the 4th order of Newton's interpolating polynomial by hand but the results weren't that good, so I wanted to make sure that I got it right is $$a_4 =\frac{f(x_4)-a_0 - a_1(x_4-x_0) -a_2(x_4-x_0)(x_4-x_1) -a_3 (x_4-x_0)(x_4-x_1)(x_4-x_2)}{(x_4-x_0)(x_4-x_1)(x_4-x_2)(x_4-x_3)}$$ ??

hn_gara
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  • Yes, that is correct, $f(x_4)=p_4(x_4)=p_3(x_4)+a_4(x_4-x_0)(x_4-x_1)(x_4-x_2)(x_4-x_3)$. But why not use the usual iteration to compute $a_4=f[x_0,x_1,x_2,x_3,x_4]$ from the divided differences of lower order? – Lutz Lehmann Dec 16 '22 at 14:14
  • @LutzLehmann I don't know what the usual iteration could u explain more please – hn_gara Dec 16 '22 at 17:54
  • It is where you compute $f[a,b,…,y,z]=\dfrac{f[a,…,y]-f[b,…,z]}{a-z}$ recursively. This is usually arranged in a "forward difference table" like in https://math.stackexchange.com/a/312586/115115. – Lutz Lehmann Dec 17 '22 at 08:25

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