Semester 2 / Methods of Mathematics / Numerical Methods / Newton's Interpolating Polynomials

Newton's Interpolating Polynomials

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Suppose that Pn(x)P_n (x) is the nn-th interpolating polynomial that agrees with the function ff at the distinct numbers x0,x1,…,xnx_0, x_1,\dots,x_n. PnP_n has the form:

Pn(x)=a0+a1(x−x0)+a2(x−x0)(x−x1)+⋯+an(x−x0)…(x−xn−1)P_n (x) = a_0 + a_1(x - x_0) + a_2(x - x_0)(x - x_1) + \dots + a_n(x - x_0)\dots(x-x_{n-1})

for appropriate constants a0,a1,…,ana_0, a_1,\dots,a_n. The constants can be found by setting xx to the known data points x0,x1,…,xnx_0,x_1,\dots,x_n.

Divided-Difference Notation

Divided differences are defined with respect to a set of distinct numbers.

Zeroth Divided Difference

f[xi]=f(xi)f[x_i] = f (x_i)

First Divided Difference

f[xi,xi+1]=f(xi+1)−f(xi)xi+1−xif[x_i,x_{i+1}] = \frac{f(x_{i+1}) - f(x_i)}{x_{i+1} - x_i}

Second Divided Difference

f[xi,xi+1,xi+2]=f[xi+1,xi+2]−f[xi,xi+1]xi+2−xif[x_i,x_{i+1},x_{i+2}] = \frac{f[x_{i+1} , x_{i+2}] - f [x_i,x_{i+1}]}{x_{i+2} - x_i}

k-th Divided Difference

f[xi,xi+1,xi+2,…,xi+k]=f[xi+1,xi+2,…,xi+k]−f[xi,xi+1,xi+2,…,xi+k−1]xi+k−xif[x_i,x_{i+1},x_{i+2},\dots,x_{i+k}] = \frac{f[x_{i+1},x_{i+2},\dots,x_{i+k}]-f [x_i,x_{i+1},x_{i+2},\dots,x_{i+k-1}]} {x_{i+k} - x_i}

Now Pn(x)P_n(x) can be rewritten in a form called Newton’s Divided Difference:

Pn(x)=f[x0]+∑k=1nf[x0,x1,…,xk](x−x0)(x−x1)…(x−xk−1)P_n(x)=f[x_0] + \sum_{k=1}^n f[x_0,x_1,\dots,x_k](x-x_0)(x-x_1)\dots(x - x_{k-1})

Wolfram Mathematica notebook

The summary notebook defines every function used in this module, with documentation and no examples, as a starting point for your own problems.

Written by September 29, 2026 1 min read
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