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Sahithyan's S2
Sahithyan's S2 — Methods of Mathematics

Newton's Interpolating Polynomials

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(xx0)+a2(xx0)(xx1)++an(xx0)(xxn1)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+1xif[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+2xif[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+k1]xi+kxif[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](xx0)(xx1)(xxk1)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})