Numerical Analysis/Newton form example

We'll find the interpolating polynomial passing through the points $(1,-6)=(x_0,y_0)$, $(2,2)=(x_1,y_1)$, $(4,12)=(x_2,y_2)$, using the Newton form of the interpolation polynomial.

The Newton form is given by the formula $p(x)=\sum _{j=0}^k{a}_j{n}_j(x)$, where $a_j=[y_0,\dots ,y_j]$ and $n_j(x)=\prod _{i=0}^{j-1}(x-x_i)$, with $n_0(x)=1$. We start by finding each $n_j(x)$.

$n_0(x)=1$

$n_1(x)=x-1$

$n_2(x)=(x-1)(x-2)=x^2-3x+2$

Next, we find the necessary divided differences. First, $[y_0]=-6$, $[y_1]=2$, and $[y_2]=12$. For the next level, we have:

$[y_0,y_1]=\frac{2+6}{2-1}=8$

$[y_1,y_2]=\frac{12-2}{4-2}=5$

Finally, we can find:

$[y_0,y_1,y_2]=\frac{5-8}{4-1}=-1$.

Now, we can find the coefficients $a_j$.

$a_0=[y_0]=-6$

$a_1=[y_0,y_1]=8$

$a_2=[y_0,y_1,y_2]=-1$

Substituting and simplifying, we get our interpolating polynomial:

$p(x)=-6+8(x-1)-(x^2-3x+2)=-x^2+11x-16$.

Adding a point

Now let's add the point $(3,-10)=(x_3,y_3)$ to our data set and find the new polynomial using the same method. Due to the formula for the Newton form, we only have to add the term $a_3{n}_3(x)$ to our previous interpolating polynomial.

First, we have

$n_3(x)=(x-1)(x-2)(x-4)=x^3-7x^2+14x-8$.

Now to find $a_3$ we calculate some more divided differences.

$[y_3]=-10$

$[y_2,y_3]=\frac{-10-12}{3-4}=22$

$[y_1,y_2,y_3]=\frac{22-5}{3-2}=17$

$a_3=[y_0,\dots ,y_3]=\frac{17+1}{3-1}=9$

So, our new interpolating polynomial is:

$p(x)=-x^2+11x-16+9(x^3-7x^2+14x-8)=9x^3-64x^2+137x-88$.