Numerical Analysis/Lagrange example

We'll find the interpolating polynomial passing through the points $(1,-6)$, $(2,2)$, $(4,12)$, using the Lagrange method.

We first use the formula to write the following:

$p(x)=-6\frac{(x-2)}{(1-2)}\frac{(x-4)}{(1-4)}+2\frac{(x-1)}{(2-1)}\frac{(x-4)}{(2-4)}+12\frac{(x-1)}{(4-1)}\frac{(x-2)}{(4-2)}$

After some simplification, we get:

$p(x)=-2(x-2)(x-4)-1(x-1)(x-4)+2(x-1)(x-2)$

$p(x)=-2(x^2-6x+8)-1(x^2-5x+4)+2(x^2-3x+2)$

And our answer:

$p(x)=-x^2+11x-16$.

Adding a point

Now we'll add a point to our data set, and find a new interpolating polynomial. Let us add the point $(3,-10)$ to our set. Starting over with the Lagrange formula, we write:

$p(x)=-6\frac{(x-2)}{(1-2)}\frac{(x-4)}{(1-4)}\frac{(x-3)}{(1-3)}+2\frac{(x-1)}{(2-1)}\frac{(x-4)}{(2-4)}\frac{(x-3)}{(2-3)}+12\frac{(x-1)}{(4-1)}\frac{(x-2)}{(4-2)}\frac{(x-3)}{(4-3)}-10\frac{(x-1)}{(3-1)}\frac{(x-2)}{(3-2)}\frac{(x-4)}{(3-4)}$

Simplifying, we get:

$p(x)=(x-2)(x-4)(x-3)+(x-1)(x-4)(x-3)+2(x-1)(x-2)(x-3)+5(x-1)(x-2)(x-4)$

$p(x)=(x^3-9x^2+26x-24)+(x^3-8x^2+19x-12)+2(x^3-6x^2+11x-6)+5(x^3-7x^2+14x-8)$

And our polynomial is:

$p(x)=9x^3-64x^2+137x-88$.