Newton's Divided Difference

Linear Interpolation

The simplest form of interpolation is linear – connect two data points with a straight line.

We can use slope/similar triangles to solve this. For some $x_0<x<x_1$

$$\frac{f_1(x)-f(x_0)}{x-x_0}=\frac{f(x_1)-f(x_0)}{x_1-x_0}$$

Rearranging:

$$f_1(x)=f(x_0)+\frac{f(x_1)-f(x_0)}{x_1-x_0}(x-x_0)$$

*The 1 in $f_1(x)$ refers to first-order.

Quadratic Interpolation

If there are 3 data points, we instead use a 2nd order fit:

$$\begin{array}{rl}{f}_2(x)& =b_0+b_1(x-x_0)+b_2(x-x_0)(x-x_1)\end{array}$$

Here, we basically have:

$$\begin{array}{rl}{a}_0& =b_0-b_1{x}_0+b_2{x}_0{x}_1\\ a_1& =b_1-b_2{x}_0-b_2{x}_1\\ a_2& =b_2\\ f(x)& =a_0+a_{1}x+a_2{x}^2\end{array}$$

To determine the values of the coefficients, we use:

$$\begin{array}{rl}{b}_0& =f(x_0)\\ b_1& =\frac{f(x_1)-f(x_0)}{x_1-x_0}\\ b_2& =\frac{\frac{f(x_2)-f(x_1)}{x_2-x_1}-\frac{f(x_1)-f(x_0)}{x_1-x_0}}{x_2-x_0}\end{array}$$