Ch. 11.2 Newton’s Forward and Backward Interpolation

Newton's Forward interpolation graphical presentation

Newton's forward difference interpolation is a method used to estimate values of a function at certain points based on known values of the function at other points. This method is particularly useful for polynomial interpolation when we have evenly spaced data points.

Key Concepts

Data Points: We need a set of data points \((x_0, y_0), (x_1, y_1), \ldots, (x_n, y_n)\), where \(y_i = f(x_i)\).

Forward Differences: These are used to create a difference table. The first forward difference is defined as:

\[ \Delta y_i = y_{i+1} - y_i \]

The second forward difference is:

\[ \Delta^2 y_i = \Delta y_{i+1} - \Delta y_i \]

And so on, until we reach the \(n\)-th forward difference.

Interpolation Formula: If we want to estimate the value of the function at a point \(x\) that is close to \(x_0\), we can use the interpolation polynomial:

\[ \small P(x) = y_0 + \frac{(x - x_0)}{h} \Delta y_0 + \frac{(x - x_0)(x - x_1)}{2!h^2} \Delta^2 y_0 \]

\[ \small + \frac{(x - x_0)(x - x_1)(x - x_2)}{3!h^3} \Delta^3 y_0 + \ldots \]

where \(h\) is the interval between the \(x\) values (assuming they are evenly spaced).

Example

Let's consider an example with the following data points:

\[ \begin{array}{|c|c|} \hline x & f(x) \\ \hline 0 & 1 \\ 1 & 2 \\ 2 & 3 \\ 3 & 4 \\ 4 & 5 \\ \hline \end{array} \]

Step 1: Create the Difference Table

Calculate Forward Differences:

\[ \begin{array}{|c|c|c|c|c|} \hline x & f(x) & \Delta y & \Delta^2 y & \Delta^3 y \\ \hline 0 & 1 & 1 & 0 & 0 \\ 1 & 2 & 1 & 0 & 0 \\ 2 & 3 & 1 & 0 & 0 \\ 3 & 4 & 1 & 0 & 0 \\ 4 & 5 & & & \\ \hline \end{array} \]

First forward differences \((\Delta y)\):

\[ \Delta y_0 = 2 - 1 = 1 \] \[ \Delta y_1 = 3 - 2 = 1 \] \[ \Delta y_2 = 4 - 3 = 1 \] \[ \Delta y_3 = 5 - 4 = 1 \]

Second forward differences \((\Delta^2 y)\):

\[ \Delta^2 y_0 = 1 - 1 = 0 \] \[ \Delta^2 y_1 = 1 - 1 = 0 \] \[ \Delta^2 y_2 = 1 - 1 = 0 \]

Third forward differences \((\Delta^3 y)\):

\[ \Delta^3 y_0 = 0 - 0 = 0 \] \[ \Delta^3 y_1 = 0 - 0 = 0 \]

Step 2: Use the Interpolation Formula

Now, let's estimate \(f(2.5)\):

Given \(h = 1\) (since the spacing is 1 between each \(x\)), \(x = 2.5\), \(x_0 = 2\), \(y_0 = f(2) = 3\), \(\Delta y_0 = 1\), \(\Delta^2 y_0 = 0\).

Plugging into the formula:

\[ P(2.5) = y_0 + \frac{(x - x_0)}{h} \Delta y_0 \]

\[ \small = 3 + (2.5 - 2) \cdot 1 + \frac{(2.5 - 2)(2.5 - 1)}{2!} \cdot 1^2 \cdot 0 \]

\[ \small = 3 + 0.5 \cdot 1 + 0 = 3 + 0.5 = 3.5 \]