A numerical scheme for finite differences is a mathematical way to approximate a derivative of a function using finite differences, which are small values that represent the difference between two nearby points on a grid. This allows us to approximate the derivative of a function by using a finite set of data points rather than an infinitely small or continuous curve.

There are many different numerical schemes that can be used to approximate derivatives using finite differences, and the specific scheme that is used will depend on the problem being solved and the desired accuracy of the approximation. Some common numerical schemes for finite differences include forward differences, backward differences, central differences, and higher-order differences.

To understand how these schemes work, let’s consider a simple example of approximating the derivative of a function at a single point using finite differences.

Suppose we have a function $f(x)$ and we want to approximate the derivative of this function at a point x0. One way to do this is to use a forward difference, which involves calculating the difference between the function value at x0 and the function value at x0 + h, where h is a small value known as the step size.

The forward difference formula is given by:

{df\over dx}_{x=x_0} = {(f(x_0 + h) - f(x_0))\over h}

This formula allows us to approximate the derivative of the function at $x_0$ by calculating the slope of the line connecting the two points $x_0$ and $x_0 + h$ on the function. The smaller the value of $h$, the more accurate the approximation will be.

Backward differences and central differences are similar to forward differences, but use different combinations of function values to approximate the derivative. Higher-order differences use even more function values to improve the accuracy of the approximation.

Here are examples of forward, backward, and central differences:

Forward differences:

Suppose we have a function $f(x)$ and we want to approximate the derivative of this function at a point $x_0$. One way to do this is to use a forward difference, which involves calculating the difference between the function value at x0 and the function value at $x_0 + h$, where h is a small value known as the step size.

The forward difference formula is given by:

{df \over dx}_{x=x_0} = {f(x_0 + h) - f(x_0)\over h}

For example, suppose we have the function $f(x) = x^2$ and we want to approximate the derivative of this function at $x_0 = 2$. We can use the forward difference formula with a step size of $h = 0.1$ to get the following approximation:

{df \over dx}_{x=2} = {f(2 + 0.1) - f(2) \over 0.1} = (2.1^2 - 2^2)/0.1 = 4.2 - 4 = 0.2

This means that the derivative of the function at $x_0 = 2$ is approximately $0.2$.

Backward differences:

Backward differences are similar to forward differences, but use the difference between the function value at $x_0$ and the function value at $x_0 – h$ to approximate the derivative.

The backward difference formula is given by:

{df \over dx}_{x=x_0} = {f(x_0) - f(x_0 - h) \over h}

For example, suppose we have the same function $f(x) = x^2$ and we want to approximate the derivative of this function at $x_0 = 2$. We can use the backward difference formula with a step size of $h = 0.1$ to get the following approximation:

{df \over dx}_{x=2} = {f(2) - f(2 - 0.1)\over 0.1} = (2^2 - (2 - 0.1)^2)/0.1 = 4 - 1.99)/0.1 = 2.01/0.1 = 20.1

This means that the derivative of the function at $x_0 = 2$ is approximately 20.1.

Central differences:

Central differences use the average of the forward and backward differences to approximate the derivative.

The central difference formula is given by:

{df \over dx}_{x=x_0} = {f(x_0 + h) - f(x_0 - h) \over 2h}

For example, suppose we have the same function $f(x) = x^2$ and we want to approximate the derivative of this function at $x_0 = 2$. We can use the central difference formula with a step size of $h = 0.1$ to get the following approximation:

{df \over dx}_{x=2} = {f(2 + 0.1) - f(2 - 0.1) \over 20.1} = (2.1^2 - (2 - 0.1)^2)/(20.1) = (4.41 - 1.99)/0.2 = 6.4/0.2 = 32

This means that the derivative of the function at $x_0 = 2$ is approximately $32$.