> Basic calculus > Solution of nonlinear equations

Solution of nonlinear equations

We very often need to find the zero of a complex non-linear function when working with mathematical modelling problems, perhaps most frequently when we wish to find the value (or values) of $x$ such that two functions, say $f$ and $g$ , are equal, that is $f (x) = g (x)$ . This problem is equivalent to letting $y (x) = f (x) - g (x)$ and determining the value(s) of $x$ such that $y (x) = 0$ .

Here we will use the numerical methods of interval bisection and Newton’s method (also known as Newton–Raphson) to find the zero of a function.

Summary of mathematics used

Interval bisection

Suppose $f (x)$ is a continuous function, such that $f (a) < 0$ and $f (b) > 0$ . An approximation to $f (x) = 0$ may be calculated using the following algorithm:

Let $c = \frac{(a + b)}{2}$
If $f (c) < 0$ , then the root lies in the interval $[c, b]$ , so let $a = c$ . If $f (c) > 0$ , then the root lies in the interval $[a, c]$ , so let $b = c$ . If $f (c) = 0$ then the root is at $x = c$
Repeat this process until you have a root to the desired accuracy.

Question

Do this problem with pen and paper.

Modify this algorithm for the case $f (a) > 0$ and $f (b) < 0$

Expand for solution

Solution

Let $c = \frac{(a + b)}{2}$
If $f (c) < 0$ , then the root lies in the interval $[a, c]$ , so let $b = c$ . If $f (c) > 0$ , then the root lies in the interval $[c, b]$ , so let $a = c$ . If $f (c) = 0$ then the root is at $x = c$ .
Repeat this process until you have a root to the desired accuracy.

Newton-Raphson

Suppose that $f (x)$ is a continuous function. An approximation to $f (x) = 0$ may be calculated using the following algorithm:

Start with an initial guess $x_{0}$ .
Calculate the next solution using: $x_{n + 1} = x_{n} - \frac{f (x_{n})}{f^{'} (x_{n})}$
Repeat step 2 until you have the root to the required accuracy.

For more details on Newton–Raphson see: https://mathworld.wolfram.com/NewtonsMethod.html