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.
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.
Do this problem with pen and paper.
Modify this algorithm for the case $f(a)>0$ and $f(b)<0$
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.
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