Newton's Method for Solving Equations

 

We will now see a very useful and interesting application of differentiation, which is due to Isaac Newton, the famous English scientist.

 

 

We often face a problem that requires finding the roots of (or solving) an equation that can be put in the form f(x)=0. We already know from algebra how to solve these equations when f(x) is a polynomial of the first or the second degree. Nevertheless, in most of the other equations, it is difficult, or even impossible to find the exact roots. We then have to use some approximation methods to find these roots. Newton’s method is one of these approximation methods, and maybe one of the best and the easiest.

 

 

First, Let’s consider the equation f(x)=0, where f is a differentiable function. We know that the solutions of this equation are the x intercepts of the graph of y=f(x). Suppose we have an approximation  of a root r of the equation. If some conditions exist, we can get a better approximation  of the root by the intersection of the tangent of the graph with the x-axis. Here,  is also an approximation of r. Therefore, we can get by the same way a better approximation , then …etc. Thus, we can have an approximate value of r correct to as many decimal places as we like.

 

NB

To apply this method, make sure that the following two conditions are satisfied:

1-  exists but is not equal to zero through an interval containing the root r, and the approximation  when we are finding the approximation .

2-  or  throughout this interval.

 

 

Now as we knew the idea of Newton’s method, let’s try to develop formulas to be used.

Let r be a root of the equation f(x)=0. Let  be a first approximation of r satisfying the conditions above. The equation of the tangent of the graph of the function at  can be written:

To find the x intercept, we put y=0 and . That gives us:

Similarly, we can find  by the formula:

Generally, the nth approximation , n is any positive integer greater than 1, can be found using the formula:

 

 

Example

Using Newton’s method, find an approximation for . Use  as a first approximation, and find .

Solution