Newton’s Method is traditionally used to find the roots of a non-linear equation.

**Definition 1** (**Newton’s Method**): Let *f*(*x*) = 0 be an equation. Define *x _{n}* recursively as follows:

Here *f′*(x_{n}) refers to the derivative *f*(*x*) of at *x _{n}*.

**Property 1**: Let *x _{n}* be defined from

*f*(

*x*) as in Definition 1. As long as function

*f*is well behaved and the initial guess is suitable, then

*f*(

*x*) ≈ 0 for sufficiently large

_{n}*n.*

**Observation**: Newton’s method requires that *f*(*x*) has a derivative and that the derivative not be zero. Usually the method converges quickly to a solution, but this can depend on the initial guess. Also in cases where there are multiple solutions, different initial guesses can lead to different solutions.

**Example 1**: Use Newton’s Method to find the square root of 25.

The problem is equivalent to solving the equation *f*(*x*) = 0 where *f*(*x*) = x^{2} – 25.

From calculus, *f′*(*x*) = 2*x*, and so

Suppose we start the iteration with *x _{0}* = 2, then as we see in Figure 1, the iterations converge to 5 as expected.

**Figure 1 – Newton’s Method for Example 1**

Note that if we select *x _{0}* = 0 the algorithm won’t converge to a solution since would be undefined. Also if we set

*x*= -2 (or any negative value) then the procedure iterates to -5, which is the other solution to

_{0}*f*(

*x*) = 0.

We now extend Newton’s method to *m* equations in *m* unknowns.

**Definition 2** (**Newton’s Method**): Let *X* = [*x _{i}*] be an

*m*× 1 column vector and let

*F*= [

*f*] be an

_{i}*m*× 1 column vector of functions. Our objective is to solve the set of

*m*equations given by

*F*(

*x*) =

*Ο*(here

*Ο*is the zero vector).

Let *J* = [*q _{ij}*] be the Jacobian matrix, defined by the partial derivatives

*q*= . Now define the

_{ij}*m*× 1 column vectors

*X*and the

_{n}*m*×

*m*matrices

*J*as follows

_{n}**Property 2**: Using the terminology in Definition 2, then for a well-behaved *F*, a reasonable guess and sufficiently large *n*, *F*(*X _{n}*) ≈

*Ο*.

**Example 2**: Find the solution to the equations y – *e ^{x} *= 0 and

*x*+ y = 3.

Set *f*(*x*, y) = y – *e ^{x} *and

*g*(

*x*, y) =

*x*+ y – 3. Then we need to find the solution to

*f*(

*x*, y) = 0 and

*g*(

*x*, y) = 0 , i.e.

*F*(

*X*) =

*Ο*where

*X*= and

*F*= . Now

where we use the notation *f_{x}*(

*x*,y) to mean . Starting with

*X*= , we see in Figure 2 that the procedure converges to

_{0}*x*= 0.79206 and y = 2.20794 after 3 iterations.

**Figure 2 – Newton’s Method for Example 2**

Representative formulas for the worksheet in Figure 2 (for iteration 2) are shown in Figure 3.

**Figure 3 – Formulas for 2 ^{nd} iteration**

The example was chosen so that we could check the result using Newton’s method in one variable since the problem is equivalent to *e ^{x }*–

*x*– 3 = 0 and y = 3 –

*x*. We can therefore apply Newton’s method as in Example 1, with

*f*(

*x*) =

*e*–

^{x}*x*– 3,

*f′*(

*x*) =

*e*– 1, and so

^{x}The iteration proceeds as in Figure 4.

**Figure 4 – Alternative use of Newton’s Method for Example 2**

Thus *x* = 0.79206 and y = 3 – *x *= 3 – 0.79206 = 2.20794, as before.

how I can find more roots and number off roots

Raad,

If you use Newton’s method, you need to make multiple guesses as to the initial value. Different guesses can potentially yield different roots. If the degree of the polynomial is n (i.e. the highest power of a polynomial; e.g. the polynomial 3x^4 + 13x^2 – .4x + 1 has degree 4), then there are at most n real roots. In fact there are exactly n roots, although some of them may be complex numbers.

To find all the roots, see the following webpage:

http://www.real-statistics.com/other-mathematical-topics/roots-of-a-polynomial/

Charles