An Introduction To Numerical Analysis (3): Newton's Approximation Method

The Newton method, actually Newton-Raphson method, for approximating the roots of an equation is taken up in this post. The goal in this case is to obtain the solution to some equation of form: f(x) = 0 by generating a series of approximations that approach the actual value(s).  We begin by picking some initial estimate  x_o   then the self-consistent method does the rest by moving stepwise to a point where the graph crosses the x-axis.

The method then basically uses the tangent to the curve y =  f(x), at ( x_o , f ( x_o)) to approximate the curve, calling the point where the tangent meets the axis  x₁ .  The value x₁ is then usually a better approximation than  x_o .   The point x₂  where the tangent to the curve at ( x₁ , f ( x₁))   crosses the x-axis is then the next approximation.  The method entails continuing on using each approximation to generate the next until we are close enough to the root to stop. Point of interest here: This is actually the method built into most scientific calculators.

To fix ideas, consider finding the positive root to the equation:

x ²  - 2 = 0

Solution:

With f(x) = x ²  - 2    and f'(x) = 2x

The next equation becomes:

x _n+1  =   x _n    -  (x _n ²  - 2) / 2 x _n

Then we rewrite in a simpler form which expedites further approximations:

  x _n+1  =   x _n    +  x _n /2  +   1/ x _n

=  x _n /2  +   1/ x _n

The equation:

x _n+1  =   x _n /2  +   1/ x _n

The approximation-generating formula (for any given f(x) )can be given as:

x _n+1  =   x_n  -  f(x_n) / f '(x_n)

This enables us to go from each approximation to the next with just a few keystrokes.  E.g.

Start the process using:  x_o  =  1, and compare each successive approximation with actual root:  

Ö2 = 1.41421

So, if  the estimate is  x_o  =  1, then the error is 1.41421 - 1 

=  0. 41421

Next we let:  x₁ =  1.5

For which: 

x _n+1  =   1.5 /2  +   1/ 1.5 = 1.41667

Which gets us even closer to the actual value.

Suggested Problems:

1)  Find the x coordinate of the point where the curve  x ³  - x  crosses the horizontal line y = 1.

2) Using Newton's method find at least one root approximation for the equation:

x ³   =   2