Methods of Approximation

Date: 7/31/96 at 14:7:48
From: Albert Nicorgua
Subject: Methods of Approximation

Hi,

My teacher used Newton's Method of Approximation to solve "Y to the 
power of Y = 25."  Do you know of another way to solve the 
above equation?

--Thanks, Andy

Date: 9/1/96 at 18:5:51
From: Doctor Jerry
Subject: Re: Methods of Approximation

You wish to solve the equation x^x-25=0.  By Newton's Method, I obtain
x=2.9632....

Another method is called the Bisection Method.  Suppose you want to 
find a zero of a continous function f(x). The idea of the Bisection 
Method is that if f has a xero in an interval [a,b] with midpoint m, 
then f must have a zero in either the left or right half of [a,b].  If 
we choose the first interval so that f(a)*f(b)<0 and let h=m-a, then 
if f(m-h)*f(m)>0, replace the interval [a,b] = [m-h,m+h] by [m,m+h]; 
otherwise, replace [m-h,m+h] by [m-h,m].

Repeat this algorithm until the containing interval is sufficiently 
small.

This method is not as fast as Newton's Method, but can be applied to 
any continuous function, not just differentiable functions.

For the function f(x) = x^x-25, here are a few results. Take the 
intial interval to be [2.5,3.5].  So m=3 and h=0.5.

Successive function values, values of m, and h are

-15.11
  2.75
  0.25

 -8.85
  2.875
  0.125

 -4.175
  2.9375
  0.0625

You can see that the midpoints are approaching 2.96...

-Doctor Jerry,  The Math Forum
 Check out our web site!  http://mathforum.org/dr.math/