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### 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/
```
Associated Topics:
High School Calculus

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