# False Position

1. Jul 15, 2007

I'm tying to use the false position method to find the root r of f(x)=0, between 2 points a0 and b0.
I have [a(i),b(i)] enclose in it f(x)=0.
There is something I don't get, if the number of iteration i goes to infinity, why the length of [a(i),b(i)] doesn't unnecessary goes to 0 ?
I understand that this have to do with the f(a0) et f(b0) the first 2 points chosen.
can somebody explain this ?

2. Jul 16, 2007

### HallsofIvy

Staff Emeritus
You seem to be confusing two different methods- in false position the length of the interval [ai, bi] does NOT necessairily go to 0!

That does happend with "bisection": if f(a0)< 0 and f(b0)> 0 (and f is continuous) then you know there is a root somewhere between a0 and b0. Take your next point to be the midpoint of the interval: c= (a0+ b0)/2. if f(c)> 0 then there is a root between a0 and c: let a1= a0, b1= c. If f(c)< 0 then there is a root between b0 and c: let a1= c, b1= b0.

In this case, because we are always dividing the interval in half the length of (ai, bi) is the (b0- a0)/2i which obviously goes to 0.

False position, however, starts with two points, a0 and b0, such that f(a0)< 0 and f(b0)> 0 (or vice-versa) and calculates the slope of the line between them (f(b0)- f(a0))/(b0- a0) and uses that to determine a1. In true "false position", it is always the "a" point that is replace while the b point remains equal to b0. ai approaches a root while bi= b0 for all i so the length of [ai, bi] does NOT go to 0.

You might want to look at Wikipedia's article on it:
http://en.wikipedia.org/wiki/False_position_method

3. Jul 16, 2007