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False Position

  1. Jul 15, 2007 #1
    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. jcsd
  3. Jul 16, 2007 #2


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    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:
  4. Jul 16, 2007 #3
    "necessarily" "unnecessary", that was a typo, I just noticed it.
    I know the difference between the 2 methods.
    So I can say that of if at the start f(a0) and f(b0) are of opposite signs...
    I would get a non zero length of [ai, bi]...
    I think I got it, I'm going to try the example given on wiki and several others and try them myself.
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