False Position method. How many itinerations should it take to find the root?

Hi! I am programming the false position method. However, with 100 itinerations, my program cannot find a root with an error smaller than 10e-10 nor 10e-6.

Is this normal, or is it just that my program that does not work properly?

Find attached the code.

Hope you can help me!

[Dyuman Joshi]

Please post the code, instead of posting the picture of the code.

From the looks of it, it seems 100 iterations are not enough to find a root with that tolerance.

Also note the different between

10e-10

ans = 1.0000e-09

1e-10 %10^x is equal to 1ex

ans = 1.0000e-10

[llucia]

Thanks! So the code is right, then? I mean, the problem is that it takes more itinerations to find the root right?

Answer 1

[Dyuman Joshi]

A correction is needed -

while n<maxits && error(n+1)>eps_x && abs(f(x0))>eps_f
                                  %^here

If any of the condition is met, exit the loop

%I'm running it as a script with some changes
f = @(x) x^3 + 2*x^2 + 10*x - 20;
x0=0;x1=1;
error=abs(x1-x0);
n=0;
while error > 1e-10 && abs(f(x0)) > 1e-10
    a=f(x0);b=f(x1);
    x = (x0*b-x1*a)/(b-a);
    if a*b<0
        x1=x;
    else
        x0=x;
    end
    error=abs(x1-x0);
    n=n+1;
end
n

n = 14

format long
x

x = 
   1.368808107821373

f(x)

ans = 
     0

[Torsten]

A correction is needed -

while n<maxits && error(n+1)>eps_x && abs(f(x0))>eps_f

Why ? I found the first version quite plausible. As long as not both conditions are met, the root is questionable and should be improved. And the code works as posted.

f = @(x)x.^3+2*x.^2 + 10*x -20;
a = 0;
b = 1;
eps_x = 1e-10;
eps_f = 1e-10;
maxits = 100;
[x n error] = RegulaFalsa1(f,a,b,eps_x,eps_f,maxits)

x = 1.3688

n = 14

error = 1×15

1.0000 0.5385 0.1882 0.1705 0.1697 0.1697 0.1697 0.1697 0.1697 0.1697 0.1697 0.1697 0.1697 0.1697 0

function [x,n,error] = RegulaFalsa1(f,a,b,eps_x,eps_f,maxits)
x0 = a;
x1 = b;
n = 0;
error(1) = abs(b-a);
while n < maxits && (error(n+1) > eps_x || abs(f(x0)) > eps_f)
  x = (x0*f(x1)-x1*f(x0))/(f(x1)-f(x0));
  if f(x0)*f(x1) < 0
    x1 = x;
  else
    x0 = x;
  end
  n = n+1;
  error(n+1) = abs(x1-x0);
end
end

[llucia]

Perfect! Thanks a lot!