Help with iteration counter not working or displaying wrong numbers?

Hi, I was creating some code to solve an equation using bisection and regula falsi. I wanted to see which method converges faster so I decided to make an iteration counter to count how many iterations were done for each result to meet my tolerance value (0.001). I'm not sure why the iterations number which are displayed along side my other value are the same for both methods. They also have decimal places. Is there somewhere in bisect.m or regfalsi.m where I am making a mistake or in main.m where I pull the iteration values in part B and C.

Specifically this area

function bisectioncalc = bisect(f,a,b) %Head is the num.soln


c = (a+b)/2; % calculation of midpoint
r = abs(f(c));
I=0;
while r > 0.001
  
   if  (f(a)*f(c))<0
      b = c; % c values becomes a
   elseif (f(b)*f(c))<0
      a = c; % c values becomes b
   end
   c = (a+b)/2;
   r = abs(f(c)); %relative error/tolerance formula
   I = I + 1;
end
bisectioncalc = c; %assign c back to bisectionfalsicalc
end

I want to know whether my placment of the counter is correct and if I am calling it correctly in main.m

Thank you!

[Walter Roberson]

The counter looks fine there.

Where is your else case if neither condition is satisfied?

(If you can be certain that one of the two will be satisfied then you do not need the second test.)

[Ralph Rodrigues]

In this case, at least one of two will be satisfied, thanks for letting me know! I'll remove the second test.

Is there any reason why my iteration numbers are in decimal places in the output? I feel like I'm incorrectly calling it out of the bisect/falsi functions when I call it in main.m for the fprintf display

f = @(H) v -((sqrt(19.62.*H)*tanh(sqrt(19.62.*H)*t/(8))));
%Head = fzero(f,b); % find root using fzero()
H = bisect(f,a,b);
I = bisect(f,a,b);
fprintf('\n%6.1f\t%6.1f\t%8.4f\t%6.1f',v,t,H,I)

[Walter Roberson]

What if f(c ) is exactly equal to 0?

Answer 1

[Walter Roberson]

f = @(H) v -((sqrt(19.62.*H)*tanh(sqrt(19.62.*H)*t/(8))));
%Head = fzero(f,b); % find root using fzero()
H = bisect(f,a,b);
I = bisect(f,a,b);

Why are you calling the same bisect() twice? Is it non-deterministic? You are not timing it, so it is not for the purposes of getting more accurate timing.

[Ralph Rodrigues]

Oh I thought that for calling up the H (head value), I (iteration number) respectively I needed two separate bisects. Is there a better way of doing that? There are 37 possible H values which are being generated, each of them has been iterated through until the tolerance is reached. Once that's done, then I display the values using:

fprintf('\n%6.1f\t%6.1f\t%8.4f\t%6.1f',v,t,H,I)

So I'm trying to generate the interation values and then have it display using the above fprintf statement.

The problem is both my tables for the bisection method and the falsi method seem to showing the same number of iterations for each head value when the three programs are run.

[Walter Roberson]

Your bisect() and regfalsi() have no way of returning the interation numbers at present.

Change to

function [bisectioncalc, I] = bisect(f,a,b) %Head is the num.soln

and

[H,I] = bisect(f,a,b);

[Ralph Rodrigues]

Perfect! It worked, thank you. Was the reason, I could not return the numbers initially because it was not setup as an array?

[Walter Roberson]

No, it was because you never tried to return I from the function.

The syntax

function [bisectioncalc, I] = bisect(f,a,b) %Head is the num.soln

does not have anything to do with arrays. The [] on the left is just syntax that does not really have to do with using [] for constructing arrays. MATLAB could have designed it as, for example,

function #bisectioncalc, #I = bisect(f,a,b) %Head is the num.soln

to invent a syntax.

[Ralph Rodrigues]

Sounds good, thanks for all your help.