My Modified regula falsi code has the same number of iterations and error value as the regular falsi code that I made. Why is this?

Question

Peter Phung 2017-9-25

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https://ww2.mathworks.cn/matlabcentral/answers/358159-my-modified-regula-falsi-code-has-the-same-number-of-iterations-and-error-value-as-the-regular-falsi

编辑： Jan 2017-9-25

This is my modified regula falsi code

clear all; clear clc;
g = 9.81; %acceleration m/s^2
m = 68.1; %mass, kg
t = 20; % time, seconds
f = @(c) g*m./c.*(1-exp(-c/m*t))-93.71;
c = 2:20;
plot (c, f(c))
xlabel('c = 2:20')
ylabel('Velocity formula in terms of c')
title('Plot of velocity function with respect to c')
a = 4; b = 8;
error = 1;
mold = 1;
SF = 6;
i = 1;
F = f(a);
G = f(b);
while (1)
     m = (a*G - b*F)/(G-F);
    if F*f(m) < 0
        b = m;
        G = f(m);
        if F*f(m) > 0
            F = F/2;
        end
    else
        a = m;
        F = f(m);
        if f(a)*f(m) > 0
            G = G/2;
        end
    end
    error = abs ((m-mold)/m);
    if error < 0.5*10^-SF
        break
    end
    mold = m;
    i = i + 1; %iteration counter
end
verse3 = [num2str(m(end)),' is our (c) value.'];
disp (verse3)
verse = ['There were ' num2str(i), ' iterations done'];
disp (verse)
verse2 = ['The error is ', num2str(error(end))];
disp (verse2)

Here is my regula falsi code

clear all; clear clc;
g = 9.81; %acceleration m/s^2
m = 68.1; %mass, kg
t = 20; % time, seconds
f = @(c) g*m./c.*(1-exp(-c/m*t))-93.71;
c = 2:20;
plot (c, f(c))
xlabel('c = 2:20')
ylabel('Velocity formula in terms of c')
title('Plot of velocity function with respect to c')
a = 4; b = 8;
error = 1;
mold = 1;
SF = 6;
i = 1;
while (1)
  m = (a*f(b)-b*f(a))/(f(b)-f(a));
  mbis(i) = m;
  if f(a)*f(m)<0;
      b = m;
  else
      a = m;
  end
  error = abs ((m-mold)/m);
  if error < 0.5*10^-SF;
      break
  end
  mold = m;
  i = i + 1; %iteration counter
end
verse3 = [num2str(m(end)),' is our (c) value.'];
disp (verse3)
verse = ['There were ' num2str(i), ' iterations done'];
disp (verse)
verse2 = ['The error is ', num2str(error(end))];
disp (verse2)

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Jan 2017-9-25

编辑：Jan 2017-9-25

Let me ask the other way around: Why do you assume that the number of iterations must be different?

Did you use the debugger already to compare the evaluations in the codes line by line?

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Answer 1

Jan 2017-9-25

1
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/358159-my-modified-regula-falsi-code-has-the-same-number-of-iterations-and-error-value-as-the-regular-falsi#answer_283011

编辑：Jan 2017-9-25

在 MATLAB Online 中打开

Let me ask the other way around: Why do you assume that the number of iterations must be different?

Did you use the debugger already to compare the evaluations in the codes line by line?

The only difference is inside this block (and the corresponding block in the other branch):

   if F*f(m) > 0
     F = F/2;
   end

And does this condition occur at all?

Note: Do not shadow the important function "error" by a variable, because this leads to unexpected behavior frequently.

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My Modified regula falsi code has the same number of iterations and error value as the regular falsi code that I made. Why is this?

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My Modified regula falsi code has the same number of iterations and error value as the regular falsi code that I made. Why is this?

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