While Loop won't work on iteration problem

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A K 2013-11-2

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回答： Roger Stafford 2013-11-2

采纳的回答： Azzi Abdelmalek

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Hi i don't seem to understand why my while loop doesn't work for my function i've done everything in logical steps:

      function [x]=reynolds(Re,x0)
      %x=zeros(1000,1);
      x=(2.5*log(Re*(x0)^0.5)+0.3)^-2;
      n=1;
      x(1)=x0;
      while abs(x(n+1)-x(n))>1e-6
          x(n+1)=x(n);
          x(n+1)=(2.5*log(Re*x(n)^0.5)+0.3)^-2;
          n=n+1;
      end

The function requires two inputs with one being an initial guess, but an error pops up saying that it can't find x(3) mea

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Azzi Abdelmalek 2013-11-2

How did you call your function with values of Re and x0

A K 2013-11-2

I called it from the command window such as, reynolds(5000,3) 5000 is Re and x0=3

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Azzi Abdelmalek 2013-11-2

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https://ww2.mathworks.cn/matlabcentral/answers/104695-while-loop-won-t-work-on-iteration-problem#answer_113927

编辑：Azzi Abdelmalek 2013-11-2

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Edit

function x=reynolds(Re,x0)
x=zeros(1,100)
n=1
x(n)=x0;
x(n+1)=(2.5*log(Re*(x0)^0.5)+0.3)^-2;
while abs(x(n+1)-x(n))>1e-6  
n=n+1;
x(n+1)=(2.5*log(Re*x(n)^0.5)+0.3)^-2
end
x=x(1:n+1)

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A K 2013-11-2

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I kind of fixed it but can't seem to stop the vector in the command window

function [x]=reynolds(Re,x0)
n=1;
x=zeros(100,1);
x(n)=x0;
x(n+1)=(2.5*log(Re*(x0)^0.5)+0.3)^-2;
while abs(x(n+1)-x(n))>1e-8
   n=n+1;
    x(n+1)=(2.5*log(Re*x(n)^0.5)+0.3)^-2;
end
if you scroll up the command window you can see the final result but it shows all the 0's after.

Azzi Abdelmalek 2013-11-2

编辑：Azzi Abdelmalek 2013-11-2

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Look at my previous comment. Remove x=zeros(100,1) or add after the while loop

x=x(1:n+1)

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Roger Stafford 2013-11-2

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Assuming that by 'log' you mean the natural logarithm and not the logarithm-base-ten, then if you define the variable w as:

w = 1/2.5/sqrt(x)

your equation to be solved can be expressed as:

w*exp(w) = Re/2.5*exp(.3/2.5)

If you have the 'lambertw' function in your system, you can solve for this directly without doing iteration:

 Re = 5000;
 w = lambertw(Re*exp(.3/2.5)/2.5);
 x = (2.5*w)^(-2);

Your initial estimate of 3 is very far from the actual solution which is in fact:

x = 0.00453573902634

and this may account for the trouble you experienced.

There are also two other methods you could use: 1) the Newton-Raphson method which requires a derivative, and the matlab function 'fzero'. Either one would surely be superior to the method you are using here. In some circumstances your method might not even converge to a solution at all.