While Loop won't work on iteration problem
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A K
am 2 Nov. 2013
Beantwortet: Roger Stafford
am 2 Nov. 2013
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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Akzeptierte Antwort
Azzi Abdelmalek
am 2 Nov. 2013
Bearbeitet: Azzi Abdelmalek
am 2 Nov. 2013
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)
9 Kommentare
Azzi Abdelmalek
am 2 Nov. 2013
Bearbeitet: Azzi Abdelmalek
am 2 Nov. 2013
Look at my previous comment. Remove x=zeros(100,1) or add after the while loop
x=x(1:n+1)
Weitere Antworten (1)
Roger Stafford
am 2 Nov. 2013
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.
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