Newton-Raphson Method
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I have the following Newton-Raphson method code.
Code.
I am attempting to call it using the following
Input.
I am receiving the following error
Error.
Can someone please help me with this? I know the number of iterations and friction factor, but I am expected to turn the loop I did into a function like this and it should work but there is something a little wrong.
3 个评论
Tony Rankin
2021-3-15
编辑：Tony Rankin
2021-3-15
Cris LaPierre
2021-3-15
@Tony Rankin, removing the substance of your question once it has been answered is not the point of the forum. The questions and answers are preserved to provide help to others down the road who have a similar question. Could you be kind enough to restore the details of your question?
Cris LaPierre
2021-3-16
Original context restored below from the cache.
I have the following Newton-Raphson method code.
function [R] = newton(f,df,x0,tol)
% R is an estimation of the root of f using the Newton-Raphson method
% f is colebrook equation for turbulent flow
% df is the first derivative of the colebrook equation
% x0 is the initial estimate for the root
% tol is the accepted tolerance
if abs(f(x0)) < tol
R = x0; % end loop, takes x0 as root of f
else
R = newton(f,df,x0-(f(x0)/df(x0)),tol); % make a recursive call
end
error = abs(f(x0)) % calculate the absolute error
end
I am attempting to call it using the following
error = 1;
p = 0.9*1000;
mu = 8*0.001;
e = 0;
D = 4*0.0254;
Q = (2000*42*3.785*10^-3)/(24*60*60);
A = (pi*(D.^2))/4;
Re = (p*D*Q)/(A*mu);
f = @(x) -1/sqrt(x)-2.*log10(((e./D)./3.7)+2.51/(Re*sqrt(x)));
df = @(x) -1/2*(x).^3/2*(1+((2*2.51)/log10(e./D)./3.7)+(2.51/(Re*sqrt(x)*Re)));
R = newton(f,df,0.01,1e-9)
I am receiving the following error
Out of memory. The likely cause is an infinite recursion within the program.
Error in newton (line 27)
if abs(f(x0)) < tol
Can someone please help me with this? I know the number of iterations and friction factor, but I am expected to turn the loop I did into a function like this and it should work but there is something a little wrong.
Colebrook equation
This is meant to be the derivative of the Colebrook equation. I hope I had input it correctly.
采纳的回答
Cris LaPierre
2021-3-15
编辑：Cris LaPierre
2021-3-15
From what I can see, the result of f(x0) keeps cycling between the same set of values, none of which results in your stopping condition tol.
Iter x0 f(x0)
2 -2.1477e+07 13.9626 + 1.3646i
3 -2.1477e+07 - 5.5096e-22i 13.9626 - 1.3646i
4 -2.1477e+07 - 9.4040e-38i 13.9626 - 1.3646i
5 -2.1477e+07 + 5.5096e-22i 13.9626 + 1.3646i
6 -2.1477e+07 - 9.4040e-38i 13.9626 - 1.3646i
...
I would check your equations for f and df.
1 个评论
Tony Rankin
2021-3-15
编辑：Tony Rankin
2021-3-15
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