while loop running forever during attempt at Newton Rhapson method
显示 更早的评论
This code keeps running seemingly forever, or at least a few minutes. I can't figure out why, could someone please explain what is going wrong? unless i wrote the actual functions of d wrong it should be expected to converge towards something until the while loop condition isn't fullfilled anymore
d=0.007;
dI=1.219;
Rfi=1.76*(10^(-4));
Rf0=Rfi;
hs=356;
ht=hs;
kw=60;
dTm=29.6;
Q=801368;
Aex=64.15;
a=((1/ht)+Rfi)/dI;
b=log(d/dI)/2*kw;
c=d*(a+b)+Rf0+(1/hs);
f=(c^(-1))-Q/(Aex*dTm);
df=-1*(c^(-2))*(a+b)*(dI/(d*2*kw))*(1/dI);
%d1-do ska va mindre än tol
%i slutet av iterationen
%så man kan sätta while
d=0.007;
d0=d;
d1=10;
abs(d1-d0);
while abs(d1-d0)>10^(-8);
d0=d;
d1=d0-f/df;
d=d1;
end
d1
采纳的回答
d=0.007;
dI=1.219;
Rfi=1.76*(10^(-4));
Rf0=Rfi;
hs=356;
ht=hs;
kw=60;
dTm=29.6;
Q=801368;
Aex=64.15;
a=((1/ht)+Rfi)/dI;
%d1-do ska va mindre än tol
%i slutet av iterationen
%så man kan sätta while
d=0.007;
d0=d;
d1=10;
abs(d1-d0);
while abs(d1-d0)>10^(-8);
b=log(d/dI)/(2*kw);
c=d*(a+b)+Rf0+(1/hs);
f=(c^(-1))-Q/(Aex*dTm);
df=-1*(c^(-2))*(a+b)*(dI/(d*2*kw))*(1/dI);
d0=d;
d1=d0-f/df;
d=d1;
end
d1
2 个评论
VBBV
2023-11-18
Place the lines of coe inside the while loop which are function of variables d and expression for evaluating d1 inside the while loop
b=log(d/dI)/(2*kw);
c=d*(a+b)+Rf0+(1/hs);
f=(c^(-1))-Q/(Aex*dTm);
df=-1*(c^(-2))*(a+b)*(dI/(d*2*kw))*(1/dI);
Torsten
2023-11-18
Are you sure it must read
b=log(d/dI)/(2*kw);
instead of
b=log(d/dI)/2*kw;
as in the original code ?
更多回答（1 个）
Your function doesn't seem to have a zero.
d=0.007;
dI=1.219;
Rfi=1.76*(10^(-4));
Rf0=Rfi;
hs=356;
ht=hs;
kw=60;
dTm=29.6;
Q=801368;
Aex=64.15;
a=((1/ht)+Rfi)/dI;
b=@(d)log(d/dI)/2*kw;
c=@(d)d.*(a+b(d))+Rf0+(1/hs);
f=@(d)(c(d).^(-1))-Q/(Aex*dTm);
D = 0.1:0.01:10;
plot(D,f(D))
df=@(d)-1*(c(d)^(-2))*(a+b(d))*(dI/(d*2*kw))*(1/dI);
%d1-do ska va mindre än tol
%i slutet av iterationen
%så man kan sätta while
d=0.007;
d0=d;
d1=10;
abs(d1-d0);
while abs(d1-d0)>10^(-8);
d0=d;
%df0 = (f(d0+1e-7)-f(d0))*1e7
d1=d0-f(d0)/df(d0);
% d1 = d0-f(d0)/df0;
d=d1
end
d1
类别
在 帮助中心 和 File Exchange 中查找有关 Interpolation 的更多信息
Translated by 
