USING newton method optimization (getting error in 'xx(1)=x') .

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RAHUL KUMAR
RAHUL KUMAR 2019-8-1
回答： Bobby Huxford 2019-8-1
rho_ss=8000;
rho_cop=8960;
x=[.001;.0014;.005;.007;.3];
D_i=x(1);D_io=x(2);D_o=x(3);D_oo=x(4);L=x(5);
wi=rho_ss*pi*D_io*(D_io-D_i)*L
wo=rho_cop*pi*D_o*(D_oo-D_o)*L
f=@(x) ((wi+wo)-.050)^2
fd=str2sym('((wi+wo)-.050)^2')
%f1=diff(fd,1)
%f2=diff(fd,2)
f1=@(x) (2.0*wi + 2.0*wo - 0.1)
f2=@(x) (2)
N=100; %no of iteration
err=.01; %result accuracy
xx(1)=x;
for i=1:N
x(i)=x-(f1(x)/f2(x));
j=i+1;
xx(j)=x;
Err=abs(xx(j))-abs(xx(j-1));
if Err<err,break;end
end

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回答（1 个）

Bobby Huxford
Bobby Huxford 2019-8-1
It would be easier to answer this if you wrote a question along with the code...
But the line xx(1)=x; will give you an error because you are trying to assign the 5 values in x to the first space in xx.
The line of code you may be looking for is:
However I am not sure what it is you are trying to do, so this could be wrong.
xx = x(1);

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