Stuck at Gauss-Newton Method code
์ด์ ๋๊ธ ํ์

I am trying to make a code with a formula (๐ฆ = ๐0 (1 โ ๐ ^(โ๐1๐ฅ ))).
Help
x=[0.25,0.75,1.25,1.75,2.25];
y=[0.23,0.58,0.70,0.76,0.79];
disp('iterations a0 a1 da0 da1')
tol= 10^-8;
a=[0.5 1];
error=1;
n=length(x);
itermax=100;
for iter=1:itermax
a0=a(1);
a1=a(2);
for i=1:n
f(i)=a0*(1-exp(-a1.*x(i)));
j(i,1)=1-exp(-a1.*x(i));
j(i,2)=a0*a1*exp(-a1.*x(i));
d(i)=y(i)-f(i);
end
da=(j'*j)\(j'*d');
a=a+da';
out= [iter a da']
disp(out);
if (abs(da(1))<tol && abs(da(2))<tol)
break
end
end
x1=min(x);
x2=max(x);
xx=linspace(x1,x2,100);
yy=(a0*xx).*(1-exp(-a1*xx));
p=plot(xx,yy,x,y,'o');
๋๊ธ ์: 2
KSSV
2022๋
5์ 5์ผ
What is the question? Where are you stuck?
seoung gi Sin
2022๋
5์ 5์ผ
๋ต๋ณ (2๊ฐ)
Walter Roberson
2022๋
5์ 5์ผ
1 ๊ฐ ์ถ์ฒ
your j 2 is the derivative with respect to x, but you need it to be the derivative with respect to a1 which is a0*x*exp(-a1*x)
๋๊ธ ์: 2
seoung gi Sin
2022๋
5์ 5์ผ
Walter Roberson
2022๋
5์ 5์ผ
da=(j'*j)\(j'*d');
I am clear why you do not factor out the j' giving you
da = j\d';
?
Change
j(i,2)=a0*a1*exp(-a1.*x(i));
to
j(i,2)=a0*x(i)*exp(-a1.*x(i));
and
xx=linspace(x1,x2,100);
yy=(a0*xx).*(1-exp(-a1*xx));
to
xx=linspace(x1,x2,100);
yy=(a0).*(1-exp(-a1*xx));
์นดํ ๊ณ ๋ฆฌ
๋์๋ง ์ผํฐ ๋ฐ File Exchange์์ Animation์ ๋ํด ์์ธํ ์์๋ณด๊ธฐ
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!
