Hi guys, so I have alot of experimental data in column matrix y. And I want to use the equation below to find the best fit for my data y. This means have constant b adjusted with x varying, I know that x typically ranges from 1 to 5 with y and b ranges from 0 to 2.
I'm thinking of using a double loop. Would anyone give me tips on how to solve for it effectively? Thanks in advance. The equation is:
y= (1-0.05*x^(b))/(1-0.05*x^b)^2

 채택된 답변

Star Strider
Star Strider 2015년 5월 29일

0 개 추천

I would use a nonlinear curve fitting function such as nlinfit, lsqcurvefit (or fminsearch indirectly).
However if your (x,y)>0, are not noisy, y>1, and you want an estimate of ‘b’, this might work:
b = log(20*(1-1./y))./log(x);
Otherwise, use a nonlinear parameter estimation routine.

댓글 수: 4

sarah
sarah 2015년 5월 30일
편집: sarah 2015년 5월 30일
Thanks.
What if I want to put in all the values of y into the equation, and i know the range for b and want it to estimate for x. However how can matlab tell me the error for each value?
I got the impression that ‘x’ is your independent variable, ‘y’ is your dependent variable, and you wanted to estimate the parameter ‘b’.
If you know ‘y’ and ‘b’ and you want to estimate ‘x’ I would still prefer a nonlinear parameter estimation routine. However, with the appropriate assumptions and reservations as I mentioned earlier, a bit of algebra produces this simplified expression:
x = (20-(1./y)).^(1./b);
The error for estimates of ‘x’ would be easiest to calculate using your original equation with the Statistics Toolbox nlinfit function, then nlparci, since x appears to be a parameter.
sarah
sarah 2015년 5월 30일
And may I ask how did you derive this expression? Thanks
Sure!
y = (1-0.05*x^b)/((1-0.05*x^b)^2
(1/y) = 1-0.05*x^b
(1-(1/y))*20 = x^b
x = ((1-(1/y))*20)^(1/b)
Then vectorise it to do the calculation:
x = ((1-(1./y))*20).^(1./b);

댓글을 달려면 로그인하십시오.

추가 답변 (0개)

카테고리

도움말 센터File Exchange에서 Get Started with Curve Fitting Toolbox에 대해 자세히 알아보기

질문:

2015년 5월 29일

댓글:

2015년 5월 30일

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by