error calculation perfect fit

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sarah 2015-5-30

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评论： Star Strider 2015-6-3

Hi guys, so I have experimental data in column matrix y. And I want to use the equation below to find x as a function of y. This means have constant b adjusted from 0 to 2 and hence want to obtain the ideal b value.

I'm thinking of using a double for loop however would anyone know how to calculate for error and set a desirable tolerance. Or tips on how to solve for it effectively?

Thanks in advance. The equation is:

y= (1-0.05*x^(b))/(1-0.03*x^b)^2

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Star Strider 2015-5-30

See also for context: fitting data with equation.

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Star Strider 2015-5-30

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I would use nlinfit and nlparci to estimate ‘b’, and use a few values of ‘x’ and ‘y’ to provide an initial estimate, with:

b = log(20*(1-1./y))./log(x);

Noting the assumptions and restrictions that I mentioned earlier.

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sarah 2015-6-1

编辑：sarah 2015-6-1

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Thank you. I'm getting this error for some strange reason. Looking into choosing another optimum solver now

                                                  First-Order                    Norm of 
 Iteration  Func-count    Residual       optimality      Lambda           step
     0           3         73.3319        4.54e+07         0.01
     1           6         17.9357         5.6e+05        0.001         1.0877
     2           9         2.28484            23.7       0.0001          4.576
     3          21         2.27261            32.7       100000    0.000236578
     4          25         2.27045            33.7        1e+06    3.26701e-05
     5          28         2.26815            34.6       100000     3.3644e-05
     6          32         2.26574            35.4        1e+06    3.45362e-05
     7          36         2.26549            35.5        1e+07    3.53694e-06
             No solution found.
               fsolve stopped because the relative size of the current step is less than the
default value of the step size tolerance, but the vector of function values
is not near zero as measured by the default value of the function tolerance.

criteria details

b =

   4.1255 - 3.2581i
  -0.0003 - 0.0001i

Star Strider 2015-6-1

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My pleasure.

I am not certain what the problem is. Your equation has to be in the form such that it equals zero. Since you mentioned that you cannot solve for one variable in terms of the other, and so cannot use nlinfit, but had terms of both variables and parameters on both sides of the equation, I suggested that you rearrange your equation so that it equals zero, and then use fsolve.

I know that this is not your function, but the idea would be that if you had for instance:

sin(x+y+b(1)) = cos(x*y*b(2))

you would rearrange it so that:

sin(x+y+b(1)) - cos(x*y*b(2)) = 0

then give it to fsolve in the form:

f = @(b,x,y) sin(x+y+b(1)) - cos(x.*y.*b(2));
[B,fval] = fsolve(@(b) f(b,x,y), b0);

(Include an options structure in the fsolve call if you want one.)

I am not certain what expressions you began with, so I cannot help you put your equation (or system of equations) in a form that fsolve can estimate, but this is the essence of the idea. (You mentioned that there could be errors in the equation you posted earlier, so I have not looked at it closely.)

sarah 2015-6-3

编辑：sarah 2015-6-3

Sure.

Also i wanted to ask you regarding residuals. Residuals for some selective data is very small e-8 and the system solves. However for say the whole range i.e. 70 data points the fit isn't good i.e. residuals are large 1000. However my b(1) and b(2) are satisfactory. It doesn't mean my model is wrong right i.e. my coding. I'm assuming that experimental data at certain range is wrong.

Star Strider 2015-6-3

It’s difficult to determine the reasons the residuals are very low in one region and high in another. My guess is that either (1) your model does not completely describe the process that created your data, or (2) some regions simply have more noise than others. This may not be a problem if your parameter estimates are acceptable and your model provides a good fit in your area of interest.

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John D'Errico 2015-5-30

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How many points do you have that you expect a perfect fit? (Hint: with only one parameter to solve for, only ONE point will give you a perfect fit in general. With more points than that, you won't get a perfect fit.)

Anyway, with the model you have chosen, is there some good reason why you choose to define it as

y= (1-0.05*x^(b))/(1-0.05*x^b)^2

instead of the algebraically identical

y= 1./(1-0.05*x.^b)

(Note my use of ./ and .^. This is important.)

And with that model, I seriously doubt you will get a very good fit, and I have not even seen your data. That model has very little capability of adjustment.

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sarah 2015-5-30

编辑：sarah 2015-5-30

my y data ranges from 1-6 but yeah i have alot of points.

so say if my experimental data is y. And say if I want to solve for b ie. the most ideal value of b. So there is no way i could do that? I think even a point of intersection which shows one value of b is good for me. As long as i get a good estimate of b - i was initially thinking of a surface plot. I do know that x ranges from 2-40. haha

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