i try to estimate an unknown parameter "a" of a non-linear model by curve fitting toolbox. the non linear model must fit the real data.
i have write my program ans i have run it but i found the fit so bad. please can you help me and show me where is the problem. it's my first time that i make a program. so my knowledge is limited.
the program is the follow:
General model:
     f(x) = (0.1374-(a*0.0118/0.051356)-((0.0118^2)/(2*(0.051356^2))))+(0.05-
                    (0.1374-(a*0.0118/0.051356)-((0.0118^2)/(2*(0.051356^2)
                    ))))*((1-exp(-0.1374*x))/(0.1374*x))+(((0.0118^2)/(4*(0.1374^3)
                    *x))*(1-exp(-0.1375*x))^2)
Coefficients (with 95% confidence bounds):
       a =     -0.1382  (-0.4263, 0.1499)

Goodness of fit:
  SSE: 0.002049
  R-square: -846.7
  Adjusted R-square: -846.7
  RMSE: 0.01848

  Warning: A negative R-square is possible if the model
           does not contain a constant term and the fit
           is poor (worse than just fitting the mean).
           Try changing the model or using a different StartPoint.

x is x = [0.5 1 2 3 5 7 10]
y = [0.079 0.0782 0.0791 0.0793 0.0791 0.0802 0.0798]
i rely so much on your help.

> i try to estimate an unknown parameter "a" of a
> non-linear model by curve fitting toolbox. the non
> linear model must fit the real data.
> i have write my program ans i have run it but i found
> the fit so bad. please can you help me and show me
> where is the problem. it's my first time that i make
> a program. so my knowledge is limited.
> the program is the follow:
> General model:
> f(x) =
> (x) =
> (0.1374-(a*0.0118/0.051356)-((0.0118^2)/(2*(0.051356^2
> ))))+(0.05-
>  (0.1374-(a*0.0118/0.051356)-((0.0118^2)/(2*(0.051356^
> 2)
>  ))))*((1-exp(-0.1374*x))/(0.1374*x))+(((0.0118^2)/(4*
> (0.1374^3)
>                     *x))*(1-exp(-0.1375*x))^2)
> Coefficients (with 95% confidence bounds):
>        a =     -0.1382  (-0.4263, 0.1499)
> 
> Goodness of fit:
>   SSE: 0.002049
>   R-square: -846.7
>   Adjusted R-square: -846.7
>   RMSE: 0.01848
> 
> Warning: A negative R-square is possible if the
> he model
> does not contain a constant term and the
> erm and the fit
> is poor (worse than just fitting the
> fitting the mean).
> Try changing the model or using a
> or using a different StartPoint.
> 
> x is x = [0.5 1 2 3 5 7 10]
> y = [0.079 0.0782 0.0791 0.0793 0.0791 0.0802 0.0798]
> i rely so much on your help.

If you write y = f(x) as
y = a*f1(x) + f2(x),
then the optimum parameter a is given by
sum_{i=1}^{N} f1(x_i)*(y_i-f2(x_i)) / 
sum_{i=1}^{N} (f1(x_i))^2

(in your case, N=7).

Best wishes
Torsten.

Torsten Hennig <Torsten.Hennig@umsicht.fhg.de> wrote in message <1842133775.51542.1264493670957.JavaMail.root@gallium.mathforum.org>...
> > 
> 
> If you write y = f(x) as
> y = a*f1(x) + f2(x),
> then the optimum parameter a is given by
> sum_{i=1}^{N} f1(x_i)*(y_i-f2(x_i)) / 
> sum_{i=1}^{N} (f1(x_i))^2
> 
> (in your case, N=7).
> 
> Best wishes
> Torsten.
first of all that you very much for your reply. how can i convert my equation to the form that you have give it to me? which form will get the function f(1) (X)and f(2)(X)