How to create a fit like 1/(x^n)

Hello,

I am trying to fit a curve to three data points. The data points are measurements from a field weakening characteristic of an electric mashine. The characteristic should be direct proportional to 1/(x^n)

datapoint1=[2250 34.4599];
datapoint2=[3500 18.3993];
datapoint3=[6000 8.3306];
% Generate a matrix wiht all datapoints

data=[datapoint1; datapoint2; datapoint3];
xdata=data(:,1);
ydata=data(:,2);

%create the new fittype (n=1)
ft=fittype({'1/(x)'})

% fit the three datapoints with the given fittype
fitobject=fit(data(:,1),data(:,2),ft);


% 
% Plot the results
plot(fitobject,data(:,1),data(:,2),'o')

Result for n=1

Result for n =1.4

So my question is: Is it possible to create a fittype like 1/(x^n) and let matlab search for the best solution for n?

I tried

fo = fitoptions('Method','NonlinearLeastSquares', 'Normalize', 'on', 'Robust', 'LAR','StartPoint',[1]);
ft = fittype('C0*x^n','problem','n','options',fo);

from a former question, but I dont know how to handle the problem parameter. Thank you very much for any suggestions :-)

[dpb]

"Generate a matrix wiht all datapoints..."

data=[2250 34.4599; 3500 18.3993; 6000 8.3306];

Answer 1

[dpb]

Of course, the above can be linearized by log transform...

b=polyfit(log10(x),log10(y),1);

>> b(1), 10^b(2)
ans =
   -1.4484
ans =
   2.4810e+06
>> 

b(1)=-b(1); b(2)=10^b(2);   % for initial start point in nonlinear model

fo=fitoptions('Method','NonlinearLeastSquares','StartPoint',flip(b));
ft=fittype(@(a,n,x) a./x.^n,'options',fo);
f1=fit(x,y,ft);

>> f1
f1 = 
     General model:
     f1(x) = a./x.^n
     Coefficients (with 95% confidence bounds):
       a =   2.317e+06  (-1.344e+06, 5.978e+06)
       n =        1.44  (1.239, 1.641)
>> 

ADDENDUM:

I forgot there is a power model built in...simplest is to use it:

>> fit(x,y,'power1')
ans = 
     General model Power1:
     ans(x) = a*x^b
     Coefficients (with 95% confidence bounds):
       a =    2.24e+06  (-1.159e+06, 5.639e+06)
       b =      -1.436  (-1.629, -1.243)
>> 

Returns slightly different coefficients as it fits x instead of 1/x; I didn't try to compare the two.

You can, of course, pass 1/x if is really important to do it that way...

>> fit(1./x,y,'power1')
ans = 
     General model Power1:
     ans(x) = a*x^b
     Coefficients (with 95% confidence bounds):
       a =    2.24e+06  (-1.159e+06, 5.639e+06)
       b =       1.436  (1.243, 1.629)
>> 

but it returns same coefficients either way using the internal power model formulation.

[Alexander Grimm]

Hello dpd,

thank you very much for your reply. The power model is exactly what i was looking for. I will keep in mind how you generated the starting points for your first suggestion. Maybe i need this for other applications in the future.

The result is amazing :-)

Thank you again and I wish you a nice day!