Nonlinear regression in case of many fits

I have tried to fit certain data points using equation of the form y=ax^-1/3. My first question is , Is it possible to linearise the equation as I do in the code below? The other problem is that how can I get the coefficients in case of many regression fits as indicated in the for loop. Here is the code I tried. The loop doesn't give me as many coefficients as equal to w. Only one coefficient is obtained.

 for w= 1:3684
 z(:,w) = R(1:12,w).^-1/3;
 y(:,w)=S(1:12,w);
 p(w,:) = polyfit(z,y,1);
 %yy = p(1) * z + p(2);
 f = polyval(p,z(:,w));
 plot(R(1:12,w),y(:,w),'o',R(1:12,w),f(:,w),'-')
  end

Answer 1

[Jos (10584)]

First, note this

x = [1 6 9]
x.^-1/3
x.^(-1/3)
(x.^-1)/3

Then, you do not need to store z and w. Here is a more efficient solution:

idx = [1 10 100 3684] ;
N = numel(idx) ;
p = zeros(N,2) ; % pre-allocation

for k = 1:N
   w = idx(k) ;
   x = R(1:12,w) ;
   xz = x.^-1/3;
   y = S(1:12,w);
   p(k,:) = polyfit(xz,y,1);
   f = polyval(p,xz) ;
   plot(x, y, 'o', x, f, '-')
end

If you want to store the fitted values as well, pre-allocate f

f = zeros(12,N) ;
...
f(:,k) = polyval(p,xz) ;

[Ede gerlderlands]

Thank you it's really helpful. But am getting an error which says 'p must be a vector' . More over, final values of x,y and xz should be 12x3684. What am getting is in the form of 12x1? the loop isn't working? what is my error here . Thank you

[Ede gerlderlands]

Thank you. This take my burden away ..

[Jos (10584)]

I meant

f = polyval(p(k,:), xz)

but you probably already figured that out yourself.