Regression curve fitting for a given equation

Find the best fit by given equation

x = [0.0191051892041436	0.0199064802088661	0.0205144445903776	0.0210029746368803	0.0216434799932356	0.0226870689634767	0.0238694334820173	0.0247271126862428	0.0255324218699147	0.0266869614901355];
y = [0.726909090909091	0.731030303030303	0.730909090909091	0.709818181818182	0.664424242424242	0.621454545454545	0.606848484848485	0.597151515151515	0.595939393939394	0.583333333333333];
plot(x, y, 'b.-');
grid on;

where y = (x/a)*ln((m-1)/b) +y0

Answer 1

[Matt J]

p=polyfit(x,y,1);

From this, you immediately obtain y0=p(2). For the remaining parameters, you can choose any of the infinite solutions to ln((m-1)/b)/a=p(1).

[Arvind]

i have want all values like m, b and a separetly

[the cyclist]

@Arvind, the equation you want to fit to,

y = (x/a)*ln((m-1)/b) + y0

is equivalent to

y = x*c + y0

where

c = (1/a)*ln((m-1)/b)

The solution that @Matt J provided is solving for y0 and what I called c.

It is not possible (without other restrictions) to determine what a, m, and b are. As Matt stated, there are literally an infinite combination of possibilities.