I want to fit some data x and y to a specific equation f(x) and then estimate this fitted data as g(x)=1/f(x) ?

x=[0,0.5,1,1.5,2,2.5];

y=[1,0.8,0.6,0.4,0.2,0];

F(x)= exp[(a*x^2)+(b*x)+c], where a,b,c are not known to me

after fitting this data to this equation i want to get g(x)=1/f(x). The picture attached below demonstrates what we are trying to do. Figure A shows x and y fitted to f(x) and figure B shows estimated g(x)=1/f(x).

Answer 1

[Ameer Hamza]

Using Curve Fitting toolbox , it is quite easy to fit the curve.

fitFunction = fittype('exp(a*x.^2+b*x+c)');
coff = fit(x', y', fitFunction);

a = coff.a;
b = coff.b;
c = coff.c;

[j.D]

And what about getting g(X)

[Walter Roberson]

Once you have the coefficients you can construct f(x) with a simple anonymous function and then define g = @(x)1./f(x)

[j.D]

Thanks for your guidance after your comments this is what i am doing but still getting some error, please guide

fitFunction = fittype('exp(a*x.^2+b*x+c)');

x=[0,0.5,1,1.5,2,2.5];

y=[1,0.8,0.6,0.4,0.2,0];

coff = fit(x', y', fitFunction);

a = coff.a;

b = coff.b;

c = coff.c;

Z=exp(-0.3365*x.^2-0.1660*x+-0.0150);

F=exp(a*x.^2+b*x+c);

G(x)= Z/F;

[Torsten]

Z=@(x)exp(-0.3365*x.^2-0.1660*x+(-0.0150));

F=@(x)exp(a*x.^2+b*x+c);

G=@(X) Z(X)./F(X);

[Walter Roberson]

 Z = @(x) exp(-0.3365*x.^2-0.1660*x+-0.0150);
 F = @(x) exp(a*x.^2+b*x+c);
 G = @(x) Z(x)./F(x);
 fplot(G, [x(1), x(end)])

[Ameer Hamza]

F = exp(-0.3365*x.^2-0.1660*x+(-0.0150))

is actually the result of curve fitting. I think OP is trying to take the inverse of F but forgot to include @(x) sign.

@j.D please use this line exactly as given below,

F = @(x) exp(a*x.^2 + b*x + c);
G = @(x) 1./F(x);

[Walter Roberson]

Ameer Hamza: if you look at the sample calibration function, it is pretty linear. We might suspect that something that is pretty close to 1:1 is expected.

For example it could be the situation that the model was defined at being exactly exp(-0.3365*x.^2-0.1660*x+-0.0150), but that actual data is then obtained, and a correction function needs to be calculated based upon the data.

[j.D]

Thanks guys for your help, both the approaches are good but ameer hamza anwser suits my scenario, thanks again walter and ameer hamza

[Ameer Hamza]

@Walter Good point, I didn't notice it earlier.

[j.D]

Walter you might be right but in your case I am not getting the inverse of data can i some how get invert of it with your logic ?

[Ameer Hamza]

@j.D, Walter equation does not invert the data. He was mentioning that you might want to compare the estimated model with an actual model which you already know. If all you want is value of f and g, then the equation i wrote will be sufficient for you.

[Ameer Hamza]

@j.D's comment from answer moved here

Thanks ameer..can you tell me does your code not give me the estimated calibration function ?and thanks for sharing your knowledge with me

[Ameer Hamza]

My code does gives you an estimate of the F function, and its inverse G.

[Walter Roberson]

 Z = @(x) exp(-0.3365*x.^2-0.1660*x+-0.0150);
 F = @(x) exp(a*x.^2+b*x+c);
 recipF = @(x) 1./F(x);
 G = @(x) Z(x) .* recipF(x);
 fplot(G, [x(1), x(end)])

This is based upon your earlier definition

G(x)= Z/F;

but breaks out the 1/F step that you had asked for originally.

We cannot tell whether your estimated calibration function for your purposes is the 1/F or is the Z/F .

[j.D]

it is 1/F

[j.D]

@Walter Roberson how can i multiply this obtained function G with a tif image ?

[Walter Roberson]

Flimg = double(YourMatrix) ;
Out = cast(Flimg.*G(Flimg), class(YourMatrix)) ;

[j.D]

Yeah that works , thanks.