Coefficient determination for polyfit

2 visualizaciones (últimos 30 días)
FL
FL el 9 de Oct. de 2018
Editada: FL el 9 de Oct. de 2018
Hi Can somebody please give some tips on below:
I have a Matrix (3x3601) and I have a vector (1x3) defined as below
gamma = 1.31; %initial values
k = -3.6; %initial values
A = (gamma-1)*(k);
B = gamma;
VECTOR = [1,A,B];
DATA = VECTOR * MATRIX;
Now, I want to use a range of values for gamma and k as below
gamma = (1.1:0.001:1.40);
k = (-3.6:0.01:-2);
k = [k, zeros(1, length(gamma) - length(k))]; %to make it of the same length
I want to use the optimum value from the range of gamma and k so that it minimizes the squared error
How do I go about it?
Thanks
  2 comentarios
Star Strider
Star Strider el 9 de Oct. de 2018
‘I want to use the optimum value from the range of gamma and k so that it minimizes the squared error’
The squared error between what and what?
What model are you regressing?
(Pardon me for this.) Your post makes absolutely no sense, especially since you define gamma initially as a scalar and later as a vector.
Please describe what you want to do in a bit more detail. Note that polyfit takes vector — not matrix — arguments for its one independent and one dependent variables, so it is likely not appropriate here.
FL
FL el 9 de Oct. de 2018
Editada: FL el 9 de Oct. de 2018
Apologies for not putting it perfectly !
I have an equation, in which I have splitted the variables (as MATRIX) and the coefficient as (VECTOR).
The equation is DATA = VECTOR*MATRIX
When I solve the equation, I get a residual vector. Now, I have to look into a range of coefficients (gamma and k, which are defined as vectors now to represent the range), which will give me a minimum residual vector. Not sure how to approach this problem. Hope it now make sense.

Iniciar sesión para comentar.

Respuestas (0)

Categorías

Más información sobre Gamma Functions en Help Center y File Exchange.

Productos


Versión

R2017b

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by