Hello , I am dealing with linear least square regression
1 Ansicht (letzte 30 Tage)
Ältere Kommentare anzeigen
Charitha Heshan
am 16 Mär. 2018
Kommentiert: Charitha Heshan
am 18 Mär. 2018
.I successfully found the coefficients for the quadratic function as follows.
function [a2,a1,a0] = QuadraticRegression (x,y)
x=[10 15 25 40 50 55];
y=[94 116 147 183 215 220];
n = length (x)
sx = sum(x)
sy = sum(y)
sx2 = sum(x.^2)
sx3 = sum(x.^3)
sx4 = sum(x.^4)
sxy = sum(x.*y)
sx2y= sum(x.*x.*y)
A = [ n sx sx2; sx sx2 sx3;sx2 sx3 sx4]
b = [sy;sxy;sx2y]
w= A\b
a2 = w(3)
a1 = w(2)
a0 = w(1)
% code
end*
func(y) = -0.0170 x.^2 + 3.8870 x + 59.0062
now i'm suppose to fit the above data into a power function as follows : y = Q (x.^M) Where Q and M are coefficients.
I went through the book https://books.google.com/books/about/Numerical_Methods_for_Engineers_and_Scie.html?id=gMrMBQAAQBAJ&printsec=frontcover&source=kp_read_button#v=onepage&q&f=false
and went through Youtube and google before i posted my question here. I see no method mentions in the any of the resources. Please give me some clues. Thanks
1 Kommentar
David Goodmanson
am 16 Mär. 2018
Hi Charitha,
try taking the logarithm of both sides and using the rule for log(C*x^m) in terms of log(C), m and log(x). That should give you an equation you can fit for two new arrays, logy = log(y) vs. logx = log(x).
Akzeptierte Antwort
Torsten
am 16 Mär. 2018
Bearbeitet: Torsten
am 16 Mär. 2018
y = Q*x^M => log(y) = log(Q) + M*log(x)
Thus take "log" of your x- and y-data and fit them by a linear function
y~ = a+b*x~
where
y~=log(y) and x~=log(x).
Once you have determined a and b, Q=exp(a) and M=b.
Another possibility is to use "lsqcurevfit" directly on the nonlinear function y = Q*x^M.
This will give slightly different fit parameters for Q and M because taking the log of x and y and making a linear fit as suggested above somehow distorts the data.
Best wishes
Torsten.
Weitere Antworten (0)
Siehe auch
Kategorien
Mehr zu Linear and Nonlinear Regression finden Sie in Help Center und File Exchange
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!