Least Square Linear Approximation
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Hi,
Based on the least square linear approximation, use this to fit a the function y = aln(x) + bcos(x) + ce^x to the data below. Find a, b, and c.
x=[0.24, 0.65, 0.95, 1.24, 1.73, 2.01, 2.23, 2.52, 2.77, 2.99];
x=x'; %Transposes x%
y=[0.23, -0.26, -1.10, -0.45, 0.27, 0.10, -0.29, 0.24, 0.56, 1.00];
y1=[log(x) cos(x) exp(x)]
When I run this code it gives me a 3x10 matrix, how do I find the value of a, b and c (each one should have one value)?
Thank you so much in advance!
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Star Strider
2017-2-11
You’re almost there! Please let me introduce you to the mldivide,\ (link) function and operator to get you the rest of the way:
The Code —
x=[0.24, 0.65, 0.95, 1.24, 1.73, 2.01, 2.23, 2.52, 2.77, 2.99];
y=[0.23, -0.26, -1.10, -0.45, 0.27, 0.10, -0.29, 0.24, 0.56, 1.00];
DesignMatrix = [log(x(:)) cos(x(:)) exp(x(:))];
Coefficients = DesignMatrix\y(:);
a = Coefficients(1)
b = Coefficients(2)
c = Coefficients(3)
The Results —
a =
-1.04103221690367
b =
-1.26131878469978
c =
0.030734825739463
The colon index reference (:) creates column vectors out of whatever they reference, from vectors to multi-dimensional matrices. That also eliminates ambiguities such as having to know if a vector is a row or column vector before transposing it.
2 个评论
(Unknown)
2017-2-11
Star Strider
2017-2-11
My pleasure!
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