How to use least square fit in matlab to find coefficients of a function?

consider the following 10 data points

x = 7.38, 5.86, 2.46, 6.66, 0.83, 6.26, 6.61, 7.29, 8.91, 9.82
y = 11.89, 2.01, 4.54, 7.26, 1.61, 3.99, 7.16, 11.17, 10.44, 1.97

the two variables x and y have a relationship of the following form:

y(x) = c1x + c2x^(2/3) + c3xsin(x)

how can i apply the least square fit algorithm for a linear combination of functions to find the optimal parameters c1,c2,and c3?

Also, i need to plot the data points and the cit curve for 0<x<10

any idea? Can't figure this one out

[Thalia Putri]

how if i have 3 variables? for example x1, x2 and y. the equation is y = c1 x1^2 x2 + c2 x1 x2 +c3 x1 x2^2 how can i get the coefficient c1 c2 c3 ?? please help me. thankyou

Answer 1

[dpb]

Matlab to the rescue!!! :) One of the most usefulest features, indeed!

>> x = [7.38, 5.86, 2.46, 6.66, 0.83, 6.26, 6.61, 7.29, 8.91, 9.82];
   y = [11.89, 2.01, 4.54, 7.26, 1.61, 3.99, 7.16, 11.17, 10.44, 1.97];
>> z=x.^2/3; w=x.*sin(x);  % define transformed variables in x
>> A=[x.' z.' w.'];        % build system matrix
>> b=A\y.'                 % solve the system
b =
  1.0146
 -0.1297
  1.0349
>> plot(x,y,'*')
>> xh=[1:10]  ;      % define the desired output points
>> zh=[xh; xh.^2/3; xh.*sin(xh)];  % and the transformed variable array from them
>> yh=b.'*zh         % evaluate the fit at those points
yh =
1.8423  3.7385  3.0929  0.2338 -0.9699  2.7963  9.7436 13.5416  9.4685  0.1927
>> hold on
>> plot(xh,yh,'b-')
>>

Actually, somewhat to my surprise after the first plot, the results aren't too bad...

ADDENDUM After the transformation, can use any of the curve fitting tools that solve the OLS problem; specifically depending on which Toolboxen you have installed, but the above is in base product and the "left divide" operator is worth the price of Matlab alone at times like this...and was particularly so before there were other alternatives readily available without "roll you own".

[nick arbogast]

thank you!

[nick arbogast]

I still cant find out how to get a c1, c2, and c3 value, any idea?

[dpb]

What you mean? The coefficients? They're the b array from

 >> b=A\y.'                 % solve the system
b =
  1.0146
 -0.1297
  1.0349

Using named (and particularly numbered) variables instead of arrays is generally counterproductive in Matlab. For example instead of

yh=b.'*zh;  % evaluate fit over vector

you'd have to write at a minimum

    yh=[c1 c2 c3]*zh;  % evaluate fit over vector

which is much more effort than needs be..

[Thalia Putri]

how if i have 3 variables? for example x1, x2 and y. the equation is

y = c1 x1^2 x2 + c2 x1 x2 +c3 x1 x2^2 how can i get the coefficient c1 c2 c3 ?? please help me. thankyou

[dpb]

That's

y = (c1 x1^2 + c2 x1 +c3 x1 x2)*x2

In Wilkinson notation that fitlm uses, that can be written as

y ~ (A + A:B + A^2) * B

Since all the Stat Toolbox functions incorporate the intercept term by default, you'll have to explicitly remove it--

y ~ (A + A:B + A^2) * B - 1

I believe should estimate the model altho I've not used the toolset much; I was so ingrained in using the backslash operator with explicitly-formed models I've just not gotten practice since TMW introduced the later tools.

[pooja sudha]

what if we don't have a linear combination of terms?

or if we have a function which includes constant in them and we can't seperate them & write them in a linear combination.