Exponential
Regression  calculate with Matlab
We’ll
work this time with exponential
regression in a curve fitting example. The
following codes find the coefficients of an equation for an exponential
curve.

The equation is in the
following form
where a
and b are the calculated
coefficients.
The
equation coefficients, coefficient of determination, coefficient of
correlation
and standard error of estimate are also displayed.

You must provide the x
and y coordinates for known data
points. Once the curve has been fitted
you may predict values of y = f(x)
for given values of x.
We’re
going to experiment with
three different methods to cope with our exponential
regression.
The first method is a classical computation using known formulas.
The second method deals with strategic optimization techniques and
gives
another example of the simplex method implemented by the NelderMead
algorithm
used in the Matlab function fminsearch. The third method just
uses
interpolation capabilities in Matlab, it doesn’t calculate any
coefficient, but
can solve interpolation values, and can even extrapolate to a certain
degree.
Example:
The
table below shows the number of strange particles present in a
scientific
experiment at various points in time. We have to fit an exponential
curve to
the data and estimate the number of particles after 7 hours.
number
of hours = [0 1 2 3 4 5 6]
number
of particles = [25 38 58 89 135 206 315]
First
method: using known formulas
We know
that coefficients a and b are determined by
where
n = number of points in given
data
We also
know that
where
Then,
this is easy to code...
clear,
clc, close all, format compact, format
long
%
enter data
x = 0 :
6;
y = [25
38 58 89 135 206 315];
%
calculate coefficients a and b
n =
length(x);
y2 =
log(y);
j =
sum(x);
k =
sum(y2);
l =
sum(x.^2);
m =
sum(y2.^2);
r2 =
sum(x .* y2);
b = (n *
r2  k * j) / (n * l  j^2)
a =
exp((kb*j)/n)
%
calculate coeff. of determination, coeff. of correlation
%
and standard error of estimate
c = b *
(r2  j * k / n);
d = m 
k^2 / n;
f = d 
c;
cf_dt =
c/d
corr =
sqrt(cf_dt)
std_err
= sqrt(f / (n  2))
%
We can calculate any point along the curve
x
=
input('Interpolation:
');
y = a *
exp(b * x)
The
results are:
b
= 0.42237507956752
a = 24.96166337377739
cf_dt =
0.99999355156242
corr = 0.99999677577601
std_err
= 0.00253817142571
Interpolation:
7
y =
4.800867130774966e+002
Second
method: optimization techniques with fminsearch
This is
a very powerful technique that can be applied in curve fitting,
exponential
regressions, or nonlinear problems in general. We can obtain the
coefficients
without
knowing the formulas above!
What we
do is set the problem in such a way that it can be minimized. The term
‘minimization’ means that we’re comparing our guesses against the given
data. The
least error, after several automatic iterations, is going to be the
solution to
our problem. Matlab builtin function fminsearch tries different
coefficients
and stops when the least error has been found.
Let’s
say that we prepare this objectivefunction:
function U =
OF_exp_regr(c)
global x y
%
try coefficients
y2 =
c(1) * exp(c(2) * x);
%
try to match original data, and return difference
U
=
norm(y2  y, 1);
We can
now use that objective with the following code:
clear,
clc, close all, format compact
global x y
%
enter data and plot it
x = 0 :
6;
y = [25
38 58 89 135 206 315];
plot(x,
y, 'go')
hold on
%
enter starting point: before optimization
c0 = [1
1];
y0 =
c0(1) * exp(c0(2) * x);
plot(x,
y0, 'ro')
legend('measured
data', 'initial
guess')
%
launch optimization method
fx = 'OF_exp_regr';
[c, f,
EF, out] = fminsearch(fx, c0)
yf =
c(1) * exp(c(2) * x);
%
plot results after optimization
figure
plot(x,
y, 'go', x, yf, 'bo')
legend('measured
data', 'final fit')
The results are:
c
= 24.88778378033923 0.42303260089823
f = 1.08673354178606
EF = 1
out =
iterations: 122
funcCount:
226
You
see? We started with a
= 1 and b
= 1 (as in c0 = [1 1]), and we got the initial
graphic. The green line is the target, the red line is our initial
guess. After
the optimization process we got c(1)= a = 24.8877
and c(2)=
b
= 0.4230, and those
coefficients produced the second plot, with green target and blue
points showing our
fit. The results are quite similar to those found by the method using
formulas,
right? It works!
Third
method: inter/extrapolation
This
method doesn’t provide results for coefficients, but it an example of
interpolation and extrapolation capabilities in Matlab.
We’ll
use builin function interp1, whose basic syntax is:
yi = interp11(X,
Y, XI, METHOD, 'extrap') uses the specified method for
extrapolation
for any elements of XI outside the interval spanned by X.
Methods,
among others, can be
'spline'
piecewise cubic spline interpolation
'cubic'
shapepreserving
piecewise cubic
interpolation
The
following code
clear,
clc, close all, format compact
%
plot initial data
x = 0 :
6;
y = [25
38 58 89 135 206 315];
plot(x,
y, 'g.o')
hold on
%
add more points using the coeff. obtained before
xi =
[1.25 3.2 5.7 7 9];
a =
24.961663;
b =
0.422375;
ye = a *
exp(b * xi);
plot(xi,
ye, '*g')
%
inter/extrapolate with 'cubic' method
yi =
interp1(x, y, xi, 'cubic', 'extrap');
plot(xi,
yi, 'ro')
%
inter/extrapolate with 'spline' method
yi =
interp1(x, y, xi, 'spline', 'extrap');
plot(xi,
yi, 'bo')
legend('Original
data', 'Extra data', 'Cubic', 'Spline')
produces
this plot
we can
see that interpolations work pretty well, but extrapolations diverge
from the
expected results. We can see that a spline method works better than a
cubic one
in this case, just for extrapolations.
From 'Exponential
Regression' to home
From
'Exponential Regression' to 'Matlab Programming'
