Matlab to find proper coefficients [duplicate] - matlab

Question

This question already has answers here:
MATLAB - Fit exponential curve WITHOUT toolbox
(2 answers)
Closed 7 years ago.
I have such x vectors and y vectors described in below :
x = [0 5 8 15 18 25 30 38 42 45 50];
y = [81.94 75.94 70.06 60.94 57.00 50.83 46.83 42.83 40.94 39.00 38.06];
with these values how can i find an coefficients of y = a*(b^x) ??
I've tried this code but it finds for y = a*e^(b*x)
clear, clc, close all, format compact, format long
% enter data
x = [0 5 8 15 18 25 30 38];
y = [81.94 75.94 70.06 60.94 57.00 50.83 46.83 42.83];
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((k-b*j)/n)
y = a * exp(b * 35)
result_68 = log(68/a)/b
I know interpolation techniques but i couldn't implement it to my solutions...
Many thanks in advance!

Answer 1

Since y = a * b ^ x is the same as log(y) = log(a) + x log(b) you can do
>> y = y(:);
>> x = x(:);
>> logy = log(y);
>> beta = regress(logy, [ones(size(x)), x]);
>> loga = beta(1);
>> logb = beta(2);
>> a = exp(loga);
>> b = exp(logb);
so the values of a and b are
>> a, b
a =
78.8627588780382
b =
0.984328823937827
Plotting the fit
>> plot(x, y, '.', x, a * b .^ x, '-')
gives you this -
NB the regress function is from the statistics toolbox, but you can define a very simple version which does what you need
function beta = regress(y, x)
beta = (x' * x) \ (x' * y);
end

Answer 2

As an extension to the answer given by Chris Taylor, which provides the best linear fit in the logarithmic-transformed domain, you can find a better fit in the original domain by solving directly the non-linear problem with, for example, the Gauss-Newton algorithm
Using for example the solution given by Chris as a starting point:
x = [0 5 8 15 18 25 30 38 42 45 50];
y = [81.94 75.94 70.06 60.94 57.00 50.83 46.83 42.83 40.94 39.00 38.06];
regress = #(y, x) (x' * x) \ (x' * y);
y = y(:);
x = x(:);
logy = log(y);
beta = regress(logy, [ones(size(x)), x]);
loga = beta(1);
logb = beta(2);
a = exp(loga)
b = exp(logb)
error = sum((a*b.^x - y).^2)
Which gives:
>> a, b, error
a =
78.862758878038164
b =
0.984328823937827
error =
42.275290442577422
You can iterate a bit further to find a better solution
beta = [a; b];
iter = 20
for k = 1:iter
fi = beta(1)*beta(2).^x;
ri = y - fi;
J = [ beta(2).^x, beta(1)*beta(2).^(x-1).*x ]';
JJ = J * J';
Jr = J * ri;
delta = JJ \ Jr;
beta = beta + delta;
end
a = beta(1)
b = beta(2)
error = sum((a*b.^x - y).^2)
Giving:
>> a, b, error
a =
80.332725222265623
b =
0.983480686478288
error =
35.978195088265906

Answer 3

One more option is to use MATLAB's Curve Fitting Toolbox which lets you generate the fit interactively, and can also emit the MATLAB code needed to run the fit.