Least squares Exponential fit using polyfit

Let's say I'm given x=[11,60,150,200] and y=[800,500,400,90] These are just random numbers (but imagine the solution is in the form of y=a*exp(b*t)

Now, I want to find what 'a' and 'b' are. This is what I'm thinking to do, but I'm not sure if it's correct:

So, if I take ln of both sides of the above equation, I'll get ln(y)= ln(a) +bx. This is in the form of y=mx+b (linear equation).

x= [10, 55, 120, 180]
y= [750, 550, 300, 100]

yPrime= log(y)%take natural logarithm of y data values
pPrime=polyfit(t,yPrime,1)%
aPrime=pPrime(1)
bPrime=pPrime(2)

so now I found the constants for my above LINEAR equation. To find 'a' and 'b' from 'y=a*exp(b*t)', should I now raise the linear constants I found to e? (e^aPrime = a, e^bPrime= b) ?

Is this how I find 'a' and 'b'?

Answer 1

[Star Strider]

Since you are starting with:

y = a * exp(b * t)

and linearising it yields:

log(y) = log(a) + b*t

however ‘aPrime’ and ‘bPrime’ are reversed with respect to the way polyfit works.

So polyfit returns:

bPrime = pPrime(1)
aPrime = pPrime(2)

you need to transform only ‘aPrime’. So:

a = exp(aPrime)

If you want to plot a line-of-fit, you could either use your originally log-transformed equation with log-transformed variables:

log(y) = aPrime + bPrime*t

or:

yfit = exp(log(aPrime)) * exp(b*t)

with your original data.

In code:

t = [11,60,150,200];
y = [800,500,400,90];

yPrime= log(y)%take natural logarithm of y data values
pPrime=polyfit(t,yPrime,1)%
aPrime=pPrime(2)
bPrime=pPrime(1)

figure(1)
plot(t, log(y), 'p',    t, polyval(pPrime, t), '-r')

figure(2)
plot(t, y, 'p',    t, exp(aPrime)*exp(t*bPrime), '-r')

figure(3)
semilogy(t, y, 'p',    t, exp(aPrime)*exp(t*bPrime), '-r')

[Rachel Dawn]

Star Strider,

Thank you. I ended up figuring it out, except for the plotting part. Thank you for that! I really appreciate it.

[Star Strider]

As always, my pleasure.

If my Answer helped you solve your problem please Accept it!

[Rachel Dawn]

Star, I accepted it.

I have one more question. I'm just curious, what if it were y=a*(x^b)? I could do "ln(y) = ln(a) + a*ln(x)" but then what would separate this from the exponential function (when writing the code)?

I'm thinking the only difference would be that I would add "xprime= log(x)".

[Star Strider]

Thank you.

That’s a power function. I’ll restate it to make it a bit clearer:

ln(y) = ln(b) + a*ln(x)

In that polyfit result, ‘bPrime’ would be ‘a’ and ‘aPrime’ would be ‘log(b)’. So to get the original value of ‘b’, calculate it as:

b = exp(aPrime);

The polyfit function returns the coefficients in descending powers of the independent variable, so in a two-parameter linear problem will return:

y = bPrime*x + aPrime

[Tamir Suliman]

didnt you also have to ployfit for log(t) values ?

yPrime= log(y)%take natural logarithm of y data values
tPrime = log(t)
pPrime=polyfit(tPrime,yPrime,1)%

[kainat rasheed]

can you write code for power function ? i am facing a problem