Do you mean? Because, if so, I can think of at least two ways to solve this problem, one of them probably easier than the other.Try fitting the points above...
In a given experiment, measurements taken showed that the number of
elements of a bacterial population decreased with time, so
approximately exponential:
time - 0 - 1 - 2 - 3 - 4 - 5 - 6 - 7 -8 - 9 - 10
amount - 419 - 274 - 181 - 122 - 85 - 61 - 46 - 36 - 30 - 26 - 24
Try adjusting the points above by a function of type:
My solution:
I find z(x) = ln(x) (*ln of each amount*)
g1(x) = 1; g2(x) = x;
I find:
a11 = 11
a12=a21 = 55
a22 = 385
b1 = 47,4564
b2 = 205,1156
Matrix x Vector(beta1 and beta2) = Vector (b1 and b2)
a1 = expoent(beta1) = 322,37
a2 = beta2 = -0,2923
Is that right?
Sorry, I use fit by exponential functions
I need to find:
and
I rode the table:
time - 0 - 1 - 2 - 3 - 4 - 5 - 6 - 7 -8 - 9 - 10
amount - 419 - 274 - 181 - 122 - 85 - 61 - 46 - 36 - 30 - 26 - 24
ln(amount) - 6,0379 - 5,6131 - 5,1985 - 4,8040 - 4,4426 - 4,1109 - 3,8286 - 3,5835 - 3,4012 - 3,2581 - 3,1780
;
;
I find:
Matrix = (a11, a12, a21, a22) x Vector(Beta1, Beta2) = Vector(b1,b2)
It's correct ?
Your numbers are fine for a linear fit to the logarithm data. However, the plot of the logarithm of the data looks a lot more like a quadratic than a straight line. That is, you can write
with an value of That's good, but not great. On the other hand, if you go with a quadratic fit to the logarithm data, that is, you assume
then
with an value of which is an incredibly good fit.
Just a thought.
The method of least squares can be applied to many more situations than just a linear fit, though that's probably the most common least squares. The fit I just described would be found using the same fundamental method.