Question

Possibly Mathematica specific.
Here f[x] is a mathematica least squares fit to a dataset.

"Use f[x]=E^(5 + .5 x)​ to get a good fit with a solution of the logistic differential equation"
GIVEN
Solution to the logistic differential equation
y[x] -> (b*E^(r*x + b*k))/(-1 + E^(r*x + b*k))​

Now somehow, I am supposed to reconcile F[x] into y[x], and create some kind of logistic model. Is it just a case of identifying the relative variables between the equations, and then use substitution, or is there more to this, requiring some manipulation?

Answer 1

beyond my mathematicabilities

Answer 2

@Miracrown

Answer 3

Do you have an a dataset to work from? Or do you need need to come up with a mathematica command line?

Answer 4

I ahve a dataset

Answer 5

Should I show the full data and conversion that led up to the creation of an f[x] =?

Answer 6

no I don't think that would be helpful you want to take the assumption that your solution function is log based they even say "solution to the logistic differential equation"

Answer 7

Right you can practice on a simpler function, then once you work through it you can use the data set you have

Answer 8

probably better to keep with the simplified version ?

Answer 9

cool

Answer 10

Right they told mathematic to fit a function to the data set

that is the line fitter[t_] = E^Fit[logdata,{1,t},t]

Answer 11

wow, you're actually going to answer this.. this is awesome.

Answer 12

LOL

Answer 13

or at least get me clued in.

Answer 14

this looks like it is the solution

Answer 15

so mathematica did it already? the fit command actually did a solution to the logistic differential equation.

Answer 16

so this is a semi log plot it should give you a straight line since it is a population time graph
so the function should look like f(t) = A e^rt where r is the slope of the semi log plot
t is in years and A is the y intercept of the plot

Answer 17

\(\color{blue}{\text{Originally Posted by}}\) @hughfuve
so mathematica did it already? the fit command actually did a solution to the logistic differential equation.
\(\color{blue}{\text{End of Quote}}\)
It should have, it should give you either an equation or the values of A and r

Answer 18

I think it should kick out an equation from what is here

Answer 19

okay hmmm.. so
fitter[t_] = E^Fit[logdata,{1,t},t]
gave me
E^(4.382968909053888 + 0.01300265909581666*x)​

and that should probably be
A e^rt
where
r is the slope of the semi log plot
t is in years and
A is the y intercept of the plot

Answer 20

so that constant on the front might be A
E^(4.382968909053888) E^(0.01300265909581666*x)​

Answer 21

well A = e^4.38 and R = .013

Answer 22

sweet.. concur

Answer 23

not quite we raise e to the 4.38 to get A then r is .0130026

Answer 24

gotcha.. we are on the same page there then.

Answer 25

Mhm

Answer 26

so as for this equation..
y[x] -> (b*E^(r*x + b*k))/(-1 + E^(r*x + b*k))​
we are basically already in a simplified version of this?

Answer 27

right we get the simplified form

Answer 28

or rather, we are now in a simplified version of this.

Answer 29

it looks like the way they have it A = b but r is the same and I think k is 0

Answer 30

and x is t

Answer 31

gotcha

Answer 32

k might be something very very small it might be 1 x 10^-12 depending on how far mathematica calculates it out

Answer 33

a limit of it's precision you think?

Answer 34

well miracrown.. that has put me in a much clearer vision.. thank you.

Answer 35

right k should go to 0 for a perfect fit then all the b*k terms go away :)

Answer 36

and our equation becomes b*e^rt or b* e^rx

Answer 37

awesome.. thank you.. thank you.

Answer 38

Yw yw :)