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?

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- nincompoop

beyond my mathematicabilities

- nincompoop

- Miracrown

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

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## More answers

- anonymous

I ahve a dataset

- anonymous

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

- Miracrown

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"

- Miracrown

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

- anonymous

probably better to keep with the simplified version ?

- anonymous

cool

- Miracrown

Right they told mathematic to fit a function to the data set
that is the line fitter[t_] = E^Fit[logdata,{1,t},t]

- anonymous

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

- Miracrown

LOL

- anonymous

or at least get me clued in.

- Miracrown

this looks like it is the solution

- anonymous

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

- Miracrown

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

- Miracrown

\(\color{blue}{\text{Originally Posted by}}\) @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

- Miracrown

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

- anonymous

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

- anonymous

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

- Miracrown

well A = e^4.38 and R = .013

- anonymous

sweet.. concur

- Miracrown

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

- anonymous

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

- Miracrown

Mhm

- anonymous

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?

- Miracrown

right we get the simplified form

- anonymous

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

- Miracrown

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

- Miracrown

and x is t

- anonymous

gotcha

- Miracrown

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

- anonymous

a limit of it's precision you think?

- anonymous

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

- Miracrown

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

- Miracrown

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

- anonymous

awesome.. thank you.. thank you.

- Miracrown

Yw yw :)

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