## anonymous one year ago 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?

1. nincompoop

beyond my mathematicabilities

2. nincompoop

@Miracrown

3. Miracrown

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

4. anonymous

I ahve a dataset

5. anonymous

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

6. 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"

7. Miracrown

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

8. anonymous

probably better to keep with the simplified version ?

9. anonymous

cool

10. 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]

11. anonymous

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

12. Miracrown

LOL

13. anonymous

or at least get me clued in.

14. Miracrown

this looks like it is the solution

15. anonymous

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

16. 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

17. 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

18. Miracrown

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

19. 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

20. anonymous

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

21. Miracrown

well A = e^4.38 and R = .013

22. anonymous

sweet.. concur

23. Miracrown

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

24. anonymous

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

25. Miracrown

Mhm

26. 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?

27. Miracrown

right we get the simplified form

28. anonymous

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

29. Miracrown

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

30. Miracrown

and x is t

31. anonymous

gotcha

32. Miracrown

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

33. anonymous

a limit of it's precision you think?

34. anonymous

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

35. Miracrown

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

36. Miracrown

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

37. anonymous

awesome.. thank you.. thank you.

38. Miracrown

Yw yw :)