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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?

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  1. nincompoop
    • one year ago
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    beyond my mathematicabilities

  2. nincompoop
    • one year ago
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    @Miracrown

  3. Miracrown
    • one year ago
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    Do you have an a dataset to work from? Or do you need need to come up with a mathematica command line?

  4. anonymous
    • one year ago
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    I ahve a dataset

  5. anonymous
    • one year ago
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    Should I show the full data and conversion that led up to the creation of an f[x] =?

  6. Miracrown
    • one year ago
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    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
    • one year ago
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    Right you can practice on a simpler function, then once you work through it you can use the data set you have

  8. anonymous
    • one year ago
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    probably better to keep with the simplified version ?

  9. anonymous
    • one year ago
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    cool

  10. Miracrown
    • one year ago
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    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
    • one year ago
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    wow, you're actually going to answer this.. this is awesome.

  12. Miracrown
    • one year ago
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    LOL

  13. anonymous
    • one year ago
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    or at least get me clued in.

  14. Miracrown
    • one year ago
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    this looks like it is the solution

  15. anonymous
    • one year ago
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    so mathematica did it already? the fit command actually did a solution to the logistic differential equation.

  16. Miracrown
    • one year ago
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    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
    • one year ago
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    \(\color{blue}{\text{Originally Posted by}}\) @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
    • one year ago
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    I think it should kick out an equation from what is here

  19. anonymous
    • one year ago
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    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
    • one year ago
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    so that constant on the front might be A E^(4.382968909053888) E^(0.01300265909581666*x)​

  21. Miracrown
    • one year ago
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    well A = e^4.38 and R = .013

  22. anonymous
    • one year ago
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    sweet.. concur

  23. Miracrown
    • one year ago
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    not quite we raise e to the 4.38 to get A then r is .0130026

  24. anonymous
    • one year ago
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    gotcha.. we are on the same page there then.

  25. Miracrown
    • one year ago
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    Mhm

  26. anonymous
    • one year ago
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    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
    • one year ago
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    right we get the simplified form

  28. anonymous
    • one year ago
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    or rather, we are now in a simplified version of this.

  29. Miracrown
    • one year ago
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    it looks like the way they have it A = b but r is the same and I think k is 0

  30. Miracrown
    • one year ago
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    and x is t

  31. anonymous
    • one year ago
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    gotcha

  32. Miracrown
    • one year ago
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    k might be something very very small it might be 1 x 10^-12 depending on how far mathematica calculates it out

  33. anonymous
    • one year ago
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    a limit of it's precision you think?

  34. anonymous
    • one year ago
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    well miracrown.. that has put me in a much clearer vision.. thank you.

  35. Miracrown
    • one year ago
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    right k should go to 0 for a perfect fit then all the b*k terms go away :)

  36. Miracrown
    • one year ago
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    and our equation becomes b*e^rt or b* e^rx

  37. anonymous
    • one year ago
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    awesome.. thank you.. thank you.

  38. Miracrown
    • one year ago
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    Yw yw :)

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