anonymous
  • anonymous
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?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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nincompoop
  • nincompoop
beyond my mathematicabilities
nincompoop
  • nincompoop
@Miracrown
Miracrown
  • 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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anonymous
  • anonymous
I ahve a dataset
anonymous
  • anonymous
Should I show the full data and conversion that led up to the creation of an f[x] =?
Miracrown
  • 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
  • Miracrown
Right you can practice on a simpler function, then once you work through it you can use the data set you have
anonymous
  • anonymous
probably better to keep with the simplified version ?
anonymous
  • anonymous
cool
Miracrown
  • 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
  • anonymous
wow, you're actually going to answer this.. this is awesome.
Miracrown
  • Miracrown
LOL
anonymous
  • anonymous
or at least get me clued in.
Miracrown
  • Miracrown
this looks like it is the solution
anonymous
  • anonymous
so mathematica did it already? the fit command actually did a solution to the logistic differential equation.
Miracrown
  • 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
  • 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
  • Miracrown
I think it should kick out an equation from what is here
anonymous
  • 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
  • anonymous
so that constant on the front might be A E^(4.382968909053888) E^(0.01300265909581666*x)​
Miracrown
  • Miracrown
well A = e^4.38 and R = .013
anonymous
  • anonymous
sweet.. concur
Miracrown
  • Miracrown
not quite we raise e to the 4.38 to get A then r is .0130026
anonymous
  • anonymous
gotcha.. we are on the same page there then.
Miracrown
  • Miracrown
Mhm
anonymous
  • 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
  • Miracrown
right we get the simplified form
anonymous
  • anonymous
or rather, we are now in a simplified version of this.
Miracrown
  • Miracrown
it looks like the way they have it A = b but r is the same and I think k is 0
Miracrown
  • Miracrown
and x is t
anonymous
  • anonymous
gotcha
Miracrown
  • Miracrown
k might be something very very small it might be 1 x 10^-12 depending on how far mathematica calculates it out
anonymous
  • anonymous
a limit of it's precision you think?
anonymous
  • anonymous
well miracrown.. that has put me in a much clearer vision.. thank you.
Miracrown
  • Miracrown
right k should go to 0 for a perfect fit then all the b*k terms go away :)
Miracrown
  • Miracrown
and our equation becomes b*e^rt or b* e^rx
anonymous
  • anonymous
awesome.. thank you.. thank you.
Miracrown
  • Miracrown
Yw yw :)

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