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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?
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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nincompoop
 one year ago
Best ResponseYou've already chosen the best response.0beyond my mathematicabilities

Miracrown
 one year ago
Best ResponseYou've already chosen the best response.11Do you have an a dataset to work from? Or do you need need to come up with a mathematica command line?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Should I show the full data and conversion that led up to the creation of an f[x] =?

Miracrown
 one year ago
Best ResponseYou've already chosen the best response.11no 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
 one year ago
Best ResponseYou've already chosen the best response.11Right you can practice on a simpler function, then once you work through it you can use the data set you have

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0probably better to keep with the simplified version ?

Miracrown
 one year ago
Best ResponseYou've already chosen the best response.11Right they told mathematic to fit a function to the data set that is the line fitter[t_] = E^Fit[logdata,{1,t},t]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0wow, you're actually going to answer this.. this is awesome.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0or at least get me clued in.

Miracrown
 one year ago
Best ResponseYou've already chosen the best response.11this looks like it is the solution

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so mathematica did it already? the fit command actually did a solution to the logistic differential equation.

Miracrown
 one year ago
Best ResponseYou've already chosen the best response.11so 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
 one year ago
Best ResponseYou've already chosen the best response.11\(\color{blue}{\text{Originally Posted by}}\) @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
 one year ago
Best ResponseYou've already chosen the best response.11I think it should kick out an equation from what is here

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0okay 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
 one year ago
Best ResponseYou've already chosen the best response.0so that constant on the front might be A E^(4.382968909053888) E^(0.01300265909581666*x)

Miracrown
 one year ago
Best ResponseYou've already chosen the best response.11well A = e^4.38 and R = .013

Miracrown
 one year ago
Best ResponseYou've already chosen the best response.11not quite we raise e to the 4.38 to get A then r is .0130026

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0gotcha.. we are on the same page there then.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so 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
 one year ago
Best ResponseYou've already chosen the best response.11right we get the simplified form

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0or rather, we are now in a simplified version of this.

Miracrown
 one year ago
Best ResponseYou've already chosen the best response.11it looks like the way they have it A = b but r is the same and I think k is 0

Miracrown
 one year ago
Best ResponseYou've already chosen the best response.11k might be something very very small it might be 1 x 10^12 depending on how far mathematica calculates it out

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0a limit of it's precision you think?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0well miracrown.. that has put me in a much clearer vision.. thank you.

Miracrown
 one year ago
Best ResponseYou've already chosen the best response.11right k should go to 0 for a perfect fit then all the b*k terms go away :)

Miracrown
 one year ago
Best ResponseYou've already chosen the best response.11and our equation becomes b*e^rt or b* e^rx

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0awesome.. thank you.. thank you.
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