What Is the Long-Term Fish Population in This Infinite Series Problem?

Burjam
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Homework Statement



A fishery manager knows that her fish population naturally increases at a rate of 1.4% per month, while 119
fish are harvested each month. Let Fn be the fish population after the nth month, where F0 = 4500 fish. Assume that that process continues indefinitely. Use the infinite series to find the long-term (steady-state) population of the fish exactly.

Homework Equations

The Attempt at a Solution



My issue is that I can't seem to set up an expression to evaluate the series. I know that the expression will involve subtracting 119 and use 0.014 to represent the percent increase. If it were only the percent increase, I would be able to set up an expression. But the -119 is really throwing me off.
 
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Burjam said:

Homework Statement



A fishery manager knows that her fish population naturally increases at a rate of 1.4% per month, while 119
fish are harvested each month. Let Fn be the fish population after the nth month, where F0 = 4500 fish. Assume that that process continues indefinitely. Use the infinite series to find the long-term (steady-state) population of the fish exactly.

Homework Equations

The Attempt at a Solution



My issue is that I can't seem to set up an expression to evaluate the series. I know that the expression will involve subtracting 119 and use 0.014 to represent the percent increase. If it were only the percent increase, I would be able to set up an expression. But the -119 is really throwing me off.
The question does not make clear whether the Fn represent the population just after a harvest or just before. I would take it as just after.
If the population is Fn after the nth month what will it be after one more month?
 
Fn+1 = Fn(1 + 0.014) - 119?
 
Burjam said:
Fn+1 = Fn(1 + 0.014) - 119?
Right. Do you know a way to solve such equations? If not, an easy thing to try is to see if you can add a constant to each Fn so that it reduces to a simple geometric progression.
 
haruspex said:
Right. Do you know a way to solve such equations? If not, an easy thing to try is to see if you can add a constant to each Fn so that it reduces to a simple geometric progression.

I don't know how to write this equation without Fn being in terms of Fn+1 or Fn-1.
 
Burjam said:
I don't know how to write this equation without Fn being in terms of Fn+1 or Fn-1.
It will be the same equation, but written in the form (Fn+1+c)=a(Fn+c) for some pair of constants a and c.
 
haruspex said:
It will be the same equation, but written in the form (Fn+1+c)=a(Fn+c) for some pair of constants a and c.

How will adding the c to both sides eliminate the Fn+1? None of the problems I have done or have examples of with infinite series so far have anything like this, so I don't really have anything to go by.
 
Burjam said:
How will adding the c to both sides eliminate the Fn+1?
I did not suggest it would.
You have this equation: Fn+1 = Fn(1 + 0.014) - 119
and I am suggesting this form of it: (Fn+1+c)=a(Fn+c)
What do you get if you combine them?
 
Burjam said:
Fn+1 = Fn(1 + 0.014) - 119?

You have ##F_{n+1} = 1.014 F_n - 119## with ##F_0 = 4500##. Try calculating ##F_1, F_2, F_3## (keeping ##F_0## symbolic instead of 4500). In fact, it might make everything much clearer if you keep all parameters symbolic, so that ##F_{n+1} = r F_n - k##. Using symbols like that instead of numbers helps keep separate the different effects.

However, I think there is something very wrong with the original problem statement: for ##r > 1## (for example, for ##r = 1.014##) you must have a very special relationship between ##F_0,r,k## in order to obtain a finite limit; otherwise you will either have ##F_n \to +\infty## as ##n \to \infty## (for some combinations of ##F_0##, ##r##, and ##k##) or else ##F_n \to -\infty## for for other combinations. Of course, the latter case really means that ##F_n## hits zero at some finite ##n## and so the fish population dies out completely and the problem ends; ##F_n## does not actually go to ##-\infty##.
 
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  • #10
haruspex said:
I did not suggest it would.
You have this equation: Fn+1 = Fn(1 + 0.014) - 119
and I am suggesting this form of it: (Fn+1+c)=a(Fn+c)
What do you get if you combine them?

By combine them, do you mean take a Fn+1 in the second equation as Fn+1 = Fn(1+0.014) - 119 and then try to solve for a and C?
 
  • #11
Ray Vickson said:
I think there is something very wrong with the original problem statement
I assumed it was intended that:
Ray Vickson said:
you will either have ##F_n \to +\infty## as n→∞... [or] ... the fish population dies out completely
 
  • #12
Burjam said:
By combine them, do you mean take a Fn+1 in the second equation as Fn+1 = Fn(1+0.014) - 119 and then try to solve for a and C?
Yes. You will have one equation with two unknowns, but remember that the equation has to be true for all Fn.
 

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