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

[Burjam]

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 F_n 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.

 

[haruspex]

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 F_n 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 F_n represent the population just after a harvest or just before. I would take it as just after.
If the population is F_n after the n^th month what will it be after one more month?

 

[Burjam]

F_n+1 = F_n(1 + 0.014) - 119?

 

[haruspex]

F_n+1 = F_n(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 F_n so that it reduces to a simple geometric progression.

 

[Burjam]

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 F_n so that it reduces to a simple geometric progression.

I don't know how to write this equation without F_n being in terms of F_n+1 or F_n-1.

 

[haruspex]

I don't know how to write this equation without F_n being in terms of F_n+1 or F_n-1.

It will be the same equation, but written in the form (F_n+1+c)=a(F_n+c) for some pair of constants a and c.

 

[Burjam]

It will be the same equation, but written in the form (F_n+1+c)=a(F_n+c) for some pair of constants a and c.

How will adding the c to both sides eliminate the F_n+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.

 

[haruspex]

How will adding the c to both sides eliminate the F_n+1?

I did not suggest it would.
You have this equation: F_n+1 = F_n(1 + 0.014) - 119
and I am suggesting this form of it: (F_n+1+c)=a(F_n+c)
What do you get if you combine them?

 

[Ray Vickson]

F_n+1 = F_n(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 $-\infty$.

 

[Burjam]

I did not suggest it would.
You have this equation: F_n+1 = F_n(1 + 0.014) - 119
and I am suggesting this form of it: (F_n+1+c)=a(F_n+c)
What do you get if you combine them?

By combine them, do you mean take a F_n+1 in the second equation as F_n+1 = F_n(1+0.014) - 119 and then try to solve for a and C?

 

[haruspex]

I think there is something very wrong with the original problem statement

I assumed it was intended that:

you will either have $F_n \to +\infty$ as n→∞... [or] ... the fish population dies out completely

 

[haruspex]

By combine them, do you mean take a F_n+1 in the second equation as F_n+1 = F_n(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 F_n.