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JH
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Problem 93 Hard Difficulty

The size of an undisturbed fish population has been modeled by the formula
$ p_{n + 1} = \frac {bp_n}{a + p_n} $
where $ p_n $ is the fish population after $ n $ years and $ a $ and $ b $ are positive constants that depend on the species and its environment. Suppose that the population in year 0 is $ p_0 > 0. $
(a) Show that if $ \{ p_n \} $ is convergent, then the only possible values for its limit are 0 and $ b - a $.
(b) Show that $ p_{n + 1} < (b/a)p_n $.
(c) Use part (b) to show that if $ a > b, $ then $ \lim_{n \to \infty} p_n = 0 $; in other words, the population dies out.
(d) Now assume that $ a < b $. Show that if $ p_0 < b - a $, then $ \{ p_n \} $ is increasing and $ 0 < p_n < b - a $. Show also that if $ p_0 > b - a $, then $ \{ p_n \} $ is decreasing and $ p_n > b - a $. Deduce that if $ a < b $, then $ \lim_{n \to \infty} p_n = b - a $.

Answer

a. $p=b-a$
b. $1+\frac{p_{m}}{a}>1$
c. $r=\frac{b}{a} \in(0,1)$
d. it is convergent by the Monotonic Sequence Theorem. It then follows from part (a) that $\lim _{n \rightarrow \infty} p_{n}=b-a$

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Video Transcript

here. PN represents a fifth fish population after and years. And that population is modelled by this formula over here. Okay, so for part A, we will show the following. If the sequence converges, then the limit it's and goes to infinity is either zero or be minus a. So let's go ahead and verify this fact solution. So let's just go ahead and supposed So we're supposing that it converges delicious. Call that limit l now using the given formula for piano up here. Let's take a limit on both sides of this equation as n goes to infinity. So we're applying the limit on both sides. So this becomes and this can be simplified now, too. The quadratic. Okay, so now at this that l equals zero is a solution to this equation over here. However, if l is not equal to zero, then we can divide it to obtain. And there are two solutions to this quadratic zero for B minus. And that was what we wanted to establish for parting. So if it converges the limit, it either dies off at zero or it stabilizes that B minus a, which is a constant. Okay with that said, Let's go on to the next page for a heartbeat. So this is where we like to show the following inequality. So it's quite and give a solution For this we have P n plus one, by definition or by the Riker JH in formula Given for p n. We can write this now. Let's divide top and bottom by a A number is just one, and then we get P n over a. Now I can just go ahead and ignore that denominator. The reason I can do this, we're dealing with positive numbers. Here's when the beginning A was a positive fish Population PM has to be not negative. So this is bigger than or equal to one or in this case, just equal to one. So that justifies this inequality here and that resolves the party because this is what we wanted PM plus one and then be over a PM And here this was You justify this. It's okay. Let me go on. I'll need some more room here, so let me go on to parse even on the next page. So here we like to use party to show that if a is bigger than be then Lim api and Ghost zero as n goes to infinity. Okay, so one way to show this is the following. So if the Siri's PN converges, then by the test for diversions we have the limited pn zero. So let's just go ahead and try to establish this fact here and then we'LL finish the problem. So let's try the ratio test for this using part see in mind because in part of seeing remember, she sees me from part B. We had key to the n plus one. It was less than be over, eh, Tien? So now we try the ratio test here. Since we're just dealing with positive numbers, we could drop the absolute value. This is B over a incense now and part see, we're assuming is larger than be that will imply be over in less than one so that the limit of Ki n plus one over p n is less than one. So that means that the Siri's converges by the ratio test and as we mentioned the beginning by the test for divergence, that will imply that the limit of P and zero and that's also what we what we wanted to establish a policy. Now there's one more party here, so we'LL go on to the next page for party. Now we're assuming is less than a B and there's a few things to show here. So first we want to show that if peanut is less than a B minus a, then we'd like to show that the sequence is increasing. And we'd like to also show that Pien is between zero on B minus eight Then on the other hand, if peanuts bigger than be minus a then we'd like to show these things The PM is decreasing and dance p and is larger than B minus a And then finally we'LL make a conclusion at the end that it is less than B and the limit of PN is b minus. Okay, so what's that? Improving this so and black thes two parts appear this worldview to parts and then two parts for the green as well. So this will take a few moments. So let's go ahead and start proving this. So we were getting so we're assuming this. But we were also given that peanut was bigger than zero, so we actually have zero less than Pena and then Liston d minus a. So now we would like to go ahead and let's show this one first. Let's do that by induction. We already have the bass case here. Bass case corresponding to an equal zero. Now, let's go ahead and uses for the induction. So we'd like to go ahead and suppose that it's true for PN and then show that would be induction. Okay, so now let's recall the following. We have tea and plus one by definition. Now, I'm just gonna do a trick up here in the numerator. And then I could rewrite this as the following. So now going on to the next page. So this was our assumption here for the inductive step at eight to both sides. Go ahead and divide. Bye, baby, he reckons, multiply both sides by the reciprocal zor football sides. And here let me put a negative motivated by what we had in the previous page. So this gives us again going back to the previous page here and then using our newest information. And that's exactly what we wanted. We wanted to show that if it was true for n that It was also true for en plus one and let's see here and by definition we know that zero will also be less than just by definition, of p m in the fact that a, B, t and R all positive. So this shows one of the parts we still have another parts ago. So now we have the following we now we want to show that it's increasing. So let's just recall that definition again. And then Now let's you not the family when we have what we already showed. So using the same type of trick is before and there we go. We see that PM is increasing sequence. So this takes care of the first part that we mentioned that was in black. And now similarly, there's two parts to show for the green steps. So this was if now we assume peanuts bigger than be minus a and so this is what we'd like to show. And then we'LL do it by induction again. Already we have the bass case up here. So now let's go ahead and recall the definition of P n plus one using our trick that we've used several times. So now since PM is bigger than be minus a. This would be our inductive step inductive hypothesis. It's at eight of both sides. Divide by eighty football sides and then multiply both sides by a negative on the next page. And so this will give us by previous work and then buy our latest inequalities. And once again, there we go. Now we have it for N plus one. Now we want to show that it's decreasing. So this was also supposed to be in green, but I moved on to a new page, so we'd like to show this. So we have add that aids of the other side using some algebra here. So this will give us that p n plus one bp on over a plus pn the shrink we wants than PM And so that's what it means to be decreasing. And so we had one more party here, and this one is to show that the following So if so, I'm getting a little sloppy. You're sorry about that. There was three parts. The party. That's why I'm using these down to you. So if peanuts less than a B minus in, then by what we showed this is positive, increasing and bounded above bye. Part by the first part of Heartbeat. We showed these things, and so his conversion bye theorems won't monotone sequence there and then also since the key and is bigger than zero and it's increasing. It can converse zero because it's increasing. So by partying we showed that the limit was either zero or B minus, and if it can be zero, it has to be B minus eight. Similarly, if peanut is bounded below, then buy this party. What we also showed is that pee and still positive. But this time it's decreasing. This is what we did in green and bounded below. Might be money, so it's converging also, by the twelve months a sequence there and then sense, um, Tien is bigger than be minus a. We have to have that the limit of P and bigger than equal city minus ing and which is bigger than zero so that the limit of PN is not equals zero and again by party. The limited it's non zero has to be a B minus, and that resolves the problem