5

Problem

A pair of birds can produce between 3 to 4 birds each year. If we start with a pair of birds and if a bird usually lives for three years, then how many birds will we have after $n$ years?

What I have done

I assumed that the pair of birds that we started with were 1 year olds.

If we let $m\in\mathbb{N}$ and $3\leq m\leq 4$; $k_n$ be the number of birds after $n$ years, then

$$ \begin{align*} k_1&=m+2\\ k_2&=\frac{1}{2}(m+2)^2-2\\ k_3&=\frac{1}{4}(m+2)^3-(m+2)\\ k_4&=\frac{1}{8}(m+2)^4-\frac{1}{2}(m+2)^2-m\\ k_5&=\frac{1}{16}(m+2)^5-\frac{1}{4}(m+2)^3-m(m+1) \end{align*} $$ and so on. The problem is that I can only see the pattern of the first term of $k_n$ and therefore can not find a general formula for $k_n$. Could you please help?

  • Does the number that pair produces per year stay always fixed ($3$ or $4$) or how do you otherwise decide how many each pair produces? – ploosu2 Feb 10 '15 at 00:42
  • I am not really sure and that is why I used $m$ instead of 3 or 4. –  Feb 10 '15 at 00:45
  • I guess you can get bounds on the number of birds by first using $m=3$ and then $m=4$. The average $m=3.5$ could give somewhat accurate behaviour (?). – ploosu2 Feb 10 '15 at 00:48
  • I have solved this with mathematica but the formula is horrendous... please bare with me while I figure out if I can make it simpler. – Alex Feb 10 '15 at 00:56
  • @Alex: What is intriguing is for $m=4$ and how the cubic $P_{67} = y^3-2y^2-2y-2=0$ pops up. I recognize this equation. The real root $y_1$ is the exact value of an eta quotient involving $\mathbb Q(\sqrt{-67})$. Hence the close approximation, $$e^{\pi\sqrt{67}} \approx y_1^{24}-24.0000000018\dots$$ Interesting that $P_{67}$ appears in a problem involving birds. – Tito Piezas III Feb 10 '15 at 02:28
  • @Alex: In fact, the sequence, $$\tfrac{1}{\color{blue}2\cdot134}( x_1,y_1^n+x_2,y_2^n+x_3,y_3^n) = 1, 3, 8, 24, 70, 204, 596,\dots$$ has a known OEIS entry A078055 – Tito Piezas III Feb 10 '15 at 02:45

2 Answers2

3

We know $$k_n=k_{n-1}+\underbrace{\frac{k_{n-1}}{2}\cdot m}_{\text{birds born this year}}-\underbrace{\frac{k_{n-4}}{2}\cdot m}_{\text{birds born 3 years ago}}$$ and also $k_{-1}=0,k_0=2,k_1=2+m,k_2=\frac{m^2}{2}+2m$.

I used Mathematica to solve this but unfortunately the result is way too long to post it here... (latex formula has 50000 characters).

Fortunately it simplifies a bit if we plug in the values $m=3,4$.

For $m=3$:

$$k_n =\frac{1}{219}(x_1\, y_1^n + x_2\, y_2^n + x_3\, y_3^n)=2,5,\tfrac{21}{2},\tfrac{105}{4},\tfrac{501}{8},\dots$$

where the $x_i$ are,

$$x_i = 88y_i^2 - 40y_i + 12$$

and the $y_i$ the roots of,

$$2y^3 - 3y^2 - 3y - 3 = 0$$

Explicitly,

$$\small k_n = \frac{1}{219} \left(\left(73+\sqrt[3]{73 \left(5767-520 \sqrt{73}\right)}+\sqrt[3]{73 \left(5767+520 \sqrt{73}\right)}\right) 2^{1-n} \left(1+\sqrt[3]{10-\sqrt{73}}+\sqrt[3]{10+\sqrt{73}}\right)^n+\left(146+i \left(\sqrt{3}+i\right) \sqrt[3]{73 \left(5767-520 \sqrt{73}\right)}+\left(-1-i \sqrt{3}\right) \sqrt[3]{73 \left(5767+520 \sqrt{73}\right)}\right) \left(\frac{1}{4} \left(2+\left(-1-i \sqrt{3}\right) \sqrt[3]{10-\sqrt{73}}+i \left(\sqrt{3}+i\right) \sqrt[3]{10+\sqrt{73}}\right)\right)^n+\left(146+\left(-1-i \sqrt{3}\right) \sqrt[3]{73 \left(5767-520 \sqrt{73}\right)}+i \left(\sqrt{3}+i\right) \sqrt[3]{73 \left(5767+520 \sqrt{73}\right)}\right) \left(\frac{1}{4} \left(2+i \left(\sqrt{3}+i\right) \sqrt[3]{10-\sqrt{73}}+\left(-1-i \sqrt{3}\right) \sqrt[3]{10+\sqrt{73}}\right)\right)^n\right)$$

For $m=4$:

$$k_n=\frac{1}{134}(x_1\, y_1^n + x_2\, y_2^n + x_3\, y_3^n) = 2, 6, 16, 48, 140, 408,\dots$$

where the $x_i$ are,

$$x_i = 38y_i^2 - 24y_i + 4$$

and the $y_i$ the roots of,

$$y^3 - 2y^2 - 2y - 2 = 0$$

Explicitly,

$$\small \frac{1}{134} 3^{-n-1} \left(2 \left(134+\sqrt[3]{134 \left(18559-909 \sqrt{201}\right)}+\sqrt[3]{134 \left(18559+909 \sqrt{201}\right)}\right) \left(2+\sqrt[3]{53-3 \sqrt{201}}+\sqrt[3]{53+3 \sqrt{201}}\right)^n+\left(268+i \left(\sqrt{3}+i\right) \sqrt[3]{134 \left(18559-909 \sqrt{201}\right)}+\left(-1-i \sqrt{3}\right) \sqrt[3]{134 \left(18559+909 \sqrt{201}\right)}\right) \left(\frac{1}{2} \left(4+\left(-1-i \sqrt{3}\right) \sqrt[3]{53-3 \sqrt{201}}+i \left(\sqrt{3}+i\right) \sqrt[3]{53+3 \sqrt{201}}\right)\right)^n+\left(268+\left(-1-i \sqrt{3}\right) \sqrt[3]{134 \left(18559-909 \sqrt{201}\right)}+i \left(\sqrt{3}+i\right) \sqrt[3]{134 \left(18559+909 \sqrt{201}\right)}\right) \left(\frac{1}{2} \left(4+i \left(\sqrt{3}+i\right) \sqrt[3]{53-3 \sqrt{201}}+\left(-1-i \sqrt{3}\right) \sqrt[3]{53+3 \sqrt{201}}\right)\right)^n\right)$$

Tito Piezas III
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Alex
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  • Thanks Alex. I am really grateful that you did this. –  Feb 10 '15 at 01:10
  • some patience you have :p – Shobhit Feb 10 '15 at 01:13
  • @user17797 You're welcome. I admit it's not the most helpful answer. If you have a copy of Mathematica and you want to fiddle with the full solution, execute the command: Simplify[ToRadicals[RSolve[{k[n] == k[n - 1] + k[n - 1]*m/2 - k[n - 4]*m/2, k[-1] == 0, k[0] == 2, k[1] == 2 + m, k[2] == m^2/2 + 2 m}, k[n], n]]] – Alex Feb 10 '15 at 01:14
  • @ILUA Mathematica has a nice function called TeXForm, which let me convert the result into LaTeX code ;) – Alex Feb 10 '15 at 01:16
  • This is a linear homogenous equation, which can be solved in general. In particular, it will be the sum of four exponential functions, with bases equal to the roots of $x^4-(1+\frac{m}2)x^{n-1}+\frac{m}2$. That's where the nasty expression comes from. – Milo Brandt Feb 10 '15 at 01:17
  • @Alex oh ok, thanks :D – Shobhit Feb 10 '15 at 01:17
  • @Meelo Yes, one root was okay ($x=1$), but the other three roots... :D If the birds died after two years, rather than three, the equation would be much nicer. – Alex Feb 10 '15 at 01:18
  • @Alex: I've simplified your expression. I hope it's ok. :) (Given two related cubics, a Tschirnhausen transformation can express the roots of one in terms of the other.) – Tito Piezas III Feb 10 '15 at 03:43
2

If you assume that each pair of birds produces either $3$ or $4$ baby birds with equal probability $1/2$, then I'm pretty sure you cannot simply use a fixed value of $m$, like $m = 3.5$. To see this, suppose you start with $2$ birds. Then with probability $1/2$ each you will have either $5$ or $6$ birds after one generation. If $5$ birds, then you only have $2$ pairs. But if $6$ birds, then you have $3$ pairs. Thus there is a bias toward having more than $3.5$ times more birds per generation per pair of birds as time goes on. Yet the bias should eventually equalize so that the number of new birds is approximately $3.5/2$ times the number of birds in the previous generation that live, due to the central limit theorem. But you cannot ignore the effects in the very early generations. It seems like you want the EXPECTED number of birds per generation, and this seems like a very hard problem. You can get bounds by fixing $m = 3$ or $m = 4$, as noted in another answer.

user2566092
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  • Thanks for the answer and my apologies that I do not have enough reputation to up-vote your answer. –  Feb 10 '15 at 01:13