Porisma

Worked solution

Unfold a geese-population recurrence in two directions

A poultry farmer raises chickens, ducks and geese. Allowing for conditions such as sales and deaths from infection, the geese population at the end of nn months is modelled by unu_n, with un+1=0.75un+80u_{n+1}=0.75u_n+80 and initial population u0u_0. Calculate u1,u2,u3,u4u_1,u_2,u_3,u_4 for (i) u0=576u_0=576 and (ii) u0=320u_0=320, commenting briefly on the results. Find unu_n in terms of n,u0n,u_0; deduce that un→320u_n\to320 for large nn, and use the formula to justify the earlier comments.

[Adapted from MEI Structured Mathematics, Pure Mathematics 2, 2001 June Exam Paper 2602/1, Q1(modified)]

CJC · Sequences and Series · Example 30

(i) Initial population 576

Answer

u1=512,u2=464,u3=428,u4=401u_1=512,u_2=464,u_3=428,u_4=401; the sequence appears to decrease.

Generate the requested four updates

u1=0.75(576)+80=512,u2=0.75(512)+80=464,u3=428,u4=401u_1=0.75(576)+80=512,\quad u_2=0.75(512)+80=464,\quad u_3=428,\quad u_4=401

The calculator starts at nMin=0 with u(0)=576 and u(n+1)=0.75u(n)+80.

Distinguish an observation from a proof

The lecturer emphasises “appears to be”: a comment based on the first five values, including u0, does not prove behaviour for every index.

How did this part go?

(ii) Initial population 320

Answer

u1=u2=u3=u4=320u_1=u_2=u_3=u_4=320; the sequence appears constant.

Use the same update with the other initial value

0.75(320)+80=320⟹u1=u2=u3=u4=3200.75(320)+80=320\quad\Longrightarrow\quad u_1=u_2=u_3=u_4=320

The second source calculator setup uses nMin=0,u(0)=320. Its displayed table remains 320.

How did this part go?

General term: two unfolding methods

Answer

un=0.75n(u0−320)+320u_n=0.75^n(u_0-320)+320 for integer n≥0n\ge0.

Method 1: work backwards from u_n

un=0.75un−1+80=0.75(0.75un−2+80)+80=0.752un−2+80(0.75)+80=0.752un−2+80(1+0.75)=0.752(0.75un−3+80)+80(1+0.75)=0.753un−3+80(1+0.75+0.752).\begin{aligned}u_n&=0.75u_{n-1}+80\\&=0.75(0.75u_{n-2}+80)+80\\&=0.75^2u_{n-2}+80(0.75)+80\\&=0.75^2u_{n-2}+80(1+0.75)\\&=0.75^2(0.75u_{n-3}+80)+80(1+0.75)\\&=0.75^3u_{n-3}+80(1+0.75+0.75^2).\end{aligned}

Repeatedly substitute the preceding recurrence; after k substitutions the accumulated constants form a geometric sum.

Collect all n contributions

un=0.75nu0+80∑j=0n−10.75j=0.75nu0+801−0.75n1−0.75=0.75nu0+320(1−0.75n)=0.75nu0+320−320(0.75n)=0.75n(u0−320)+320.\begin{aligned}u_n&=0.75^nu_0+80\sum_{j=0}^{n-1}0.75^j\\&=0.75^nu_0+80\frac{1-0.75^n}{1-0.75}\\&=0.75^nu_0+320(1-0.75^n)\\&=0.75^nu_0+320-320(0.75^n)\\&=0.75^n(u_0-320)+320.\end{aligned}

For n≥1 the bracket is a GP with first term 1, ratio 0.75 and n terms. The resulting closed formula also holds at n=0.

Method 2: work forwards from u_0

u1=0.75u0+80,u2=0.75u1+80=0.75(0.75u0+80)+80=0.752u0+0.75(80)+80,u3=0.75u2+80=0.75[0.752u0+0.75(80)+80]+80=0.753u0+0.752(80)+0.75(80)+80.\begin{aligned}u_1&=0.75u_0+80,\\u_2&=0.75u_1+80\\&=0.75(0.75u_0+80)+80\\&=0.75^2u_0+0.75(80)+80,\\u_3&=0.75u_2+80\\&=0.75[0.75^2u_0+0.75(80)+80]+80\\&=0.75^3u_0+0.75^2(80)+0.75(80)+80.\end{aligned}

The source also expands successive early terms before recognising the same pattern.

Sum the forward pattern

un=0.75nu0+80∑j=0n−10.75n−1−j=0.75nu0+801−0.75n1−0.75=0.75nu0+320(1−0.75n)=0.75n(u0−320)+320.\begin{aligned}u_n&=0.75^nu_0+80\sum_{j=0}^{n-1}0.75^{n-1-j}\\&=0.75^nu_0+80\frac{1-0.75^n}{1-0.75}\\&=0.75^nu_0+320(1-0.75^n)\\&=0.75^n(u_0-320)+320.\end{aligned}

The same n contributions now appear in descending order of their powers.

How did this part go?

Limit and justification of the initial comments

Answer

un→320u_n\to320. For u0=576u_0=576, it strictly decreases toward 320; for u0=320u_0=320, it equals 320 for every n≥0n\ge0.

Let the decaying factor vanish

0.75n→0⟹un=0.75n(u0−320)+320→3200.75^n\to0\quad\Longrightarrow\quad u_n=0.75^n(u_0-320)+320\to320

For every fixed finite initial value, 0.75^n tends to zero.

Justify decreasing behaviour for the larger initial value

un=256(0.75n)+320,un+1−un=−64(0.75n)<0u_n=256(0.75^n)+320,\qquad u_{n+1}-u_n=-64(0.75^n)<0

The positive distance 256·0.75^n shrinks each month. This formula justifies the earlier observation for every index, rather than only the calculator rows.

Justify the constant case

un=0.75n(320−320)+320=320(n∈Z, n≥0)u_n=0.75^n(320-320)+320=320\qquad(n\in\mathbb Z,\ n\ge0)

With u0=320 the coefficient of the decaying power is zero.

How did this part go?

Solution notes

The Student copy has a stray u₄=320 below the general derivation. It applies only when u₀=320; the other requested initial value gives u₄=401. Calculator values later in the source are rounded displays of the model, not intermediate rounding instructions.