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Worked solution

Unfold a loss-then-top-up recurrence in two directions

A storage tank loses 10%10\% of its water each day, then receives 9090 litres at the end of the day. Let xnx_n be the volume after n days and x0x_0 its initial volume.

(i) Give a recurrence. mark

(ii) Show xn=0.9n(x0−900)+900x_n=0.9^n(x_0-900)+900. marks

(iii) Deduce the limiting volume. mark

[2008/MJC/I/8(a)]

CJC · Review problems · Question 16

Answer

xn=0.9xn−1+90=0.9n(x0−900)+900x_n=0.9x_{n-1}+90=0.9^n(x_0-900)+900; limit 900900 litres.

(i) Apply the loss before the addition

xn=0.9xn−1+90x_n=0.9x_{n-1}+90, n≥1n\ge1. The added 9090 litres is not reduced by that same day’s loss.

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(ii) Method 1: unfold backwards from x_n

xn=0.92xn−2+90(1+0.9)=0.93xn−3+90(1+0.9+0.92)=⋯=0.9nx0+90∑j=0n−10.9jx_n=0.9^2x_{n-2}+90(1+0.9)=0.9^3x_{n-3}+90(1+0.9+0.9^2)=\cdots=0.9^nx_0+90\sum_{j=0}^{n-1}0.9^j. The geometric sum gives xn=0.9nx0+900(1−0.9n)=0.9n(x0−900)+900x_n=0.9^nx_0+900(1-0.9^n)=0.9^n(x_0-900)+900.

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(ii) Method 2: generate forward from the initial state

x1=0.9x0+90x_1=0.9x_0+90, x2=0.92x0+90(0.9+1)x_2=0.9^2x_0+90(0.9+1), and x3=0.93x0+90(0.92+0.9+1)x_3=0.9^3x_0+90(0.9^2+0.9+1). Continuing yields the same n-term geometric sum and the required formula. This retains the source’s separate forward-pattern derivation.

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(iii) Take the limit of the geometric factor

Since 0.9n→00.9^n\to0, xn→900x_n\to900 litres for any finite initial volume. If x0=900x_0=900, it is already constant; otherwise the deviation from 900900 is multiplied by 0.90.9 each day.

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Common mistakes

Multiplying the whole expression x_(n−1)+90 by 0.9 would model a different order of events.

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