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Worked solution

Solve a daily retention recurrence

Celine takes 500500 mg of Vitamin C once daily. The model assumes 20%20\% remains immediately before the next day’s pill. Let xnx_n be the amount immediately after dose nn, with x1x_1 immediately after the first dose.

(i) Write a recurrence. (ii) Show xn=(0.2)n−1(x1−625)+625x_n=(0.2)^{n-1}(x_1-625)+625. (iii) Deduce the limit.

CJC · Discussion · Question 13

Answer

xn=0.2xn−1+500x_n=0.2x_{n-1}+500, x1=500x_1=500; xn=(0.2)n−1(x1−625)+625→625x_n=(0.2)^{n-1}(x_1-625)+625\to625 mg.

(i) Combine retained amount and the new dose

x1=500x_1=500 and xn=0.2xn−1+500x_n=0.2x_{n-1}+500 for n≥2n\ge2.

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(ii) Unroll and sum the geometric contributions

x2=0.2x1+500x_2=0.2x_1+500, x3=(0.2)2x1+0.2(500)+500x_3=(0.2)^2x_1+0.2(500)+500, and x4=(0.2)3x1+(0.2)2(500)+0.2(500)+500x_4=(0.2)^3x_1+(0.2)^2(500)+0.2(500)+500. Hence xn=(0.2)n−1x1+500[1+0.2+⋯+(0.2)n−2]=(0.2)n−1x1+625[1−(0.2)n−1]x_n=(0.2)^{n-1}x_1+500[1+0.2+\cdots+(0.2)^{n-2}]=(0.2)^{n-1}x_1+625[1-(0.2)^{n-1}], proving the stated form (also valid at n=1n=1).

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(iii) Take the post-dose limit

As n→∞n\to\infty, (0.2)n−1→0(0.2)^{n-1}\to0, so xn→625x_n\to625 mg. This is the amount immediately after dosing in the stated recurrence.

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Common mistakes

The sequence records the amount after each dose, not just before the next one.

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