一名住院患者在中午 12 点服用了一颗含有 50 单位抗生素的药丸。到下午 1 点，患者体内的抗生素单位数下降了 12%。到下午 2 点，下午 1 点时残留在血液中的抗生素又减少了 12%。已知其减少速率使得抗生素量每小时下降 12%，求下午 6 点时血液中残留的抗生素量。 医生考虑制定一个疗程，要求患者在很长一段时间内每 6 小时服用一次该药丸。已知 $x_{n+1}=50+(0.88)^6 x_n,$ 其中 $x_n$ 表示服用第 $n$ 颗药丸后体内的抗生素单位数。 当 $n \to \infty$ 时，$x_n \to L$。确定 $L$ 的值。 证明：若 $x_n<L$，则 $x_{n+1}>x_n$； $x_{n+1}<x_n\text{ if }x_n>L.$

1. Determine the time elapsed and the constant decay multiplier.

The time elapsed from 12pm to 6pm is $6$ hours.
The amount of antibiotic drops by $12\%$ every hour, so the multiplier is $1 - 0.12 = 0.88$.

2. Formulate and evaluate the amount remaining after 6 hours.

$\text{Amount remaining} = 50 \times (0.88)^6$
$= 23.220... \approx 23.2 \text{ units}$

1. Apply the limit as $n \to \infty$ to the recurrence relation

As $n \to \infty$, both $x_n \to L$ and $x_{n+1} \to L$.
Substituting this into the given recurrence relation:
$L = 50 + (0.88)^6 L$

2. Solve the linear equation for $L$

$L - (0.88)^6 L = 50$
$L(1 - 0.88^6) = 50$
$L = \frac{50}{1 - 0.88^6}$
$L \approx 93.3539$
$L \approx 93.4 \text{ (to 3 s.f.)}$

1. Form an expression for the difference $x_{n+1} - x_n$.

$x_{n+1} - x_n = 50 + (0.88)^6 x_n - x_n$
$x_{n+1} - x_n = 50 - x_n(1 - (0.88)^6)$

2. Substitute the exact relation for the limit $L$ into the difference expression.

From part (ii)(a), as $n \to \infty$, $x_n \to L$, so the limit satisfies:
$L = 50 + (0.88)^6 L \implies 50 = L(1 - (0.88)^6)$
Substitute this equivalent of 50 into the expression for $x_{n+1} - x_n$:
$x_{n+1} - x_n = L(1 - (0.88)^6) - x_n(1 - (0.88)^6)$
$x_{n+1} - x_n = (L - x_n)(1 - (0.88)^6)$

3. Analyze the sign of the difference to deduce the inequalities.

Since $(0.88)^6 < 1$, the constant term $(1 - (0.88)^6) > 0$.

• If $x_n < L$, then $L - x_n > 0$.
  $(L - x_n)(1 - (0.88)^6) > 0 \implies x_{n+1} - x_n > 0 \implies x_{n+1} > x_n$

• If $x_n > L$, then $L - x_n < 0$.
  $(L - x_n)(1 - (0.88)^6) < 0 \implies x_{n+1} - x_n < 0 \implies x_{n+1} < x_n$

(Both cases are shown)