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减少了 12%。到下午 2 点，下午 1 点时血液中残留的抗生素又流失了 12%。已知抗生素含量的减少速率为每小时降低 12%，求下午 6 点时血液中残留的抗生素量。 一位医生考虑制定一个疗程，让患者在很长一段时间内每 6 小时服用一次药丸。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：当 $x_n<L$ 时，$x_{n+1}>x_n$； $\text{当 } x_n>L \text{ 时，} x_{n+1}<x_n\,.$

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

The number of hours from 12pm to 6pm is $6$.
Since the antibiotic decreases by $12\%$ every hour, the multiplier is $1 - 0.12 = 0.88$.

2. Calculate the amount of antibiotic remaining after 6 hours.

Amount remaining $= 50 \times (0.88)^6$
$= 23.220\dots$
$\approx 23.2 \text{ units}$ (to 3 s.f.)

1. Apply the limit to the recurrence relation by setting $x_{n+1}$ and $x_n$ to $L$.

As $n \to \infty$, $x_n \to L$. Since $x_{n+1}$ also approaches the same limit as $n$ grows indefinitely, we substitute $L$ into the given recurrence relation:
$L = 50 + (0.88)^6 L$

2. Solve the linear equation for $L$.

$L - (0.88)^6 L = 50$
$L(1 - 0.88^6) = 50$
$L = \frac{50}{1 - 0.88^6}$
$L \approx 93.354$
$L = 93.4 \text{ (to 3 s.f.)}$

1. Express the difference $x_{n+1} - x_n$ in terms of $L$ and $x_n$.

Consider the difference between consecutive terms:
$x_{n+1} - x_n = 50 + (0.88)^6 x_n - x_n$
$x_{n+1} - x_n = 50 - (1 - 0.88^6)x_n$
From part (ii)(a), the steady-state limit $L$ satisfies $L = 50 + (0.88)^6 L$, which means $50 = L(1 - 0.88^6)$. Substituting this into the equation gives:
$x_{n+1} - x_n = L(1 - 0.88^6) - x_n(1 - 0.88^6)$
$x_{n+1} - x_n = (1 - 0.88^6)(L - x_n)$

2. Deduce the first inequality when $x_n < L$.

Since $0.88 < 1$, the constant term $(1 - 0.88^6) > 0$.
If $x_n < L$, then $(L - x_n) > 0$.
Therefore, the product is positive:
$(1 - 0.88^6)(L - x_n) > 0$
$x_{n+1} - x_n > 0$
$x_{n+1} > x_n \text{ (Shown)}$

3. Deduce the second inequality when $x_n > L$.

Similarly, if $x_n > L$, then $(L - x_n) < 0$.
Therefore, the product is negative:
$(1 - 0.88^6)(L - x_n) < 0$
$x_{n+1} - x_n < 0$
$x_{n+1} < x_n \text{ (Shown)}$