Hey there! Let's figure out how to find that steady-state lower limit for your recursive sequence—great catch noticing the initial value $x_1$ doesn't affect the end state, that's a key observation for these kinds of geometric-like recursions.

Step 1: Understand the steady-state assumption #

When the sequence reaches its lower limit (or steady state), as $n$ gets really large, $x_n$ and $x_{n-1}$ will be practically equal. Let's call this steady-state value $L$.

Step 2: Substitute into the recursive formula #

Plug $L = x_n = x_{n-1}$ into your original equation:
$$
L = \left( L + p \right) \times \left( 1 - \frac{t}{100} \right)
$$

Step 3: Solve for $L$ #

Let's expand and rearrange terms to isolate $L$:

1. Expand the right-hand side:
   $$
   L = L \times \left(1 - \frac{t}{100}\right) + p \times \left(1 - \frac{t}{100}\right)
   $$
2. Move all terms with $L$ to the left side:
   $$
   L - L \times \left(1 - \frac{t}{100}\right) = p \times \left(1 - \frac{t}{100}\right)
   $$
3. Factor out $L$ on the left:
   $$
   L \times \left[ 1 - \left(1 - \frac{t}{100}\right) \right] = p \times \left(1 - \frac{t}{100}\right)
   $$
4. Simplify the bracket term:
   $$
   L \times \frac{t}{100} = p \times \frac{100 - t}{100}
   $$
5. Cancel out the $\frac{1}{100}$ from both sides and solve for $L$:
   $$
   L = p \times \frac{100 - t}{t}
   $$

Step 4: Verify with your test cases #

Let's check your examples to make sure this works:

• For $p=10$, $t=20$: $L = 10 \times \frac{100-20}{20} = 10 \times 4 = 40$ — matches your manual calculation!
• For $p=11$, $t=20$: $L = 11 \times \frac{80}{20} = 44$ — perfect, that's exactly your rough collapsed value.

Why the initial value $x_1$ doesn't matter #

The term $\left(1 - \frac{t}{100}\right)$ is a positive number less than 1 (since $t>0$). Every time you iterate the formula, the difference between the current $x_n$ and the steady-state $L$ gets multiplied by this factor, which shrinks it more and more. Eventually, that initial difference becomes negligible, so the sequence converges to $L$ no matter what $x_1$ you start with.

备注：内容来源于stack exchange，提问作者MrShoe