Find limit of Decrementing Recursive Series

Question

I want to find a formula to find the lower limit part of this recursive or geometric series $$ x_{n} = \left( x_{n-1} + p \right) \times \left( 1 - \frac{t}{100} \right) $$ I was just wondering what method I should use to find the lower limit of this equation with a given value of $p$ and $t$ as the value $x_1$ doesn't seem to effect the end state of the formula

When calculating by hand I found that, for example, when using $$ x_{1}=100\ \ \ \ p=10\ \ \ \ t=20 $$ I find it hits a lower limit of 40

here is what I found when doing some simple tests as what they collapsed to

p	t	rough collapsed value
10	20	40
11	20	44
12	20	48
10	30	90
10	10	~23.33333
11	10	99

So far I haven't been able to find a formula or equation that charts this out, can anyone help?

Answer 1

If you assume a limit exists, then you can solve for it by setting $x_{n-1}=x_n=z$. I'm going to use $k= \left( 1 - \frac{t}{100} \right)$ for shorthand: $$ x = \left( x + p \right) \times k \\ x(1-k) = pk \\ x = \frac{pk}{1-k} \\ x = \frac{pk}{\frac{t}{100}}=\frac{100pk}{t} $$

A lower limit will definitely exist for $0 < t < 200$. For $t=0$, a limit only exists if $p=0$ (in that case, the limit is just $x_1$). For $t \ge 200$ or $t<0$, things are a little complex; if you set $x_1$ to exactly the right value, you can still have a fixed point (or a cycle like 1,-1,1,-1).

Answer 2

If the limit exists, it is found by solving the equation

$$x=(x+p)\left(1-\frac{t}{100}\right)$$

$$x=p\left(\frac{100}{t}-1\right)$$

To prove when this is indeed the limit, note that for $t\neq 0$ (where the recursion obviously diverges if $p\neq 0$ and converges for $p=0$) we can define

$$y_n= \left|x_n-p\left(\frac{100}{t}-1\right)\right|$$

Then the recursion becomes

$$y_n=\left|x_n-p\left(\frac{100}{t}-1\right)\right|=\left|(x_{n-1}+p)\left(1-\frac{t}{100}\right)-p\left(\frac{100}{t}-1\right)\right|$$

$$=\left|\frac{(-100 + t) (-100 p + p t + t x_{n-1})}{100 t}\right|=\left|\frac{t-100}{100}\right|\left| x_{n-1}-p\left(\frac{100}{t}-1\right)\right|$$

$$=\left|\frac{t-100}{100}\right| y_{n-1}$$

For $x_n$ to converge, $y_n$ must converge to $0$: ie

$$|t-100|<100 \text{ or } y_{n-1}=0$$

$$0<t<200\text{ or }y_1=0$$

We already talked about what happens at $t=0$ (diverges unless $p=0$). At $t=200$ the recursion becomes

$$x_n=-x_{n-1}-p$$

which diverges unless $x_1=-\frac{p}{2}$ (in this case then $y_1=0$). If $t<0$ or $t>200$ then the sequence diverges unless $y_1=0$: that is

$$x_1=p\left(\frac{100}{t}-1\right)$$

Summarizing:

$$x_n\to p\left(\frac{100}{t}-1\right)\text{ if }t\in (0,200)$$

$$x_n\to x_1\text{ if }t,p\in\{0\}$$

$$x_n\to x_1\text{ if }t\in(-\infty,0)\cup [200,\infty)\text{ and }x_1=p\left(\frac{100}{t}-1\right)$$

$$x_n\text{ diverges otherwise}$$