Solve a recurrence relation

Question

The sequence x is defined as follows:
$x_{0} = 1, x_{t} =\sqrt{0.2x_{t-1}+0.9x_{t-1}^{2}}$
I want to know what is t when $x_{t} = 2$.
I use a spreadsheet to calculate it. When t is 104, $x_{t} = 2$, cor. to 2 d.p. But, is there a more mathematical way to do it?
More generally, can we define $x_{t}$ in terms of t?

Answer 1

In the image below, the blue line is the function $y=\sqrt{0.2 x+0.9x^2}$. The purple line is $y=x$. When you evaluate the function at $x<2$, you go up to the blue line (that's the next $x$ value). Then, you go to the right to the purple line, to get $x[t+1]=y=f(x[t])$. From the purple line, you go up again to the blue line, to get $f(x[t+1])$. This continues forever. The intersection of the two lines is at $(2,2)$. When $x[t]<2$, you will always get that $x[t+1]>x[t]$, because $f(x)>x$. Also, you will never get past $2$, because $f(x)<2$ when $0<x<2$. Therefore, your sequence will converge to $2$, but never actually reach it.

Answer 2

Using the ratio test, we get

$$\frac{\sqrt{0.9x_{t-1}^2+0.2x_{t-1}}}{x_{t-1}}=\sqrt{0.9+\frac{0.2}{x_{t-1}}}$$

With this, it is immediately clear that whenever $x_{t-1}\lt 2, x_t\gt x_{t-1}$ and also that whenever $x_{t-1}\gt 2, x_t\lt x_{t-1}$. Consider the value of $x_t$ when $x_{t-1}=2-\epsilon$ for $\epsilon\gt0$:

$$x_t=\sqrt{0.9(2-\epsilon)^2+0.2(2-\epsilon)}=\sqrt{4-3.8\epsilon+0.81\epsilon^2}$$

If $\epsilon\le 1$, then $3.8\epsilon\gt 0.81\epsilon^2$ and since $\lim_{\epsilon\to0}\sqrt{4-3.8\epsilon+0.81\epsilon^2}=2$ and $\forall t,x_t\gt x_{t-1}$ by choice of $x_0=1$, the limit of the recurrence is $x_{\infty}=2$.