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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?

Chris Cheung
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  • It looks like $x_t=2$ is a limit of the sequence... If I say that $x_{t-1}=2$ and consider $x_t$, I get $2$ again... – abiessu Dec 17 '13 at 14:21
  • @abiessu Does this mean x can never reach 2? If this is the case, is there a mathematical way to prove it? – Chris Cheung Dec 17 '13 at 14:29
  • @abiessu: It is not because the image of 2 is 2 that it is a limit. I agree there is a chance but that's it. – user88595 Dec 17 '13 at 14:48
  • @user88595: true, I was only looking at it as "what happens if..." and considering what that might mean. – abiessu Dec 17 '13 at 15:12
  • I am now wondering can we know what is $x_{t}$ given t? Is it impossible to do it without using a spreadsheet as it is hard to define $x_{t}$ in terms of t since this recurrence relation is non-linear? – Chris Cheung Dec 18 '13 at 01:25

2 Answers2

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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.

image of given function

Ragnar
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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 $x_{\infty}=2$.

abiessu
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  • Another good answer to explain the convergence of the sequence to 2 algebraically. I am just curious can we find the limit of the sequence directly, i.e. not setting $x_{t-1} = 2-ϵ$ and prove 2 as the limit of the sequence? – Chris Cheung Dec 17 '13 at 16:16
  • I'm not sure why using $x_t=2-\epsilon$ is not algebraic... I'm working on a more expanded version that uses a $\epsilon_t=2-\sqrt{4-3.8\epsilon+0.81\epsilon^2}$ transform, and I'm just running into minor detail difficulty with showing that $\epsilon_t\to 0$. – abiessu Dec 17 '13 at 16:19
  • Surely, using $x_{t} = 2-ϵ$ is algebraic. My meaning is to find the limit of the sequence directly without plotting a graph as Ragnar has done or setting $x_{t} = 2-ϵ$ deliberately to conclude 2 as the limit. Sorry for the confusion. – Chris Cheung Dec 17 '13 at 16:25
  • The ratio test with the ratio of $$\sqrt{0.9+\frac{0.2}{x_{t-1}}}$$ demonstrates that $x_t=2$ is the unique point where the ratio of one term to the next is $1$, thereby showing that $2$ is an interesting point in the sequence. Further finding that the ratio is $\gt1$ for $x_t\lt 2$ and that the sequence is bounded above by $2$ when $x_0=1$ is enough to demonstrate it fully, in my estimation... – abiessu Dec 17 '13 at 18:12