Recursions, I have no idea how to do these

Question

Find the equilibria of the difference equation and classify them as stable or unstable. (Enter your answers as comma-separated lists. If an answer does not exist, enter DNE.)

x_t+1=1/3x_t²

Answer 1

The sequence represented by the recurrence relation diverges if |x₀| > 1.

It converges to 0 if |x₀| < 1

If x₀ = 1, it becomes a convergent geometric sequence.

To see why these answers are so, simply write out the first few terms of the sequence.

OK, please follow along

x₁ = (1/3)x₀²

x₂ = (1/3)x₁² = (1/3)(1/9)x₀⁴

x₃ = (1/3)x₂² = (1/3)(1/27)²x₀⁸ and so on

You can see that if |x₀|>1, the terms of the sequence will just get bigger and bigger, i.e. they will diverge.

If |x₀| < 1 then the terms will get smaller and smaller but positive, i.e. they will approach 0.

If x₀ = 1 the terms will just be come powers of (1/3), i.e. a geometric series.

12/4/2019 9:12 AM UPDATE

I am sorry; the above explanation is close, but not correct.

I believe this is the correct answer.

The general term of the sequence is 3(x₀/3)^2n.

^{Thus the sequence converges for |x}0^{| < 3, diverges for |x}0^{| > 3.}

^{For x}₀ = 1 it is the geometric sequence with first term 1 and ratio 1/9.

Again I apologize for the incorrect answers given initially!