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There is a sequence given by:$\ \ h(0) = \frac{2}{3} ,\ \ h(n) = h(n-1)^2 - 2 \ \ $for $n \in\mathbb{N^{+}}$.

Let's have a look a the first values: $$h(0)=\ \ \frac{2}{3}\\ \ \ \ h(1) =-\frac{14}{9}\\ \ \ h(2) = \ \ \frac{34}{81}\\ \ \ \ \ \ \ h(3) =-\frac{11966}{6561} \\ \ \ \ \ \ \ \ \ \ h(4)=\frac {57091714}{43046721}\\ $$ $...$

Terms are growing fast, I would not post any further.

Interesting for me is the change in the signs. At first glance, do you expect alternating sign changes?

The signs I determined for $ h(n) $ for $n=0..31 \ \ $:

$$ [1, -1, 1, -1, 1, -1, -1, 1, 1, -1, -1, 1, -1, 1, 1, -1, -1, -1, -1, 1, 1, 1, 1, -1, -1, -1, -1, 1, -1, 1, -1, -1]$$

...then my computer was running out of memory. He needs almost 4 GByte for the last operation.

Question: Is it possible to determine if $h(n)$ is positive or negative for increasing values of $n$? For example for $n=100$ or even $n=1000000$?

Question: Is it true that for all $h(0)$ with $0not$ \sqrt{2}$, the signs of $h(n)$ change infinitely often?

Thanks in advance!

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    There are infinitely many initial values for which we get a fixed sign, in fact a fixed function value, after only finitely many terms. Plug in $h(0)=2\cos (\pi/(2^k))$, where $k$ is a natural numver $\ge 2$ to make the initial value positive, then apply the chronologicaly first answer below. – 2017-01-08
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    I see, it is about an appropriate trigonometric identity. Solve $ \frac{2}{3}=2 \cos(\pi/(2^{k}))$ for $k$. right? – 2017-01-08
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    No, for $2/3$ you have no integer solution for $k$. If $\theta_0 = arc \cos(1/3)$ is not a rational multiple if $\pi$, the sign changes in your sequence will keep happening forever (but not at regular intervals). – 2017-01-08
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    @MartinHopf: A related question, dealing with the case $h_0 > 2$: http://math.stackexchange.com/questions/2073920/explicit-solution-of-the-recursion-x-n-x-n-12-2-with-x-02 – 2017-01-10

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Terms are not growing fast, only our attempted representations using primitive operations is.

Assume that $h(k) = 2\cos (\theta(k))$. This is certainly true for the initial term $k=0$. Use the appropriate trigonometric identity to show that:

$h(k+1) = 2\cos (2\theta(k))$

and so:

$\theta(n)=2^n\theta(0)$

$h(n) = 2\cos (2^{n}\theta(0))$

Now you can work out the sign pattern easily given any initial value for $h(0)$ or $\theta(0)$.

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    Thank you for the quick answer. btw what does $\theta ()$ mean? I'm a bit clueless. – 2017-01-08
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    A variable I introduce for this analysis. If a term is between -2 and +2, it's twice the cosine of some argument- which I call $\theta$. We have proof that $\theta$ doubles from one term to the next the terms $h=2\cos(\theta)$ have to follow suit. – 2017-01-08