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This maybe a question answered in some textbooks, I had trouble figure out how to solve it. Here is a function defined recursively,

$ F_1(x) = x $

$ F_2(x) = x^2 -2 $

$ F_n(x) = F_{n-1}^{2}(x) -2 $

Is there a way to express $F_n(x) $ in term of $x$ directly?

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Hint: $\;F_n = F_{n-1}^2 - 2 \iff \dfrac{F_n}{2} = 2 \left(\dfrac{F_{n-1}}{2}\right)^2 - 1\,$, then think of $\cos 2a = 2 \cos^2 a - 1$.


[ EDIT ]   Let $\,\dfrac{F_n}{2}=\cos a_n\,$, then the recurrence gives $\,\cos a_n = 2 \cos^2 a_{n-1} - 1 = \cos 2 a_{n-1}\,$ which has the solution$\,a_n = 2 a_{n-1}\,$. By telescoping the GP $\,a_n = 2^{n-1} a_1\,$, and $\,a_1=\arccos \dfrac{x}{2}\,$ from the initial condition $\,F_1 = x\,$, so in the end $\,F_n(x) = 2 \cos \left(2^{n-1} \arccos\left(\dfrac{x}{2}\right)\right)\,$.

The above works for $\,|x| \le 2\,$, in the other cases you'll need to resort to hyperbolic functions like it's done for the trigonometric definition of the Chebyshev polynomials, which are in fact related to your $\,F_n\,$ by the relation $\,F_n(2x) = 2 T_{2^{n-1}}\left(x\right)\,$.

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