Problem 2

Let

P_1(x) = x^2 - 2

and

P_j(x) = P_1(P_{j-1}(x))

for

j = 2, 3, \ldots

. Prove that, for any positive integer

n

, the roots of the equation

P_n(x) = x

are all real and distinct.

Step 3 of 6: Translate the equation

P_n(x)=x

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In plain words

The original algebraic fixed-point equation is now a purely trigonometric one, which is solvable by elementary means — this is the whole point of the substitution.

2\cos(2^n\theta) = 2\cos\theta \iff \cos(2^n\theta) = \cos\theta

Detailed analysis

Substituting $x=2\cos\theta$ into $P_n(x)=x$ and using Step 2 gives $2\cos(2^n\theta)=2\cos\theta$ , i.e. $\cos(2^n\theta)=\cos\theta$ .