MathLabs

Problem 2

Let P1(x)=x2−2P_1(x) = x^2 - 2 and Pj(x)=P1(Pj−1(x))P_j(x) = P_1(P_{j-1}(x)) for j=2,3,…j = 2, 3, \ldots. Prove that, for any positive integer nn, the roots of the equation Pn(x)=xP_n(x) = x are all real and distinct.
Step 1 of 6: Substitute and apply the double-angle formula
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In plain words

The map x↦x2−2x\mapsto x^2-2 looks exactly like the double-angle formula for cosine in disguise; recognizing this hidden trigonometric structure is the key insight that makes iterating P1P_1 tractable.

x=2cos⁡θ  ⟹  P1(x)=4cos⁡2θ−2=2cos⁡(2θ)x = 2\cos\theta \implies P_1(x) = 4\cos^2\theta - 2 = 2\cos(2\theta)
Point at angle θ\theta on the unit circle, whose horizontal coordinate is cos⁡θ\cos\theta
Unit circle with a radius drawn at angle $\theta$ from the positive $x$-axis, illustrating the substitution $x=2\cos\theta$ used to linearize the doubling map.
Detailed analysis

Let x=2cos⁡θx=2\cos\theta. Then P1(x)=x2−2=4cos⁡2θ−2=2(2cos⁡2θ−1)=2cos⁡(2θ)P_1(x)=x^2-2=4\cos^2\theta-2=2(2\cos^2\theta-1)=2\cos(2\theta), using the double-angle identity cos⁡(2θ)=2cos⁡2θ−1\cos(2\theta)=2\cos^2\theta-1.