MathLabs

Problem 6

A sequence (un)(u_n) is defined by u0=2u_0=2, u1=52u_1=\frac52, and un+1=un(un−12−2)−u1u_{n+1}=u_n(u_{n-1}^2-2)-u_1 for n=1,2,…n=1,2,\ldots. Prove that for every positive integer nn, ⌊un⌋=22n−(−1)n3\lfloor u_n\rfloor=2^{\frac{2^n-(-1)^n}{3}}, where ⌊x⌋\lfloor x\rfloor denotes the greatest integer less than or equal to xx.
Step 4 of 6: Establish the alternating exponent identity
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In plain words

The exponent difference is always 11 or −1-1; the expression 2r+2−r2^r+2^{-r} cannot distinguish those two signs, so the correction term is constantly a1a_1.

xn−1−2xn−2=(−1)n,n≥2x_{n-1}-2x_{n-2}=(-1)^n,\quad n\ge2
Detailed analysis

Subtracting twice the recurrence gives xn−2xn−1=−(xn−1−2xn−2)x_n-2x_{n-1}=-(x_{n-1}-2x_{n-2}), so the quantity changes sign when the index advances: directly, x1−2x0=1x_1-2x_0=1, x2−2x1=−1x_2-2x_1=-1, and hence xn−1−2xn−2=(−1)nx_{n-1}-2x_{n-2}=(-1)^n for n≥2n\ge2. Therefore 2xn−1−2xn−2+2−xn−1+2xn−2=2+12=52=a12^{x_{n-1}-2x_{n-2}}+2^{-x_{n-1}+2x_{n-2}}=2+\frac12=\frac52=a_1.