Problem 6

A sequence

(u_n)

is defined by

u_0=2

u_1=\frac52

, and

u_{n+1}=u_n(u_{n-1}^2-2)-u_1

for

n=1,2,\ldots

. Prove that for every positive integer

n

\lfloor u_n\rfloor=2^{\frac{2^n-(-1)^n}{3}}

, where

\lfloor x\rfloor

denotes the greatest integer less than or equal to

x

Step 5 of 6: Verify the recurrence for

a_n

Previous step Next step

In plain words

The complicated recurrence collapses after expanding products of exponentials; the alternating term is exactly what the subtraction by $u_1$ removes.

a_{n-1}(a_{n-2}^2-2)-a_1=a_n

Detailed analysis

Using $(2^r+2^{-r})^2-2=2^{2r}+2^{-2r}$ and $x_n=x_{n-1}+2x_{n-2}$ , expand: $a_{n-1}(a_{n-2}^2-2)=2^{x_n}+2^{-x_n}+2^{x_{n-1}-2x_{n-2}}+2^{-x_{n-1}+2x_{n-2}}$ . Step 4 says the last two terms equal $a_1$ , so subtracting $a_1$ leaves $a_n$ . Thus $a_n$ and $u_n$ have identical initial values and recurrence, hence $u_n=a_n$ for all $n$ .