HMMT November, 2016

Algebra Difficulty 4.8 AIME Prove it United States

Problem:
Let the sequence $\{a_{i}\}_{i=0}^{\infty}$ be defined by $a_{0}=\frac{1}{2}$ and $a_{n}=1+(a_{n-1}-1)^{2}$ . Find the product
$\prod_{i=0}^{\infty} a_{i} = a_{0} a_{1} a_{2} \ldots$

Solution

Solution:
Let $\{b_{i}\}_{i=0}^{\infty}$ be defined by $b_{n}=a_{n}-1$ and note that $b_{n}=b_{n-1}^{2}$ . The infinite product is then
$(1+b_{0})(1+b_{0}^{2})(1+b_{0}^{4}) \ldots (1+b_{0}^{2^{k}}) \ldots$
By the polynomial identity
$(1+x)(1+x^{2})(1+x^{4}) \ldots (1+x^{2^{k}}) \cdots = 1+x+x^{2}+x^{3}+\cdots = \frac{1}{1-x}$
Our desired product is then simply
$\frac{1}{1-(a_{0}-1)} = \frac{2}{3}$

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.