A difficult problem

Define a real sequence : $a_{n+1}=2a_{n}^{2}-1,\forall n\in \mathbb{N^{+}}$

and $T_{n}=2^{n}a_{1}a_{2}\cdots a_{n}$

If $\exists A \in \mathbb{R}$ : $\forall n\in \mathbb{N^{+}}, |T_{n}|<A$ , then what values can $a_{1}$ take?

Any idea that will help?

#Algebra

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Remember to wrap math in `$` ... `$` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
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Comments

Hint: $\cos(2x) = 2\cos^2 x - 1$ .

Hint 2: Prove that $a_{n+m} = \cos(2^m \theta)$ .

Hint 3: Multiply $T_n$ by $\frac{\sin \theta}{\sin \theta}$ , apply $\sin(2x) = 2\sin x \cos x$ repeatedly.

Pi Han Goh - 2 years, 11 months ago

Really cool ! Hence $a_{1}\in (-1,1)$ is the answer. Thank you a lot.

Haosen Chen - 2 years, 11 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$