Nested Roots

Consider the set $S_n$ of all the $2^n$ numbers of the type $2\pm\sqrt{2\pm\sqrt{2\pm\dots}}$ , where the number $2$ appears $n+1$ times.

(a) Show that all members of $S_n$ are real.

(b) Find the product $P_n$ of all elements of $S_n$ .

Source: Austria 1989

#Olympiad

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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}`	$\boxed{123}$

Comments

(a)Assume, up to a number n, that all values are positive and real.These values are bounded by 4, since the infinite radical $\sqrt{2+\sqrt{2+...}}=2$ .So the n+1st number is bigger than $2-\sqrt{2+\sqrt{2+...}}=0$ .Thus, by induction, every such number is real and positive (the base case is obvious)

(b)I will show that the product is the same for every n.Let $\alpha_n$ be a number $\sqrt{2\pm\sqrt{2\pm...}}$ .That means that $P_n=\displaystyle\prod_{}{}(2+\alpha_n)(2-\alpha_n)=\prod 4-\alpha_n^2=\prod (2+\alpha_{n-1})(2+\alpha_{n-1})=P_{n-1}$ .Since $P_1=2$ , they are all 2.

Bogdan Simeonov - 6 years, 2 months ago

Bravo !

Cody Johnson - 6 years, 1 month ago

Thanks :D .By the way when is the next Proofathon?

Bogdan Simeonov - 6 years, 1 month ago

@Bogdan Simeonov – See the schedule on the website: http://proofathon.org/ongoing_contest.php

Cody Johnson - 6 years, 1 month 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}`	$\boxed{123}$