### Identity NO. 4 : Square root “sequence” of 2

30Dec07

Reading some of the Lev Kourliandtchik books, I’ve found a tricky observation how to write this gigantic expression in an extremely simple,
$\sqrt{2+\sqrt{2 + \ldots + \sqrt{2} }}$
with $n$ number of square roots.
It’s easy to check that $2 + 2 \cos{2x} = 4\cos^2{x}$
and using that equality we have
$\sqrt{2+\sqrt{2 + \ldots + \sqrt{2} }} = \newline = \sqrt{2 + \ldots + \sqrt{2 + 2 \cos{\frac{\pi}{4}} } } = \newline = \sqrt{2 + \ldots + \sqrt{2 + 2 \cos{\frac{\pi}{8}} } } = \newline \ldots \newline = \sqrt{2+2\cos{\frac{\pi}{2^n} } } = 2\cos{\frac{\pi}{2^{n+1}}}$
cool, huh? 🙂

#### 5 Responses to “Identity NO. 4 : Square root “sequence” of 2”

1. 1 ecik

Ya, cool 😀

2. And then as $n \to \infty$ , the limit is 2, as you’d expect using induction. 🙂

3. 3 ecik

btw. you have an error here.
You should have written (cos x)^2 instead of cos x^2. Currently it looks as it’s an argument what’s squared but it’s not true 🙂

4. 4 Will M

Bryce Sheppard is a genius..whoever he is.

5. ozjdgcoesrrssayulxcv, Locksmiths Mount Pleasant, tDeDBBv.

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