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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