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 2009-12-31, 21:26 #1 Damian     May 2005 Argentina BA16 Posts Square root of 3 The square root of 2 can be written as an infinite product in the form: $\sqrt{2} = \left(1+\frac{1}{1} \right)\left(1-\frac{1}{3} \right)\left(1+\frac{1}{5} \right)\left(1-\frac{1}{7} \right) \ldots$ I wonder if there is an analogous infinite product for $\sqrt{3}$, does anyone know it, if there it is one? Thanks, Damián
 2009-12-31, 22:00 #2 cheesehead     "Richard B. Woods" Aug 2002 Wisconsin USA 22·3·641 Posts If you come up with some sine-cosine identity involving √3 (analogous to the cos(pi/4) = sin(pi/4) = 1/√2 identity shown in the Wikipedia article for √2), you've got it made. For instance, http://upload.wikimedia.org/math/a/6...daed90ca43.png or, more simply, cos(pi/6) = sin(pi/3) = √3/2 - - http://en.wikipedia.org/wiki/Exact_t...tric_constants and http://mathworld.wolfram.com/TrigonometryAngles.html have others. Last fiddled with by cheesehead on 2009-12-31 at 22:15
2009-12-31, 23:30   #3
Damian

May 2005
Argentina

2728 Posts

Quote:
 Originally Posted by cheesehead If you come up with some sine-cosine identity involving √3 (analogous to the cos(pi/4) = sin(pi/4) = 1/√2 identity shown in the Wikipedia article for √2), you've got it made. For instance, http://upload.wikimedia.org/math/a/6...daed90ca43.png or, more simply, cos(pi/6) = sin(pi/3) = √3/2 - - http://en.wikipedia.org/wiki/Exact_t...tric_constants and http://mathworld.wolfram.com/TrigonometryAngles.html have others.
Thank you!
Following your advice, and using the productory for the sine function, I could derive the following identities:

$\sqrt{2} = \frac{\pi}{2} \Pi_{n=1}^{\infty} \left( 1 - \frac{1}{16n^2} \right)$

$\sqrt{3} = \frac{2\pi}{3} \Pi_{n=1}^{\infty} \left( 1 - \frac{1}{9n^2} \right)$

$\displaystyle \sqrt{5} = 1 + \frac{2\pi}{5} \Pi_{n=1}^{\infty} \left( 1 - \frac{1}{100 n^2} \right)$

I'll see if I can find a way to generalize those to a productory for $\sqrt{n}$ for any integer $n$

Any hint on that?
Thanks,
Damián.

 2010-01-01, 01:56 #4 Batalov     "Serge" Mar 2008 Phi(4,2^7658614+1)/2 232108 Posts You will probably get to the same n's as in triangulation

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