mersenneforum.org > Math Square root of 3
 User Name Remember Me? Password
 Register FAQ Search Today's Posts Mark Forums Read

 2009-12-31, 21:26 #1 Damian     May 2005 Argentina 2·3·31 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 1E0C16 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

2×3×31 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(3,3^1118781+1)/3 902510 Posts You will probably get to the same n's as in triangulation

 Similar Threads Thread Thread Starter Forum Replies Last Post chris2be8 Msieve 7 2017-02-06 03:34 davar55 Lounge 0 2016-03-16 20:19 paul0 Factoring 10 2015-01-19 12:25 bsquared Msieve 17 2010-04-09 14:11 davar55 Puzzles 3 2007-09-05 15:59

All times are UTC. The time now is 07:56.

Sun May 31 07:56:00 UTC 2020 up 67 days, 5:29, 1 user, load averages: 1.43, 1.66, 1.63