Square root of 3

Damian · 2009-12-31, 21:26

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

cheesehead · 2009-12-31, 22:00

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.

Damian · 2009-12-31, 23:30

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.

Batalov · 2010-01-01, 01:56

You will probably get to the same n's as in triangulation

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

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Batalov "Serge" Mar 2008 Phi(4,2^7658614+1)/2 3⁶·13 Posts	You will probably get to the same n's as in triangulation