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Old 2009-12-31, 21:26   #1
Damian
 
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Default 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
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Old 2009-12-31, 22:00   #2
cheesehead
 
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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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Old 2009-12-31, 23:30   #3
Damian
 
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Quote:
Originally Posted by cheesehead View Post
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.
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Old 2010-01-01, 01:56   #4
Batalov
 
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You will probably get to the same n's as in triangulation
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