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Wallis' product exploration
02-09-2020, 01:57 PM
Post: #10
RE: Wallis' product exploration
Using identity 5.36 ( p. 186 of Concrete Mathematics )

\( \binom{n-\frac{1}{2}} {n} = \binom{2n} {n} ÷ 2^{2n} \)

I think we can also reduce the limit to:
\( \pi = \lim\limits_{n\to\infty} \frac {1} {n+\frac {1}{2}} ÷ \binom{n-\frac{1}{2}} {n}^{2} \)

(I could have messed something up there..)

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Messages In This Thread
Wallis' product exploration - pinkman - 02-07-2020, 10:55 PM
RE: Wallis' product exploration - pinkman - 02-08-2020, 02:42 PM
RE: Wallis' product exploration - Allen - 02-08-2020, 07:13 PM
RE: Wallis' product exploration - Allen - 02-08-2020, 08:58 PM
RE: Wallis' product exploration - pinkman - 02-09-2020, 05:44 AM
RE: Wallis' product exploration - Allen - 02-08-2020, 08:13 PM
RE: Wallis' product exploration - Allen - 02-09-2020, 01:08 PM
RE: Wallis' product exploration - Allen - 02-09-2020 01:57 PM
RE: Wallis' product exploration - Allen - 02-09-2020, 03:08 PM
RE: Wallis' product exploration - pinkman - 02-09-2020, 02:14 PM
RE: Wallis' product exploration - EdS2 - 02-10-2020, 10:35 AM
RE: Wallis' product exploration - pinkman - 02-11-2020, 10:02 AM
RE: Wallis' product exploration - pinkman - 02-12-2020, 10:01 PM



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