Wallis' product exploration

02092020, 03:08 PM
(This post was last modified: 02092020 03:59 PM by Allen.)
Post: #12




RE: Wallis' product exploration
Nice, Thank you for checking in CAS!
of course at infinite limits the extra +1/2 is irrelevant, and since \( \binom{n\frac{1}{2}} {n}^{2} = \binom{1/2} {n}^{2} \), we can remove some more symbols. as we can see on a discrete limit Wolfram alpha (or with continuous n) \( \pi = \lim\limits_{n\to\infty} \frac {1} {n \binom{\frac{1}{2}} {n}^{2}} \) or circumference \(c \) is \( c = \lim\limits_{n\to\infty} 2 \binom{\frac{1}{2}} {n}^{2} \) 17bii  32s  32sii  41c  41cv  41cx  42s  48g  48g+  48gx  50g  30b 

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