. . . It is about me
Fetches 1 by pi to 16 terms in 4 iterations.
Added
Just my ((n!)(2n!)(6/85))/((n!)^(n+2)) fetches 1 by pi upto 5 places in 4 iterations. If I use (1+0.000041206)^n also then 1 by pi upto 9 places in 4 iterations. I claim I used Wolfram alpha resources sensibly.
There is tremendous scope in tuner concept. Because 41206 by 54 has repeated decimals. And others too.
54 and 22628 can be expressed as (0.41206)^n into (0.318309)^(n(n+1))/2 . So my formula can be made iterative.
Just my ((n!)(2n!)(6/85))/((n!)^(n+2)) fetches 1 by pi upto 5 places in 4 iterations.
ReplyDeleteIf I use (1+0.000041206)^n also then 1 by pi upto 9 places in 4 iterations.
I claim I used Wolfram alpha resources sensibly.
There is tremendous scope in tuner concept. Because 41206 by 54 has repeated decimals. And others too.
ReplyDelete54 and 22628 can be expressed as (0.41206)^n into (0.318309)^(n(n+1))/2 .
ReplyDeleteSo my formula can be made iterative.