Re: Quantum Pi

Robert J. Kolker wrote :
> magicvsmagick@xxxxxxxxx wrote:
>> I have decided that I would rather define Pi as 3, instead of using the
>> other definitions. I have a circle I created in photoshop that measures
>> 8 pixels across, however you measure it. The circle's circumfrence is
>> 24 pixels around. If you can accept this as a circle then Pi is 3.
>
> Why not use a square. The ratio of the circumference to the diameter is
> 4/sqrt(2) which is the sqare root of 8 which is 2 plus a fraction.
>
> The real pi has some advantages. For example you get equalities
> like exp(i*pi) = -1. I will bet you can't do that with your pixel thingy.
>
> Bob Kolker

Depends on value of e may be...
or of value of -1 (what is -1 pixel eh ?)
Sure OP would get some other very fundamental relation using his new
pi.

BTW I just (re-)read some paper about Gauss method to find the number
of grid points inside a circle (a true circle, not a special one)
that is the number of integers x,y with x^2 + y^2 < r^2
Just from the number of decompositions of n in sum of two squares being
r(n) = 4*d(n) - 4*d'(n) where d(n) is the number of divisors of n of
the form 4k+1 and d'(n) the number of 4k+3 divisors.
and finishing with the interesting formula :

Number of points = 1 + 4( [r^2/1] - [r^2/3] + [r^2/5] - [r^2/7]...)
with [x] being integer part (floor) of x
The sum being finite because after some rank, the integer part becomes
0

with r = 4 (diameter 8), this gives 49 points inside,
resulting in pi = 49/4^2 = 3.0625, what OP considers to be 3

r = 10 results into 317 points (summing 50 terms)
r = 100 gives 31417 points (sum of 5000 terms, program required)

Regards.

--
philippe
mail : chephip at free dot fr
site : http://chephip.free.fr/


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