[math-fun] 2 surprising approximations and 1 exact formula for pi.

21 Jan 2011

      Hello everybody,

  these days I am working on exponential sums and I
have found something of interest,
like
                             infinity
                              -----          3
                               \            n
                                )     ---------------
                               /          2 Pi n
                              -----   exp(------) - 1
                              n = 1         13

= 119.00000000000000000000000000000009593745851025547335588584913...

the precision is 31 digits.

  Another one is
                             infinity
                              -----          3
                               \            n
                                )     ---------------
                               /          2 Pi n
                              -----   exp(------) - 1
                              n = 1         7

  = 
10.0000000000000001901617678886626755843593058554453334802548978434061099438\
The precision here is 15 digits.

For an argument of 2*Pi*n/163, the precision is 435 digits! I
don't know why I took 163... (of course).

in general, these sums will be near an integer if the argument
is 2*Pi*n/k and k is NOT a multiple of 2,3 or 5, strange isn't ?
These are the simplest I could find,
if the exponent is of the form 4m-1 then the sum is often an
integer with the same conditions. This, I believe extends
a little bit the known formula of Ramanujan/Berndt/etc.

  A good question is : does someone has a simple explanation
of this ? I don't. I am preparing a paper on these results.

Because, I do have another one which is EXACT, namely for pi:
In general these sums with a fractional exponent are very
close to integers or pi, I have a couple of series for 1/pi, 1/pi^2,
etc.

here is the  EXACT formula with fractional arguments mixed with
integer exponents.

    /infinity                  \      /infinity                    \
    | -----                    |      | -----                      |
    |  \              1        |      |  \               1         |
10 |   )     -----------------| - 40 |   )     -------------------|
    |  /      n (exp(Pi n) - 1)|      |  /      n (exp(2 Pi n) - 1)|
    | -----                    |      | -----                      |
    \ n = 1                    /      \ n = 1                      /

           /infinity                    \      /infinity                  \
           | -----                      |      | -----                    |
           |  \               1         |      |  \              1        |
      + 10 |   )     -------------------| - 10 |   )     -----------------|
           |  /      n (exp(4 Pi n) - 1)|      |  /        /    Pi n     \|
           | -----                      |      | -----   n |exp(----) - 1||
           \ n = 1                      /      \ n = 1     \     5       //

           /infinity                    \      
/infinity                    \
           | -----                      |      | 
-----                      |
           |  \               1         |      |  \               
1         |
      + 40 |   )     -------------------| - 10 |   )     
-------------------|
           |  /        /    2 Pi n     \|      |  /        /    4 Pi 
n     \|
           | -----   n |exp(------) - 1||      | -----   n |exp(------) 
- 1||
           \ n = 1     \      5        //      \ n = 1     \      
5        //
...
evalf(%);
3.14159265358979323846264338327950288419716939937510582097494459230781640628\
     62089986280348253421170679821480865132823066470938442...
well, I verified up to 1000 digits and it holds.
This is trivial, ? I do not see how.

If someone has a piece of information on why this exist, I would
be glad to ear from it, references, known results, etc.
Me, I never saw these kind of formulas before,

  have a good day,
  Simon Plouffe