As for calculating pi...
My favourite rational approximation is 355/113 by Tsu Ch'ung-Chi has a nice 
pattern of digits and is accurate to 6 decimal places (better than the 2 of 
22/7 and better than any other below 103993/33102). Others are
>223/71 and <22/7 (Archimedes)
3123/994 (the average of the above two, and better than either)
377/120  (Ptolemy)
1019514486099146/324521540032945 (Johann Lambert)
Indeed, there are an infinite number of approximations. It just depends on how 
approximate you want to be.
3?
31/10?
157/50?
3141/1000?
6283/2000?

In 1700, Jacob Marcelis gave this approximation (he claimed exact):
    1008449087377541679894282184894
3 + -------------------------------
    6997183637540819440035239271702

Are you taking notes? There will be a test at the end... :-)

As for pi itself...

Which method you you want? There are (literally) an infinite number. Since pi 
is transcendental, there is no finite polynomial with integer coefficients 
(like 6x^13+2264448x^7+32123x^2+654) that you can just take a root of and out 
pops pi. Plenty of iterative methods though:
The first was
pi = 
2/(
     1       1 1     1        1 1     1 1     1
sqrt(-) sqrt(-+-sqrt(-)) sqrt(-+-sqrt(-+-sqrt(-)))...
     2       2 2     2        2 2     2 2     2
  )
...where "sqrt" is the square-root function, of course. I hope I managed to 
make the pattern clear enough.

pi   2   2   4   4   6   6
-- = - x - x - x - x - x -...
2    1   3   3   5   5   7

pi       1   1   1   1
-- = 1 - - + - - - + - ...
4        3   5   7   9

...but don't use this one in practice, it takes ages.

sqrt(2)pi       1   1   1   1   1    1    1
--------- = 1 + - - - - - + - + -- - -- - -- + ...
    4           3   5   7   9   11   13   15

pi-3     1       1       1
---- = ----- - ----- + ----- - ...
  4    2x3x4   4x5x6   6x7x8

pi^2        1     1     1
---- = 1 + --- + --- + --- + ...
 6         2^2   3^2   4^2

pi^2        1     1     1     1
---- = 1 + --- + --- + --- + --- + ...
 8         3^2   5^2   7^2   9^2

  1      1      1      1          2^24x76977927xpi^26
---- + ---- + ---- + ---- + ... = -------------------
1^26   2^26   3^26   4^26                 27!

Now just the primes!
 2^2    3^2       5^2     7^2    11^2          pi^2
----- x ----- x ----- x ----- x ------ x ... = ----
2^2-1   3^2-1   5^2-1   7^2-1   11^2-1          6

Srinivasa Ramanujan gave some stunning approximations:

(9^2+19^2/22)^(1/4) = 3.14159265262...

63(17+15sqrt(5))
----------------
25( 7+15sqrt(5))

    99^2
-----------
2206sqrt(2)

Some methods for calculating pi might be in order. I don't suppose you have 
the time or patience to work through an infinite sum or product to the number 
of terms or factors needed to get even a reasonable approximation (like I 
said, some of them take ages to get anywhere). So here it goes with a few 
methods. 

    Step 0: Set a = x = 1, b = 1/sqrt(2) and c = 1/4.
/-->Step 1: Set y = a
|           Set a = (a+b)/2
|           Set b = sqrt(by)
|           Set c = c-x(a-y)^2
|           Set x = 2x
|           Set p = (a+b)^2/4c.
\---p is your approximation. To get a better one, go back to Step 1 and go 
through the loop again.

After 3 times around the loop, the approximation is correct to 5 decimal 
places.

Some others, which get more accurate faster, at the expense of extra 
calculation, are as follows:

Step 0: Set y = 1/sqrt(2), c = 0 and p = 1/2
Step 1: Set c = c+1
        Set a = sqrt(1-y^2)
        Set y = (1-a)/(1+a)
        Set p = p(1+y)^2-y2^c
1/p is your approximation. Loop back to Step 1 to improve it.
This is accurate to 8 decimal places after three loops.

Step 0: Set y = sqrt(2)-1, c = 0 and p = 6-4sqrt(2)
Step 1: Set c = c+1
        Set a = (1-y^4)^(1/4)
        Set y = (1-a)/(1+a)
        Set p = p(1+y)^4-y(1+y+y^2)2^(2c+1)
1/p is your approximation. Loop back to Step 1 to improve it.
This is accurate to 171 decimal places after three loops and 694 after four.

The best I know of:
Step 0: Set s = 5(sqrt(5)-2), c=0 and p=1/2
Step 1: Set p = ps^2-5^c((s^2-5)/2+sqrt(s(s^2-2s+5)))
        Set x = 5/s-1
        Set y = (x-1)^2+7
        Set z = (x(y+sqrt(y^2-4x^3))/2)^(1/5)
        Set c = c+1
        Set s = 25/(s(z+x/z+1)^2)
1/p is still your approximation. Loop back to Step 1 to improve it.
This is accurate to 166 decimal places after three loops, and 848 after four.


1/pi = 
                                    n
  inf                           (-1) (6n)!
12SUM(545140134n+13591409)-----------------------
  n=0                           3n+1/2          3
                          640320       (3n)!(n!)

Where n! is the factorial function n!=1x2x3x...xn.

4/pi =
1+1/(2+9/(2+25/(2+49/(2+...))))

  2
pi  =
                 n
  inf (2-sqrt(3))
72SUM ------------
  n=1     2 2n
         n    C
               n
with 2nCn being the choose function 2nCn = (2n)!/(n!)^2

pi = 
         1
3+----------------
          1
  7+--------------
            1
    15+-----------
             1
       1+---------
               1
         292+-----
                1
             1+---
               ...

although I admit there's not much pattern there.

Step 0: s=sqrt(2); p=2+s; y=sqrt(s)

Step 1: Set x = (sqrt(x)+1/sqrt(x))/2
        Set p = p(x+1)/(y+1)
        Set y = (ysqrt(x)+1/sqrt(x))/(y+1)
And p is your approximation. Repeat Step 1 as necessary. After four
times through the loop, this procedure gives 
 p=3.141592653589793238462643383279502884197...
and only after the "97" do the digits go haywire.

I think I got those procedures right.

Morgan L. Owens (himself)
.sig under construction. Protective clothing MUST be worn.