Posted by Geeks around the world will be celebrating Pi day, a shameless excuse to put the month before the day like the Americans do so we can talk Mathematics on “3.14” day 
So what better way to show how cool SQL can be with some expressions to approach Pi with continuous series
All we really need to know by way of background is the CONNECT BY LEVEL trick to generate arbirtary sequences of row, eg
SQL> select level from dual connect by level <= 5;
LEVEL
----------
1
2
3
4
5
Now we can get some approximations to Pi using the some of the standard SQL functions SQRT and POWER
SQL> select sqrt(8*sum(1/power((level*2-1),2))) pi
2 from dual
3 connect by level <= 100000
4 order by 1 desc;
PI
----------
3.14158947
1 row selected.
SQL>
SQL>
SQL> select sqrt(6*sum(1/(level*level))) pi
2 from dual
3 connect by level <= 100000
4 order by 1 desc;
PI
----------
3.1415831
1 row selected.
SQL>
SQL> select 8*sum(1/((4*(level-1)+1)*(4*(level-1)+3))) pi
2 from dual
3 connect by level <= 100000;
PI
----------
3.14158765
1 row selected.
SQL>
SQL>
SQL> select
2 4*sum(
3 power(-1,level-1)/(level*2-1)*
4 ( 12*power(1/18,level*2-1)+
5 8*power(1/57,level*2-1)-
6 5*power(1/239,level*2-1))) pi
7 from dual
8 connect by level <= 100;
PI
----------
3.14159265
1 row selected.
SQL>
Some methods to calculate Pi need factorials, but there isn’t a native SQL function for that. But in 12c, that’s no problem, we can define SQL functions on the fly directly inside our SQL statement !
SQL>
SQL> with
2 function factorial(n int) return int is
3 f int := 1;
4 begin
5 for i in 1 .. n loop
6 f := f * i;
7 end loop;
8 return f;
9 end;
10 select 2*sum(
11 power(2,level-1)*power(factorial(level-1),2)/factorial(2*level-1)
12 ) pi
13 from dual
14 connect by level <= 17;
15 /
PI
----------
3.1415864
1 row selected.
But what if you’re not on 12c yet ? Well, you should be!
But whilst you’re on 11g, there is still plenty of cool SQL options to play with Pi. Here we can use recursive SQL to generate Pi, based on the formula:
SQL> with term(numerator,product,seq) as
2 ( select sqrt(2) numerator, sqrt(2)/2 product , 1 seq from dual
3 union all
4 select sqrt(2+numerator), sqrt(2+numerator)*product/2 , seq+1
5 from term, dual
6 where term.seq <= 16
7 )
8 select 2/product pi
9 from term
10 where seq = 16;
PI
----------
3.14159265
1 row selected.
SQL…still the most awesome language out there!
Learning is not a spectator sport 
select
sum(
(
4 / (8 * (level – 1) + 1) –
2 / (8 * (level – 1) + 4) –
1 / (8 * (level – 1) + 5) –
1 / (8 * (level – 1) + 6)
) / power(16, level – 1)
)
from
dual
connect by
level <= 10
probably is the most performant way to get a very good approximation
I forget where I came across this, but acos(-1) also gives π. (Don’t ask me what an arc cosine is.)
In case the boring old 3.1416 isn’t the π you’re looking for: http://oracle-wtf.blogspot.co.uk/2017/10/mmm.html