20 replies to this topic

[guest3456]

here ya go

x := 1

Loop
{
   Send {%x%}

   if (x == 10)
      x := 1
   else
      x := x + 1
}

return

[Tilter_of_Windmills]

Would not getting the exact value of pi be more desirable?

MsgBox pi = π

[Tilter_of_Windmills]

Would not getting the exact value of pi be more desirable?

Loop
{
MsgBox, 4,, pi = π.  Close enough?
IfMsgBox, Yes
     Break
}
ExitApp

Here's an algorithm for you...I just don't know how to code it. It'd be interesting to see if another could.

pi = sqrt(6/1 + 6/2^2 + 6/3^2 + ...)

So loop that enough times and the square root of that series should nudge up pretty close to π. A billion digits? Can AHK handle that?

Here's another one:

pi = 3 + (1/(6 + 3^2/(6 + 5^2/ (6 + 7^2/(6 + 9^2/(6 + 11^2/(....

[Morpheus]

<!-- m -->http://www.optimnem.co.uk/pi.php<!-- m -->

[Ohnitiel]

pi = sqrt(6/1 + 6/2^2 + 6/3^2 + ...)

pi := 6/1
Loop, % 1000000000 - 1
	pi += 6/(A_Index^2)
pi := sqrt(pi)

...?

Not right... that formula is wrong. This one works though 3+(1/(6+3**2/(6+5**2/(6+7**2..., but Autohotkey is not capable of making it for too long. My guess was to try and make it in parts, but I'm not in the mood to think about that right now.

[Tilter_of_Windmills]

pi = sqrt(6/1 + 6/2^2 + 6/3^2 + ...)

pi := 6/1
Loop, % 1000000000 - 1
	pi += 6/(A_Index^2)
pi := sqrt(pi)

...?

Not right... that formula is wrong. This one works though 3+(1/(6+3**2/(6+5**2/(6+7**2..., but Autohotkey is not capable of making it for too long. My guess was to try and make it in parts, but I'm not in the mood to think about that right now.

The first formula is right. It's Euler's solution to the Basel problem, solved for pi. Euler proved that the summation as n goes from 1 to infinity of 1/n^2 is pi^2/6. Multiplying through by 6 would give us that pi^2 = the summation as n goes from 1 to infinity of 6/n^2. Taking the square root of both sides would produce pi = the square root of (6/1 +6/2^2 + 6/3^2...). It's the code that "Guest" wrote that's wrong.

[Tilter_of_Windmills]

How does that code differ from your posted algo (1st one)?

That code is calculating a bunch of nested square roots rather than the square root of the sum.

Edit: The code below works pretty good. It produces 3.141018. I tried 100000 loops and it locked up on me.

Loop, 10000
{	
	pi += 1/(A_Index**2)
}
pi := sqrt(6*pi)
MsgBox, 4,, pi = %pi%...Close enough?
IfMsgBox, Yes
return

[Tilter_of_Windmills]

Is there a limit on how many loops AHK will run?

[Ohnitiel]

Well... Autohotkey just finished making 100000000 (yup 8 0's) and it return only 16 decimal places... Took over half a hour for that and returned this 3.1415926449823899. Anyone willing to try a greater amount of loops?

PS: I've used setformat for it to show 50 decimal places.