I'm interested in using Python, how do I get started?

Easy to understand code. I like that example because it really shows that pi is real. But it has a veeerryyy slow convergence!

I don't use python myself, I tend to use php, javascript and actionscript (for flash).

I made it as far as here: http://www.python.org/download/ but I'm not sure which to download. I'm using Windows XP.

I know a little C++, but mostly use Actionscript.

Wow, I need an update. I'm currently using Python 2.6...
And the programming tutorial needs only a little programming experience. I think you should be able to go through it with not much trouble.

Yeah, I normally only use betas that have been around for a few weeks or so, I'm not much into new ones when I downloaded Firefox 3.1 Beta and it decided to crash every time it started.

Ahh, help!

I was going through this tutorial: http://docs.python.org/tutorial/index.html

And I quickly realized that I was using 3.0 and the tutorial is written for 2.6. I didn't think that was that big of a deal, but as it turns out, it is. (One example I came across was that the tutorial told me to use the print function without parenthesizes, while the 3.0 required that I type print())

As I get deeper and deeper into the tutorial, it gets worse and worse.

So I guess my question is, should I find a different tutorial, or download a different version?

Thanks in advance.

Last edited by bossk171 (2009-05-20 05:49:00)

simron wrote:

(Oh, and I'd recommend installing Python to a flash drive.)

Why is that? Is that some kind of coder's trick I'm not familiar with, or is it simply for portability?

Hey im probably a little old but i wanna learn more math and i've a passing familiarity with python.
I gave it a shot and it's really slow, can anyone else resolve pi faster?

def seq(n):
    n = n+2
    return [n, n+2]

def findpi():
    pi = 0
    k = -1
    while True:
        i = seq(k)
        a = (4.0/i[0])-(4.0/i[1])
        pi = pi + a
        k= k + 4
        print pi
findpi()

Im pretty sure 3.0 is "the future", aka an experimental version. All the interesing modules I've found are for 2.5+

added more simple

Last edited by gwar (2009-06-22 17:19:23)

It was the easiest the use though. Im not sure how to translate this http://en.wikipedia.org/wiki/Bellard%27s_formula
To me it just looks like this, it's obviously not right.

n = 0
pi = 1/2^6
while True:
        a= -1*n/2^(10*n)*(-2^5/4*(n+1)-1/4*(n+3)+2^8/10*(n+1)-2^6/10*(n+3)-2^2/10*(n+5)-2^2/10*(n+7)+1/10*(n+9))/10*(n+9))
        pi+=a
        n+=1
        print pi

Last edited by gwar (2009-06-23 16:41:35)

Hi gwar;

Even though mathsyperson has corrected that proggie.
Your choice of algorithm is not right.
That formula I believe is a spigot type algorithm.
It will not evaluate pi. It is designed to get the nth binary digit of pi.

Last edited by bobbym (2009-06-23 23:20:08)

ok well that makes sense.

Last edited by gwar (2009-06-23 23:30:59)

bobbym wrote:

Your choice of algorithm is not right.
That formula I believe is a spigot type algorithm.
It will not evaluate pi. It is designed to get the nth binary digit of pi.

It does calculate pi, it's just that it does so by calculating its binary digits, then summing them.