[Tutor] square root
Salim
salim@nstp.com.my
Fri, 02 Feb 2001 15:38:23 +0800
Please make a correction:
x**1/2==the square root of x
please note there is no bracket...
tq
At 09:21 AM 2/2/01 +0800, you wrote:
>There's no operator for squareroot (unless I've been misleading myself
>for this long).
>
>Use the power rule of x**(1/y). Here's the relavent rules:
>
>x**y == x to the power of y,
>x**(1/y) == x to the root of y,
>x**(z/y) == x to the root of y - all raised to the power of z,
>
>So x**(1/2) == the square root of x
>
>I don't think it was a dumb question, but then, I'm a newbie script
>kiddie! =)
>
>Andrew Wilkins
>
> > -----Original Message-----
> > From: tutor-admin@python.org
> > [mailto:tutor-admin@python.org]On Behalf Of
> > W.W. van den Broek
> > Sent: Thursday, 1 February 2001 5:13
> > To: tutor@python.org
> > Subject: [Tutor] square root
> >
> >
> > How do you use the
> > square root as operator
> > in python?
> > Dumb question, but
> > thanks anyway,
> > walter
> > --
> > W.W. van den Broek
> > e-mail:
> > vandenbroek@psyd.azr.nl
> > AZR-Dijkzigt fax:
> > 010-4633217
> > afdeling psychiatrie
> > tel: 010-4639222
> > Postbus 2040 e-mail
> > vdbroekw@wxs.nl (thuis)
> > 3000 CA Rotterdam
> > homepage:
> > http://home.planet.nl/~vdbroekw
> >
> > _______________________________________________
> > Tutor maillist - Tutor@python.org
> > http://mail.python.org/mailman/listinfo/tutor
> >
>
>
>
>_______________________________________________
>Tutor maillist - Tutor@python.org
>http://mail.python.org/mailman/listinfo/tutor