On Tue, Jul 5, 2011 at 12:15 PM, Steven Schveighoffer <schveiguy@yahoo.com> wrote:
On Tue, 05 Jul 2011 04:31:09 -0400, James Fisher <jameshfisher@gmail.com> wrote:

Hopefully this won't be taken as frivolous.  I (and possibly some of you)
have been convinced by the argument at http://tauday.com/.  It's very
convincing, and I won't rehash it here.

The use of τ instead of π will only become really convenient when one does
not have to preface everything with "let τ = 2π".

For example, in D, in order to think in terms of τ instead of π, one must
define `enum real TAU = std.math.PI * 2;`, and possibly also TAU_2, TAU_4,
etc.

As well as being a typing inconvenience, I also think things are not that
easy due to loss of precision (though I'm far from an expert on intricacies
of floating point).

There is an initiative to add TAU to the Python standard library:
http://www.python.org/dev/peps/pep-0628/

To this end, I suggest adding the constant TAU to std.math, and possibly
also TAU_2 as an alias for PI, TAU_4 as an alias for PI_2, TAU_8 as PI_4.

In any case, I'd like to know what's necessary in order for me to define
these constants without loss of precision.
d

I read an article about this recently, it's definitely interesting.  The one place where I haven't seen it mentioned is what happens when you want the area of a circle, since that necessarily involves the radius.  I'd guess you'd have to use τ/2 * r^2, but even then, that's one formula vs. the rest.  It's probably a good tradeoff.  I can definitely see the advantage when using radians.  Never thought I'd have to re-learn trig again...

One thing I like about Pi vs Tau is that it cannot be mistaken for a normal character.

I'm not a floating point expert, but I would expect since floating point is stored in binary, dividing or multiplying by 2 loses no precision at all.  But I could be wrong...

Sorry, I didn't state this very clearly.  Multiplying the approximation of PI in std.math should yield the exact double of that approximation, as it should just involve increasing the exponent by 1.  However, [double the approximation of the constant] is not necessarily equal to [the approximation of double the constant].  Does that make sense?