Re: modulo 2 pi?

[Johan Kullstam]

Søren Hauberg <address@hidden> writes:

> ons, 21 01 2009 kl. 14:23 +0100, skrev Francesco Potortì:
>> >If I have two angles 'theta' and 'psi', and I want to compute their
>> >difference I can simply do
>> >
>> >  d = theta - psi;
>> >
>> >However, I'd like this result to be in the range [0, 2*pi]. Can I simply
>> >do
>> >
>> >  d = mod (theta - psi, 2*pi);
>> >
>> >or will this fail in some cases?
>> 
>> It should not.  It would fail if you used 'rem', but using 'mod' you
>> should have no problems.
>
> Thanks. I've been using 'mod' for some time now, but I wanted to double
> check.

You can also use cos, sin and atan2 as in atan2(sin(theta),cos(theta)).

Sometimes it is more convenient to use the complex number
  z = exp(I*theta)
instead of the real number theta.

Use multiplication of z instead of addition (and division for
subtraction).  The angle wraps naturally.  Welcome to the Lie group
U(1,C).

For your case:
t = exp(I*theta);
p = exp(I*psi);
d = t/p;
delta = arg(d);

delta will be between -pi and +pi.



Looking at the help information
~:> help arg
 -- Mapping Function:  arg (Z)
 -- Mapping Function:  angle (Z)
     Compute the argument of Z, defined as THETA = `atan (Y/X)'.  in
     radians.

The above is wrong as it should be atan2(Y,X) not atan(Y/X).  How to
file a bug against the documentation?

-- 
Johan KULLSTAM