Hi Andries!
Nice to hear from you again!
On Thu, Sep 15, 2011 at 6:24 PM, Andries E. Brouwer
<Andries.Brouwer@xxxxxx> wrote:
> On Fri, Aug 05, 2011 at 11:34:54PM +0200, Richard B. Kreckel wrote:
>> Hi everybody,
>>
>> On 11/27/2010 08:37 AM, Michael Kerrisk wrote:
>> >Hi Andries,
>> >
>> >Since you are the mathematician, can you comment?
>
> Hi everybody,
>
> Now that I see this discussion again:
> Mathematically speaking there is no unique correct definitions.
> The function is multiple-valued and some arbitrary agreement
> fixes one of the values.
>
> Richard's point of view is that the actual behavior of the library functions
> should be documented. Not a bad idea, but perhaps one should document what
> the standard says, and then file a bug report in case the implementation
> does something else.
>
> I lost most of my old docs, but just rediscovered a copy of
> ISO-C-FDIS.1999-04.pdf, I don't know whether that still is the
> appropriate source to consult.
>
> (i)
> It says in 7.3.1: cacos: branch cuts outside [-1,1] along the real axis
> returns arccos z where Re arccos z lies in [0,pi].
> So, if z = cos w then w = arccos z, but because cos (w + 2pi) = cos w
> and cos (-w) = cos w, we can pick w with real part in [0,pi].
>
> It says in 7.3.7.2: clog: branch cut along the negative real axis
> returns log z where Im log z lies in [-I*pi,I*pi].
>
> It says in 7.3.8.3: csqrt: branch cut along the negative real axis
> returns sqrt z where Im sqrt z >= 0.
>
> We want to check whether it is true that
> cacos(z) = -I* clog(z + I*csqrt(1 - z * z))
>
> Suppose w = cacos(z). Then w is defined by cos w = z and Re w in [0,pi].
> The question is whether iw = log(cos w + i sqrt(sin^2 w)).
> If sin w has nonnegative imaginary part, then sqrt(sin^2 w) = sin w
> and the question is whether iw = log(e^{iw}). This logarithm is
> chosen with imaginary part in [-pi,pi] but w has imaginary part
> in [0,pi], so this equality holds.
> If sin w has negative imaginary part, then sqrt(sin^2 w) = -sin w
> and the question is whether iw = log(e^{-iw}), and again this holds.
>
> I agree with Richard's equation.
>
> (ii)
> Claim: catan(z) = (clog(1 + iz) - clog(1 - iz)) / 2i
>
> 7.3.5.3: catan: branch cuts outside [-i,i] along the imaginary axis
> returns atan z where Re atan z lies in [-pi/2, pi/2].
>
> Suppose w = catan(z). Then w is defined by tan w = z and Re w in [-pi/2, pi/2].
> Up to a multiple of 2pi*i the RHS is
> (1/2i)log((cos w + i sin w)/(cos w - i sin w)) = (1/2i) log e^{2iw},
> looks good. The two multiples of 2pi*i partially cancel, so after
> dividing by 2i this is in [-pi/2, pi/2].
>
> I agree with Richard's equation.
>
> (iii)
> Claim: catanh(z) = (clog(1 + z) - clog(1 - z))/2
>
> 7.3.6.3: catanh: branch cuts outside [-1,1] along the real axis
> returns atanh z where Im atanh z lies in [-pi/2,pi/2].
>
> Suppose w = catanh(z).
> Then w is defined by tanh w = z and Re w in [-pi/2,pi/2].
> Up to a multiple of 2pi*i the RHS is correct, and just like before
> we get the right value.
>
> I agree with Richard's equation.
>
> So, if I make no mistake, Richard is correct on all counts.
Thanks very much for the above analysis. I'm happy to have a second
mathematician confirm what the first one says,
Best regards,
Michael
--
Michael Kerrisk
Linux man-pages maintainer; http://www.kernel.org/doc/man-pages/
Author of "The Linux Programming Interface"; http://man7.org/tlpi/
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