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. Andries -- To unsubscribe from this list: send the line "unsubscribe linux-man" in the body of a message to majordomo@xxxxxxxxxxxxxxx More majordomo info at http://vger.kernel.org/majordomo-info.html