I have to agree with you on most counts that Numerical Recipes in C=20 is not a full-blown encyclopedia on all subtleties of doing numerical=20 computations. However it does its job greatly for a big class of stuff I am=20 interested in: minimization, non-linear system of equations solving=20 (The newton routine given there is good, accurate and fast.) There are errors and problems as in most other numerical books. In=20 truth, I don't think there is anything fully correct out there. When trying to make computations you have to do a lot of testing and=20 a lot of thought even if the algorithm seems perfect. That applies to=20 all books, recipes et al. I have just discovered that tan(pi/2) =3D 1.63317787284e+16 in=20 numerical python. And we all know this is bad. It should be infinity=20 period. We should define a symbol called infinity and put the correct=20 definition of tan, sin .... for all known angles then interpolate if=20 needed for the rest, or whatever is used to actually compute those thing= s. Peace <snip> - ------- End of Forwarded Message I believe you're actually talking about the math module, not the numeric module (I'm not aware of tan or pi definitions in numeric, but I haven't bothered to double check that). Never the less, I think it has relevance here as Numeric is all about doing serious number crunching. This problem is caused by the lack of infinite precision in python. Of course, how is it even possible to perform infinite precision on non-rational numbers? The obvious solution is to allow the routine (tan() in this case) to recognize named constants that have relevance in their domain (pi in this case). This would fix the problem: math.tan(math.pi) = -1.22460635382e-16 but it still doesn't solve your problem because the named constant would have the mathematical operation performed on it before it's passed into the function, ruining whatever intimate knowledge of the given named constant that routine has. Perhaps you could get the routine to recognize rational math on named constants (the problem with that solution is how do you not burden other routines with the knowledge of how to process that expression). Assuming you had that, even for your example, should the answer be positive or negative infinity? Another obvious solution is just to assume that any floating point overflow is infinity and any underflow is zero. This obviously won't work because some asymptotic functions (say 1/x^3) will overflow or underflow at values for x for which the correct answer is not properly infinity or zero. It is interesting to note that Matlab's behaviour is the same as Python's, which would indicate to me that there's not some easy solution to this problem that Guido et. al. overlooked. I haven't really researched the problem at all (although now I'm interested), but I'd be interested if anyone has a proposal for (or reference to) how this problem can be solved in a general purpose programming language at all (as there exists the distinct possibility that it can not be done in Python without breaking backwards compatibility). Terry -----BEGIN PGP SIGNATURE----- Version: PGPfreeware 5.0i for non-commercial use Charset: noconv iQA/AwUBOL2OUqfuGVwXgOQkEQJTpQCggOuFT2ZVavzMhy+jZgoehnrK5uIAoMzO D5OOdLtBvT97ee7vkckO+0Qt =SmqL -----END PGP SIGNATURE-----