Do not rudely accuse me of writing tautologies! Compare with: * A function A(x) admits a hypergeometric form whenever a_0=1, a_{n+1}=poly1(n)/poly2(n)*a_{n}. * The function A(s) admits a product form whenever each a_n obeys the previously stated identity. In both of these statements a deciding criterion is given for dividing the set of all functions into has and has not classes with regard to some apparently arbitrary property. That is distinctly different from the tautological form “A is A”! We also have a lot of cases where “B is not A”. Honestly, if you are hoping to get a better response, you might want to do some self-critique and try to improve on communicating what you want to hear. As far as I can tell, I answered your question, and then you attacked me with a fallacious argument, meaning to imply that I am some sort of idiot. Wrong! Cheers, Brad
On May 9, 2019, at 6:39 AM, Mike Stay <metaweta@gmail.com> wrote:
Of course, but that's a condition on a, not directly on A. It's like saying "exp(z) equals its conjugate when exp(z) is real" rather than "exp z equals its conjugate when Im(z) is an integer multiple of pi".
On Wed, May 8, 2019 at 8:32 PM <bradklee@gmail.com> wrote:
a_{p1^m1 p2^m2 ... p^m} = a_{p1^m1} a_{p2^m2} ... a_{p^m}
?? —Brad