There has been a data type error here. A derivation is a linear operator on the space of functions, not a function. If 1 denotes the constant function, then d(1) = d(1.1) = d(1).1 + 1.d(1) = 2.d(1), so d(1) = 0, and thus constant functions are in the null space of all derivations. On the other hand, I seen no reason why d(x) must be 1. If d(x) = q (i.e. the function q(x)), then for any function f, it looks like d(f) = q.D(f), where D is the standard differentiation operator. Gene ----- Original Message ---- From: "mcintosh@servidor.unam.mx" <mcintosh@servidor.unam.mx> To: math-fun@mailman.xmission.com Cc: mcintosh@servidor.unam.mx Sent: Tuesday, June 19, 2007 10:57:40 AM Subject: Re: [math-fun] Continuous derivations on the reals Quoting David Wilson <davidwwilson@comcast.net>:

Let [1] d(uv) = d(u) v + u d(v)

Then [2] d(u^2) = 2u d(u)

not when the quantities are noncommutative, as is typically the case when the object of discussion is a Lie Algebra. As for this implying a vanishing Maclaurin series, what does this say about ordinary functions, say polynomials, which do indeed have (non-zero) derivatives? - hvm ------------------------------------------------- www.correo.unam.mx UNAMonos Comunicándonos _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun ____________________________________________________________________________________ No need to miss a message. Get email on-the-go with Yahoo! Mail for Mobile. Get started. http://mobile.yahoo.com/mail