User talk:Ed g2s/Maths

Firstly, I apologize that I didn't take the time to read the whole thing (ironically, I'm a math major and yet I have a hard time followubg math articles on Wikipedia), but I do see one "flaw" in there (not really a flaw, the whole thing seems makes sense, but what I noticed invalidates your purpose). In the Order 1 section, you say that the gradient of the line is a which is also the derivative. Gradients are just expanding a derivative into multiple dimensions, still using the basic definition (which is the difference quotient). Your comment that this doesn't involve the difference quotient thus becomes false.

Also, in Order 2 Method 2, you say that f(x)-g(x) has a double root at x=x1. You should provide a proof of that. It's not obvious to me that that is the case (while other proofs in calculus usually are obvious to me), and although I'm sure it'll only take me a few minutes to get that result, others may not see it. --ZeromaruTC 14:39, 30 May 2006 (UTC)

What I have pretty much repeats Zeromaru above; you define a derivative formally to be the expression you expect for polynomials over any ring or more generally for power series, which in turn looks like your expression in line 1. And then it has all the nice algebraic properties. Here because the definition is algebraic one can only deduce algebraic results from it. To define a derivative of more general functions using this one can use Weierstrass preparation theorem but again that is using limits. What you did is define gradient or slope of ax+b to be a. Then used the fact that there is a unique line passing through any point on a curve defined by a polynomial which has multiple intersection. And then defined the gradient to be the gradient of this line. You can do the same with power series over complex numbers. And you have found an expression for the function which takes value this gradient at each point. Of course it has to agree with derivative of the polynomial. Mornington 15:56, 27 March 2007 (UTC)