Recent content by Josh Swanson

It occurred to me there are some details missing yet. Hawkeye's statement, essentially on the annulus of convergence, requires some justification. The nth term test only gives \lim_{n \rightarrow \infty} c_n z^n + c_{-n} z^{-n} = 0 \mbox{ for } r < |z| < R \quad \mbox{ (*)} which is not a...

I don't see how this is true, though you made me consider the fact that the nth term goes to 0 on this annulus, which gives the result anyway. Outside the radius of convergence of a power series, the nth terms do not go to 0 (which is how divergence is proven there), so each r < z < R is not...

Suppose we have f(z) = \sum_{n=-\infty}^\infty c_n z^n; \quad U(z) = \sum_{n=0}^\infty c_n z^n; \quad L(z) = \sum_{n=1}^\infty c_{-n} z^{-n} where f(z) converges* in the interior of some annulus with inner radius r and outer radius R > r. Further suppose U(z) has radius of convergence R_0...

Ah, wonderful, thanks for the link! I was hoping the result was still true outside of inner product spaces, and so it is. That completes the proof then. P.S. Non-degeneracy can also be deduced easily without recourse to the existence of f from (c_1, ..., c_i, ..., c_n) . (0, ..., 0, 1...

Let V be a vector space over a field h and let n be a positive integer. Let f:V -> h^n be a linear map given by f(v) = (f1(v), f2(v), ..., fn(v)). Call two vectors (g1, ..., gn) and (h1, ..., hn) in h^n "orthogonal" if g1 h1 + ... + gn hn = 0 Suppose the only vector orthogonal to every...

I agree. "Restricting" is more proper. Adding a circle at infinity is a much more elegant approach than my ad-hoc notation, even if they amount to the same thing. I had considered using projective points instead of vectors but I wanted +\infty and -\infty to be distinct. Right now I'm...

Well, I again disagree that defining it as 0 at 0 won't work just because it has an essential singularity there. But I understand that the specific property of continuity is maintained in the narrower case while it's lost in the more general case, so to maintain the more general version of...

I've been using the term "indeterminate form" in its technical sense (eg. see the http://en.wikipedia.org/wiki/Indeterminate_form" on the topic for a more formal description of what I mean). In that context, sin(x)/x's behavior is irrelevant, though of course it provides a nice example that 0/0...

@micromass: If I understand correctly, you would define \text{^}(x, y): \mathbb{R} \times \mathbb{Z}^+_0 \to \mathbb{R}~\text{with}~\text{^}(0, 0) = 1 and \text{^}(x, y): \mathbb{R} \times \mathbb{R} \setminus \{0\} \times \mathbb{Z}^-_0 \to \mathbb{R}? [It occurred to me that x^y is...

Thank you for discussing this with me. I find it very interesting. That a function can't be made continuous at a point isn't enough to prevent me from defining it at that point anyway. The argument "x^y cannot be made continuous at (0, 0) so it shouldn't be defined at (0, 0)" is incomplete...

@Micromass: Actually, I don't agree. If we already have 0^0 = 1 in the ring of integers, it's very natural to say 0^0 = 1 in the ring of rationals. This can be extended to the ring of reals naturally as well. Take the reals as the usual completion of the rationals, and take the real number 0...

I'm sorry if I was unclear. I'm aware of that argument (which boils down to the fact that 0^0 is an indeterminate form) but I don't accept it. Coming at it from another direction, the binomial theorem holds in general in unital commutative rings and (if 0^0 is defined) it implies 0^0 = 1. Why...

0^0 (that is, if ^ is the exponentiation function, ^(0, 0)) is sometimes undefined. The only argument I've seen for this is that 0^0 is an indeterminate form, though I don't accept that as an argument against defining ^(0, 0). It seems that 0^0 could be interpreted as an indeterminate form or as...

Thank you all for your input. Sorry for the delay in getting back to this thread. Some day I'll have to start a study of set theory, since it's clear to me that there are far more technicalities in the ideas discussed in this thread than can reasonably be included in it. But, my original...

The only relevant result I'm aware of is the http://en.wikipedia.org/wiki/Lindemann%E2%80%93Weierstrass_theorem" (actually, Lindemann's half is sufficient) which says that e^a where a != 0 is any algebraic number is transcendental. [Briefly, an algebraic number is a root of some polynomial...