Recent content by Diophantus

Thanks I'll take a look.

I am a complete novice of philosophy but my imagination was recently captured by the notion of existentialism - in particular the notion that humans have complete responsibility for attaching meaning to their lives. I'd like to read more and was told that Sartre might be a good place to start...

Ok to simplify the situation a little, it has been claimed to me that \sum_{\gamma > 0}\frac{1}{\gamma^2} is convergent, where, as usual, the gamma are just the imaginary parts of the Riemann zerosin the critical strip. I think this is false based on the following heuristic arguament...

I've just been reading through my number theory lecture notes and I noticed this line of reasoning: ***Assuming the Rieman hypothesis*** \sum_{\gamma > 0}\sin(\gamma \log x) \frac{-(1/2)^2}{\gamma((1/2)^2 + \gamma^2)} = O(1) by comparisson with \sum_{n \ge 1}\frac{1}{n^2}. Let...

I haven't seen that one but I'll look it up in the library if you recommend it. The two I have been using are Miles Reid's 'Undergraduate Algebraic Geometry', and Cox, Little and O'Shea's 'Ideals, Varieties and Algorithms'. The former isn't that great but is usefull in that it follows my course...

I was just wondering what the more fundamental definition of a conic in complex projective 2 space is. Is it better to say that it is a curve such that the dehomogenisation of its defining equation is a represents a conic in R^2; OR simply a curve defined by a homogeneous degree two polynomial...

I think we may have gone off on a bit of a tangent. Thank you for trying to expain Gib Z but I think a few of us do not understand mathwonk's motives for his method and hence don't understand your motives for limiting the number of solutions. Not a clue what's going on there. I myself have only...

Try n = \frac{2\pi i}{\log 2} \not = 0.

I don't understand why you said when it is clearly wrong. And what's all this talk about it not being a function? I have only ever alluded to 2^n being 1 whose imaginary part, if I am not mistaken, lies between + and - pi.

Oh I see now, the primary branch of the complex log function. I forgot about that. Still, do we have poles here or not?

I'm sorry you've lost me now. Is it or is it not true that the function 2^(1-z) has poles at z=1 + \frac{2\pi in}{log(2)}? What do you mean by primary branch?

What about z=1 + \frac{2\pi in}{log(2)} for any integer n?

Hmmm, it's actually 1/log2 but thanks anyway. I'm really intrigued to know whether there is a reasonable way of calculating the Laurant series though. Surely there must be a bit of trickery that will work.

OK so I've opted for the Laurant Series route. Where am I going wrong here: \frac{1}{1-2^{1-z}} = \displaystyle\sum_{n=0}^{\infty}(2^{1-z})^n = \displaystyle\sum_{n=0}^{\infty}(e^{(1-z)log2})^n = \displaystyle\sum_{n=0}^{\infty}(\displaystyle\sum_{m=0}^{\infty}\frac{(1-z)^m(log2)^m}{m!})^n...

How exactly would one go about proving that \frac{1}{1-2^{1-z}} has a simple pole at z=1? I've tried writing 2^{1-z} in terms of e to get a Taylor series for the denominator but can't quite figure out where to go from there.