Recent content by xiavatar

Acceptances have been sent out for SMALL already.

Its really hard to say whether or not this schedule is too much to handle because of how greatly the difficulty of a class could vary based on the professor. That being said I noticed that you have taken a foundation of higher mathematics course. I am guessing it introduced you to some basic set...

I am currently looking at grad schools, and I am wondering if anyone knew who are the leading researchers in differential geometry. I know that question is a little vague considering how diverse differential geometry is, but I was hoping that something could direct me in the right direction...

Okay, so we can choose an analytic branch, but your above equality with the logarithm is only correct up to a multiple of $$2\pi i$$. What allows you to assume that the multiple vanishes in this case?

But the problem is that complex logarithms don't preserve the properties were used to with real valued logarithms. In order for them to be holomorphic we have to choose an analytic branch for the logarithm, and once we've done that we can't just rewrite the log as you have written. Or have I...

I've posted a proof on http://math.stackexchange.com/questions/1199285/infinite-product-converges-to-meromorphic-function/1200018#1200018. I'm wondering if anyone can take a look at it and critique it.

I did take logarithms, but the problem is bounding it and applying Weirstrass M-test. So what were reduced to showing is that $$\sum_{n=k}^\infty \text{log}(\frac{n}{z+n}(\frac{n+1}{n})^z)\leq \sum_{n=k}^\infty M_n<\infty$$ on a disk $$D_r$$ centered at zero where $$k>r$$.

How do you show that $$\frac{1}{z}\prod_{n=1}^\infty \frac{n}{z+n}(\frac{n+1}{n})^z$$ is meromorphic? Any hints would be helpful, I'm having trouble bounding the functions and their logarithms. This is exercise XIII.3 problem 15 in Gamelin's Complex Analysis.

I was just stating that if ##[x+y]=[x]+[y] \rightarrow [x+y]-[y]=[x]+[y]-[y]=[x]\rightarrow [x+y]-[y]-[x]=0##. But this isn't important since everything were working with is abelian. I understood the rest of your explanation Terandol. Thanks.

Why do we write it as [x+y]-[x]-[y], shouldn't it be [x+y]-[y]-[x] since your adding inverses to the right? Or are we assuming the fact that the representatives are abelian, so we can write [x+y]=[x]+[y]=[y]+[x].

Are we basically factoring through the commutator of the free group, ##F_{ab}(M)##.

In Lang's book,page 39-40, he factorizes ##F_{ab}(M)## with respect to the subgroup generated by all elements of type ##[x+y]-[x]-[y]##. I don't quite understand why he does this. I know that he is trying to create inverse elements, but I don't see why that factorization necessarily satisfies...

If I create a bijective map between the open balls of two metric spaces, does that automatically imply that this map is a homemorphism?

Use pegging and method of loci. You can also use acronyms and acrostics to strengthen these memories.

Remember that when you integrate a function you have find the constant of integration. He just combined the two steps of finding the constant and integrating into one step, essentially.