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 Tumbleweed
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Aug
13
comment Show $\int_{0}^{1} \frac{\ln x}{1-x}dx$=$\sum_{1}^{\infty}\frac{1}{n^2}$ and converges
And what about $\lim_{x \rightarrow 1} \frac{\log(x)}{1-x}$?
Aug
13
comment Show $\int_{0}^{1} \frac{\ln x}{1-x}dx$=$\sum_{1}^{\infty}\frac{1}{n^2}$ and converges
But how unbounded? $\frac{\log(x)}{1-x} \sim \log(x)$ for $x \sim 0$. And we know $\int_0^\epsilon \log(x) dx$ to be convergent...
Aug
3
comment $\mathbb{C}$-points on a $\mathbb{Z}$-scheme
Regarding the morphism $\mathbb{F}_p \rightarrow \mathbb{Z}$ I have only been sloppy and I didn't write $\mathrm{Spec}$... ...but for my claim that $\mathbb{C}\otimes_{\mathbb{Z}}\mathbb{F}_p \simeq \mathbb{C}\backslash p\mathbb{Z}$ I don't have such an easy excuse! Thank you very much again!
Aug
3
accepted $\mathbb{C}$-points on a $\mathbb{Z}$-scheme
Aug
3
comment $\mathbb{C}$-points on a $\mathbb{Z}$-scheme
Still, if I understand correctly then the following should be true. Let $p: \mathbb{F}_p \rightarrow \mathbb{Z}$ be a point in $\mathrm{Spec}(\mathbb{Z})$. Base changing gives the scheme $\mathrm{Spec}(\mathbb{C} \otimes_{\mathbb{Z}} \mathbb{F}_p)$ defined over $\mathbb{C}$, but $\mathbb{C} \otimes_{\mathbb{Z}} \mathbb{F}_p \simeq \mathbb{C} / p \mathbb{Z}$ admits no ring-morphism to $\mathbb{C}$, thus this scheme has no complex points. If this is correct then I probably grasped a bit what´s going on here. Thanks again and $+1$!
Aug
3
comment $\mathbb{C}$-points on a $\mathbb{Z}$-scheme
Thanks! In general it is not clear to me why the lack of $\mathbb{C}$-point over $\mathcal{X}$ implies the lack of a $\mathbb{C}$-point over $\mathcal{X} \otimes \mathbb{C}$. I mention this because I thought that base changing to $\mathbb{C}$ could have changed the situation of the existence or not of a "complexification" of $P$, but you seem not to consider it important.
Aug
3
asked $\mathbb{C}$-points on a $\mathbb{Z}$-scheme
Jul
7
awarded  Tumbleweed
Jun
30
asked Is the structure constant additive on connected components?
Jun
16
comment What is the order of a cusp form at a cusp?
Thanks! I've been a bit confused by the general sloppiness that seems to permeate this issue...
Jun
16
asked What is the order of a cusp form at a cusp?
Apr
27
comment When are heat kernels only dependent on the distance?
Thanks for your answer! You address a point I'm eager to explore, but unfortunately I do not fully understand your answer. Specifically, I do not see the statement "for a same amount of time, at least for small times, the heat from q2 in the flat part of M will be more spread out than the heat from q1 in the thin part". This seems to be in contradiction with Varadhan's large deviation formula (on a complete manifold): $$\lim_{t \rightarrow 0} -4t \log(K(t;x,y)) = d(x,y)^2.$$ In order to mark your answer as accepted, could I ask you to elaborate a but more please? Thanks again and +1!
Apr
24
revised When are heat kernels only dependent on the distance?
edited title
Apr
23
revised When are heat kernels only dependent on the distance?
added 1 character in body
Apr
23
asked When are heat kernels only dependent on the distance?
Apr
20
awarded  Yearling
Apr
2
answered If $a,b,c$ are integers such that $4a^3+2b^3+c^3=6abc$, is $a=b=c=0$?
Apr
2
awarded  Popular Question
Mar
27
comment Closed form for $\sum_{k=1}^\infty \frac{k+1}{q^{k(k+n)}}$
@nayrb Thanks, actually I tried to complete the squares, isolating then the $k^2$-factor, but the question I linked doesn't seem to provide any closed form for the series I'm interested into. Or am I missing something?
Mar
27
revised Closed form for $\sum_{k=1}^\infty \frac{k+1}{q^{k(k+n)}}$
deleted 6 characters in body