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1d
comment Meromorphic functions on the unit disk
Almost certainly not. For starters, this class includes all biholomorphisms from $\mathbb D$ to sufficiently nice connected subsets of $\mathbb D$.
1d
comment Prove that if $n \equiv 7 \pmod 8$, then $n$ cannot be expressed as the sum of three squares.
Hint: Find all the values squares can take mod 8. Can any sum of these three be 7?
Apr
14
awarded  Socratic
Apr
14
answered All the zeroes of analytic $f$ in $A$ are isolated, or $f \equiv 0$ on $A$.
Apr
14
comment All the zeroes of analytic $f$ in $A$ are isolated, or $f \equiv 0$ on $A$.
Use the fact that holomorphic functions have power series expansions.
Apr
13
awarded  Popular Question
Apr
12
awarded  Good Question
Apr
7
comment Proving completeness of $L^p$
@VincentBoelens It's not a definition of completeness, it's a definition of convergence. In a metric space, a sequence $x_n$ converges to $x$ if $d(x,x_n)\rightarrow 0$ as $n\rightarrow \infty$.
Apr
6
comment Proving completeness of $L^p$
Yes. This is just the definition of convergence in $L^p$: $f_n \rightarrow f$ if $\|f-f_n\|\rightarrow 0$.
Apr
2
asked Connections between probability theory and algebraic topology?
Mar
25
reviewed Close How do you solve $f(x)=4$ for $x=2$
Mar
25
reviewed Close How to compute linear recurrence of a sum of binomial-multiplied linear recurrences
Mar
25
reviewed Leave Open 19 points on a hexagon
Mar
25
reviewed Leave Open discrete math-Complexity of algorithms
Mar
25
reviewed Close are these propositions equivalent, case xor and then?
Mar
25
reviewed Leave Open Why is this laplace identity true $\int_{\Bbb{R}^+}\frac{f(t)}{t}\,dt = \int_{\Bbb{R}^+}\mathcal{L}\{f\}$?
Mar
25
reviewed Close Quaternion ^ Quaternion
Mar
23
comment Why is the differentiation of $e^x$ is $e^x$?
I think you need to justifying switching the limit and the differentiation.
Mar
23
comment $H^{n}(M)$ where $M$ is compact, orientable and connected manifold
@cactus Fix some form $\alpha$ that has nonzero integral. We may assume by scaling that $\int_M \alpha =1$. Then if $[\beta]$ is some element of $H^n$ with $\int \beta = c$, $\int c\alpha -\beta =0$, so $\beta-c\alpha$ is exact and $[\beta]=[c\alpha]$. Hence $\alpha$ generates $H^n$ and $H^n\cong\mathbb R$.
Mar
23
comment $H^{n}(M)$ where $M$ is compact, orientable and connected manifold
@cactus Look up the definition of $H^n$ again. Elements of $H^n$ are equivalence classes of forms, not forms.