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16h
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1d
revised Classification of conformally equivalent annuli via periods
added 86 characters in body
1d
answered Classification of conformally equivalent annuli via periods
1d
asked Classification of conformally equivalent annuli via periods
2d
comment Proof of Weierstrass Preparation Theorem
I'm not sure if it will answer this question, but the book by Daniel Huybrechts on Complex Geometry covers much of the same material as the early chapters of G&H. It might be a helpful reference.
2d
answered Index of essential singularity
May
17
comment If $f:[0,1]\to\mathbf{R}$ is continuous, and that $\int_0^1 f(x)x^k \, dx = 0$ for all integers $k>2004$, show that $f = 0$.
This won't work. The function $f$ my not be representable as a Taylor series.
May
15
comment problems in the book A Course in Functional Analysis - John B. Conway
@letran.dn95 The $n$th coordinate is the value at $n$.
May
14
answered If $A$ is a square matrix and $Ax = b$ has a unique solution for some $b$, is $A$ necessarily invertible?
May
14
comment problems in the book A Course in Functional Analysis - John B. Conway
@letran.dn95 It's the norm defined in question 2 (integral of the absolute value of a function).
May
14
comment problems in the book A Course in Functional Analysis - John B. Conway
Does whoever downvoted care to comment? Is there an error?
May
14
answered problems in the book A Course in Functional Analysis - John B. Conway
May
13
comment Integral relations in Fricke and Klein
@glebovg That's just the fundamental theorem of calculus, basically. (Stoke's theorem implies the integral of an exact form over a closed path is zero.)
May
13
comment Integral relations in Fricke and Klein
@glebovg I get a reasonable translation of the parts I tried. Presumably a German-English math dictionary would help with the trouble spots.
May
10
comment Integral relations in Fricke and Klein
A quick and dirty way to get an answer: type the text into Google translate.
May
10
answered Prove that the complement of an open ball in $\mathbb{R^n}$ has exactly one unbounded component
May
9
comment To Find Eigenvalues
@mathers101 To say that $p(x)$ is annihilating polynomial just means that $p(A)=0$. Here, we have $A^2-I=0$.
May
9
comment $Im(A+B) \subset ImA + ImB$
@mathcraze Actually, that seems less clear to me. In any case, the asker had no trouble understanding what I meant.
May
9
comment $Im(A+B) \subset ImA + ImB$
@mathcraze It's correct as written. The variables $x$ and $y$ are implicitly quantified over all elements of the vector space (separately).