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Complex analysis, holomorphic functions, automorphic group actions and forms, pseudoconvexity, complex geometry, analytic spaces, analytic sheaves.
21
votes
Integral of $\log|e^{it}-1|$
Here's another elementary proof, again assuming one has checked that
the integral converges absolutely despite the singularity at $t=0$.
Let $I$ be the average of the $2\pi$-periodic function
$\log \l …
16
votes
Accepted
A statement on complex polynomials
The conjecture is easily seen to be true for $n<3$.
We give a counterexample for $n=3$.
Let $p_i = z^2 - \omega_i$ where the $\omega_i$ are the cube roots of unity.
These are linearly dependent becaus …
8
votes
Accepted
Modular form not meromorphic at $\infty$
[expanding on my comment to convert it to an answer]
An example is $e^j \varphi$ where $j$
is the $j$-invariant and $\varphi$ is any nonzero form of weight $k$.
In general a holomorphic (or meromorphi …
4
votes
Accepted
A generalization of polynomial algebra on a Riemann surface
Counterexample to the first question ("is $A$ an algebra of functions?"):
Let $M$ be a vertical strip such as {$x + iy : 0 < x < 1$},
and define $f_1,f_2$ as the restriction to $M$ of the
entire func …
4
votes
Accepted
Sums of entire surjective functions
One expects there to be no such $a_n$ in general, because the
"typical" entire functions is surjective (those that aren't are of the
special form $z \mapsto c + \exp g(z)$). An explicit example is
$ …
15
votes
Accepted
Conformal mappings that preserve angles and areas but not perimeters?
No.
Either condition implies $|\,f'(z)| = 1$ for all $z$ in the domain of $\,f$,
which in turn implies that $\,f'(z)$ is locally constant
(for instance using the open mapping theorem) and thus that
$ …
2
votes
Zeroes of trigonometric-like function
Robert Israel's calculations suggest that if $0<a<b$ then
$f(z)$ has infinitely many zeros near but not on the imaginary axis,
at least on certain rational rays $b=ra$ such as Robert's
example with $ …
3
votes
What is $\sum (x+\mathbb{Z})^{-2}$?
For the record, one can prove the product formula for the sine without complex analysis (and without the Gamma function), from which (as David Speyer noted) one can recover $\sum_{i \in \bf Z} (x+i)^{ …
65
votes
Absolute value inequality for complex numbers
In general, once you've proven an inequality like this in ${\bf R}$
it holds automatically in any Euclidean space (including ${\bf C}$)
by averaging over projections. ("Inequality like this" = inequa …
11
votes
Accepted
Abscissa of convergence of Dirichlet series
That's not true even for power series with real coefficients.
Let
$$
D(s) = \sqrt{1-2^{-s}}
= 1 - \frac12 2^{-s} - \frac18 4^{-s} - \frac1{16} 8^{-s} - \frac5{128} 16^{-s}
- \cdots .
$$
Then $P(s) = …
4
votes
construct a power series with infinitely many zeros in the complex plane, bounded coefficien...
It's been noted already in the comments that the problem is still
too easy as stated, because one can easily find functions
$f(z) = \sum_{n=0}^\infty c_n z^n$ such as $f(z) = \sin(1 / (r-z))$,
with in …
4
votes
on completeness of R_mn, the set of all rational functions of type (m,n)
At the risk that I'm doing somebody's homework...
Yes, $R_{m,n}[a,b]$ is complete with respect to the uniform metric
$\left\| \cdot \right\|$.
Let $f \in {\cal C}([a,b])$ be a uniform limit of ratio …
1
vote
The integral inequality
No, such an inequality need not hold: one can construct $f$ of
exponential type and a sequence $\{a_n\}$ of real numbers such that
$$
\frac1{f(a_n)} \int_{a_n - \frac12}^{a_n - \frac12}
\left|\phant …
1
vote
Complex Zeroes of Stirling functions of the second kind
Alexandre Eremenko already showed that for each $n>1$ the function
$$
S_n(x) := \frac1{n!} \sum_{k=1}^n {n \choose k} (-1)^{n-k} k^x
$$
has infinitely many zeros $x \in {\bf C}$.
One can still say mor …
24
votes
Rational functions with a common iterate
Over ${\bf C}$, An easy counterexample to question 3 is
$f(x) = x^2$, $g(x) = cx^2$ where $c$ is a nontrivial cube root of unity.
Then $f(f(x)) = g(g(x)) = x^4$ but $f$ and $g$ do not commute.
There a …