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Complex analysis, holomorphic functions, automorphic group actions and forms, pseudoconvexity, complex geometry, analytic spaces, analytic sheaves.
25
votes
Asymptotic approximation of $x^\alpha$ by entire functions
As a matter of fact, real entire functions (that is, entire functions that map the real line into itself, or equivalently, functions represented by a power series centered in 0, with real coefficient …
16
votes
Accepted
Analytic functions with integer coefficients
We may assume w.l.o.g $0<\beta<1$. Write $\beta^{-1}$ in the $ \beta^{-2}$ expansion as:
$$\beta^{-1}=\sum_{k=0}^\infty d_k\beta^{2k}$$
with integer digits $0\le d_k<\beta^{-2}$, and define
$$f(x):=-1 …
12
votes
Demystifying complex numbers
If they have a suitable background in linear algebra, I would not omit the interpretation of complex numbers in terms of conformal matrices of order 2 (with nonnegative determinant), translating al …
11
votes
Roots of $x^n-x^{n-1}-\cdots-x-1$
Let's prove the equivalent claim:
The only complex solutions to $2z=z^{n+1} +1$ with $|z|\le 1$ are $z={1\over\alpha}$ and $z=1$.
Remark: The number $\beta:={1\over\alpha}$ is the minimum fixed …
11
votes
Why are lacunary series so badly behaved?
"Objection, the question assumes facts not in evidence!"
Talking about the general question as in the title, I wonder in what measure can we say that lacunary series are particularly badly behaved. M …
10
votes
Accepted
Bernoulli number formula involving roots of Taylor polynomial of $\exp-1$
Start by the definition of the Bernoulli numbers via their generating function: we have
$$\big(e^x-1\big)\sum_{k=0}^\infty B _ k\frac{x^k}{k!}=x\, .$$
Therefore, writing with your notation $ t _ n(x)= …
9
votes
Is the exponential function the sole solution to these equations?
Here's a proof that also uses Bernstein's representation theorem for totally monotone functions (I wondered if that result was needed, and Alexander Eremenko's answer now makes me think, that it is). …
8
votes
Complete metric on the space of Jordan curves?
I fear this is possibly a bit too abstract for your purposes, but tells that there is such a complete distance topologically equivalent to the uniform distance, and can be made more concrete (check th …
8
votes
Uniqueness of Neumann series
edit: This was meant to recall a first elementary but relevant fact that was not mentioned at all, that is orthogonality.
Bessel functions $\{J_n\}_{n\in\mathbb N_+}$ are orthogonal on $\mathbb R$ w.r …
7
votes
Accepted
complex polynomials and inequalities
For $n\ge 3$, let's take $z_j$, for $1\le j\le n$, be the $n$-th roots of unity, and $w_j:=z_j+4\bar{z_j}$, so that the assumption are satisfied by $$f(z):=z+4z^{n-1}$$ $$g(z):=4z+z^{n-1}.$$ Howev …
7
votes
Intersection of two Jordan curves lying in the rectangle
Alternatively, wlog the rectangle $R$ is $[0,1]\times[0,1]$, and denoting $L_1$, $L_2$ the corresponding parametrization curves, one may also consider the map $$[0,1]\times[0,1]\ni (t,s)\mapsto L_1(t) …
7
votes
Real integral which cannot be evaluated without complex analysis?
I'm quite sure one could camuflage any complex analysis technique into a real calculus computation, leaving no appeal even to the notion of complex numbers. For instance, the Cauchy's formula on a cir …
6
votes
Accepted
If $f(x)+f(2x)$ is quasianalytic, is $f(x)$ necessarily quasianalytic?
Assuming $V:=U=(-\alpha,+\alpha)$, we want the sequence of real numbers
$$B_n:=\bigg( \frac{\| f^{(n)} \|_{\infty,U}} {n!M_n} \bigg) ^{1/n}$$
to be bounded (in which case $B:=\sup_{n\in\mathbb{N}}B_n$ …
6
votes
Complex structures on $R^{2N}$ with complex annulus
Consider, for $N=1$, the case where $M$ is the open unit disk $B_1$: diffeomeorphic to $\mathbb{R}^2$, not biholomorphic to $\mathbb{C}$. The compact $K:=\{0\}$ is actually such that $M\setminus K$ …
6
votes
Fourier expansion of Takagi-function (everywhere non differentiable function).
Yes. The triangle wave $s(x):=\min_{k\in\mathbb{Z}}\big|x-k\big|$ has an absolutely convergent Fourier series
$$s(x)=\frac{1}{4}-\frac{2}{\pi^2}\sum_{k=0}^\infty\frac{1}{(2k+1)^2}\cos\big(2\pi (2k+1 …