Timeline for Integrality of complex infinite series
Current License: CC BY-SA 3.0
33 events
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Nov 24, 2016 at 13:03 | history | edited | T. Amdeberhan | CC BY-SA 3.0 |
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Nov 24, 2016 at 7:42 | comment | added | Anthony Quas | I have edited the post to include the information that you withheld in the original posting. I believe this fixes the mistake that you have acknowledged making (I would have expected you to do this yourself). | |
Nov 24, 2016 at 7:41 | history | edited | Anthony Quas | CC BY-SA 3.0 |
Revealed crucial information that was deliberately withheld in the original formulation of the question
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Nov 24, 2016 at 7:34 | history | edited | Anthony Quas | CC BY-SA 3.0 |
Revealed crucial information that was deliberately withheld in the original formulation of the question
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Nov 24, 2016 at 3:07 | review | Close votes | |||
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Nov 24, 2016 at 2:53 | history | edited | T. Amdeberhan | CC BY-SA 3.0 |
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Nov 24, 2016 at 2:37 | comment | added | Anthony Quas | It is disingenuous to say that no proof has been furnished yet | |
Nov 24, 2016 at 0:43 | comment | added | T. Amdeberhan | Please take a look at Christian's "answer" below for the level of existing proofs. | |
Nov 24, 2016 at 0:31 | comment | added | fedja | OK, "the most elementary" was a bad choice of words. It should be "the least non-elementary". I just do not want to waste my and your time by posting something that is more complicated or uses more advanced tools than an argument you know already. | |
Nov 24, 2016 at 0:26 | comment | added | T. Amdeberhan | @fedja: I do not have any suggestion for the elementary proof. Really. | |
Nov 24, 2016 at 0:20 | comment | added | fedja | Then you'd better give us the initial approximation: what is the most elementary proof you know? (there is no need to present full details, a general outline will suffice) | |
Nov 24, 2016 at 0:14 | comment | added | T. Amdeberhan | We should try an "elementary as possible", leaving the gauging up to you. | |
Nov 23, 2016 at 23:49 | comment | added | fedja | How "elementary" do you want it to be? You can get away with basic theory of Fourier series and Lebesgue integration, but I'm not so sure about "elementary complex analysis" alone... | |
Nov 23, 2016 at 21:49 | history | edited | T. Amdeberhan | CC BY-SA 3.0 |
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Nov 23, 2016 at 4:31 | history | edited | T. Amdeberhan | CC BY-SA 3.0 |
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Nov 23, 2016 at 0:43 | answer | added | Christian Remling | timeline score: 1 | |
Nov 22, 2016 at 23:56 | comment | added | T. Amdeberhan | @AnthonyQuas: Path from $f=1$ to $f=z$? Uhmm ... beside existence of such "paths", we might need homotopy invariance of a sort. BTW, let's hope to keep the proof (if true) to basic complex analysis. | |
Nov 22, 2016 at 23:22 | comment | added | T. Amdeberhan | The smooth functions are dense (even polynomials), so we're not short of approximating functions. As Christian pointed out, the obvious approximants may not do the job. Yet, is it possible to choose a "good" set of estimating functions maintaining the "integrality" property? It's a fair question, though not sure if it easy to achieve this. | |
Nov 22, 2016 at 23:06 | comment | added | user78249 | @T.Amdeberhan Aww yes, how silly of me. I don't know why I went from constant modulus to constant function. | |
Nov 22, 2016 at 23:04 | comment | added | Anthony Quas | @ChristianRemling: It's clear that you can do an approximation as you suggest. The difference between the approximation and the original function is small in the Sobolev sense. But is it, for example, uniformly small? If it is, you can make it satisfy $|f|=1$ by a real scaling that is close to 1. | |
Nov 22, 2016 at 22:59 | comment | added | Christian Remling | @AnthonyQuas: They can certainly be approximated just fine, for example by truncated sums $\sum_{|n|\le N} a_n z^n$, or by $\sum a_nr^{|n|}z^n$, but the problem seems to be that the approximating functions won't satisfy $|f|=1$, so it's not obvious (to me) how to pass to the limit in your argument. | |
Nov 22, 2016 at 22:50 | comment | added | Anthony Quas | We agreed that we're talking about functions satisfying $|f|=1$. If the function is differentiable, the result holds. Question: what are continuity properties of functions in your Sobolev space? Can these functions be approximated in the space by differentiable functions? If so, maybe there is a simple proof by approximation? | |
Nov 22, 2016 at 22:34 | answer | added | Pat Devlin | timeline score: 0 | |
Nov 22, 2016 at 21:20 | comment | added | T. Amdeberhan | @james.nixon: As per your notation, let $f(\theta)=\sum_{\mathbb{Z}}a_ne^{in\theta}$ then condition (2) can be encapsulated as saying $\int_{\mathbb{S}^1}(\vert f(\theta)\vert^2-1)e^{ik\theta}d\theta=0$ for all $k$. This implies $\vert f(\theta)\vert=1$ almost everywhere. | |
Nov 22, 2016 at 21:10 | comment | added | T. Amdeberhan | @james.nixon: your argument suggests $g=1$ or $\vert f\vert=1$ (see Anthony's comment above, this is okay). But, $\vert f\vert=1$ does not imply $f=1$ for complex functions. Take $f(\theta)=e^{i\theta}$. | |
Nov 22, 2016 at 20:14 | comment | added | user78249 | If $f(x) = \sum_{n=-\infty}^{\infty }a_n e^{2\pi i n x}$ then your second condition is troubling me. Essentially it reads $\int_0^1 f(x)\overline{f(x)}e^{-2\pi i k x} \,dx = \delta_0(k)$. But then if $g(x) = |f(x)|^2$ it satisfies $\int_0^1 g(x)e^{2\pi i k x}\,dx = 0$ for all $k \neq 0$. I'm no pro at this but doesn't this imply $g$ and therefore $f$ is constant? particularly $f = 1$ by your normalization conditions? | |
Nov 22, 2016 at 20:09 | comment | added | Anthony Quas | So this is quite an interesting question. Your conditions basically say that $f$ satisfies $|f|=1$ and the quantity you are asking about is $(1/2\pi)\int f'(z)\bar f(z)\,dz$, which at least for differentiable functions is the number of times that the argument of $f(z)$ goes around the circle. I'm not familiar enough with Sobolev spaces to see quickly if this extends to $W^{1/2,2}$. In particular, is there a continuous (in the Sobolev space) path of functions from $f(z)=1$ to $f(z)=z$ with range in the circle? | |
Nov 22, 2016 at 19:16 | history | edited | T. Amdeberhan | CC BY-SA 3.0 |
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Nov 22, 2016 at 19:15 | comment | added | T. Amdeberhan | You're right. I fixed that. Thank you. | |
Nov 22, 2016 at 19:15 | history | edited | T. Amdeberhan | CC BY-SA 3.0 |
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Nov 22, 2016 at 19:13 | comment | added | Joe Silverman | When you say "$a_n$ is a sequence of infinite series", do you mean that each $a_n$ is an infinite series? Or did you just mean to say that $(a_n)_{n\in\mathbb Z}$ is a (double-sided) sequence? | |
Nov 22, 2016 at 18:59 | history | edited | T. Amdeberhan | CC BY-SA 3.0 |
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Nov 22, 2016 at 18:53 | history | asked | T. Amdeberhan | CC BY-SA 3.0 |