Timeline for Building representation of an arbitrary umbral calculus
Current License: CC BY-SA 4.0
15 events
| when toggle format | what | by | license | comment | |
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| Feb 13 at 22:30 | comment | added | Anixx | @SamHopkins I think, it is not productive to edit the question after each comment or partial answer. | |
| Feb 13 at 22:27 | comment | added | Sam Hopkins | Can you refine your question? If it is "does every sequence $a_n$ arise in this way" the answer is clearly no as explained in the comments (there are many restrictions). But: can we specifically get the Bernoulli numbers seems more interesting. So focus on that question. | |
| S Feb 13 at 7:23 | history | suggested | CommunityBot | CC BY-SA 4.0 |
format and other copyediting
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| Feb 13 at 7:07 | review | Suggested edits | |||
| S Feb 13 at 7:23 | |||||
| Feb 13 at 4:57 | history | edited | Sam Hopkins | CC BY-SA 4.0 |
edited body
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| Feb 13 at 0:22 | comment | added | Anixx | @paulgarrett I just tried to include motivation. Also, the umbra with moments $\frac1{n+1}$ corresponts to the operator $\Delta D^{-1}$ while Bernoulli umbra corresponds to $D \Delta^{-1}$, so it is anti-Bernoulli umbra in a sense. | |
| Feb 13 at 0:19 | comment | added | paul garrett | There would be less overhead (possibly irrelevant) to the question if it were more direct: "given a sequence $a_n$, is there a function $f$ on the unit interval such that $\int_0^1f^n =a_n$?" ? | |
| Feb 13 at 0:16 | comment | added | Anixx | @TomCopeland tried function $(2 x-1)^2-1$ as an example, got sequence $\frac{(-4)^n \Gamma (n+1)^2}{\Gamma (2 n+2)}$ with moments $\left\{-\frac{2}{3},\frac{8}{15},-\frac{16}{35},\frac{128}{315},-\frac{256}{693},...\right\}$ | |
| Feb 13 at 0:11 | comment | added | Tom Copeland | I scanned very quickly two papers that would likely have shown such a function for the Bernoullis, but did not find one. | |
| Feb 13 at 0:04 | comment | added | Tom Copeland | My bad, you're right. $\int_{0}^x f(t) dt = b_0 x + b_1 x^2/2! + ...$ if $f(x) = b_0 + b_1 x+b_2x^2/2! + \cdots$. | |
| Feb 13 at 0:04 | comment | added | Christian Remling | There are many restrictions, for example $a_1^2\le a_2$, by Cauchy-Schwarz. | |
| Feb 13 at 0:02 | comment | added | Anixx | @TomCopeland but I agree that a sequence with zeros at even positions cannot be represented this way. | |
| Feb 12 at 23:58 | comment | added | Anixx | @TomCopeland hmm, no. $a_1=1$ does not determine $f(x)=1$. | |
| Feb 12 at 23:54 | comment | added | Tom Copeland | Consider the simplest cast $a_0 =1$, $a_1 = 1$, and $a_k =0$ otherwise. $a_1 =1$ determines $f(x) = 1$, but then this determines $a_k=1$ for $k >1$. | |
| Feb 12 at 23:34 | history | asked | Anixx | CC BY-SA 4.0 |