Timeline for Transformation converting power series to Bernoulli polynomial series
Current License: CC BY-SA 4.0
19 events
| when toggle format | what | by | license | comment | |
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| Sep 6 at 19:32 | comment | added | Tom Copeland | Eqn. 37 on pg. 43 of "A universal formula for representing Lie algebra generators as formal power series with coefficients in the Weyl algebra" by Durov, Meljanac, Samsarov, and Skoda is a statement of the umbral compositional inverse relation between the Bernoulli polynomials $B_n(x)$ and the umbral inverse Bernoulli polynomials $\hat{B}_n(x)$. | |
| Sep 3 at 17:32 | history | edited | Tom Copeland | CC BY-SA 4.0 |
Added example of inversion for a polylogarithm
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| Sep 3 at 4:42 | history | edited | Tom Copeland | CC BY-SA 4.0 |
Mislabeled index corrected and Mellin transform rep presented
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| Sep 2 at 23:33 | history | edited | Tom Copeland | CC BY-SA 4.0 |
Gave more robust rep of Bernoulli transformation op
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| Jan 13, 2021 at 18:30 | comment | added | Tom Copeland | Also “Exact Euler–Maclaurin formulas for simple lattice polytopes” by Karshon, Sternberg, and Weitsman. | |
| Jan 2, 2021 at 8:45 | vote | accept | Anixx | ||
| Aug 20, 2018 at 18:56 | history | edited | Tom Copeland | CC BY-SA 4.0 |
Added a link on relation to physics
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| Jun 20, 2017 at 22:38 | comment | added | Tom Copeland | See also the Todd op in Computing the Continuous Discretely by Beck and Robins. | |
| Jun 20, 2017 at 20:17 | history | edited | Tom Copeland | CC BY-SA 3.0 |
Introduced standard name fo the operator and ref
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| Apr 19, 2017 at 16:16 | comment | added | Anixx | Is it possible you also answer this question? mathoverflow.net/questions/267648/… | |
| Apr 9, 2017 at 21:11 | comment | added | Tom Copeland | @Anixx, not sure, but another way to rep the Bernoulli series for a function $f(x)$ is as $f(R) 1$ where $R $ is the raising op for the Bernoulli series. See my website for a description of diff reps for $R$. In particular cases, this might allow you to simplify the results. | |
| Apr 8, 2017 at 21:55 | comment | added | Anixx | @Tom Copeland is there expression for this operator not using formal power series? I mean, more traditional form? | |
| Apr 8, 2017 at 21:28 | comment | added | Tom Copeland | @Anixx, Roman in The Umbral Calculus calls such an operator a transfer operator. (I don't use Mathematica, but it just amounts to a simple substitution.) | |
| Apr 8, 2017 at 10:35 | comment | added | Anixx | Can I write it somehow in Mathematica? | |
| Apr 8, 2017 at 10:34 | comment | added | Anixx | Is there a name for such operator anywhere? | |
| Apr 8, 2017 at 10:05 | comment | added | Anixx | In this notation, what would be the same but applied to the integral of a function? Also, I wonder what would be non-operator expression for this transform. | |
| Apr 2, 2017 at 18:02 | comment | added | Tom Copeland | The inverse transformation is simply $(e^D-1)/D=e^{R.(0)D}$ with $R_n(0)= 1/(n+1)$ and associated Appell sequence $R_n(x)=(R.(0)+x)^n$; therefore, $R_n(B.(x))=x^n=B_n(R.(x))$, and the two Appell sequences form an umbral compositional inverse pair. | |
| Apr 1, 2017 at 1:40 | comment | added | T. Amdeberhan | This is also Umbral Calculus. | |
| Apr 1, 2017 at 0:47 | history | answered | Tom Copeland | CC BY-SA 3.0 |