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Sep 5 at 22:14 history edited Tom Copeland CC BY-SA 4.0
Removed erroneous statement about a signed, masked Pascal triangle
Jan 16, 2021 at 21:39 history edited Tom Copeland CC BY-SA 4.0
Elaborated on raising op
Jan 16, 2021 at 21:27 comment added Tom Copeland Uncorrect is incorrect, in both senses.
Jan 16, 2021 at 21:26 history edited Tom Copeland CC BY-SA 4.0
Corrected notation, added dJ(x)/dx
Jan 11, 2021 at 8:44 history edited Tom Copeland CC BY-SA 4.0
Added another operator and link to applications of the Todd operator
Sep 17, 2019 at 23:12 comment added reuns It is uncorrect
Sep 17, 2019 at 22:59 comment added Tom Copeland I@reuns I mean exactly what I derive in my blog post that I linked to. The pdf file does present Dirichlet series as well, but they aren't central to the discussion per se. If there is a spot in the analysis you don't follow, please point it out. The Mellin transform and the diff op derivations are pretty straightforwardly based on manipulations of Euler's product formula.
Sep 17, 2019 at 20:13 comment added reuns You meant $\exp(-D_x\log n) \delta(x) = \sum_{k=0}^\infty \frac{(-\log n)^k}{k!} \delta^{(k)}(x)=\delta(x-\log n)$ in the sense of analytic functionals, with a convergent Dirichlet series $F(s)=\sum_{n=1}^\infty a_n n^{-s}$ then $ F(D_x)\delta = \sum_{n=1}^\infty a_n \exp(-D_x\log n)\delta$ converges in the sense of analytic functionals acting on bounded analytic functions and it is $ = \sum_{n=1}^\infty a_n \delta(x-\log n)$ which is the inverse Laplace transform of $F(s)$ in the sense of distributions.
Sep 17, 2019 at 19:42 history edited Tom Copeland CC BY-SA 4.0
Added raising operators
Sep 17, 2019 at 2:51 history asked Tom Copeland CC BY-SA 4.0