Timeline for Iterated sums--something like a differsum
Current License: CC BY-SA 3.0
11 events
when toggle format | what | by | license | comment | |
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Nov 26, 2016 at 6:02 | history | edited | user78249 | CC BY-SA 3.0 |
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Nov 26, 2016 at 5:17 | vote | accept | CommunityBot | ||
Nov 26, 2016 at 5:16 | comment | added | user78249 | @JulianRosen I feel like such a tool right now. I can't believe I missed that there was a $k$ in the upper argument. I can't even understand how I missed that. Everything is as clear as day now. | |
Nov 26, 2016 at 0:49 | answer | added | Pat Devlin | timeline score: 0 | |
Nov 25, 2016 at 23:18 | comment | added | Julian Rosen | Should it be $1\times\ldots\text{($k$ times)}\ldots\times 1 = {n+k-1\choose k-1}$? | |
Nov 25, 2016 at 20:50 | comment | added | Pat Devlin | And this even fails for positive integers $s$ and $q$. Right? | |
Nov 25, 2016 at 20:45 | comment | added | Pat Devlin | This fails when $n=0,$ right? | |
Nov 25, 2016 at 19:58 | comment | added | user78249 | @PietroMajer I'm well aware of that, I guess I didn't make myself clear enough. I was more interested in the properties of the iterated sum, and then the complex iterated sum. If this construction exists in literature somewhere. | |
Nov 25, 2016 at 19:56 | comment | added | Pietro Majer | This is the discrete convolution, en.wikipedia.org/wiki/Convolution#Discrete_convolution In the particular case of sequences $u:\mathbb{Z}\to\mathbb{C}$ supported on $\mathbb{Z}_+$, it is also called "one-side convolution", and corresponds to the Cauchy product of power series. | |
Nov 25, 2016 at 19:42 | history | edited | user78249 | CC BY-SA 3.0 |
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Nov 25, 2016 at 19:37 | history | asked | user78249 | CC BY-SA 3.0 |