Timeline for Are all discrete-analytic funtions as defined here also natural?
Current License: CC BY-SA 3.0
19 events
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| Nov 7, 2013 at 4:38 | vote | accept | Anixx | ||
| Nov 7, 2013 at 4:37 | comment | added | Aaron Meyerowitz | You are right, now I see it. As I said above, the value at $x$ is the limit on $N$ of $(-1)^N\binom{x-N}{N}=\prod_{j=1}^N(1-\frac{x}{j})$ so $x=0$ is easy, for any $x \gt 0$ the terms are positive and decrease fast enough to have limiting product $0$, and for any $x\lt 0$ the terms are positive and decrease to $1$ slowly enough that the product diverges to $\infty$ like $\ln{N}$. | |
| Nov 7, 2013 at 4:31 | history | edited | Anixx | CC BY-SA 3.0 |
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| Nov 7, 2013 at 4:23 | comment | added | Anixx | @Aaron Meyerowitz now I think it does not converge at -1<x<0 | |
| Nov 7, 2013 at 3:53 | comment | added | Anixx | @Aaron Meyerowitz if f(x) indeed 1 in x=0 and 0 in x>0, then it just cannot converge for -1<x<0. So this fact is important. Either it converges for -1<x<0 and then it is non-zero for positive non-integer x or it does not converge and comes to infinity, in that case it approaches the Г-shaped limit. | |
| Nov 7, 2013 at 3:46 | comment | added | Aaron Meyerowitz | Yes I said it. Now I say that we should ignore that fact. It would be fair game if we could shift to $f(x-1)$ but it diverges at $-1.$ So start at $x=0.$ | |
| Nov 7, 2013 at 2:52 | history | edited | Anixx | CC BY-SA 3.0 |
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| Nov 7, 2013 at 2:41 | history | edited | Anixx | CC BY-SA 3.0 |
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| Nov 7, 2013 at 2:29 | comment | added | Anixx | @Aaron Meyerowitz you yourself said it converges for all x>-1. If so, it is non-zero. If it diverges for -1<x<0, then it is indeed zero. | |
| Nov 6, 2013 at 4:18 | comment | added | Aaron Meyerowitz | Your graphic is ( I claim) not the graph of the infinite Newton expansion. That expansion should be ignored for negative $x$ but is $0$ for all positive $x$, integral or not. | |
| Nov 6, 2013 at 3:45 | comment | added | Anixx | @Andy Putman i posted it there after a day since i posted it here because of littele number of answers here. Is it you who downvoted? Just because i crossposted it? | |
| Nov 6, 2013 at 3:39 | review | Close votes | |||
| Nov 6, 2013 at 7:26 | |||||
| Nov 6, 2013 at 3:29 | comment | added | Aaron Meyerowitz | I claim (to my initial amazement) that the true function $g(x)$ is $1$ at $0$ but $0$ for all positive real $x$! The convergence can be slow. You may confirm that the $k$th partial sum ( of the Newton expansion ) for $g(x)$ is $(-1)^k\binom{x-k}{k}$ i.e. $\prod_{j=1}^k(1-\frac{x}{j})$. Not that that leaves me much more confident. | |
| Nov 6, 2013 at 3:20 | comment | added | Andy Putman | Crossposted to math.se : math.stackexchange.com/questions/553795/… | |
| Nov 6, 2013 at 2:09 | history | edited | Anixx | CC BY-SA 3.0 |
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| Nov 5, 2013 at 5:39 | answer | added | Aaron Meyerowitz | timeline score: 1 | |
| Nov 5, 2013 at 4:15 | history | edited | Anixx | CC BY-SA 3.0 |
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| Nov 5, 2013 at 4:08 | comment | added | Anixx | I wonder why the first horizontal line is not visible and the formulas collide with the text. Please feel free to correct it you can. | |
| Nov 5, 2013 at 4:00 | history | asked | Anixx | CC BY-SA 3.0 |