Timeline for Series expansion with criteria on the coefficients
Current License: CC BY-SA 4.0
13 events
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| Apr 22, 2021 at 0:03 | history | bumped | CommunityBot | This question has answers that may be good or bad; the system has marked it active so that they can be reviewed. | |
| Mar 23, 2021 at 23:54 | comment | added | Daniel Lichtblau | Also: I suspect this is an X/Y problem. It seems like a combinatorial counting problem is the actual thing of interest? If so, some background on that could be useful. | |
| Mar 23, 2021 at 14:13 | comment | added | Daniel Lichtblau | You can remove the numerator and just ask for the constant (with respect to the x_j) term. Now consider the expansion the 1/(1-term) in the usual way as 1+term+term^2... There is a product of these. You get a constant by selecting the 1 from each factor. Suppose there is another way, using some of the nontrivial terms. Then you can use their squares and also get a constant term. Same with cubes, etc. So if I am seeing this correctly, you have an infinite sum of such terms. | |
| Mar 22, 2021 at 22:14 | answer | added | yarchik | timeline score: 1 | |
| Mar 22, 2021 at 22:11 | comment | added | CA Trevillian | Alessandro, you did not add any useable Mathematica code of the new example...can you, please, do this? | |
| Mar 22, 2021 at 22:03 | comment | added | yarchik | No, you do not need to loop over all of them. You can set $x_i=\xi^{10i}$. This is just a possibility. You do not provide the more complicated example in MA form, I cannot test. | |
| Mar 22, 2021 at 21:52 | comment | added | Alessandro Mininno | As you can see, if I replace the example I gave without the numerator and asking for Coefficient 0 for all the x's, first of all I need to loop over all the x's, so I add a loop, and then it's very RAM consuming the first time I do it. I hope I'm explaining myself. | |
| Mar 22, 2021 at 21:50 | comment | added | Alessandro Mininno | Yes, exactly, I understand, but they must be free from all the x's and this is slow and RAM consuming. I was asking if there was a way to select the coefficients that are free from x while expanding, so that I don't need to drop the ones that contains x's only afterwards. | |
| Mar 22, 2021 at 21:31 | comment | added | yarchik | But that's simple, if you set numerator to 1, you need to pick up terms in the series expansion that are free from $x$. | |
| Mar 22, 2021 at 21:27 | comment | added | Alessandro Mininno | I added a very complicated example that I cannot do on my laptop but I'm doing very slowly computing it on a cluster. | |
| Mar 22, 2021 at 21:26 | history | edited | Alessandro Mininno | CC BY-SA 4.0 |
added 1032 characters in body
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| Mar 22, 2021 at 19:28 | comment | added | CA Trevillian | Can you, please, provide a more complicated example that you’d want to deal with? | |
| Mar 22, 2021 at 15:48 | history | asked | Alessandro Mininno | CC BY-SA 4.0 |