Timeline for Why does Mathematica not recognize a convergent series?
Current License: CC BY-SA 4.0
31 events
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| Dec 4, 2023 at 8:15 | comment | added | drer |
@ВалерийЗаподовников It was already clarified in comments that the function Sum with an Infinity as limit of summation and no other additional options means computing series.
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| Dec 4, 2023 at 5:59 | history | edited | drer | CC BY-SA 4.0 |
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| Dec 4, 2023 at 5:21 | comment | added | Валерий Заподовников | @drer Series is not an infinite sum, because there is no such thing as infinite summation. The operation defined in school MUST BE FINITE. By definition, series is a sequence of partial sums of another sequence. That way limit is the "sum" of "infinite summation" operation. Infinite summation is sometimes talking about regularization, but there are many types of that, not only the standard Series way. | |
| Dec 4, 2023 at 5:06 | history | edited | bmf | CC BY-SA 4.0 |
edited the bugs header
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| Dec 4, 2023 at 5:04 | comment | added | Daniel Lichtblau | It has something to do with Floor being in the input. Beyond that I don’t really know the specifics. The change was not arbitrary though; it came about in the process of fixing a bug. | |
| Dec 3, 2023 at 7:27 | comment | added | drer |
@DanielLichtblau Could you tell more about the bug? Is it indeed caused by the factor (-1)^Floor [k/2]?
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| Dec 1, 2023 at 23:07 | comment | added | Daniel Lichtblau | To clarify (or confound) nomenclature, a power series is an infinite sum. The result from Mathematica's `Series function is not. | |
| Dec 1, 2023 at 23:05 | comment | added | Daniel Lichtblau | Reported as a bug. This appears to have regressed beginning with version 12.1. | |
| Dec 1, 2023 at 23:04 | history | edited | Daniel Lichtblau |
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| Dec 1, 2023 at 21:57 | history | edited | drer | CC BY-SA 4.0 |
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| Dec 1, 2023 at 20:17 | history | became hot network question | |||
| Dec 1, 2023 at 17:44 | answer | added | Bob Hanlon | timeline score: 3 | |
| Dec 1, 2023 at 16:54 | comment | added | user64494 | @drer: Sorry, I find "infinite sum" only in the historic remarks to the article in Encyclopedia of Mathematics. You words are only wishful thinking . | |
| Dec 1, 2023 at 16:22 | comment | added | drer | @user64494 From your reference I conclude that another word for series is infinite sum and I am quite sure that a lot of mathematicians use both terms interchangably. | |
| Dec 1, 2023 at 16:03 | comment | added | user64494 | @drer: Wiki sometimes is too lightweight in contrast to other encyclopedias. Also see Encyclopedia of Mathematics. | |
| Dec 1, 2023 at 15:46 | comment | added | drer | @user64494 en.wikipedia.org/wiki/Series_(mathematics) | |
| Dec 1, 2023 at 15:38 | answer | added | chyanog | timeline score: 6 | |
| Dec 1, 2023 at 15:30 | comment | added | user64494 | @drer: What I mean is stated, for example, here. | |
| Dec 1, 2023 at 14:25 | comment | added | drer | @user64494 Do you mean there is a special command for computing series? | |
| Dec 1, 2023 at 14:15 | answer | added | Roman | timeline score: 7 | |
| Dec 1, 2023 at 14:12 | answer | added | bmf | timeline score: 8 | |
| Dec 1, 2023 at 13:52 | answer | added | user64494 | timeline score: 5 | |
| Dec 1, 2023 at 13:48 | comment | added | user64494 | Series is not an infinite sum. | |
| Dec 1, 2023 at 13:43 | history | edited | user64494 | CC BY-SA 4.0 |
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| Dec 1, 2023 at 12:49 | comment | added | drer | @Syed I assume you mean: mathematician = developer of Mathematca. :) | |
| Dec 1, 2023 at 12:46 | comment | added | Syed | You will have to wait for a knowledgeable answer from a mathematician. Things change (for the better mostly) between versions. | |
| Dec 1, 2023 at 12:44 | comment | added | drer | @Syed Thank you very much! I remember I successfully computed the same sum several years ago without any additional options. What has changed? | |
| Dec 1, 2023 at 12:42 | comment | added | drer |
@Nasser The Sum[(-1)^k/(k + 1), {k, 0, Infinity} is computed fast and correct.
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| Dec 1, 2023 at 12:37 | comment | added | Nasser |
I think because it is oscillating, it does not converge. Here is illustration: data = Table[{k, (-1)^(Floor[k/2])/(k + 1)}, {k, 0, 300}]; data = Table[{data[[k, 1]], Total[data[[1 ;; k]][[All, 2]]]}, {k, 1, Length@data}]; p = ListLinePlot[data]; Show[p, Plot[Pi/4 + Log[2]/2, {x, 0, Length[data]}, PlotStyle -> Red], PlotRange -> All] screen shot !Mathematica graphics so on average it does converge, but because it never settles due to +-, it is considered not to converge?
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| Dec 1, 2023 at 12:35 | comment | added | Syed |
Sum[(-1)^(Floor[k/2])/(k + 1), {k, 0, Infinity}, Regularization -> "Abel"]
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| Dec 1, 2023 at 12:16 | history | asked | drer | CC BY-SA 4.0 |
