Timeline for power series of the reciprocal... does a recursive formula exist for the coefficients [closed]
Current License: CC BY-SA 4.0
27 events
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| Sep 28, 2021 at 10:39 | comment | added | qifeng618 | The best answers are posted at math.stackexchange.com/a/4262414/945479 and math.stackexchange.com/a/4262417/945479. | |
| Jun 29, 2021 at 5:56 | comment | added | Rainb | Wow, this question helped me, how come it's too localized, very very weird!, If it should be closed, change the reason at least! | |
| Dec 17, 2019 at 20:55 | review | Reopen votes | |||
| Dec 18, 2019 at 12:27 | |||||
| S Dec 17, 2019 at 20:03 | history | suggested | IV_ | CC BY-SA 4.0 |
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| Dec 17, 2019 at 18:55 | comment | added | IV_ | This question has answers here: math.stackexchange.com/questions/1264615/… and here: math.stackexchange.com/questions/710252/… and here: mathoverflow.net/questions/53384/… . | |
| Dec 17, 2019 at 18:45 | review | Suggested edits | |||
| S Dec 17, 2019 at 20:03 | |||||
| Jul 6, 2017 at 16:58 | comment | added | fairytale | "This question is unlikely to help any future visitors"? NO, this question helped me a lot. | |
| Feb 1, 2017 at 19:20 | comment | added | Tom Copeland | See oeis.org/A263633, oeis.org/A133314, and oeis.org/A049019 | |
| May 6, 2016 at 22:21 | comment | added | Tom Copeland | Related to mathoverflow.net/questions/238186/… | |
| Jul 31, 2014 at 1:08 | review | Reopen votes | |||
| Jul 31, 2014 at 19:14 | |||||
| S Jan 27, 2011 at 1:22 | vote | accept | AUK1939 | ||
| Jan 27, 2011 at 1:12 | comment | added | Pietro Majer | Actually I don't see the need of closing this question, for these reasons: (1) even a simple question may be of interest to other professional mathematicians not in that very field; and (2) sometimes simple or naive questions here gave rise to wonderful answer by our best users. That said, I would recommend aukm not to feel offended, and to avoid quarreling --for some reasons it's considerd umpolite. | |
| Jan 27, 2011 at 1:05 | vote | accept | AUK1939 | ||
| S Jan 27, 2011 at 1:22 | |||||
| Jan 27, 2011 at 0:12 | comment | added | AUK1939 | And Willie Wong!! the reference you pointed me to simply suggests you didnt even read or understand the question ... so you foolishly closing the question has annoyed because I was having a good discussion about the solution with other members. | |
| Jan 27, 2011 at 0:11 | comment | added | AUK1939 | Yuan, how can you close this question on the basis that its a staightforward application of the Faa di Bruno equation ... since 1) this equation is not known by many (read the third line of this paper, romanpress.com/MathArticles/FaaDiBruno.pdf). and 2) Even if the Faa di Bruno equation does provide an answer, its equivalence to the answer posted at functions.wolfram.com/GeneralIdentities/7 is far from trivial ... and finally before you stupidly voted to close this question look at the responses below. The problem may not be hard but certainly not trivial!! | |
| Jan 26, 2011 at 22:04 | history | closed |
Willie Wong Gjergji Zaimi Andrés E. Caicedo Qiaochu Yuan Gerry Myerson |
too localized | |
| Jan 26, 2011 at 21:26 | comment | added | Qiaochu Yuan | This is a straightforward application of Faa di Bruno (en.wikipedia.org/wiki/Fa%C3%A0_di_Bruno's_formula). Voting to close. | |
| Jan 26, 2011 at 21:22 | answer | added | Ira Gessel | timeline score: 24 | |
| Jan 26, 2011 at 20:56 | answer | added | Pietro Majer | timeline score: 19 | |
| Jan 26, 2011 at 20:11 | comment | added | AUK1939 | Zen, $d_n+1$ is found recursively by a computer but I am looking for a recursive or non recursive expression for $d_n+1$. A computer wont do this for me. I dont know how to make this any clearer. The formula in the link does answer my question, however, I would like to know how it was derived, who derived it ... basically where it came from. Note that it is a recursive formula. As far as I know a non recursive formula simply just doesn't exist. | |
| Jan 26, 2011 at 19:48 | comment | added | Zen Harper | aukm, I agree: finding an explicit closed form formula for $d_n$ is not easy. However, by the method I outlined above, knowing the values of $d_0, d_1, \ldots, d_n$ (and the $b_j$), you can easily find $d_{n+1}$. Then you could easily programme a computer to calculate as many $d_j$ as you wish. As I understand it, that falls under the category of "recursive" formulae for $d_n$, which you say is allowed. But then you refer to a website listing explicit formulae. So that answers your question, doesn't it? If you want a proof or derivation instead, you should say this in your question! | |
| Jan 26, 2011 at 19:43 | history | edited | AUK1939 | CC BY-SA 2.5 |
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| Jan 26, 2011 at 19:31 | comment | added | AUK1939 | A homework question?? I dont think you read the question properly, or I lack your higher intellect. its one thing to solve a system on equations its quite another to write down a formula for the solution of the nth variable, n being arbitrary. I found the solution at functions.wolfram.com/GeneralIdentities/7 ... I wonder how it was derived... Possibly somebody smart expanded enough terms found a pattern and used induction. That pattern is quite encryptic though. Any refs found would be appreciated. | |
| Jan 26, 2011 at 19:28 | comment | added | Willie Wong | See also: en.wikipedia.org/wiki/Power_series#Multiplication_and_division | |
| Jan 26, 2011 at 19:19 | history | edited | AUK1939 | CC BY-SA 2.5 |
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| Jan 26, 2011 at 18:59 | comment | added | Zen Harper | There's no $a_n$ in your functions, only $b_n$. And this really, really does look like homework, and is not research level. But anyway, before this gets closed: just use $f(x) \cdot \frac{1}{f(x)} = 1 + 0 \cdot x + 0 \cdot x^2 + \ldots$ and the formulae for multiplication of power series. | |
| Jan 26, 2011 at 18:52 | history | asked | AUK1939 | CC BY-SA 2.5 |