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Warn moderators| Author : | Topic: Topic: Basic series | Bottom |
| saucer admin Posts : 673 A Good Tautology is Hard to Find!  |
- http://tutorial.math.lamar.edu/AllBrowsers/2414/Series_Basics.asp citing. ...a sequence is a list of numbers while a series is a single number, provided it makes sense to even compute the series. Students will often confuse the two and try to use facts pertaining to one on the other. However, since they are different beasts this just won’t work. There will be problems where we are using both sequences and series, but we’ll always have to remember that they are different. - |
| ferme Posts : 85 |
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| zee Posts : 115 |
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| 360pan gold Posts : 14  |
series http://thesaurus.maths.org/mmkb/entry.html?action=entryById&id=1045 Lagrange inversion formula From: Robin Chapman <rjc@maths.ex.ac.uk> Subject: Re: Power Series With Binomials Date: Tue, 07 Sep 1999 09:14:21 GMT Newsgroups: sci.math Keywords: Application of Lagrange inversion formula In article <erh4lvu9oix3@forum.swarthmore.edu>, qqquet@hotbot.com (Leroy Quet) wrote: > I know that sum_{m=0}^infinity[binomial(2m,m)x^m]= > 1/sqrt(1-4x). > But,in general, what is sum_{m=0}^infinity[binomial(rm,m)x^m]? > (r is a positive integer.) Consider the equation x = f(x) - f(x)^r. By the Lagrange inversion formula this has a unique power series solution f(x) with f(0) = 0 and this is f(x) = sum_{k=0}^infinity binomial(rk,k) x^{1+k(r-1)}/[1+k(r-1)]. Hence f'(x) = sum_{k=0}^infinity binomial(rk,k) x^{k(r-1)}. Now f'(x)[1 - rf(x)^{r-1}] = 1 so f'(x) = 1/[1 - rf(x)^{r-1}]. Should we think the effort worthwhile we could get a degree r equation for f' from that for f. 360 --Last edited by 360pan gold on 2007-05-22 16:44:32 -- |
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