Re: [R] Infinite Series

but you get a different hide with sum vs. cumsum. David is right if you want the sum of n terms. > sum(1/(1:100)^2) - pi^2/6 [1] -0.009950167 > sum(1/(1:1000)^2) - pi^2/6 [1] -0.0009995002 etc. > On Jul 24, 2015, at 7:24 PM, Janh Anni wrote: > > Wow! So many (simpler) ways to skin a cat. Th

Re: [R] Infinite Series

Wow! So many (simpler) ways to skin a cat. Thanks! On Fri, Jul 24, 2015 at 8:07 PM, David Winsemius wrote: > > On Jul 24, 2015, at 4:37 PM, Janh Anni wrote: > > > Hello Jeff, > > > > Thanks a lot. I tried it and see that it prints out the entire 100 > partial > > sums, so I can take the last v

Re: [R] Infinite Series

Thanks Bert! On Fri, Jul 24, 2015 at 7:57 PM, Bert Gunter wrote: > Janh: > > It sounds like you really need to go through an R tutorial or two > before posting further, as this is a pretty basic query. Or am I wrong > about this? > > An answer: Just use indexing > > cumsum(1/seq_len(100)^2)[seq(

Re: [R] Infinite Series

cumsum(1/(1:100)^2)[100] > On Jul 24, 2015, at 4:37 PM, Janh Anni wrote: > > Hello Jeff, > > Thanks a lot. I tried it and see that it prints out the entire 100 partial > sums, so I can take the last value as the partial sum for the first 100 > terms. Would there be any way cumsum can print on

Re: [R] Infinite Series

On Jul 24, 2015, at 4:37 PM, Janh Anni wrote: > Hello Jeff, > > Thanks a lot. I tried it and see that it prints out the entire 100 partial > sums, so I can take the last value as the partial sum for the first 100 > terms. Would there be any way cumsum can print only the nth partial sum, > i.e.

Re: [R] Infinite Series

Janh: It sounds like you really need to go through an R tutorial or two before posting further, as this is a pretty basic query. Or am I wrong about this? An answer: Just use indexing cumsum(1/seq_len(100)^2)[seq(10, to = 100,by = 10)] ## keeps every 10th [1] 1.549768 1.596163 1.612150 1.62024

Re: [R] Infinite Series

Hello Jeff, Thanks a lot. I tried it and see that it prints out the entire 100 partial sums, so I can take the last value as the partial sum for the first 100 terms. Would there be any way cumsum can print only the nth partial sum, i.e. the last value in the array, instead of printing the entire

Re: [R] Infinite Series

Please reply-all so the mailing list stays in the loop. cumsum(1/(1:100)^2) gives you the partial sums up through i=100. --- Jeff NewmillerThe . . Go Live... DCN:Basics: ##.#

Re: [R] Infinite Series

?cumsum --- Jeff NewmillerThe . . Go Live... DCN:Basics: ##.#. ##.#. Live Go... Live: OO#.. Dead: OO#.. Playing Research Enginee

[R] Infinite Series

Dear All, Does anyone know of any R functions that compute partial sums of series? Thanks in advance! Janh [[alternative HTML version deleted]] __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/

[R] Infinite series

Is there any routine/trick in R to compute a convergence series? __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal