similar to: exponent confusion

Why does the expression "(-8)^(1/3)" return NaN, instead of -2? This is not answered by http://cran.r-project.org/doc/FAQ/R-FAQ.html#Why-are-powers-of-negative-numbers-wrong_003f Thanks, Dave [[alternative HTML version deleted]]

Dear all, Has anyone worked on coding for calculating Lyapunov Exponent for a time series data? or any package is available for computing Lyapunov? Please advice and many thanks in advance. Catherine X Wang

Dear All, I am a novice when it comes to time-series analysis and at the moment I am actually interested in calculating the Hurst exponent of a time series. This question has already been asked quite some time ago http://bit.ly/98dZsi and I trust some progress has been made ever since. I was able to find some functions in the packages http://cran.r-project.org/web/packages/Rwave/index.html

Hello, I'm using fArma package to estimate the value of Hurst exponent using R/S method. However, for a certain set of data I get H ~ 1.8. How do I interpret this? Following are the output that I get for this set: > mean(data[,2]) [1] 400.5433 > sd(data[,2]) [1] 1139.786 > > rsFit(data[,2], levels = 64) Title: Hurst Exponent from R/S Method Call: rsFit(x = data[, 2], levels

Hello, Does anybody know if there is somewhere in R a function to calculate the Lyapunov exponent in a time series? Thanks, Philippe Grosjean .......................................................<??}))><.... ) ) ) ) ) ( ( ( ( ( Prof. Philippe Grosjean \ ___ ) \/ECO\ ( Numerical Ecology of Aquatic Systems /\___/ ) Mons-Hainaut University, Pentagone / ___ /( 8, Av. du

Dear R experts, I really have difficulty when I try to deal with this question. suppose X is a square symmetric matrix. The exponent of X is defined by the matrix limit as following: exp(X) = lim (I + X/n)^n, note: the limit is from n to infinite. How can I write R function for the above? Thank you very much -- View this message in context:

Hello, I'm using the fArma package to estimate Hurst exponent by R/S method. I've some measurements in the following format: Call_number Call_duration I'm using the following steps. Am new to R, so it would help if someone could please confirm if my steps are correct. Further, this method seems to give a value H ~= 0.8 (> 0.5). Is the use of R/S method to estimate H OK? >

Dear R users, The function lmacfPlot( ) in fBasics, returns a list including the Hurst Exponent. I get sometimes a value for Hurst Exponent using lmacfPlot bigger than 1 which is theoretically incorrect. What could be reason ? From the related documents, I could not find how the Hurst Exponent in lmacfPlot is estimated. Could you please say how it is estimated here? Thank you so

I have the following bug report to submit. OpenSSH 2.5.1p1 and 2.5.1p2 HP-UX 11.00 Dart 51 64bit (32bit compile) OpenSSL 0.9.6 Zlib 1.1.3 Cflags: -Ae I keep getting "rsa_public_encrypt() exponent too small or not odd" with the SSH 1 or 1.5 protocols. I can't get OpenSSH to communicate with itself with any protocal other than SSH 2. I have searched everywhere. Google, OpenBSD,

I'm trying to make a function that will output the exponent... so f2(2,2) = 4 and f2(2,3)=8. But I don't want to just use the x^n function, I want to do it another way, and without a recursion. I did the follow code but for some reason it doesn't work. Help please? f2 <- function(x, n) #without recursion { y <- 1 if (n==0) {return(1)} else { if (length(y) < (n+1) { y

I am attempting to deploy OpenSSH. The trouble is I keep getting the rsa_public_encrypt() exponent too small or not odd with the SSH 1 or 1.5 protocols. I can't get OpenSSH to communicate with itself with any protocal other than SSH 2. Platform notes: HP-UX 11.00 Dart 51 64bit OpenSSL 0.9.6 Zlib 1.1.3 Cflags: -Ae I have tried with and without optimizations. I noticed that this problem

Has anyone writen R code for estimating Hurst exponent with R/S method or other methods? or any other source of R code available? Many thanks Catherine Wang

Hello, I'm an absolute beginner with R. I'm hoping to do some time-series analysis on my data. The data looks like #time value 18 153 20 426 70 7 83 130 84 7 and so on where time could be in seconds or hours or days (not all at the same time). How could I import such a file to R and do some simple stuff (say plot the values)? As per the tutorials on time series, I could use the ts()

Dear R Users, I am using code from the following links to do Hurst exponent estimation. link: http://r-forge.r-project.org/plugins/scmsvn/viewcvs.php?rev=1&root=rmetrics&view=rev file: LongRangeDependence.R Take a look at the following run: > x<-(cos((1:200)/10)) > rsFit(diff(x,15))@hurst$H [1] 1.027420 > aggvarFit(diff(x,15))@hurst$H [1] 0.02301331 First of all, can anyone

Hi all, I want to label an axis with exponents, but can't get it done with expression. Any hints would be very welcome! # simulated data, somewhat similarly distributed to my real data: set.seed(12); d <- rbeta(1e6, 0.2,2)*150 ; d <- d[d>1e-8] hist( d? , breaks=100) # now on a logarithmically scaled axis: hist(log10(d), breaks=100, xaxt="n") abline(v=

Dear all, I would like to (i) produce boxplot graphs with axis in logarithm in base 10 and (ii) showing the values on the axis in 10^exponent format rather than 10E+exponent. To illustrate with an example, I have some widely spread data that I chart plot using boxplot() [figure on the left]; the log="y" option of boxplot() I obtained the natural logarithm conversion of the data and

plot(1,2, ylab= paste("insects", expression(m^2), sep=" ")) I get insects m^2 I would like m to the 2 what is the problem? -- Let's not spend our time and resources thinking about things that are so little or so large that all they really do for us is puff us up and make us feel like gods. We are mammals, and have not exhausted the annoying little problems of being

Hello, I have compiled OpenSSL-0.9.7d - the lastest version and when OpenSSH-3.7.1p2 is compiled with this ssl library [0.9.7d], I am getting the following error when SSH-1 connection is done. I am using HP-UX IPF box and I am doing 32 bit compilation only. Even I have changed the optimization level for OpenSSL and no use. Any clue why this problem is occuring? Advance thanks, Kumaresh

I agree there is no reason they _need_ to be the same precedence, but I think SPECIALS are already have the proper precedence for both unary and binary calls. Namely higher than all the binary operators (except for `:`), but lower than the other unary operators. Even if we gave unary specials their own precedence I think it would end up in the same place. `%l%` <- function(x) tail(x, n =

R has long supported user defined binary (infix) functions, defined with `%fun%`. A one line change [1] to R's grammar allows users to define unary (prefix) functions in the same manner. `%chr%` <- function(x) as.character(x) `%identical%` <- function(x, y) identical(x, y) %chr% 100 #> [1] "100" %chr% 100 %identical% "100" #> [1] TRUE