similar to: PR#751

Dear R Development Team, I have encountered the following difficulty in using the function polyroot under either NT4.0 (R version 1.2.1) or linux (R version 0.90.1). In the provided example, the non-zero root of c(0,0,0,1) depends on the results of the previous call of polyroot. R : Copyright 2001, The R Development Core Team Version 1.2.1 (2001-01-15) R is free software and comes with

I have found that the polyroot() function in R-1.1.1(both solaris and Win32 version) gives totally incorrect result. Here is the offending code: # Polyroot bug report: # from R-1.1.1 > sort(abs(polyroot(c(1, -2,1,0,0,0,0,0,0,0,0,0,-2,5,-2,0,0,0,0,0,0,0,0,0,1,-2,1)))) [1] 0.8758259 0.9486499 0.9731015 1.5419189 1.7466214 1.7535362 1.7589484 [8] 2.0216317 2.4421509 2.5098488 2.6615572

Is there an easy way to compare complex numbers? Here is a small example: > (z1=polyroot(c(1,-.4,-.45))) [1] 1.111111-0i -2.000000+0i > (z2=polyroot(c(1,1,.25))) [1] -2+0i -2+0i > x=0 > if(any(identical(z1,z2))) x=99 > x [1] 0 # real and imaginary parts: > Re(z1); Im(z1) [1] 1.111111 -2.000000 [1] -8.4968e-21 8.4968e-21 > Re(z2); Im(z2) [1] -2

Hi, All: Is there a simple way to detect complex conjugates in the roots returned by 'polyroot'? The obvious comparison of each root with the complex conjugate of the next sometimes produces roundoff error, and I don't know how to bound its magnitude: (tst <- polyroot(c(1, -.6, .4))) tst[-1]-Conj(tst[-2]) [1] 3.108624e-15+2.22045e-16i

In a bug report from Nov.28 2000, Li Dongfeng writes: ----- I have found that the polyroot() function in R-1.1.1(both solaris and Win32 version) gives totally incorrect result. Here is the offending code: # Polyroot bug report: # from R-1.1.1 > sort(abs(polyroot(c(1,-2,1,0,0,0,0,0,0,0,0,0,-2,5,-2,0,0,0,0,0,0,0,0,0,1,-2)))) [1] 0.8758259 0.9486499 0.9731015 1.5419189 1.7466214 1.7535362

El problema del módulo es que pierde el signo. En tu caso sale igual porque has invertido el signo del coeficiente en el polinomio (en realidad se me pasó a a mí advertir que el término independiente debe ir con signo negativo): .> polyroot(z=c(0.5,0,0,0,0,1)) [1] 0.7042902+0.5116968i -0.2690149+0.8279428i -0.2690149-0.8279428i [4] 0.7042902-0.5116968i -0.8705506+0.0000000i .> .>

Hi, I'm responding to a post about finding roots of a cubic or quartic equation non-iteratively. One obviously could create functions using the explicit algebraic solutions. One post on the subject noted that the square-roots in those solutions also require iteration, and one post claimed iterative solutions are more accurate than the explicit solutions. This post, however, is about

I'm attempting to compile and install R version 1.7.1 for my statistical geneticists. It seems to compile correctly -- that is, it compiles without errors -- but the regression test is failing in the following manner: =-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-= > ## log > stopifnot(all.equal(log(1:10), log(1:10, exp(1)))) > stopifnot(all.equal(log10(30), log(30, 10))) >

Mirando los comentarios, realmente lo que deseo es encontrar la raíz real de (-0.5)^(1/5) la cual debería ser -0.87055056329. José me hace caer en cuenta que además de no encontrar la raiz real, tampoco da todas las raiz complejas. Habría alguna manera de que tuviera en cuenta? > ------------------------------ > > Message: 6 > Date: Thu, 15 Oct 2015 11:25:39 +0200 > From: José

Dear useRs, I need to compute zero of polynomial function fitted by lm. For example if I fit cubic equation by fit=lm(y~x+I(x^2)+i(x^3)) I can do it simply by polyroot(fit$coefficients). But, if I fit polynomial of higher order and optimize it by stepAIC, I get of course some coefficients removed. Then, if i have model y ~ I(x^2) + I(x^4) i cannot call polyroot in such way, because there is

Hi, I want find all roots for the following polynomial : a <- c(-0.07, 0.17); b <- c(1, -4); cc <- matrix(c(0.24, 0.00, -0.08, -0.31), 2); d <- matrix(c(0, 0, -0.13, -0.37), 2); e <- matrix(c(0.2, 0, -0.06, -0.34), 2) A1 <- diag(2) + a %*% t(b) + cc; A2 <- -cc + d; A3 <- -d + e; A4 <- -e fn <- function(z) { y <- diag(2) - A1*z - A2*z^2 - A3*z^3 - A4*z^4

I noticed on the mailing list archives and through Google searches that a couple of months ago there was discussion of maintaining rpms for suse 9.1 on the x86_64 architecture. Are these rpms still planning on being released? It would help me a great deal. I've tried compiling from source using the R-base spec files provided on CRAN. Using that, I am able to produce a running version of

If I have matrics as follows: > a <- c(1,1,0,0) > b <- c(4,4,0,0) > c <- c(3,5,5,6) How can I use R code to solve the equation ax^2+bx+c=0. thanks! yuying shi [[alternative HTML version deleted]]

Dear R people: Is there a way to perform simple polynomial multiplication; that is, something like (x - 3) * (x + 3) = x^2 - 9, please? I looked in poly and polyroot and expression. There used to be a package that had this, maybe? thanks, Erin -- Erin Hodgess Associate Professor Department of Computer and Mathematical Sciences University of Houston - Downtown mailto: erinm.hodgess at

To R-helpers, R offers the polyroot function for solving mentioned equations iteratively. However, Dr Math and Mathworld (and other places) show in detail how to solve mentioned equations non-iteratively. Do implementations for R that are non-iterative and that solve mentioned equations exists? Regards, Mads Jeppe

Here is a relatively simple script (with comments as to the logic interspersed): # Some of these libraries are probably not needed here, but leaving them in place harms nothing: library(tseries) library(xts) library(quantmod) library(fGarch) library(fTrading) library(ggplot2) # Set the working directory, where the data file is located, and read the raw data

[I thought I'd submitted this bug report some time ago, but it's never showed up on the bug tracking system, so I'm submitting again.] qr.solve() gives incorrect results when dealing with complex matrices or with qr objects that have been computed with LAPACK=TRUE, whenever the b argument has more than one column. This bug flows from qr.coef(), which has a similar problem. I believe

No sé si he entendido bien la pregunta, pero creo que lo que quieres obtener es esto: (as.complex(-0.5)^(1/5)) Saludos,Salva > To: r-help-es en r-project.org > From: canadasreche en gmail.com > Date: Thu, 15 Oct 2015 10:45:10 +0200 > Subject: Re: [R-es] potencia fracional de un número negativo > > Hola. > No sé si va por aquí, pero prueba a quitar el paréntesis a (-0.5) >

Hi, with small matrices eigen works as expected: > eigen(cbind(c(1,4),c(4,7)), only.values = TRUE) $values [1] 9 -1 $vectors NULL > eigen(cbind(c(1,4),c(4,7))) $values [1] 9 -1 $vectors [,1] [,2] [1,] 0.4472136 -0.8944272 [2,] 0.8944272 0.4472136 > eigen(cbind(c(1,-1),c(1,-1))) $values [1] -3.25177e-17+1.570092e-16i -3.25177e-17-1.570092e-16i $vectors

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]]