search for: polyroot

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 1.7...

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 ABSOL...

I'd just like to report a possible R bug--or rather, confirm an existing one (bug #751). I have had some difficulty using the polyroot() function. For example, in Win 98, R 1.1.1, > polyroot(c(2,1,1)) correctly (per the help index) gives the roots of 1 + (1*x) + (2*x^2) as [1] -0.5+1.322876i -0.5-1.322876i However, > polyroot(c(-100,0,1)) gives the roots of [1] 10+0i -10+0i which corresponds to -100 + x^2, not 1 -...

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 1.7535...

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 abs(tst[-1]-Conj(tst[-2]))/abs(tst[-1]) 1...

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$coefficients). But, if I fit polynomial of higher order and optimize it by stepAIC, I get of course some coefficients removed. Then, if i have model y ~ I(x^2) + I(x^4) i cannot call polyroot in such way, because there is a need to call polyroot(c(0,0,fit$coefficients[1],0,fit$coefficients...

.... 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 comparative accuracy of (1) the R solve functionused in the included post, (2) the R polyroot function, (3) the Matlab roots function, (4) the SAS IML polyroot function. I tried the posted polynomial: -8 + 14*x - 7*x^2 + x^3 = 0 and a repeating-roots example: 8 - 36*x + 54*x^2 - 27*x^3 = 0 I used Mathematica solutions as the reference: (* Posted example *) Roots[-8 + 14 x - 7 x...

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 .> .> z_all <- polyroot(z=c(-0.5,0,0,0,0,-1)) .> z_all [1] 0.7042902+0.5116968i -0.2690149+0.8279428i -0.2690149-0.8279428i [4] 0.7042902-0.511696...

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]...

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 gmail.com

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

...so that the current record has rt = 0 d$rt <- seq(1-nrow(d),0) # apply the regression to fit a 4th degree polynomial in rt polyfit <- lm(Close ~ poly(rt,4),d) # get the coefficients p <- coef(polyfit) # get the roots of the first derivative of the fitted polynomial pr <- polyroot(c(p[2],2*p[3],3*p[4],4*p[5])) # define a function that evaluates the second derivative as a function of x dd <- function(x) { rv = 2*p[3]+6*p[4]*x+12*p[5]*x*x;rv;} # evaluate the second derivative at the ith root, and print the result r <- dd(pr[i]) r } rollRegFun(alpha,1) r...

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

...lo to all I've to calculate an arima model and I need only the first and 365 th parameter and also the sar1 and the intercept, so I'm traing with: arima(X,order=c(365,0,0),seasonal=list(order=c(1,0,0),..),fixed=c(NA,rep(0,363),NA,NA,NA),transform.pars=F) but the error answer is: Error in polyroot(z) : polynomial degree too high (49 max) also there are problems in allocating memory (I've 512 mb ram) may be somebody could help me? can R do it? thanks michele ______________________________________________________________________ http://it.yahoo.com/mail_it/foot/?http://it.mail.yahoo.c...

...ncxreg <- ncxreg + 1 } xm <- if (narma == 0) drop(as.matrix(newxreg) %*% coefs) else drop(as.matrix(newxreg) %*% coefs[-(1:narma)]) } else xm <- 0 if (arma[2] > 0) { ma <- coefs[arma[1] + 1:arma[2]] if (any(Mod(polyroot(c(1, ma))) < 1)) warning("MA part of model is not invertible") } if (arma[4] > 0) { ma <- coefs[sum(arma[1:3]) + 1:arma[4]] if (any(Mod(polyroot(c(1, ma))) < 1)) warning("seasonal MA part of model is not invertible")...

R friends, I need to find the determinant of this matrix x 1 0 0 1 x 1 0 0 1 x 1 0 0 1 x det yields x^4-3x^2+1 I can then use polyroot to find the roots of the coefficients. The question is about the use of "x", which is what I'm solving for. thanks in advance, and this is a back-burner question. Apologies if I have posted this incorrectly/to the wrong place, I'm a newbie to this list... -- Robert Gotwals,...

Is there a way to find all roots of a polynomial equation? Lets say x^5+a*x^4+b*x^3+c*x^2+d*x+e=0 how to find its all roots? [[alternative HTML version deleted]]

...he bug is that the C code (ver 2.4.0) assumes *npsi is the length of the psi vector (which is n+p), whilst the predict.ar function in R passes out as.integer(npsi), where npsi <- n-1. Some R code following reproduces the error. Let p=4, n=6, then > ar<-c(-0.5,0.25,-0.125,0.0625) > Mod(polyroot(c(1,-ar))) # ar model is stationary [1] 1.037580 2.444163 2.444163 2.581298 > #Pass in values as predict.ar does > cumsum(c(1,.C("artoma",as.integer(4),as.double(ar),double(9),as.integer(5))[[3]][1:5]^2)) [1] 1.000000 1.250000 1.312500 1.328125 1.332031 1.332031 > #Pass in value...

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 there, Does anyone know how I solve for x from a given y in a polynomial function? Here's some example code: ##example file a<-1:10 b<-c(1,2,2.5,3,3.5,4,6,7,7.5,8) po.lm<-lm(a~b+I(b^2)+I(b^3)+I(b^4)); summary(po.lm) (please ignore that the model is severely overfit- that's not the point). Let's say I want to solve for the value b where a = 5.5. Any thoughts? I did