similar to: solving cubic/quartic equations non-iteratively

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

The programming language mosml comes with foldr that 'accumulates' a function f over a list [x1,x2,...,xn] with initial value b as follows foldr f b [x1,x2,...,xn] = f(x1,...,f(xn-1,f(xn,b))...) Observe that "list" should have same elements so in R terminology it would perhaps be appropriate to say that the accumulation takes place over a 'vector'. I wonder if R

Hello Everyone This samba server was working perfectly without problems. Running as an Domain member vs Win2K ADS One day it stopped working. All that happened 5 days ago was a change of the administrator/root password We adjusted the wbinfo -set-auth-user towards the new password. But nothing have worked since. install:/ # wbinfo -V Version 3.0.13-1.1-SUSE What might be wrong

Hi there, I would like to solve the following equation in R to estimate 'a'. I have the amp, d, x and y. amp*y^2 = 2*a*(1-a)*(-a*d+(1-a)*x)^2 test data: amp = 0.2370 y= 0.0233 d= 0.002 x= 0.091 Can anyone suggest how I can set this up? Thanks, Diviya [[alternative HTML version deleted]]

Hi All R-helpers This is my first (but probartly not last ;-) mail to R-help, so hello to everybody. My problem: Is there a way to give colors to the labels (sample labels) in plots for a hclust object for better visualization? I have looked through plot, points, hclust and more but cannot find anything on label color. Anybody know if this is doable? Best regards Jeppe

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

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

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

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

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

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

R-users E-mail: r-help@r-project.org My understanding is that package "mgcv" is based on "Generalized Additive Models: An Introduction with R (by Simon N. Wood)". On the page 126 of this book, eq(3.4) looks a quartic equation with respect to "x", not a cubic equation. I am wondering if all routines which uses cubic splines in mgcv are based on this quartic

re: Cubic splines in package "mgcv" I don't have access to Gu (2002) but clearly the function R(x,z) defined on p126 of Simon Wood's book is piecewise quartic, not piecewise cubic. Like Kunio Takezawa (below) I was puzzled by the word "cubic" on p126. As Simon Wood writes, this basis is not actually used by mgcv when specifying bs="cr". Maybe the point is

Upon fresh installation of R-patched and rimage on openSUSE 11.3 box, simple fftw on renown lena image cause memory free failure. Quick: Go to the end of this mail and read the error message from R Thoruogh: Find detailed step-by-step how sopprt library fftw2 was installed and how rimage was installed. Regards Mads Jeppe VERSIONS ======== openSUSE -------- # cat /etc/SuSE-release

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

Hello, I want to solve some two dimensional equation system with R. Some systems are not solvable analytically. Here is an example: (I) 1/n*sum{from_i=1_to_n}(Xi) = ln lambda + digamma(c) (II) mean(X) = x / lambda I want to find lambda and c, which R-function could do that task? Carsten [[alternative HTML version deleted]]

Hello all, In the course of trying to implement the CDF of an InverseGammaDistribution, I have run across the need for an igamma() function. Several others have needed this function but the answers I have found so far are not totally clear to me. I'm writing for three reasons: 1) to present a small error in the docs 2) to clarify the approach we are expected to take 3) to request,for the

Hello Everyone This samba server was working perfectly without problems. Running as an Domain member vs Win2K ADS One day it stopped working? All that happened 5 days ago was a change of the administrator/root password We adjusted the wbinfo ?set-auth-user towards the new password. But nothing have worked since. install:/ # wbinfo -V Version 3.0.13-1.1-SUSE What might be wrong when the following

Hi, all, Does anybody got the most general expression of a unitary matrix? I found one in the book, four entries of the matrix are: (cos\theta) exp(j\alpha); -(sin\theta)exp(j(\alpha-\Omega)); (sin\theta)exp(j(\beta+\Omega)); (cos\theta) exp(j\beta); where "j" is for complex. However, since for any two unitary matrices, their product should also be a unitary matrix. When I

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 .> .>