similar to: How to find a root for a polynomial between [-inf, -3]?

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

Hello, I have 3 columns : a, b and a*b I would like to find the pair (a,b) so that a*b is the minimum but not from the points I measured but from the fit of the curve (I have more points that the ones given below but I fit only on this part because I know that the minimum a*b is in this interval). I thought doing it this way : - to fit a*b=f(a) abfit<-lm(ab ~ poly(a,8,raw=T)) - to use the

? Mon, 6 Nov 2023 17:53:49 +0100 Troels Ring <tring at gvdnet.dk> ?????: > Hence I wonder if I could somehow have non linear regression to find > the 3 pK values. Below is HEPESFUNC which delivers charge in the > fluid for known pKs, HEPTOT and SID. Is it possible to have > root-finding in the formula with nls? Sure. Just reformulate the problem in terms of a function that

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

Thanks a lot! This was amazing. I'm not sure I see how the conditiion pK1 < pK2 < pK3 is enforced? - it comes from the derivation via generalized Henderson-Hasselbalch but perhaps it is not really necessary. Anyway, the use of Vectorize did the trick! Best wishes Troels Den 06-11-2023 kl. 19:19 skrev Ivan Krylov: > ? Mon, 6 Nov 2023 17:53:49 +0100 > Troels Ring <tring at

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

Dear list, Did anybody have something like multiroot(), similar to uniroot() only for several varibles/function? I know the difficulty several varibles bring. But for nice functions, it can be useful. We are thinking of writing one....I thought to ask first. Mai Zhou -.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.- r-help mailing list -- Read

Hi all, This is probably a blindingly obvious question: Why does it matter in the uniroot function whether the f() values at the end points that you supply are of the same sign? For example: f <- function(x,y) {y-x^2+1} #this gives a warning uniroot(f,interval=c(-5,5),y=0) Error in uniroot(f, interval=c(-5, 5), y = 0) : f() values at end points not of opposite sign #this doesn't give a

Dear friends - occurring in Windows R2.6.2 I am modeling physical chemistry in collaboration with a friend who has preferred working in Excel. I used uniroot, and find a solution to a two buffer problem in acid-base chemistry which I believe is physiologically sensible. Using "goal seek" in Excel my friend found another plausible root, quite close to zero, and a plot of the function

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]]

A recent message on ansari.test() prompted me to play with the examples. This doesn't work for me in R version 2.4.1 R> ansari.test(rnorm(100), rnorm(100, 0, 2), conf.int = TRUE) Error in uniroot(ab, srange, tol = 1e-04, zq = qnorm(alpha/2, lower = FALSE)) : object "ab" not found It looks like there's a small typo in ccia() inside ansari.test.default() in which

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 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

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

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

Dear friends - I have a function for the charge in a fluid (water) buffered with HEPES and otherwise only containing Na and Cl so that [Na] - [Cl] = SID (strong ion difference) goes from -1 mM to 1 mM. With known SID and total HEPES concentration I can calculate accurately the pH if I know 3 pK values for HEPES by finding the single root with uniroot Now, the problem is that there is some

I am trying to understand what is the difference between linear and polynomial kernel: linear: u'*v polynomial: (gamma*u'*v + coef0)^degree It would seem that polynomial kernel with gamma = 1; coef0 = 0 and degree = 1 should be identical to linear kernel, however it gives me significantly different results for very simple data set, with linear kernel

Hi everybody! I'm looking some way to do in R a polynomial fit, say like polyfit function of Octave/MATLAB. For who don't know, c = polyfit(x,y,m) finds the coefficients of a polynomial p(x) of degree m that fits the data, p(x[i]) to y[i], in a least squares sense. The result c is a vector of length m+1 containing the polynomial coefficients in descending powers: p(x) = c[1]*x^n +

Hi, I would like to know whether there exist algorithms to compute the coefficients or, at least, the degree of the minimal polynomial of a square matrix A (over the field of complex numbers)? I don't know whether this would require symbolic computation. If not, has any of the algorithms been implemented in R? Thanks very much, Ravi. P.S. Just for the sake of completeness, a

Hello All First - a warning. I'm not very R or programming savvy. I am trying to do something without much luck, and have scoured help-pages, but nothing has come up. Here it is: I have a matrix (m) of approx 40,000 rows and 3 columns, filled with numbers. I would like to convert the contents of this matrix into another matrix (m_p), where the numbers of (m) have been coerced into a