similar to: solving x in a polynomial function

Displaying 20 results from an estimated 2000 matches similar to: "solving x in a polynomial function"

2008 Oct 15
4
a really simple question on polynomial multiplication
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
2009 Jan 11
4
How to get solution of following polynomial?
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
2002 Oct 09
5
polynomial
Any better (more efficient, built-in) ideas for computing coef[1]+coef[2]*x+coef[3]*x^2+ ... than polynom <- function(coef,x) { n <- length(coef) sum(coef*apply(matrix(c(rep(x,n),seq(0,n-1)),ncol=2),1,function(z)z[1]^z[2])) } ? Ben -- 318 Carr Hall bolker at zoo.ufl.edu Zoology Department, University of Florida http://www.zoo.ufl.edu/bolker
2010 Jan 05
4
solving cubic/quartic equations non-iteratively
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
2005 Aug 08
2
coefficient of polynomial expansion
Hi, I would like to get the coefficient of polynomial expansion. For example, (1+ x)^2 = 1 + 2x + x^2, and the coefficients are 1, 2 and 1. (1 + x + x^2)^3 = 1 + 3*x + 6*x^2 + 7*x^3 + 6*x^4 + 3*x^5 + x^6, and the coefficients are 1, 3, 6, 7, 6, 3, and 1. I know that we can use polynom library. Is there any other way to do it without loading a library. Thanks a lot for your help. Peter
2004 Dec 03
3
Computing the minimal polynomial or, at least, its degree
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
2010 Oct 14
1
Fw: Problem to create a matrix polynomial
Awaiting some suggestion. Was my question not very understandable? Please let me know how can I offer more elaborate clarification. Additionally, I would like to solve the determinant of "p1" for the values of "z" (I am working with some multivariate time series modelling). When I use det() function, it am getting error that, that function is not for objects with class
2006 Nov 13
1
wishlist: xlim in lines.polynomial (PR#9362)
Full_Name: Tamas K Papp Version: 2.4.0 OS: linux Submission from: (NULL) (140.180.166.160) I was using the lines.polynomial method for plotting piecewise polynomials (parts of splines). I needed a feature to limit the range of plotting using a parameter given to the function (as opposed to par("usr")). I think that the following changes would be a nice addition: lines.polynomial
2012 Apr 18
3
Is there a way to find all roots of a polynomial equation in R?
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]]
2006 Oct 09
1
Coefficients of a factorized polynomial
Hi all, starting from a vector "v[1:n]" I would like to compute the coefficients of the polynomial (1+x^v[1])*(1+x^v[2])*...*(1+x^v[n]). The following code works but is extremely slow for a large "n" due to, I believe, the polynomial being factorized. I wanted to try the package "polynom" command "unclass" but I could not figure out how to input a
2010 Oct 05
2
Using as.polynomial() over a matrix
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
2009 May 14
1
automated polynomial regression
Dear all - We perform some measurements with a machine that needs to be recalibrated. The best calibration we get with polynomial regression. The data might look like follows: > true_y <- c(1:50)*.8 > # the real values > m_y <- c((1:21)*1.1, 21.1, 22.2, 23.3 ,c(25:50)*.9)/0.3-5.2 > # the measured data > x <- c(1:50) > # and the x-axes > > # Now I do the following:
2004 Aug 09
1
returns the value of a polynomial of degree n evaluated at x.
> Background: > OS: Linux Mandrake 9.1 > release: R 1.9.0 > editor: Xemacs 21.4 > frontend: ESS 5.1.23 > --------------------------------- > > Colleagues > Is there a function in R that is similar to polyval in matlab? (y = polyval(p,x) returns the value of a polynomial of degree n evaluated at x. The input argument p is a vector of length n+1 whose elements are the
2010 Jan 08
0
solving cubic/quartic equations non-iteratively -- comparisons
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
2001 Jul 09
1
polynomial regression and poly
When doing polynomial regression I believe it is a good idea to use the poly function to generate orthogonal polynomials. When doing this in Splus there is a handy function (transform.poly I think) to convert the coefficients produced by regression with the poly function back to the original scale. Has somebody written something similar for R ? Robert
2006 Feb 22
1
Gram-Charlier series
Good day everyone, I want to use the Gram-Charlier series expansion to model some data. To do that, I need functions to: 1) Calculate 'n' moments from given data 2) Transform 'n' moments to 'n' central moments, or 3) Transform 'n' moments to 'n' cumulants 4) Calculate a number of Hermite polynomials Are there R-functions to do any of the above?
2007 Nov 06
1
How to find the zero (only the real solution) with the package polynom ?
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
2004 Jun 25
4
Bug in parse(text = <long polynom>) (PR#7022)
Merci beaucoup, Jean, for the bug report -- which I'm no "completeing" to R-bugs >>>>> "Jean" == Jean Coursol <coursol@cristal.math.u-psud.fr> >>>>> on Thu, 24 Jun 2004 15:22:37 +0200 (CEST) writes: Jean> I was exploring the polynom library with students: <and found a segmentation fault from parsing a long expression>
2005 Aug 19
1
Using lm coefficients in polyroot()
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
2001 Jul 16
1
polyroot() (PR#751)
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