similar to: Fitting Data with errors to non-polynomial Linear Model

Dear Thomas, Unfortunately, I do not know if any packages implement this functionality. Though, it is a topic that interests me. Unlike the "classic discriminant", I prefer to work with the reduced polynomial. This "discriminant" is generalizable to a superset of Chebysev polynomials (which I called Cardano-polynomials). x^3 - 3*c*x - 2*d = 0 x^5 - 5*c*x^3 + 5*c^2*x - 2*d =

Hello, i'm fairly familiar with R and use it every now and then for math related tasks. I have a simple non polynomial function that i would like to approximate with a polynomial. I already looked into poly, but was unable to understand what to do with it. So my problem is this. I can generate virtually any number of datapoints and would like to find the coeffs a1, a2, ... up to a given

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 +

Hello, I have a question regarding fitting a polynomial to a data set, then constructing a polynom from the coefficients so that I can integrate it. I first use lm to fit the polynomial setting the coefficients to raw=TRUE - this appears to work fine. I plot the model and it is a true representation of the data. I then take the coefficients vector and construct a polynom from the

Dear R helpers     Suppose I have a following data   y  <- c(9.21, 9.51, 9.73, 9.88, 10.12. 10.21)   t  <- c(0, 0.25, 1, 3, 6, 12)   I want to find out the polynomial which fits y in terms of t i.e. y = f(t) some function of t.   e.g.   y = bo + b1*t + (b2 * t^2) + (b3 * t^3) + ...... and so on.   In Excel I have defined y as independent variable, then defined t, t^2 and t^3 and using

I'm trying to find a way to fit a polynomial of degree n in x and y to a set of x, y, and z data that I have and obtain the coefficients for the terms of the fitted polynomial. However, when I try to use the surf.ls function I'm getting odd results. > x <- seq(0, 10, length=50) > y <- x > f <- function (x, y) {x^2 + y} > library(spatial) > test <-

I'm looking for a way to solve a linear equation system where the factors are polynomials: Here is a simplified example (To solve my problem, I have to deal with dimensions larger than 2): ( s + 2) x1 + (s - 3) x2 = 2 ( s^2 + 2s - 1) x1 - 2 x2 = 1 Theoretically the solution is easy: By performing polynomial multiplications, divisions and sums. I found out, that R is able to

Hey all, first time poster here. I'm new to R and working on my first real programming and forecasting asignment. I'm using unemployment data from 1948-2012. I successfully completed part a and the linear fit for part b, but i am really struggling fitting a polynomial with a power greater than 2 to my forecast. I'll upload my R code at the bottom. Any help is very much appreciated!

I wonder how one in R can fit a 3rd degree polynomial to some data? Say the data is: y <- c(15.51, 12.44, 31.5, 21.5, 17.89, 27.09, 15.02, 13.43, 18.18, 11.32) x <- seq(3.75, 6, 0.25) And resulting degrees of polynomial are: 5.8007 -91.6339 472.1726 -774.2584 THanks in advance! -- Jonas Malmros Stockholm University Stockholm, Sweden

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

Hello All, This might seem elementary to everyone, but please bear with me. I've just spent some time fitting poly functions to time series data in R using lm() and predict(). I want to analyze the functions once I've fit them to the various data I'm studying. However, after pulling the first function into Octave (just by plotting the polynomial function using fplot() over

Hi all, I am looking to fit a logistic regression using the lrm function from the Design library. I am interested in this function because I would like to obtain "pseudo-R2" values (see http://tolstoy.newcastle.edu.au/R/help/02b/1011.html). Can anyone help me with the syntax? If I fit the model using the stats library, the code looks like this: model <- glm(x$trait ~ x$PC1 +

igraph is a package for graphs/networks. It has a C core and uses a simple and fast graph representation allowing millions of vertices and edges. LINKS Release notes for the 0.5.2 version: http://igraph.sourceforge.net/relnotes-0.5.2.html Release notes for the 0.5.1 version: http://igraph.sourceforge.net/relnotes-0.5.1.html Complete list of changes: http://igraph.sourceforge.net/news.html The

igraph is a package for graphs/networks. It has a C core and uses a simple and fast graph representation allowing millions of vertices and edges. LINKS Release notes for the 0.5.2 version: http://igraph.sourceforge.net/relnotes-0.5.2.html Release notes for the 0.5.1 version: http://igraph.sourceforge.net/relnotes-0.5.1.html Complete list of changes: http://igraph.sourceforge.net/news.html The

Hi. Is there an analog to abline() that can be used to plot a polynomial fit? For example, I can draw the straight-line fit fit <- lm(y ~ x) via abline(coef=fit$coef) but I'm not sure how to draw the polynomial fit fit <- lm(y ~ poly(x,2)) I do see the function curve(), but not how to prepare an expr for curve() based on the coefficients returned by the polynomial

I was looking for suggestions as to how to smooth a timeseries and, having accomplished that, how to find the fitted curve values for intermediate points. ? I've tried numerous examples of possible approaches in R that I've found on the web, but when applied to my simple data, R returns an error message in numerous cases. ? The main problem seems to be that I have monthly data, starting

I would like to experiment with testing the fit of a loess model against the fit from an ordinary linear regression. The 1988 JASA paper by Cleveland and Devlin *appears* to indicate that this can be done, at least ``approximately''. They, as I read it, advocate the use of an ANOVA type test with degree of freedom chosen to make the ``F ratio'' have an approximate F distribution

Hi, I can not comprehend the linear fitting results of polynoms. For example, given the following data (representing y = x^2): > x <- 1:3 > y <- c(1, 4, 9) performing a linear fit > f <- lm(y ~ poly(x, 2)) gives weird coefficients: > coefficients(f) (Intercept) poly(x, 2)1 poly(x, 2)2 4.6666667 5.6568542 0.8164966 However the fitted() result makes sense: >

Hi R experts, Could you please help me to fit a linear, cubic and quadratic curve in a figure? I was trying to show all these three fitting curves with different colour in one figure. I spent substantial time to figure it out, but I could not. I have given here a example and what I did for linear, but no idea for cubic and quadratic fitting curve > dput(test) structure(list(sp = c(4L, 5L,

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