similar to: polynomial regression and poly

Hi, I am curious about how to interpret the results of a polynomial regression-- using poly(raw=TRUE) vs. poly(raw=FALSE). set.seed(123456) x <- rnorm(100) y <- jitter(1*x + 2*x^2 + 3*x^3 , 250) plot(y ~ x) l.poly <- lm(y ~ poly(x, 3)) l.poly.raw <- lm(y ~ poly(x, 3, raw=TRUE)) s <- seq(-3, 3, by=0.1) lines(s, predict(l.poly, data.frame(x=s)), col=1) lines(s,

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 Folks! Im using the function poly to generate orthogonal polynomials, but Id like to see the actual polynomials so that I could convert it to a polynomial in my original variable. Is that possible and if so how do I do it? /E

Dear all, I am using poly() in lm() in the following form. 1> DelsDPWOS.lm3 <- lm(DelsPDWOS[,1] ~ poly(DelsPDWOS[,4],3)) 2> DelsDPWOS.I.lm3 <- lm(DelsPDWOS[,1] ~ poly(I(DelsPDWOS[,4]),3)) 3> DelsDPWOS.2.lm3 <- lm(DelsPDWOS[,1]~DelsPDWOS[,4]+I(DelsPDWOS[,4]^2)+I(DelsPDWOS[,4]^3)) 1 and 2 lead to identical but wrong results. 3 is correct. Surprisingly (to me) the residuals

Dear All, I have found in the poly help this sentence: The orthogonal polynomial is summarized by the coefficients, which can be used to evaluate it via the three-term recursion given in Kennedy & Gentle (1980, pp. 343–4), and used in the predict part of the code. My question: which type of orthogonal polynomials are used by this function? Hrmite, legendre.. TIA Giovanni [[alternative HTML

Dear all, Can any one tell me how can i perform Orthogonal Polynomial Regression parameter estimation in R? -------------------------------------------- Here is an "Orthogonal Polynomial" Regression problem collected from Draper, Smith(1981), page 269. Note that only value of alpha0 (intercept term) and signs of each estimate match with the result obtained from coef(orth.fit). What

What does the warning message "1: Singular precision matrix in level -1, block 1" mean? I get this warning 50+ times when I try to fit the following model lme( response ~ covariateA + poly(covariateB,3), ~poly(covariateB,3)|group ) It's not a small dataset - a set of up to 20 blood pressure readings on just over 2000 people, and I don't get the error message when I try to fit

In S-Plus, I can obtain polynomial contrasts for an ordered factor with contr.poly(). The function also exists in R, however is limited to factors where the levels are equally spaced. In S-Plus, one can obtain the contrasts for a set of numeric values representing unequally spaced ordered factors. Has anyone implemented this in R? I see that the S-Plus function calls another function (poly.raw())

How can ordinary polynomial coefficients be calculated from an orthogonal polynomial fit? I'm trying to do something like find a,b,c,d from lm(billions ~ a+b*decade+c*decade^2+d*decade^3) but that gives: "Error in eval(expr, envir, enclos) : Object "a" not found" > decade <- c(1950, 1960, 1970, 1980, 1990) > billions <- c(3.5, 5, 7.5, 13, 40) > #

Here's my little discussion example for a quadratic regression: http://pj.freefaculty.org/R/WorkingExamples/regression-quadratic-1.R Students press me to know the benefits of poly() over the more obvious regression formulas. I think I understand the theory on why poly() should be more numerically stable, but I'm having trouble writing down an example that proves the benefit of this. I

Full_Name: Renaud Lancelot Version: Version 2.3.0 (2006-04-24) OS: MS Windows XP Pro SP2 Submission from: (NULL) (82.239.219.108) I think there is a bug in predict.lme, when a polynomial generated by poly() is used as an explanatory variable, and a new data.frame is used for predictions. I guess this is related to * not * using, for predictions, the coefs used in constructing the orthogonal

I have a simple X, Y data frame that I am trying to run regression analysis on. The linear regression looks great, but when I use lm(formula = y ~ poly(x, degree = 5)) I get the same coeffecients. So for example if I use degree =3 my formula would look like y = 4.2 x^3 + 3.2x^2 + 2.1x + 1.0 and my degree 5 would look like y = 6.5x^5+ 5.4x^4 + 4.2 x^3 + 3.2x^2 + 2.1x + 1.0, which doesn't make

Hi all, I want to fit data called "metal" with a polynominal function as dP ~ a.0 + a.1 * U0 + a.2 * U0^2 + a.3 * U0^3 + a.4 * U0^4 The data set includes, the independant variable U0 and the dependant variable dP. I've seen that the combination of lm() and poly() can do that instead of using the nls() function. But I don't get how to interpret the results from the linear

By any chance is anyone aware of an R function that duplicates Octave's poly function? Here is a description of Octave's poly function: Function File: poly (A) If A is a square N-by-N matrix, `poly (A)' is the row vector of the coefficients of `det (z * eye (N) - a)', the characteristic polynomial of A. As an example we can use this to find the eigenvalues

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

Often in practical situations a predictor has missing values, so that poly crashes. For instance: > x<-1:10 > y<- x - 3 * x^2 + rnorm(10)/3 > x[3]<-NA > lm( y ~ poly(x,2) ) Error in poly(x, 2) : missing values are not allowed in 'poly' > > lm( y ~ poly(x,2) , subset=!is.na(x)) # This does not help?!? Error in poly(x, 2) : missing values are not allowed in

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 <-

Full_Name: Russell Lenth Version: 2.6.2 OS: Windows XP Pro Submission from: (NULL) (128.255.132.36) The poly() function allows a higher-degree polynomial than it should, when raw=FALSE. For example, consider 5 distinct 'x' values, each repeated twice. we can fit a polynomial of degree 8: ===== R> x = rep(1:5, 2) R> y = rnorm(10) R> lm(y ~ poly(x, 8)) Call: lm(formula = y ~

Hey all, I'm performing polynomial regression. I'm simulating x values using runif() and y values using a deterministic function of x and rnorm(). When I perform polynomial regression like this: fit_poly <- lm(y ~ poly(x,11,raw = TRUE)) I get some NA coefficients. I think this is due to the high correlation between say x and x^2 if x is distributed uniformly on the unit interval

Full_Name: Jens Keienburg Version: 2.3.0 OS: Windows XP Submission from: (NULL) (193.174.53.122) I used the function lm() to calculate the coefficients of a polynome. If I used the function poly(t,2) to denote a polynome of form 1 + x + x^2, the coefficients are wrong. I appended an excerpt below: > t=1:100 > p=-20 - 10 * t + 2 * t^2 > p [1] -28 -32 -32 -28 -20 -8 8