Hi Terry, If I understood your problem you would estimate trend and seasonal (as sum of sin and cos) in a ts.
If t is time, Y is your ts, T=f(t) is trend function of time (it could be linear, quadratic, etc. as better is for your data), e=errors/residuals Your model to fit will'be: Y(t)=T(t)+a*cos(2*pi*t/12)+b*sin(2*pi*t/12)+e(t) using lm() function to estimate a linear/polinomial trend and sin/cos seasonal: cos.t <- cos(2*pi*t/12) sin.t <- sin(2*pi*t/12) gfit<-lm(y~t+cos.t+sin.t, data=yourdf) see this example: > t<-seq(1:48) > t [1] 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 [26] 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 y<-10+5*t+0.5*cos(2*pi*t/12)+0.2*sin(2*pi*t/12)+rnorm(48) > cos.t <- cos(2*pi*t/12) > sin.t <- sin(2*pi*t/12) > gfit<-lm(y~t+cos.t+sin.t) > summary(gfit) Call: lm(formula = y ~ t + cos.t + sin.t) Residuals: Min 1Q Median 3Q Max -2.10222 -0.62184 -0.09387 0.50586 2.74299 Coefficients: Estimate Std. Error t value Pr(>|t|) (Intercept) 10.30466 0.30009 34.339 <2e-16 *** t 4.98987 0.01071 465.793 <2e-16 *** cos.t 0.30207 0.20604 1.466 0.150 sin.t 0.08699 0.20961 0.415 0.680 --- Signif. codes: 0 `***' 0.001 `**' 0.01 `*' 0.05 `.' 0.1 ` ' 1 Residual standard error: 1.008 on 44 degrees of freedom Multiple R-Squared: 0.9998, Adjusted R-squared: 0.9998 F-statistic: 7.525e+04 on 3 and 44 DF, p-value: < 2.2e-16 I hope I helped you. Best Vito ===== Diventare costruttori di soluzioni Became solutions' constructors "The business of the statistician is to catalyze the scientific learning process." George E. P. Box Visitate il portale http://www.modugno.it/ e in particolare la sezione su Palese http://www.modugno.it/archivio/palese/ ______________________________________________ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html