# Extrapolation with exponential curve

I would like to extrapolate time series using exponential curve while getting the parameters via linear regression.

Exponential curve is given as $g=e^{~a + b \cdot t}$. Since I want to use linear regression, I transformed it as $ln(g) = a + b \cdot t$.

My observations are (written as R vectors)

t <- c(3,4,5,6,7,8,9,10,11)
g <- c(1.038459504,1.019448815,1.017729187,1.010076583,1.00895011,1.007841198,+
+1.006566597,1.009939696,1.003751382)


I omitted values for t = 1, 2 as they were way too different. It should be smooth.

Here comes my solution. I should probably regress $ln(g)$ against $t$. $ln(g)$ is

lng<-c(0.037738369, 0.019262104, 0.017573858, 0.010026152, 0.008910295+
+0.007810615, 0.006545131, 0.009890622, 0.003744363)


The output of linear model lm(lng~t) in R is

lm(formula = lng ~ t)

Coefficients:
(Intercept)            t
0.035476    -0.003139


Now, I would expect to simply put ln(g) = 0.037571 - 0.003477*t, which means g = e^(0.037571 - 0.003477*t) for t > 11 in order to get the extrapolated values, but it's weird. lm(g~t) is giving me better output.

I believe that I need to get $a$ and $b$ firstly and then I can predict values. The result is a=-2.390289,b=-0.326016. I just don't know how to get it.

• Don't know if this will help much, but I don't get those values for $\ln g$ at all. My first few numbers are 0.0377384, 0.0192621, 0.0175739.... – tabstop Feb 15 '14 at 18:07
• Because in r you use the lm function as lm(thing you want to predict ~ variables) also if it is a log fit then why not use ln(g)~ intercept + t since r will do the rest for you? Ps I am using a phone hence no formatting ;) – Chinny84 Feb 15 '14 at 18:08
• tabstop: my bad, it's edited now.___ @Chinny84: what do you mean? – emilio Feb 15 '14 at 18:47 When I plot your data after taking the log, I get $ln g = 0.0355-0.0031 t$ as the best fit. It doesn't look too good. The chart is from Excel. Your values for $\ln g$ are not correct. 