Facing a simple, yet frustrating formula like this

and the task to solve it for x left me googling around for hours until I found salvation in Wolfram Alpha, Wikipedia, and a nice blog post with R-syntax to solve a similar equation.

Using the results from Wolfram Alpha I was able to find the solution with the ‘gsl’ library

# install.packages("gsl") library(gsl) # create some example data dat <- data.frame(a = 0.109861, x = 10) # a is set so that b is roughly 30. # Lazy as I am I used Excel and its solver ability to find numbers # to check if b is close to 30. Using the initial formula dat$b <- dat$x * exp(dat$a * dat$x) dat # solve for x2 and see if x and x2 are similar and close to 10 dat$x2 <- lambert_W0(dat$a * dat$b)/dat$a dat # a x b2 x2 #1 0.109861 10 29.99993 10.00001 # Hurray!

Sometimes life can be so easy (after a long time searching for the right results….).

# Appendix: Improvements

After revisiting this article some time later, I wondered what the speed is compared to Dan Kelley’s (see comment below) alternative. After firing up some repetitions using microbenchmark I got the following:

library(gsl) library(rootSolve) library(microbenchmark) library(ggplot2) dat <- data.frame(a = 0.109861, x = 10) dat$b <- dat$x * exp(dat$a * dat$x) f <- function(x, a, b) x*exp(a*x) - b autoplot(microbenchmark( lambertW = dat$x2 <- lambert_W0(dat$a * dat$b)/dat$a, uniroot = dat$x3 <- uniroot.all(f, interval = c(0, 100), a = dat$a, b = dat$b), times = 10000))

[…] Getting that X with the Glog function and Lambert’s W Facing a simple, yet frustrating formula like this xe^{ax}=b and the task to solve it for x left me googling around for hours until I found salvation in Wolfram Alpha, Wikipedia, and a nice blog post with R-syntax to solve a similar equation. […]

LikeLike

Is the following helpful?

> a b f library(rootSolve)

> uniroot.all(f, interval=c(0,100))

[1] 10

LikeLike

Hi Dan,

thanks for the comment, I haven’t looked into the rootSolve library. Can you post a minimum working example?

Thanks!

LikeLike