CODEWITHSUNDEEP
r
user2005253
user2005253

Reputation:

Solving transcendental equations in R

Is there a function for solving transcendental equations in R?

For example, I want to solve the following equation

x = 1/tan(x)

Any suggestions? I know the solution has multiple roots so I also want to be able to recover all the answers for a given interval

Upvotes: 2

Views: 1631

Answers (2)

Ben Bolker
Ben Bolker

Reputation: 226637

You can use uniroot to find roots of any 1D equations within a given range. However, getting multiple roots seems like a very hard problem in general (e.g. see the relevant chapter of Numerical Recipes for some background: chapter 9 at http://apps.nrbook.com/c/index.html ). Which root is found when there are multiple roots is hard to predict. If you know enough about the problem to subdivide the space into subregions with zero or one roots, or if you're willing to divide it into lots of regions and hope that you found all the roots, you can do it. Otherwise I look forward to other peoples' solutions ...

In this particular case, as shown by @liuminzhao's solution, there's (at most? exactly?) one solution between n*pi and (n+1)*pi

y = function(x) x-1/tan(x)
curve(y,xlim=c(-10,10),n=501,ylim=c(-5,5))
abline(v=(-3:3)*pi,col="gray")
abline(h=0,col=2)

enter image description here

This is a bit of a hack, but it will find roots of your equation (provided they are not too close to a multiple of pi: you can reduce eps if you like ...). However, if you want to solve a different multi-root transcendental equation you might need another (specialized) strategy ...

f <- function(n,eps=1e-6) uniroot(y,c(n*pi+eps,(n+1)*pi-eps))$root
sapply(0:3,f)
## [1] 0.8603337 3.4256204 6.4372755 9.5293334

Upvotes: 5

liuminzhao
liuminzhao

Reputation: 2455

I would plot the function curve and look at it to see what it looks like:

R > y = function(x) { x - 1/tan(x) }
R > curve(y, xlim = c(-10, 10))
R > abline(h = 0, color = 'red')

enter image description here

Then I saw there is a root between 0 and 3, I would use uniroot to get the root I want:

R > uniroot(y, interval = c(0, 3))
$root
[1] 0.8603

$f.root
[1] 6.612e-06

$iter
[1] 7

$estim.prec
[1] 6.104e-05

Upvotes: 7

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