Solve simple equation in R - r

Question

I have a probably really basic question concerning the possibility to solve functions in R, but to know the answer would really help to understand R better.
I have following equation:
0=-100/(1+r)+(100-50)/(1+r)^2+(100-50)/(1+r)^3+...(100-50)/(1+r)^10
How can I solve this equation in R finding the variable r?
I tried sth. like this:
n <- c(2:10)
0 = -100/(r+1)+sum((100-50)/((1+r)^n))
But got an error message:
Error in 0 = -100/(r + 1) + sum((100 - 50)/((1 + r)^n)) :
invalid (do_set) left-hand side to assignment
What's the problem and how can I find r?

Answer 1

There are plenty of optimization and root finding libraries for R link here. But in native R:
fnToFindRoot = function(r) {
n <- c(2:10)
return(abs(-100/(r+1)+sum((100-50)/((1+r)^n))))
}
# arbitrary starting values
r0 = 0
# minimise the function to get the parameter estimates
rootSearch = optim(r0, fnToFindRoot,method = 'BFGS', hessian=TRUE)
str(rootSearch)
fnToFindRoot(rootSearch$par)
That function is very volatile. If you are willing to bracket the root, you are probably better off with uniroot:
fnToFindRoot = function(r,a) {
n <- c(2:10)
return((-100/(r+1)+sum((100-50)/((1+r)^n)))-a)
}
str(xmin <- uniroot(fnToFindRoot, c(-1E6, 1E6), tol = 0.0001, a = 0))
The a argument is there so you can look for a root to any arbitrary value.

Answer 2

Try bisection. This converges to r = 0.4858343 in 25 iterations:
library(pracma)
bisect(function(r) -100/(1+r) + sum(50/(r+1)^seq(2, 10)), 0, 1)
giving:
$root
[1] 0.4858343
$f.root
[1] 8.377009e-07
$iter
[1] 25
$estim.prec
[1] 1.490116e-08

Answer 3

Let x = 1/(1+r), so your equation should be:
0-100x + 50x^2 + 50x^3 + ... + 50x^10 = 0.
then in R:
x <- polyroot(c(0, -100, rep(50, 9)))
(r <- 1/x - 1)
Here is the answer:
[1] Inf+ NaNi 0.4858344-0.0000000i -1.7964189-0.2778635i
[4] -0.3397136+0.6409961i -0.3397136-0.6409961i -1.4553556-0.7216708i
[7] -0.9014291+0.8702213i -0.9014291-0.8702213i -1.7964189+0.2778635i
[10] -1.4553556+0.7216708i