How can I run my Newton's method in this case? - r

Question

There is a function like:
y = (e^x - 2)^n
The x is an unknown, for n = 2,3,4,...,8
Now I want to use NR method to find the root of this function(initial x is 0).
I know how to write an NR method if the n is a fixed value, here's my origin NR code:
NR <- function(f, x0, tol = 1e-5, ite = 1000){
require(numDeriv) #call the package for computing dx
k <- ite
for (i in 1:ite){
#calculate dx
dx <- genD(func = f, x = x0)$D[1]
#get the x1
x1 <- x0 - (f(x0) / dx)
k[i] <- x1
if(abs(x1 - x0) < tol){
root <- x1
re <- list('root approximation' = root, 'iteration' = length(k))
return(re)
}
x0 <- x1
}
print('Outside the upper iteration')
}
Now I rewrite my function:
f <- function(x, n){
(exp(x) - 2) ^ n
}
If I want to output every root for different n, I think I should add another loop before the loop "for (i in 1:ite)"
So I rewrite my NR function code:
NR <- function(f, x0, tol = 1e-5, ite = 1000){
require(numDeriv) #call the package for computing dx
k <- ite
for(n in 2:8){
for (i in 1:ite){
#calculate dx
dx <- genD(func = f, x = x0)$D[1]
#get the x1
x1 <- x0 - (f(x0, n) / dx)
k[i] <- x1
if(abs(x1 - x0) < tol){
root <- x1
re <- list('root approximation' = root, 'iteration' = length(k))
return(re)
}
x0 <- x1
}
print('Outside the upper iteration')
}
}
But when I run NR(f,0), R showed me the error is :
Error in func(x, ...) : argument "n" is missing, with no default
How can I figure this out?
Thank you for your help!

Answer 1

I hope you find my answer helpful:
If you try ?genD you will read this:
Usage
genD(func, x, method="Richardson",
method.args=list(), ...)
## Default S3 method: genD(func, x, method="Richardson",
method.args=list(), ...) Arguments
func a function for which the first (vector) argument is used as a
parameter vector. x The parameter vector first argument to func.
And in the bottom of the R Documentation this example:
Examples
func <- function(x){c(x[1], x[1], x[2]^2)}
z <- genD(func, c(2,2,5))
Therefore, the issue with your code is that you need to use a vector as an argument for f:
f <- function(c){ (exp(c[1]) - 2) ^ c[2] }
NR <- function(f, x0, tol = 1e-5, ite = 1000){ require(numDeriv)
#call the package for computing dx k <- ite for(n in 2:8){
for (i in 1:ite){
#calculate dx
dx <- genD(func = f, x = c(x0,n))$D[1]
#get the x1
x1 <- x0 - (f(c(x0,n)) / dx)
k[i] <- x1
if(abs(x1 - x0) < tol){
root <- x1
re <- list('root approximation' = root, 'iteration' = length(k))
return(re)
}
x0 <- x1
}
print('Outside the upper iteration') } }
NR(f,0)
If I run that my output is:
$`root approximation` [1] 0.6931375
$iteration [1] 15
Best!