I do not know if other R users have found the following problem.
Within R I do the folowing operation:
> (3/-2)^(1/3)
[1] NaN
I obtain a NaN result.
I the similar way if I set:
> w<-(3/-2)
> g<-1/3
> w^g
[1] NaN
However, if I do:
> 3/-2
[1] -1.5
> -1.5^(1/3)
[1] -1.144714
Is there anybody that can explain this contradiction?
Where do you see a problem? -1.5^(1/3) is not the same as (-1.5)^(1/3). If you have basic maths education you shouldn't expect these to be the same.
Read help("Syntax") to learn that ^ has higher precedence than - in R.
This is due to the mathematical definition of exponentiation. For the continuous real exponentiation operator, you are not allowed to have a negative base.
Begin by doing (3/2)^(1/3) and after add "-"
you can't calculate a cube root of a negative number !
If you really want the answer you can do the computation over the complex numbers, i.e. get the cube root of -1.5+0i:
complex(real=-1.5,im=0)^(1/3)
## [1] 0.5723571+0.9913516i
This is actually only one of three complex roots of x^3+1.5==0:
polyroot(c(1.5,0,0,1))
[1] 0.5723571+0.9913516i -1.1447142+0.0000000i 0.5723571-0.9913516i
but the other answers probably come closer to addressing your real question.
Related
I have looked a bit online and in the site but I did not find any solution. My problem is relatively simple so if you could point me to a possible solution, much appreciated.
test_vec <- c(2,8,709,600)
mean(exp(test_vec))
test_vec_bis <- c(2,8,710,600)
mean(exp(test_vec_bis))
exp(709)
exp(710)
# The numerical limit of R is at exp(709)
How can I calculate the mean of my vector and deal with the Inf values knowing that R could probably handle the mean value but not all values in the numerator of the mean calculation ?
There is an edge case where you can solve your problem by simply restating your problem mathematically, but that would require that the length of your vector is extremely large and/or that your large exp. numbers are close to the numeric limit:
Since the mean sum(x)/n can be written as sum(x/n) and since exp(x)/exp(y) = exp(x-y), you can calculate sum(exp(x-log(n))), which gives you a relief of log(n).
mean(exp(test_vec))
[1] 2.054602e+307
sum(exp(test_vec - log(length(test_vec))))
[1] 2.054602e+307
sum(exp(test_vec_bis - log(length(test_vec_bis))))
[1] 5.584987e+307
While this works for your example, most likely this won't work for your real vector.
In this case, you will have to consult packages like Rmpfr as suggested by #fra.
Here's one way where you qualify to only select those in your test_vec that give an answer < Inf:
mean(exp(test_vec)[which(exp(test_vec) < Inf)])
[1] 1.257673e+260
t2 <- c(2,8,600)
mean(exp(t2))
[1] 1.257673e+260
This assumes you were looking to exclude values that result in Inf, of course.
When I plug these equations into R I get:
> 1/(1+exp(-18))
[1] 1
> 1/(1+exp(-16))
[1] 0.9999999
But when I plug the same equations into Chrome I get:
1/(1+exp(-18))
0.99999998477
1/(1+exp(-16))
0.99999988746
So it seems like R is not very precise and rounds numbers up. Is it possible for me to get more digit precision with R? If so, how can I do that?
Thanks to user G5W, I realized I needed to do:
options(digits=11)
And rerun my exp() equations. B/c by default R does not display so many digits.
In the following code:
(-8/27)^(2/3)
I got the result NaN, despite the fact that the correct result should be 4/9 or .444444....
So why does it return NaN? And how can I have it return the correct value?
As documented in help("^"):
Users are sometimes surprised by the value returned, for example
why ‘(-8)^(1/3)’ is ‘NaN’. For double inputs, R makes use of IEC
60559 arithmetic on all platforms, together with the C system
function ‘pow’ for the ‘^’ operator. The relevant standards
define the result in many corner cases. In particular, the result
in the example above is mandated by the C99 standard. On many
Unix-alike systems the command ‘man pow’ gives details of the
values in a large number of corner cases.
So you need to do the operations separately:
R> ((-8/27)^2)^(1/3)
[1] 0.4444444
Here's the operation in the complex domain, which R does support:
(-8/27+0i)^(2/3)
[1] -0.2222222+0.3849002i
Test:
> ((-8/27+0i)^(2/3) )^(3/2)
[1] -0.2962963+0i
> -8/27 # check
[1] -0.2962963
Furthermore the complex conjugate is also a root:
(-0.2222222-0.3849002i)^(3/2)
[1] -0.2962963-0i
To the question what is the third root of -8/27:
polyroot( c(8/27,0,0,1) )
[1] 0.3333333+0.5773503i -0.6666667-0.0000000i 0.3333333-0.5773503i
The middle value is the real root. Since you are saying -8/27 = x^3 you are really asking for the solution to the cubic equation:
0 = 8/27 + 0*x + 0*x^2 + x^2
The polyroot function needs those 4 coefficient values and will return the complex and real roots.
I am running into an issue when I exponentiate floating point data. It seems like it should be an easy fix. Here is my sample code:
temp <- c(-0.005220092)
temp^1.1
[1] NaN
-0.005220092^1.1
[1] -0.003086356
Is there some obvious error I am making with this? It seems like it might be an oversight on my part with regard to exponents.
Thanks,
Alex
The reason for the NaN is because the result of the exponentiation is complex, so you have to pass a complex argument:
as.complex(temp)^1.1
[1] -0.002935299-0.000953736i
# or
(temp + 0i)^1.1
[1] -0.002935299-0.000953736i
The reason that your second expression works is because unary - has lower precedence than ^, so this is equivalent to -(0.005220092^1.1). See ?Syntax.
In the following code:
(-8/27)^(2/3)
I got the result NaN, despite the fact that the correct result should be 4/9 or .444444....
So why does it return NaN? And how can I have it return the correct value?
As documented in help("^"):
Users are sometimes surprised by the value returned, for example
why ‘(-8)^(1/3)’ is ‘NaN’. For double inputs, R makes use of IEC
60559 arithmetic on all platforms, together with the C system
function ‘pow’ for the ‘^’ operator. The relevant standards
define the result in many corner cases. In particular, the result
in the example above is mandated by the C99 standard. On many
Unix-alike systems the command ‘man pow’ gives details of the
values in a large number of corner cases.
So you need to do the operations separately:
R> ((-8/27)^2)^(1/3)
[1] 0.4444444
Here's the operation in the complex domain, which R does support:
(-8/27+0i)^(2/3)
[1] -0.2222222+0.3849002i
Test:
> ((-8/27+0i)^(2/3) )^(3/2)
[1] -0.2962963+0i
> -8/27 # check
[1] -0.2962963
Furthermore the complex conjugate is also a root:
(-0.2222222-0.3849002i)^(3/2)
[1] -0.2962963-0i
To the question what is the third root of -8/27:
polyroot( c(8/27,0,0,1) )
[1] 0.3333333+0.5773503i -0.6666667-0.0000000i 0.3333333-0.5773503i
The middle value is the real root. Since you are saying -8/27 = x^3 you are really asking for the solution to the cubic equation:
0 = 8/27 + 0*x + 0*x^2 + x^2
The polyroot function needs those 4 coefficient values and will return the complex and real roots.