Golf the xᵗʰ root of x

Question

While bored in high-school (when I was half my current age...), I found that \$f(x) = x^{x^{-1}}\$ had some interesting properties, including e.g. that the maximum \$f\$ for \$0 ≤ x\$ is \$f(e)\$, and that the binding energy per nucleon of an isotope can be approximated as \$6 × f(x ÷ 21)\$...

Anyway, write the shortest function or program that calculates the x^th root of x for any number in your language's domain.

Examples cases

For all languages

     -1   >       -1
   ¯0.2   >    -3125
   ¯0.5   >        4
    0.5   >     0.25
      1   >        1
      2   >    1.414
      e   >    1.444 
      3   >    1.442
    100   >    1.047
  10000   >    1.001

For languages that handle complex numbers

   -2   >        -0.7071i
    i   >            4.81         
   2i   >    2.063-0.745i
 1+2i   >   1.820-0.1834i
 2+2i   >   1.575-0.1003i

For languages that handle infinities

-1/∞   >   0    (or ∞ or ̃∞)
   0   >   0    (or 1 or ∞)
 1/∞   >   0
   ∞   >   1
  -∞   >   1

For languages that handle both infinities and complex numbers

 -∞-2i   >   1      (or ̃∞)

---

^{̃∞ denotes directed infinity.}

Answer 1

TI-BASIC, 3 bytes

Ans×√Ans

TI-BASIC uses tokens, so Ans and ×√ are both one byte.

Explanation

Ans is the easiest way to give input; it is the result of the last expression. ×√ is a function for the x'th root of x, so for example 5×√32 is 2.

Answer 2

Jelly, 2 bytes

*İ

Try it online!

How it works

*İ    Main link. Input: n

 İ    Inverse; yield 1÷n.
*     Power (fork); compute n ** (1÷n).

Answer 3

Python 3, 17 bytes

lambda x:x**(1/x)

Self-explanatory

Answer 4

Javascript (ES2016), 11 bytes

x=>x**(1/x)

I rarely ever use ES7 over ES6.

Answer 5

J, 2 bytes

^%

Try it online!.

How it works

^%  Monadic verb. Argument: y

 %  Inverse; yield 1÷y.
^   Power (hook); compute y ** (1÷y).

Answer 6

Haskell, 12 11 bytes

Thanks @LambdaFairy for doing some magic:

(**)<*>(1/) 

My old version:

\x->x**(1/x)

Answer 7

Pyth, 3 bytes

@QQ

Trivial challenge, trivial solution...

(noncompeting, 1 byte)

@

This uses the implicit input feature present in a version of Pyth that postdates this challenge.

Answer 8

JavaScript ES6, 18 bytes

n=>Math.pow(n,1/n)

Answer 9

Java 8, 18 bytes

n->Math.pow(n,1/n)

Java isn't in last place?!?!

Test with the following:

import java.lang.Math;

public class Main {
  public static void main (String[] args) {
    Test test = n->Math.pow(n,1/n);
    System.out.println(test.xthRoot(6.0));
  }
}

interface Test {
  double xthRoot(double x);
}

Answer 10

R, 19 17 bytes

function(x)x^x^-1

-2 bytes thanks to @Flounderer

Answer 11

Java, 41 bytes

float f(float n){return Math.pow(n,1/n);}

Not exactly competitive because Java, but why not?

Answer 12

MATL, 5 bytes

t-1^^

Try it online!

t       % implicit input x, duplicate
 -1     % push -1
   ^    % power (raise x to -1): gives 1/x
    ^   % power (raise x to 1/x). Implicit display

Answer 13

Mathematica, 8 7 4 7 bytes

#^#^-1&

More builtin-only answers, and now even shorter! Nope. By definition, the next answer should be 13 bytes. (Fibonacci!) The pattern is still broken. :(

Answer 14

Perl 5, 10 bytes

9 bytes plus 1 for -p

$_**=1/$_

Answer 15

Ruby, 15 Bytes

a=->n{n**n**-1}

Ungolfed:

-> is the stabby lambda operator where a=->n is equivalent to a = lambda {|n|}

Answer 16

NARS APL, 2 bytes

√⍨

NARS supports the √ function, which gives the ⍺-th root of ⍵. Applying commute (⍨) gives a function that, when used monadically, applies its argument to both sides of the given function. Therefore √⍨ x ↔ x √ x.

Other APLs, 3 bytes

⊢*÷

This is a function train, i.e. (F G H) x ↔ (F x) G H x. Monadic ⊢ is identity, dyadic * is power, and monadic ÷ is inverse. Therefore, ⊢*÷ is x raised to 1/x.

Answer 17

Python 2 - 56 bytes

The first actual answer, if I'm correct. Uses Newton's method.

n=x=input();exec"x-=(x**n-n)/(1.*n*x**-~n);"*999;print x

Answer 18

CJam, 6 bytes

rd_W##

Try it online!

How it works

rd     e# Read a double D from STDIN and push it on the stack.
  _    e# Push a copy of D.
   W   e# Push -1.
    #  e# Compute D ** -1.
     # e# Compute D ** (D ** -1).

Answer 19

dc, 125 bytes

15k?ddsk1-A 5^*sw1sn0[A 5^ln+_1^+ln1+dsnlw!<y]syr1<y1lk/*sz[si1[li*li1-dsi0<p]spli0<p]so0dsw[lzlw^lwlox/+lw1+dswA 2^!<b]dsbxp

Unlike the other dc answer, this works for all real \$x\$ greater than or equal to \$1 (\$1 ≤ x$). Accurate to 4-5 places after the decimal.

I would have included a TIO link here, but for some reason this throws a segmentation fault with the version there (dc 1.3) whereas it does not with my local version (dc 1.3.95).

Explanation

As dc does not support raising numbers to non-integer exponents to calculate \$x^\frac1x\$, this takes advantage of the fact that:

$$x^\frac1x = e^\frac{\ln x}x$$

So, to calculate \$\ln(x)\$, this also takes advantage of the fact that:

$$\int \frac1x dx = \ln(x) + c$$

whose definite integral from \$1\$ to \$b = x\$ is numerically-approximated in increments of \$10^{-5}\$ using the following summation formula:

$$\int_1^b \frac1x dx = \sum_{i=1}^{10^{5}(b-1)} \frac 1 {10^5 + i}$$

The resulting sum is then multiplied by \$\frac1x\$ to get \$\frac{\ln(x)}x\$. \$e^{\frac{\ln(x)}x}\$ is then finally calculated using the \$e^x\$ Maclaurin Series to 100 terms as follows:

$$e^x=\sum^{10^2}_{n=0}\frac{x^n}{n!}$$

This results in our relatively accurate output of \$x^\frac1x\$.

Answer 20

Seriously, 5 bytes

,;ì@^

Try it online!

Explanation:

,;ì@^
,;     input, dupe
  ì@   1/x, swap
    ^  pow

Answer 21

Pylons, 5 bytes.

ideAe

How it works.

i # Get command line input.
d # Duplicate the top of the stack.
e # Raise the top of the stack to the power of the  second to the top element of the stack.
A # Push -1 to the stack (pre initialized variable).
e # Raise the top of the stack to the power of the second to the top element of the stack.
  # Implicitly print the stack.

Answer 22

Japt, 3 bytes

UqU

Test it online!

Very simple: U is the input integer, and q is the root function on numbers.

Answer 23

C++, 48 bytes

#include<math.h>
[](auto x){return pow(x,1./x);}

The second line defines an anonymous lambda function. It can be used by assigning it to a function pointer and calling it, or just calling it directly.

Try it online

Answer 24

Milky Way 1.6.5, 5 bytes

'1'/h

---

Explanation

'      ` Push input
 1     ` Push the integer literal
  '    ` Push input
   /   ` Divide the STOS by the TOS
    h  ` Push the STOS to the power of the TOS

x**(1/x)

---

Usage

$ ./mw <path-to-code> -i <input-integer>

Answer 25

O, 6 bytes

j.1\/^

No online link because the online IDE doesn't work (specifically, exponentiation is broken)

Explanation:

j.1\/^
j.      push two copies of input
  1\/   push 1/input (always float division)
     ^  push pow(input, 1/input)

Answer 26

𝔼𝕊𝕄𝕚𝕟, 5 chars / 7 bytes

Мű⁽ïï

Try it here (Firefox only).

Trivial.

Answer 27

Pyke (commit 29), 6 bytes

D1_R^^

Explanation:

D      - duplicate top
 1_    - load -1
   R   - rotate
    ^  - ^**^
     ^ - ^**^

Answer 28

C# - 18 43 41 bytes

float a(float x){return Math.Pow(x,1/x);}

-2 byes thanks to @VoteToClose

Try it out

Note:

First actual attempt at golfing - I know I could do this better.

Answer 29

C, 23 bytes

#define p(a)pow(a,1./a)

This defines a macro function p which evaluates to the ath root of a.

Thanks to Dennis for reminding me that gcc doesn't require math.h to be included.

Thanks to @EʀɪᴋᴛʜᴇGᴏʟғᴇʀ for reminding me that the space after the first ) is not needed.

Try it online

Answer 30

PHP 5.6, 32 30 29 bytes

function($x){echo$x**(1/$x);}

or

function($x){echo$x**$x**-1;}

_{30->29, thank you Dennis!}