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

Javascript (ES2016), 11 bytes

x=>x**(1/x)

I rarely ever use ES7 over ES6.

Answer 4

Python 3, 17 bytes

lambda x:x**(1/x)

Self-explanatory

Answer 5

Haskell, 12 11 bytes

Thanks @LambdaFairy for doing some magic:

(**)<*>(1/) 

My old version:

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

Answer 6

J, 2 bytes

^%

Try it online!.

How it works

^%  Monadic verb. Argument: y

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

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

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 11

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 12

Java, 41 bytes

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

Not exactly competitive because Java, but why not?

Answer 13

R, 19 17 bytes

function(x)x^x^-1

-2 bytes thanks to @Flounderer

Answer 14

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 15

Perl 5, 10 bytes

9 bytes plus 1 for -p

$_**=1/$_

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

Ruby, 15 Bytes

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

Ungolfed:

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

Answer 18

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 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^(1/x), this takes advantage of the fact that:

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

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

.

The resulting sum is then multiplied by 1/x to get ln(x)/x. e^(ln(x)/x) is then finally calculated using the e^x Maclaurin Series to 100 terms as follows:

.

This results in our relatively accurate output of x^(1/x).

Answer 20

Seriously, 5 bytes

,;ì@^

Try it online!

Explanation:

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

Answer 21

, 5 chars / 7 bytes

Мű⁽ïï

Try it here (Firefox only).

Trivial.

Answer 22

Pyke (commit 29), 6 bytes

D1_R^^

Explanation:

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

Answer 23

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 24

Japt, 3 bytes

UqU

Test it online!

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

Answer 25

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 26

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 27

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 28

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 29

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 30

PARI/GP, 9 bytes

x->x^x^-1

This is shorter than the alternatives x->x^(1/x) and x->sqrtn(x,x). It uses the right-associativity of ^ and the strong binding of the unary -. GP isn't quite functional enough to curry (a,b)->sqrtn(a,b) into (x)->sqrtn(x,x); the best it can manage is x->call(sqrtn,[x,x]) which is far too long.

Answer 31

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!}

Answer 32

Awk, 8 bytes

$1^$1^-1

A shell command to check it

awk '{print $1^$1^-1}'

---

Longer version (+1) with

$1^(1/$1)

Answer 33

MATLAB, 11bytes

@(x)x^(1/x)

Creates an anonymous function. This will work with all test cases including complex numbers and infinities.

This also works with Octave. You can try online here. Simply enter something like:

f=@(x)x^(1/x)

At the command line of the online interpreter, then do:

f(-1)

To test.

Answer 34

C#, 25 bytes, type: double

Bare form of a lambda expression:

x=>System.Math.Pow(x,1/x)

The expression can be assigned to a function variable, e.g. (Func<double, double>), ...

System. can be omitted, if using System; is used (but more bytes).

Complete test program (143 bytes) with the input of the question:

using System;class P{static void Main(){Array.ForEach(new double[]{-.2,-.5,.5,1,2,Math.E,3,100,10000},x=>Console.WriteLine(Math.Pow(x,1/x)));}}

Result:

-3125
4
0,25
1
1,4142135623731
1,44466786100977
1,44224957030741
1,0471285480509
1,0009214583193

C#, 37 bytes, type: Complex

x=>System.Numerics.Complex.Pow(x,1/x)

Also the assembly for System.Numerics needs to be referenced. The division operator is overloaded for complex arguments.

Test program (189 bytes) with the input of the question for complex numbers:

using System;using C=System.Numerics.Complex;class P{static void Main(){Array.ForEach(new C[]{new C(-2,0),new C(0,1),new C(0,2),new C(1,2),new C(2,2)},x=>Console.WriteLine(C.Pow(x,1/x)));}}

Result:

(4,32963728535968E-17, -0,707106781186548)
(4,81047738096535, 0)
(2,0628722350809, -0,745007062179724)
(1,81984053615444, -0,183434723804562)
(1,57500291115344, -0,100274868564155)

Test program (264 bytes) with infinity:

using System;using C=System.Numerics.Complex;class P{static void Main(){var p=double.PositiveInfinity;var n=double.NegativeInfinity;Array.ForEach(new C[]{new C(-1/p,0),new C(0,0),new C(1/p,0),new C(p,0),new C(n,0),new C(n,2) },x=>Console.WriteLine(C.Pow(x,1/x)));}}

Result:

(0, 0)
(0, 0)
(0, 0)
(1, 0)
(1, 0)
(1, 0)

Answer 35

Vitsy, 4 + 1 (function declaration) = 5 bytes

Hey, look, my language actually does this. Go figure.

Di^^
D    Duplicate the top item (input)
 i   Push -1.
  ^  Pop top (-1), put the second to top to the power of the popped item (x^-1)
   ^ Same as above, except we now have the desired number.

This is a function that leaves the value requested on the stack. For view-ability, I have also made the program output on the Try it Online! link.

Try it Online!

Answer 36

Bean, 11 bytes (non-competing)

Hexdump:

00000000 26 4c a0 43 95 4c a5 3a 8e 20 43  &L C.L¥:. C
0000000b

Equivalent JavaScript:

A**(1/A)

Implicitly converts first line of input to number as A and implicitly outputs the result of the expression.

Try the demo here.

Try the test suite here.

Answer 37

Codename Dragon, 16 bytes

a=<>;dispa^(1/a)

I'm abusing the lack of a syntax error.

Ungolfed:

disp((a=<>)^(1/a))

This is a simple program that takes input through <>.

You can find an interpreter here, no github site yet.

Answer 38

Jolf, 5 bytes

Try it here!

^j/1j
 j    input
^ /1j to the 1/j pow.

Answer 39

Gnuplot, 13 bytes

f(x)=x**(1/x)

or

f(x)=x**x**-1

Answer 40

Racket, 28 22 bytes

(λ(n)(expt n(/ 1 n)))

@cat gets credit for the 6 byte save. Thanks!

Seriously, Racket is impossible to golf with :). Nothing to see here, move along!

Answer 41

Factor, 14 10 bytes

1 over / ^

Expects input on top of the stack.

Answer 42

Convex, 3 bytes

Convex is a new language that I am developing that is heavily based on CJam and Golfscript. The interpreter and IDE can be found here. Input is an integer into the command line arguments. Indexes are one-based. Uses the CP-1252 encoding. This answer in non-competing as it was created after this challenge.

_¹#

Explanation:

    Implied input
_   Duplicate top of stack
 ¹  Find reciprocal
  # Power
    Implied output

Answer 43

Brachylog, 8 7 bytes

Thanks to Fatalize for saving a byte.

^?^-1=.

Explanation

^?^-1=.
     =.   unify output with
^?^-1     input raised to the 1/input power

Answer 44

Lua, 20

n=(...)print(n^n^-1)

How it works:

n=(...)              Take input and store at n
       print(
             n^n^-1  n^(n^(-1))
                   )

Answer 45

(Noncompeting) Quetzalcoatl, 4 bytes

c.I?

Explanation:

c    # Input
 .   # Duplicate
  I  # Inverse
   ? # Power

This answer was too short to post before adding this sentence.

Answer 46

dc, 8 6 chars

Outputs to the top of the stack:

?d1r/^

Or if I can expect the input at the top of the stack, this 5 char solution works too:

d1r/^

Explanation:

?        take input from STDIN and execute it (numbers are executed by pushing to the stack)
 d1r     duplicate top of stack, push 1 and swap them
    /^   divide and exponentiate

Alas, dc does not allow for fractional exponents so this answer does not actually work for any given inputs above 1, and gives the wrong answer for 0 < input < 1. It does the same thing as many of the other programs though.

Answer 47

Python 3: 34

n=complex(input());print(n**(1/n))

Unless I'm missing something this would do nicely. I like that Python normally uses spaces, but this solution doesn't need them.

Answer 48

05AB1E, 3 bytes (non-competing)

Non-competing, because 05AB1E couldn't parse floats correctly at the time the challenge was posted. Code:

Dzm

Explanation:

D    # Duplicate the input
 z   # Inverse, 1 ÷ input
  m  # Power, calculating input ** (1 ÷ input)

Try it online!.

Answer 49

tcl, 23

I post two approaches which occupy the exact same number of bytes.

puts [expr $n**(1./$n)]

—

puts [expr $n**$n**-1.]

Available to test on http://rextester.com/EMW19680

Answer 50

SmileBASIC, 18 bytes

INPUT X?POW(X,1/X)

Answer 51

Common Lisp, 24 bytes

(lambda(x)(expt x(/ x)))

Try it online!

Answer 52

TI-Basic, 4 bytes

Ans^Ansֿ ¹

Since TI-Basic is tokenized, this code is a mere four bytes.

Answer 53

QBIC, 9 bytes

:?a^(1/a)

Gets a numeric input from the command line and does the math...

Answer 54

Mouse-2002, 13 bytes

x:x.b/x.&ROOT

Other programs assume the number's on the top of the stack and so shall I.

x:    ~ put TOS in x
x.    ~ push x
b /   ~ divide by 1
x.    ~ push x again
&ROOT ~ y to the x root

Answer 55

Scala, 33 bytes

def f=(x:Double)=>Math.pow(x,1/x)

Answer 56

Perl, 11 bytes

Added 2 bytes for -pn.

$_**=1/$_

Try it here !

Answer 57

(Non-competing) Pyth, 1 byte

@

Pyth takes implicit input, so @ is really equivalent to @QQ, where Q is the evaluated input.

Try it online!

Answer 58

Mumps, 14 Bytes

R I W I**(1/I)

R reads from input into variable I, and the W (write) statement does the mathematical calculations.

I will warn you: the output will be ugly as there are no implied newlines in Mumps, and at least the way I read the challenge, the OP didn't require them. If you did want cleaner output, add 2 bytes and add a newline:

R I W !,I**(1/I)

Answer 59

SQL (MySQL), 60 bytes

CREATE FUNCTION G(X REAL) RETURNS REAL RETURN POWER(X,1./X);

Caps lock is cruise control for cool golf.

Answer 60

Desmos, 14 bytes

a=1
\sqrt[a]{a}

Demonstration

Answer 61

Excel, 10 bytes

=A1^(1/A1)

Self explanatory.