Nth root

Task

Implement the algorithm to compute the principal n^th root ${\displaystyle {\sqrt[{n}]{A}}}$ of a positive real number A, as explained at the Wikipedia page.

11l

Translation of: Nim

<lang 11l>F nthroot(a, n)

  V result = a
  V x = a / n
  L abs(result - x) > 10e-15
     x = result
     result = (1.0 / n) * (((n - 1) * x) + (a / pow(x, n - 1)))
  R result

print(nthroot(34.0, 5)) print(nthroot(42.0, 10)) print(nthroot(5.0, 2))</lang>

Output:

2.0244
1.4532
2.23607

An example of converting integer floating-point using unnormalized short format. The 'include' file FORMAT, to format a floating point number, can be found in: Include files 360 Assembly. <lang 360asm>* Nth root - x**(1/n) - 29/07/2018 NTHROOT CSECT

        USING  NTHROOT,R13        base register
        B      72(R15)            skip savearea
        DC     17F'0'             savearea
        SAVE   (14,12)            save previous context
        ST     R13,4(R15)         link backward
        ST     R15,8(R13)         link forward
        LR     R13,R15            set addressability
        BAL    R14,ROOTN          call rootn(x,n)
        LE     F0,XN              xn=rootn(x,n)
        LA     R0,6               decimals=6
        BAL    R14,FORMATF        edit xn
        MVC    PG(13),0(R1)       output xn
        XPRNT  PG,L'PG            print buffer
        L      R13,4(0,R13)       restore previous savearea pointer
        RETURN (14,12),RC=0       restore registers from calling sav

ROOTN MVC ZN,=E'0' zn=0 ----------------------------

        MVC    ZN,N               n
        MVI    ZN,X'46'           zn=unnormalize(n)
        LE     F0,ZN              zn
        AE     F0,=E'0'           normalized         
        STE    F0,ZN              zn=normalize(n)
        LE     F6,=E'0'           xm=0
        LE     F0,X               x
        DE     F0,ZN              /zn
        STE    F0,XN              xn=x/zn

WHILEA LE F0,XN xn

        SER    F0,F6              xn-xm
        LPER   F0,F0              abs((xn-xm)
        DE     F0,XN              /xn
        CE     F0,EPSILON         while abs((xn-xm)/xn)>epsilon
        BNH    EWHILEA            ~
        LE     F6,XN                xm=xn
        LE     F0,ZN                zn
        SE     F0,=E'1'             zn-1
        MER    F0,F6                f0=(zn-1)*xm
        L      R2,N                 n
        BCTR   R2,0                 n-1        
        LE     F2,=E'1'             xm

POW MER F2,F6 *xm

        BCT    R2,POW               f2=xm**(n-1)   
        LE     F4,X                 x
        DER    F4,F2                x/xm**(n-1)
        AER    F0,F4                (zn-1)*xm+x/xm**(n-1)
        DE     F0,ZN                /zn
        STE    F0,XN                xn=((zn-1)*xm+x/xm**(n-1))/zn
        B      WHILEA             endwhile

EWHILEA LE F0,XN xn

        BR     R14                return ---------------------------
        COPY   FORMATF            format a float

X DC E'2' x <== input N DC F'2' n <== input EPSILON DC E'1E-6' imprecision XN DS E xn :: output ZN DS E zn=float(n) PG DC CL80' ' buffer

        REGEQU 
        END    NTHROOT </lang>

Output:

     1.414213

AArch64 Assembly

Works with: as version Raspberry Pi 3B version Buster 64 bits

<lang AArch64 Assembly> /* ARM assembly AARCH64 Raspberry PI 3B */ /* program nroot64.s */ /* link with gcc. Use C function for display float */

/*******************************************/ /* Constantes file */ /*******************************************/ /* for this file see task include a file in language AArch64 assembly*/ .include "../includeConstantesARM64.inc"

/*******************************************/ /* Initialized data */ /*******************************************/ .data szFormat1: .asciz "Root= %+09.15f\n" .align 4 qNumberA: .quad 1024 /*******************************************/ /* UnInitialized data */ /*******************************************/ .bss .align 4 /*******************************************/ /* code section */ /*******************************************/ .text .global main main: // entry of program

   /* root 10ieme de 1024  */
   ldr x0,qAdriNumberA                 // number address
   ldr d0,[x0]                         // load number in registre d0
   scvtf d0,d0                         // conversion in float
   mov x0,#10                          // N
   bl nthRoot
   ldr x0,qAdrszFormat1                // format
   bl printf                           // call C function !!!
                                       // Attention register dn lost !!!
   /* square root of 2   */ 
   fmov d0,2                           // conversion 2 in float register d0
   mov x0,#2                           // N
   bl nthRoot
   ldr x0,qAdrszFormat1                // format
                                       // d0 contains résult
   bl printf                           // call C function !!!

100: // standard end of the program

   mov x0, #0                          // return code
   mov x8, #EXIT                       // request to exit program
   svc 0                               // perform the system call

qAdrszFormat1: .quad szFormat1 qAdriNumberA: .quad qNumberA

/******************************************************************/ /* compute nth root */ /******************************************************************/ /* x0 contains N */ /* d0 contains the value */ /* x0 return result */ nthRoot:

   stp x1,lr,[sp,-16]!            // save  registers
   stp x2,x3,[sp,-16]!            // save  registers
   stp d1,d2,[sp,-16]!            // save float registers
   stp d3,d4,[sp,-16]!            // save float registers
   stp d5,d6,[sp,-16]!            // save float registers
   stp d7,d8,[sp,-16]!            // save float registers
   fmov d6,x0                     //
   scvtf d6,d6                    // N conversion in float double précision (64 bits)
   sub x1,x0,#1                   // N - 1
   fmov d8,x1                     // 
   scvtf d4,d8                    //conversion in float double précision
   fmov d2,d0                     // a = A
   fdiv d3,d0,d6                  // b = A/n
   adr x2,dfPrec                  // load précision
   ldr d8,[x2]

1: // begin loop

   fmov d2,d3                     // a <- b
   fmul d5,d3,d4                  // (N-1)*b

   fmov d1,1                      // constante 1 -> float
   mov x2,0                       // loop indice

2: // compute pow (n-1)

   fmul d1,d1,d3                  // 
   add x2,x2,1
   cmp x2,x1                      // n -1 ?
   blt 2b                         // no -> loop
   fdiv d7,d0,d1                  // A / b pow (n-1)
   fadd d7,d7,d5                  // + (N-1)*b
   fdiv d3,d7,d6                  // / N -> new b
   fsub d1,d3,d2                  // compute gap
   fabs d1,d1                     // absolute value
   fcmp d1,d8                     // compare float maj FPSCR
   bgt 1b                         // if gap > précision -> loop 
   fmov d0,d3                     // end return result in d0

100:

   ldp d7,d8,[sp],16              // restaur  2 float registers
   ldp d5,d6,[sp],16              // restaur  2 float registers
   ldp d3,d4,[sp],16              // restaur  2 float registers
   ldp d1,d2,[sp],16              // restaur  2 float registers
   ldp x2,x3,[sp],16              // restaur  2 registers
   ldp x1,lr,[sp],16              // restaur  2 registers
   ret                            // return to address lr x30

dfPrec: .double 0f1E-10 // précision /********************************************************/ /* File Include fonctions */ /********************************************************/ /* for this file see task include a file in language AArch64 assembly */ .include "../includeARM64.inc" </lang>

Output:

Root= +2.000000000000000
Root= +1.414213562373095

Ada

The implementation is generic and supposed to work with any floating-point type. There is no result accuracy argument of Nth_Root, because the iteration is supposed to be monotonically descending to the root when starts at A. Thus it should converge when this condition gets violated, i.e. when x_k+1≥xk. <lang Ada> with Ada.Text_IO; use Ada.Text_IO;

procedure Test_Nth_Root is

  generic
     type Real is digits <>;
  function Nth_Root (Value : Real; N : Positive) return Real;
  
  function Nth_Root (Value : Real; N : Positive) return Real is
     type Index is mod 2;
     X : array (Index) of Real := (Value, Value);
     K : Index := 0;
  begin
     loop
        X (K + 1) := ( (Real (N) - 1.0) * X (K) + Value / X (K) ** (N-1) ) / Real (N);
        exit when X (K + 1) >= X (K);
        K := K + 1;
     end loop;
     return X (K + 1);
  end Nth_Root;

  function Long_Nth_Root is new Nth_Root (Long_Float);

begin

  Put_Line ("1024.0 10th  =" & Long_Float'Image (Long_Nth_Root (1024.0, 10)));
  Put_Line ("  27.0 3rd   =" & Long_Float'Image (Long_Nth_Root (27.0, 3)));
  Put_Line ("   2.0 2nd   =" & Long_Float'Image (Long_Nth_Root (2.0, 2)));
  Put_Line ("5642.0 125th =" & Long_Float'Image (Long_Nth_Root (5642.0, 125)));

end Test_Nth_Root; </lang> Sample output:

1024.0 10th  = 2.00000000000000E+00
  27.0 3rd   = 3.00000000000000E+00
   2.0 2nd   = 1.41421356237310E+00
5642.0 125th = 1.07154759194477E+00

ALGOL 68

Translation of: C

Works with: ALGOL 68 version Standard - no extensions to language used

Works with: ALGOL 68G version Any - tested with release 1.18.0-9h.tiny

<lang algol68>REAL default p = 0.001;

PROC nth root = (INT n, LONG REAL a, p)LONG REAL: (

 [2]LONG REAL x := (a, a/n);

 WHILE ABS(x[2] - x[1]) > p DO
   x := (x[2], ((n-1)*x[2] + a/x[2]**(n-1))/n )
 OD;
 x[2]

);

PRIO ROOT = 8; OP ROOT = (INT n, LONG REAL a)LONG REAL: nth root(n, a, default p); OP ROOT = (INT n, INT a)LONG REAL: nth root(n, a, default p);

main: (

 printf(($2(" "gl)$,
        nth root(10, LONG 7131.5 ** 10, default p),
        nth root(5, 34, default p)));
 printf(($2(" "gl)$,
        10 ROOT ( LONG 7131.5 ** 10 ),
        5 ROOT 34))

)</lang> Output:

 +7.131500000000000000001144390e  +3
 +2.024397462171090138953733623e  +0
 +7.131500000000000000001144390e  +3
 +2.024397462171090138953733623e  +0

ALGOL W

<lang algolw>begin

   % nth root algorithm                                              %
   % returns the nth root of A, A must be > 0                        %
   %         the required precision should be specified in precision %
   long real procedure nthRoot( long real value A
                              ; integer   value n
                              ; long real value precision
                              ) ;
       begin
           long real xk, xd;
           integer   n1;
           n1 := n - 1;
           xk := A / n;
           while begin
               xd := ( ( A / ( xk ** n1 ) ) - xk ) / n;
               xk := xk + xd;
               abs( xd ) > precision
           end do begin end;
           xk
       end nthRoot ;
   % test cases %
   r_format := "A"; r_w := 15; r_d := 6; % set output format %
   write( nthRoot( 7131.5 ** 10, 10, 1'-5 ) );
   write( nthRoot(           64,  6, 1'-5 ) );

end.</lang>

Output:

    7131.500000
       2.000000

ARM Assembly

Works with: as version Raspberry Pi

<lang ARM Assembly> /* ARM assembly Raspberry PI */ /* program nroot.s */ /* compile with option -mfpu=vfpv3 -mfloat-abi=hard */ /* link with gcc. Use C function for display float */

/* Constantes */ .equ EXIT, 1 @ Linux syscall

/* Initialized data */ .data szFormat1: .asciz " %+09.15f\n" .align 4 iNumberA: .int 1024

/* UnInitialized data */ .bss .align 4

/* code section */ .text .global main main: @ entry of program

   push {fp,lr}                        @ saves registers

   /* root 10ieme de 1024  */
   ldr r0,iAdriNumberA                 @ number address
   ldr r0,[r0]
   vmov s0,r0                          @ 
   vcvt.f64.s32 d0, s0                 @conversion in float single précision (32 bits)
   mov r0,#10                          @ N
   bl nthRoot
   ldr r0,iAdrszFormat1                @ format
   vmov r2,r3,d0
   bl printf                           @ call C function !!!
                                       @ Attention register dn lost !!!
   /* square root of 2   */ 
   vmov.f64 d1,#2.0                    @ conversion 2 in float register d1
   mov r0,#2                           @ N
   bl nthRoot
   ldr r0,iAdrszFormat1                @ format
   vmov r2,r3,d0
   bl printf                           @ call C function !!!

100: @ standard end of the program

   mov r0, #0                          @ return code
   pop {fp,lr}                         @restaur  registers
   mov r7, #EXIT                       @ request to exit program
   swi 0                               @ perform the system call

iAdrszFormat1: .int szFormat1 iAdriNumberA: .int iNumberA

/******************************************************************/ /* compute nth root */ /******************************************************************/ /* r0 contains N */ /* d0 contains the value */ /* d0 return result */ nthRoot:

   push {r1,r2,lr}                    @ save  registers 
   vpush {d1-d8}                         @ save float registers
   FMRX    r1,FPSCR                   @ copy FPSCR into r1
   BIC     r1,r1,#0x00370000          @ clears STRIDE and LEN
   FMXR    FPSCR,r1                   @ copy r1 back into FPSCR

   vmov s2,r0                         @ 
   vcvt.f64.s32 d6, s2                @ N conversion in float double précision (64 bits)
   sub r1,r0,#1                       @ N - 1
   vmov s8,r1                         @ 
   vcvt.f64.s32 d4, s8                @conversion in float double précision (64 bits)
   vmov.f64 d2,d0                     @ a = A
   vdiv.F64 d3,d0,d6                  @ b = A/n
   adr r2,dfPrec                      @ load précision
   vldr d8,[r2]

1: @ begin loop

   vmov.f64 d2,d3                     @ a <- b
   vmul.f64 d5,d3,d4                  @ (N-1)*b

   vmov.f64 d1,#1.0                   @ constante 1 -> float
   mov r2,#0                          @ loop indice

2: @ compute pow (n-1)

   vmul.f64 d1,d1,d3                  @ 
   add r2,#1
   cmp r2,r1                          @ n -1 ?
   blt 2b                             @ no -> loop
   vdiv.f64 d7,d0,d1                  @ A / b pow (n-1)
   vadd.f64 d7,d7,d5                  @ + (N-1)*b
   vdiv.f64 d3,d7,d6                  @ / N -> new b
   vsub.f64 d1,d3,d2                  @ compute gap
   vabs.f64 d1,d1                     @ absolute value
   vcmp.f64 d1,d8                     @ compare float maj FPSCR
   fmstat                             @ transfert FPSCR -> APSR
                                      @ or use VMRS APSR_nzcv, FPSCR
   bgt 1b                             @ if gap > précision -> loop 
   vmov.f64 d0,d3                     @ end return result in d0

100:

   vpop {d1-d8}                       @ restaur float registers
   pop {r1,r2,lr}                     @ restaur arm registers
   bx lr

dfPrec: .double 0f1E-10 @ précision

</lang>

AutoHotkey

<lang autohotkey>p := 0.000001

MsgBox, % nthRoot( 10, 7131.5**10, p) "`n"

       . nthRoot(  5, 34.0      , p) "`n"
       . nthRoot(  2, 2         , p) "`n"
       . nthRoot(0.5, 7         , p) "`n"

---------------------------------------------------------------------------

nthRoot(n, A, p) { ; http://en.wikipedia.org/wiki/Nth_root_algorithm

---------------------------------------------------------------------------

   x1 := A
   x2 := A / n
   While Abs(x1 - x2) > p {
       x1 := x2
       x2 := ((n-1)*x2+A/x2**(n-1))/n
   }
   Return, x2

}</lang> Message box shows:

7131.500000
2.024397
1.414214
49.000000

AutoIt

<lang AutoIt>;AutoIt Version: 3.2.10.0 $A=4913 $n=3 $x=20 ConsoleWrite ($n& " root of "& $A & " is " &nth_root_it($A,$n,$x)) ConsoleWrite ($n& " root of "& $A & " is " &nth_root_rec($A,$n,$x))

Iterative

Func nth_root_it($A,$n,$x)

  $x0="0"
  While StringCompare(string($x0),string($x))
     ConsoleWrite ($x&@CRLF)
     $x0=$x
     $x=((($n-1)*$x)+($A/$x^($n-1)))/$n
  WEnd
  Return $x

EndFunc

Recursive

Func nth_root_rec($A,$n,$x)

  ConsoleWrite ($x&@CRLF)
  If $x==((($n-1)*$x)+($A/$x^($n-1)))/$n Then
     Return $x
  EndIf
  Return nth_root_rec($A,$n,((($n-1)*$x)+($A/$x^($n-1)))/$n)

EndFunc</lang> output :

20
17.4275
17.0104009124137
17.0000063582823
17.0000000000024
17
3 root of 4913 is 17

AWK

!/usr/bin/awk -f

BEGIN {

       # test

print nthroot(8,3) print nthroot(16,2) print nthroot(16,4) print nthroot(125,3) print nthroot(3,3) print nthroot(3,2) }

function nthroot(y,n) {

       eps = 1e-15;   # relative accuracy
       x   = 1;

do { d = ( y / ( x^(n-1) ) - x ) / n ; x += d; e = eps*x; # absolute accuracy } while ( d < -e || d > e )

return x } </lang>

Sample output:

BASIC

Works with: QBasic

Works with: FreeBASIC

Works with: PowerBASIC

Works with: Visual Basic

This function is fairly generic MS BASIC. It could likely be used in most modern BASICs with little or no change.

<lang qbasic>FUNCTION RootX (tBase AS DOUBLE, tExp AS DOUBLE, diffLimit AS DOUBLE) AS DOUBLE

   DIM tmp1 AS DOUBLE, tmp2 AS DOUBLE
   ' Initial guess:
   tmp1 = tBase / tExp
   DO
       tmp2 = tmp1
       ' 1# tells compiler that "1" is a double, not an integer
       tmp1 = (((tExp - 1#) * tmp2) + (tBase / (tmp2 ^ (tExp - 1#)))) / tExp
   LOOP WHILE (ABS(tmp1 - tmp2) > diffLimit)
   RootX = tmp1

END FUNCTION</lang>

Note that for the above to work in QBasic, the function definition needs to be changed like so: <lang qbasic>FUNCTION RootX# (tBase AS DOUBLE, tExp AS DOUBLE, diffLimit AS DOUBLE)</lang>

The function is called like so:

<lang qbasic>PRINT "The "; e; "th root of "; b; " is "; RootX(b, e, .000001)</lang>

Sample output:

The  4th root of  16 is  2

For BASICs without the ^ operator, it would be trivial to write a function to reproduce it (as is done in the C example below).

See also the Liberty BASIC and PureBasic solutions.

Basic09

Works with: OS9 operating system -- RUN nth(root,number,precision)

<lang Basic09> PROCEDURE nth

  PARAM N : INTEGER; A, P : REAL
  DIM   TEMP0, TEMP1 : REAL

TEMP0 := A TEMP1 := A/N WHILE ( abs(TEMP0 - TEMP1) > P) DO

 TEMP0 := TEMP1
 TEMP1 := (( N - 1.0) * TEMP1 + A / TEMP1 ^ (N - 1.0)) / N

ENDWHILE PRINT "Root Number Precision" PRINT N, A, P PRINT "The Root is: ";TEMP1 END </lang>

BBC BASIC

Works with: BBC BASIC for Windows

<lang bbcbasic> *FLOAT 64

     @% = &D0D
     PRINT "Cube root of 5 is "; FNroot(3, 5, 0)
     PRINT "125th root of 5643 is "; FNroot(125, 5643, 0)
     END
     
     DEF FNroot(n%, a, d)
     LOCAL x0, x1 : x0 = a / n% : REM Initial guess
     REPEAT
       x1 = ((n% - 1)*x0 + a/x0^(n%-1)) / n%
       SWAP x0, x1
     UNTIL ABS (x0 - x1) <= d
     = x0</lang>

Output:

Cube root of 5 is 1.709975946677
125th root of 5643 is 1.071549111198

bc

<lang bc>/* Take the nth root of 'a' (a positive real number).

* 'n' must be an integer.
* Result will have 'd' digits after the decimal point.
*/

define r(a, n, d) {

   auto e, o, x, y, z

   if (n == 0) return(1)
   if (a == 0) return(0)

   o = scale
   scale = d
   e = 1 / 10 ^ d

   if (n < 0) {
       n = -n
       a = 1 / a
   }

   x = 1
   while (1) {
       y = ((n - 1) * x + a / x ^ (n - 1)) / n
       z = x - y
       if (z < 0) z = -z
       if (z < e) break
       x = y
   }
   scale = o
   return(y)

}</lang>

Bracmat

Bracmat does not have floating point numbers as primitive type. Instead we have to use rational numbers. This code is not fast! <lang bracmat>( ( root

 =   n a d x0 x1 d2 rnd 10-d
   .   ( rnd       { For 'rounding' rational numbers = keep number of digits within bounds. }
       =   N r
         .   !arg:(?N.?r)
           & div$(!N*!r+1/2.1)*!r^-1
       )
     & !arg:(?n,?a,?d)
     & !a*!n^-1:?x0
     & 10^(-1*!d):?10-d
     &   whl
       ' (   ( rnd$(((!n+-1)*!x0+!a*!x0^(1+-1*!n))*!n^-1.10^!d)
             . !x0
             )
           : (?x0.?x1)
         & (!x0+-1*!x1)^2:~<!10-d   { Exit loop when required precision is reached. }
         )
     & flt$(!x0,!d)      { Convert rational number to floating point representation. }
 )

& ( show

 =   N A precision
   .   !arg:(?N,?A,?precision)
     & out$(str$(!A "^(" !N^-1 ")=" root$(!N,!A,!precision)))
 )

& show$(10,1024,20) & show$(3,27,20) & show$(2,2,100) & show$(125,5642,20) )</lang> Output:

1024^(1/10)=2,00000000000000000000*10E0
27^(1/3)=3,00000000000000000000*10E0
2^(1/2)=1,4142135623730950488016887242096980785696718753769480731766797379907324784621070388503875343276415727*10E0
5642^(1/125)=1,07154759194476751170*10E0

C

Implemented without using math library, because if we were to use pow(), the whole exercise wouldn't make sense. <lang c>#include <stdio.h>

include <float.h>

double pow_ (double x, int e) {

   int i;
   double r = 1;
   for (i = 0; i < e; i++) {
       r *= x;
   }
   return r;

}

double root (int n, double x) {

   double d, r = 1;
   if (!x) {
       return 0;
   }
   if (n < 1 || (x < 0 && !(n&1))) {
       return 0.0 / 0.0; /* NaN */
   }
   do {
       d = (x / pow_(r, n - 1) - r) / n;
       r += d;
   }
   while (d >= DBL_EPSILON * 10 || d <= -DBL_EPSILON * 10);
   return r;

}

int main () {

   int n = 15;
   double x = pow_(-3.14159, 15);
   printf("root(%d, %g) = %g\n", n, x, root(n, x));
   return 0;

} </lang>

C#

Almost exactly how C works. <lang csharp> static void Main(string[] args) { Console.WriteLine(NthRoot(81,2,.001));

       Console.WriteLine(NthRoot(1000,3,.001));
       Console.ReadLine();

}

public static double NthRoot(double A,int n, double p) { double _n= (double) n; double[] x = new double[2]; x[0] = A; x[1] = A/_n; while(Math.Abs(x[0] -x[1] ) > p) { x[1] = x[0]; x[0] = (1/_n)*(((_n-1)*x[1]) + (A/Math.Pow(x[1],_n-1)));

} return x[0]; } </lang>

C++

<lang cpp>double NthRoot(double m_nValue, double index, double guess, double pc)

  {
      double result = guess;
      double result_next;
      do
      {
          result_next = (1.0/index)*((index-1.0)*result+(m_nValue)/(pow(result,(index-1.0))));
          result = result_next;
          pc--;
      }while(pc>1);
      return result;
  };

</lang>

<lang cpp>double NthRoot(double value, double degree) {

   return pow(value, (double)(1 / degree));

}; </lang>

Clojure

<lang clojure> (ns test-project-intellij.core

 (:gen-class))

define abs & power to avoid needing to bring in the clojure Math library

(defn abs [x]

 " Absolute value"
 (if (< x 0) (- x) x))

(defn power [x n]

 " x to power n, where n = 0, 1, 2, ... "
 (apply * (repeat n x)))

(defn calc-delta [A x n]

 " nth rooth algorithm delta calculation "
 (/ (- (/ A (power x (- n 1))) x) n))

(defn nth-root

 " nth root of algorithm: A = numer, n = root"
 ([A n] (nth-root A n 0.5 1.0))  ; Takes only two arguments A, n and calls version which takes A, n, guess-prev, guess-current
 ([A n guess-prev guess-current] ; version take takes in four arguments (A, n, guess-prev, guess-current)
  (if (< (abs (- guess-prev guess-current)) 1e-6)
    guess-current
    (recur A n guess-current (+ guess-current (calc-delta A guess-current n)))))) ; iterate answer using tail recursion

</lang>

COBOL

      IDENTIFICATION DIVISION.
      PROGRAM-ID. Nth-Root.
      AUTHOR.  Bill Gunshannon.
      INSTALLATION.  
      DATE-WRITTEN.  4 Feb 2020.
     ************************************************************
     ** Program Abstract:
     **   Compute the Nth Root of a positive real number.
     **   
     **   Takes values from console.  If Precision is left
     **   blank defaults to 0.001.
     **   
     **   Enter 0 for first value to terminate program.
     ************************************************************
      
      ENVIRONMENT DIVISION.
      
      INPUT-OUTPUT SECTION.
      FILE-CONTROL.
           SELECT Root-File ASSIGN TO "Root-File"
                ORGANIZATION IS LINE SEQUENTIAL.
      
      DATA DIVISION.
      
      FILE SECTION.
      
      FD  Root-File
          DATA RECORD IS Parameters.
      01  Parameters.
          05 Root                       PIC 9(5).
          05 Num                        PIC 9(5)V9(5).
          05 Precision                  PIC 9V9(9).

      WORKING-STORAGE SECTION.
      
      01  TEMP0                         PIC 9(9)V9(9).
      01  TEMP1                         PIC 9(9)V9(9).
      01  RESULTS.
          05  Field1                        PIC ZZZZZ.ZZZZZ.
          05  FILLER                        PIC X(5).
          05  Field2                        PIC ZZZZ9.
          05  FILLER                        PIC X(14).
          05  Field3                        PIC 9.999999999.

      01  HEADER.
          05  FILLER                        PIC X(72) 
              VALUE "   Number           Root           Precision.".
      
      01  Disp-Root                         PIC ZZZZZ.ZZZZZ.
      
      PROCEDURE DIVISION.
      
      Main-Program.
          PERFORM FOREVER
          
             PERFORM Get-Input
             IF Precision = 0.0 
                 THEN MOVE 0.001 to Precision
             END-IF

             PERFORM Compute-Root

             MOVE Root TO Field2
             MOVE Num TO Field1
             MOVE Precision TO Field3
             DISPLAY HEADER
             DISPLAY RESULTS
             DISPLAY " "
             MOVE TEMP1 TO Disp-Root
             DISPLAY "The Root is: " Disp-Root
          END-PERFORM.
      
      Get-Input.
          DISPLAY "Input Base Number: " WITH NO ADVANCING
          ACCEPT Num
          IF Num EQUALS ZERO
             THEN 
                  DISPLAY "Good Bye."
                  STOP RUN
          END-IF
          DISPLAY "Input Root: " WITH NO ADVANCING
          ACCEPT Root
          DISPLAY "Input Desired Precision: " WITH NO ADVANCING
          ACCEPT Precision.

      Compute-Root.
         MOVE Root TO TEMP0
         DIVIDE Num BY Root GIVING TEMP1

         PERFORM UNTIL FUNCTION ABS(TEMP0 - TEMP1) 
                                   LESS THAN Precision 
              MOVE TEMP1 TO TEMP0
              COMPUTE TEMP1 = (( Root - 1.0) * TEMP1 + Num / 
                                       TEMP1 ** (Root - 1.0)) / Root
         END-PERFORM.
      
      END-PROGRAM.

</lang>

Output:

     
Input Base Number: 25.0
Input Root: 2
Input Desired Precision: 0.0001
   Root             Number         Precision.                           
   25.00000         2              0.000100000
 
The Root is:     5.00000
Input Base Number: 5642.0
Input Root: 125
Input Desired Precision: 
   Root             Number         Precision.                           
 5642.00000       125              0.001000000
 
The Root is:     1.07155
Input Base Number: 0
Good Bye.

CoffeeScript

<lang coffeescript> nth_root = (A, n, precision=0.0000000000001) ->

 x = 1
 while true
   x_new = (1 / n) * ((n - 1) * x + A / Math.pow(x, n - 1))
   return x_new if Math.abs(x_new - x) < precision
   x = x_new

tests

do ->

 tests = [
   [8, 3]
   [16, 4]
   [32, 5]
   [343, 3]
   [1024, 10]
   [1000000000, 3]
   [1000000000, 9]
   [100, 2]
   [100, 3]
   [100, 5]
   [100, 10]
 ]
 for test in tests
   [x, n] = test
   root = nth_root x, n
   console.log "#{x} root #{n} = #{root} (root^#{n} = #{Math.pow root, n})"

</lang> output <lang> > coffee nth_root.coffee 8 root 3 = 2 (root^3 = 8) 16 root 4 = 2 (root^4 = 16) 32 root 5 = 2 (root^5 = 32) 343 root 3 = 7 (root^3 = 343) 1024 root 10 = 2 (root^10 = 1024) 1000000000 root 3 = 1000 (root^3 = 1000000000) 1000000000 root 9 = 10 (root^9 = 1000000000) 100 root 2 = 10 (root^2 = 100) 100 root 3 = 4.641588833612778 (root^3 = 99.99999999999997) 100 root 5 = 2.5118864315095806 (root^5 = 100.0000000000001) 100 root 10 = 1.5848931924611134 (root^10 = 99.99999999999993) </lang>

Common Lisp

This version does not check for cycles in x_i and x_i+1, but finishes when the difference between them drops below ε. The initial guess can be provided, but defaults to n-1.

<lang lisp>(defun nth-root (n a &optional (epsilon .0001) (guess (1- n)))

 (assert (and (> n 1) (> a 0)))
 (flet ((next (x)
          (/ (+ (* (1- n) x)
                (/ a (expt x (1- n))))
             n)))
   (do* ((xi guess xi+1)
         (xi+1 (next xi) (next xi)))
        ((< (abs (- xi+1 xi)) epsilon) xi+1))))</lang>

nth-root may return rationals rather than floating point numbers, so easy checking for correctness may require coercion to floats. For instance,

<lang lisp>(let* ((r (nth-root 3 10))

      (rf (coerce r 'float)))
 (print (* r r r ))
 (print (* rf rf rf)))</lang>

produces the following output.

1176549099958810982335712173626176/117654909634627320192156007194483 
10.0

D

<lang d>import std.stdio, std.math;

real nthroot(in int n, in real A, in real p=0.001) pure nothrow {

   real[2] x = [A, A / n];
   while (abs(x[1] - x[0]) > p)
       x = [x[1], ((n - 1) * x[1] + A / (x[1] ^^ (n-1))) / n];
   return x[1];

}

void main() {

   writeln(nthroot(10, 7131.5 ^^ 10));
   writeln(nthroot(6, 64));

}</lang>

Output:

7131.5
2

Delphi

<lang delphi> USES

  Math;

function NthRoot(A, Precision: Double; n: Integer): Double; var

  x_p, X: Double;

begin

  x_p := Sqrt(A);
  while Abs(A - Power(x_p, n)) > Precision do
  begin
     x := (1/n) * (((n-1) * x_p) + (A/(Power(x_p, n - 1))));
     x_p := x;
  end;
  Result := x_p;

end; </lang>

E

Rather than choosing an arbitrary precision, this implementation continues until a cycle in the iterated result is found, thus producing an answer almost as precise as the number type.

(Disclaimer: This was not written by a numerics expert; there may be reasons this is a bad idea. Also, it might be that cycles are always of length 2, which would reduce the amount of calculation needed by 2/3.)

<lang e>def nthroot(n, x) {

 require(n > 1 && x > 0)
 def np := n - 1
 def iter(g) { return (np*g + x/g**np) / n }
 var g1 := x
 var g2 := iter(g1)
 while (!(g1 <=> g2)) {
   g1 := iter(g1)
   g2 := iter(iter(g2))
 }
 return g1

}</lang>

EasyLang

<lang>func power x n . r .

 r = 1
 for i range n
   r *= x
 .

. func nth_root x n . r .

 r = 2
 repeat
   call power r n - 1 p
   d = (x / p - r) / n
   r += d
   until abs d < 0.0001
 .

. call power 3.1416 10 x call nth_root x 10 r fscale 4 print r</lang>

Elixir

Translation of: Erlang

<lang elixir>defmodule RC do

 def nth_root(n, x, precision \\ 1.0e-5) do
   f = fn(prev) -> ((n - 1) * prev + x / :math.pow(prev, (n-1))) / n end
   fixed_point(f, x, precision, f.(x))
 end
 
 defp fixed_point(_, guess, tolerance, next) when abs(guess - next) < tolerance, do: next
 defp fixed_point(f, _, tolerance, next), do: fixed_point(f, next, tolerance, f.(next))

end

Enum.each([{2, 2}, {4, 81}, {10, 1024}, {1/2, 7}], fn {n, x} ->

 IO.puts "#{n} root of #{x} is #{RC.nth_root(n, x)}"

end)</lang>

Output:

2 root of 2 is 1.4142135623746899
4 root of 81 is 3.0
10 root of 1024 is 2.00000000022337
0.5 root of 7 is 48.99999999999993

Erlang

Done by finding the fixed point of a function, which aims to find a value of x for which f(x)=x: <lang erlang>fixed_point(F, Guess, Tolerance) ->

   fixed_point(F, Guess, Tolerance, F(Guess)).

fixed_point(_, Guess, Tolerance, Next) when abs(Guess - Next) < Tolerance ->

   Next;

fixed_point(F, _, Tolerance, Next) ->

   fixed_point(F, Next, Tolerance, F(Next)).</lang>

The nth root function algorithm defined on the wikipedia page linked above can advantage of this: <lang erlang>nth_root(N, X) -> nth_root(N, X, 1.0e-5). nth_root(N, X, Precision) ->

   F = fun(Prev) -> ((N - 1) * Prev + X / math:pow(Prev, (N-1))) / N end,
   fixed_point(F, X, Precision).</lang>

Excel

This will work in any spreadsheet that uses Excel-compatible expressions -- i.e. KOffice's KCells (formerly KSpread), Calligra Tables, and OpenOffice.org Calc.

Beside the obvious; <lang Excel>=A1^(1/B1)</lang>

Cell A1 is the base.
Cell B1 is the exponent.
Cell A2 is the first guess (any non-zero number will do).

In cell A3, enter this formula:

=((($B$1-1)*A2)+($A$1/(A2^($B$1-1))))/$B$1

Copy A3 down until you get 2 cells with the same value. (Once there are two visibly-identical cells, all cells below those two will also be identical.)

For example, here we calculate the cube root of 100:

	A	B
1	100	3
2	7	(first guess)
3	5.346938776
4	4.730544697
5	4.643251125
6	4.641589429
7	4.641588834
8	4.641588834

Alternately, Excel could use the BASIC example above as VBA code, deleting A2 and replacing A3's formula with something like this:

=RootX(A1,B1,.00000001)

F#

<lang fsharp> let nthroot n A =

   let rec f x =
       let m = n - 1.
       let x' = (m * x + A/x**m) / n
       match abs(x' - x) with
       | t when t < abs(x * 1e-9) -> x'
       | _ -> f x'
   f (A / double n)

[<EntryPoint>] let main args =

   if args.Length <> 2 then
       eprintfn "usage: nthroot n A"
       exit 1
   let (b, n) = System.Double.TryParse(args.[0])
   let (b', A) = System.Double.TryParse(args.[1])
   if (not b) || (not b') then
       eprintfn "error: parameter must be a number"
       exit 1
   printf "%A" (nthroot n A)
   0

</lang>

Compiled using fsc nthroot.fs example output:

nthroot 0.5 7
49.0

Factor

Translation of: Forth

<lang factor>USING: kernel locals math math.functions prettyprint ;

th-root ( a n -- a^1/n )

   a [
       a over n 1 - ^ /f
         over n 1 - *
       + n /f
       swap over 1e-5 ~ not
   ] loop ;

34 5 th-root . ! 2.024397458499888 34 5 recip ^ . ! 2.024397458499888</lang>

Forth

<lang forth>: th-root { F: a F: n -- a^1/n }

 a
 begin
   a fover n 1e f- f** f/
     fover n 1e f- f*
   f+ n f/
   fswap fover 1e-5 f~
 until ;

34e 5e th-root f. \ 2.02439745849989 34e 5e 1/f f** f. \ 2.02439745849989</lang>

Fortran

<lang fortran>program NthRootTest

 implicit none

 print *, nthroot(10, 7131.5**10)
 print *, nthroot(5, 34.0)

contains

 function nthroot(n, A, p)
   real :: nthroot
   integer, intent(in)        :: n
   real, intent(in)           :: A
   real, intent(in), optional :: p

   real :: rp, x(2)

   if ( A < 0 ) then
      stop "A < 0"       ! we handle only real positive numbers
   elseif ( A == 0 ) then
      nthroot = 0
      return
   end if

   if ( present(p) ) then
      rp = p
   else
      rp = 0.001
   end if

   x(1) = A
   x(2) = A/n   ! starting "guessed" value...

   do while ( abs(x(2) - x(1)) > rp )
      x(1) = x(2)
      x(2) = ((n-1.0)*x(2) + A/(x(2) ** (n-1.0)))/real(n)
   end do

   nthroot = x(2)

 end function nthroot

end program NthRootTest</lang>

FreeBASIC

<lang freebasic>' version 14-01-2019 ' compile with: fbc -s console

Function nth_root(n As Integer, number As Double) As Double

   Dim As Double a1 = number / n, a2 , a3

   Do
       a3 = Abs(a2 - a1)
       a2 = ((n -1) * a1 + number / a1 ^ (n -1)) / n
       Swap a1, a2
   Loop Until Abs(a2 - a1) = a3

   Return a1

End Function

' ------=< MAIN >=------

Dim As UInteger n Dim As Double tmp

Print Print " n 5643 ^ 1 / n nth_root ^ n" Print " ------------------------------------" For n = 3 To 11 Step 2

   tmp = nth_root(n, 5643)
   Print Using " ###    ###.########    ####.########"; n; tmp; tmp ^ n

   tmp = nth_root(n, 5643)
   Print Using " ###    ###.########    ####.########"; n; tmp; tmp ^ n

Output:

   n    5643 ^ 1 / n     nth_root ^ n
 ------------------------------------
   3     17.80341642    5643.00000000
   5      5.62732516    5643.00000000
   7      3.43502583    5643.00000000
   9      2.61116581    5643.00000000
  11      2.19303907    5643.00000000

  25      1.41273402    5643.00000000
  50      1.18858488    5643.00000000
  75      1.12207047    5643.00000000
 100      1.09022240    5643.00000000
 125      1.07154911    5643.00000000

FutureBasic

<lang futurebasic> include "ConsoleWindow"

def tab 8

local fn NthRoot( root as long, a as long, precision as double ) as double dim as double x0, x1

x0 = a : x1 = a /root while ( abs( x1 - x0 ) > precision ) x0 = x1 x1 = ( ( root -1.0 ) * x1 + a / x1 ^ ( root -1.0 ) ) /root wend end fn = x1

print " 125th Root of 5643 Precision .001", using "#.###############"; fn NthRoot( 125, 5642, 0.001 ) print " 125th Root of 5643 Precision .001", using "#.###############"; fn NthRoot( 125, 5642, 0.001 ) print " 125th Root of 5643 Precision .00001", using "#.###############"; fn NthRoot( 125, 5642, 0.00001 ) print " Cube Root of 27 Precision .00001", using "#.###############"; fn NthRoot( 3, 27, 0.00001 ) print "Square Root of 2 Precision .00001", using "#.###############"; fn NthRoot( 2, 2, 0.00001 ) print "Square Root of 2 Precision .00001", using "#.###############"; sqr(2) // Processor floating point calc deviation print " 10th Root of 1024 Precision .00001", using "#.###############"; fn NthRoot( 10, 1024, 0.00001 ) print " 5th Root of 34 Precision .00001", using "#.###############"; fn NthRoot( 5, 34, 0.00001 ) </lang> Output:

 125th Root of 5643 Precision .001      1.071559602191682
 125th Root of 5643 Precision .001      1.071559602191682
 125th Root of 5643 Precision .00001    1.071547591944772
  Cube Root of   27 Precision .00001    3.000000000000002
Square Root of    2 Precision .00001    1.414213562374690
Square Root of    2 Precision .00001    1.414213562373095
  10th Root of 1024 Precision .00001    2.000000000000000
   5th Root of   34 Precision .00001    2.024397458499885

Go

<lang go>func root(a float64, n int) float64 {

   n1 := n - 1
   n1f, rn := float64(n1), 1/float64(n)
   x, x0 := 1., 0.
   for {
       potx, t2 := 1/x, a
       for b := n1; b > 0; b >>= 1 {
           if b&1 == 1 {
               t2 *= potx
           }
           potx *= potx
       }
       x0, x = x, rn*(n1f*x+t2)
       if math.Abs(x-x0)*1e15 < x {
           break
       }
   }
   return x

}</lang>

The above version is for 64 bit wide floating point numbers. The following uses `math/big` Float to implement this same function with 256 bits of precision.

A set of wrapper functions around the somewhat muddled big math library functions is used to make the main function more readable, and also it was necessary to create a power function (Exp) as the library also lacks this function. The exponent in the limit must be at least one less than the number of bits of precision of the input value or the function will enter an infinite loop!

<lang go> import "math/big"

func Root(a *big.Float, n uint64) *big.Float { limit := Exp(New(2), 256) n1 := n-1 n1f, rn := New(float64(n1)), Div(New(1.0), New(float64(n))) x, x0 := New(1.0), Zero() _ = x0 for { potx, t2 := Div(New(1.0), x), a for b:=n1; b>0; b>>=1 { if b&1 == 1 { t2 = Mul(t2, potx) } potx = Mul(potx, potx) } x0, x = x, Mul(rn, Add(Mul(n1f, x), t2) ) if Lesser(Mul(Abs(Sub(x, x0)), limit), x) { break } } return x }

func Abs(a *big.Float) *big.Float { return Zero().Abs(a) }

func Exp(a *big.Float, e uint64) *big.Float { result := Zero().Copy(a) for i:=uint64(0); i<e-1; i++ { result = Mul(result, a) } return result }

func New(f float64) *big.Float { r := big.NewFloat(f) r.SetPrec(256) return r }

func Div(a, b *big.Float) *big.Float { return Zero().Quo(a, b) }

func Zero() *big.Float { r := big.NewFloat(0.0) r.SetPrec(256) return r }

func Mul(a, b *big.Float) *big.Float { return Zero().Mul(a, b) }

func Add(a, b *big.Float) *big.Float { return Zero().Add(a, b) }

func Sub(a, b *big.Float) *big.Float { return Zero().Sub(a, b) }

func Lesser(x, y *big.Float) bool { return x.Cmp(y) == -1 } </lang>

Groovy

Solution: <lang groovy>import static Constants.tolerance import static java.math.RoundingMode.HALF_UP

def root(double base, double n) {

   double xOld = 1
   double xNew = 0
   while (true) {
       xNew = ((n - 1) * xOld + base/(xOld)**(n - 1))/n
   if ((xNew - xOld).abs() < tolerance) { break }
       xOld = xNew
   }
   (xNew as BigDecimal).setScale(7, HALF_UP)

} </lang>

Test: <lang groovy>class Constants {

   static final tolerance = 0.00001

}

print

  Base   Power  Calc'd Root  Actual Root

------ ----------- -----------

def testCases = [

   [b:32.0, n:5.0, r:2.0],
   [b:81.0, n:4.0, r:3.0],
   [b:Math.PI**2, n:4.0, r:Math.PI**(0.5)],
   [b:7.0, n:0.5, r:49.0],

]

testCases.each {

   def r = root(it.b, it.n)
   printf('%7.4f  %6.4f  %11.4f  %11.4f\n',
       it.b, it.n, r, it.r)
   assert (r - it.r).abs() <= tolerance

}</lang>

Output:

   Base   Power  Calc'd Root  Actual Root
-------  ------  -----------  -----------
32.0000  5.0000       2.0000       2.0000
81.0000  4.0000       3.0000       3.0000
 9.8696  4.0000       1.7725       1.7725
 7.0000  0.5000      49.0000      49.0000

Haskell

Function exits when there's no difference between two successive values. <lang Haskell>n `nthRoot` x = fst $ until (uncurry(==)) (\(_,x0) -> (x0,((n-1)*x0+x/x0**(n-1))/n)) (x,x/n)</lang> Use:

*Main> 2 `nthRoot` 2
1.414213562373095

*Main> 5 `nthRoot` 34
2.024397458499885

*Main> 10 `nthRoot` (734^10)
734.0

*Main> 0.5 `nthRoot` 7
49.0

Or, in applicative terms, with formatted output:

<lang haskell>nthRoot :: Double -> Double -> Double nthRoot n x =

 fst $
 until
   (uncurry (==))
   (((,) <*> ((/ n) . ((+) . (pn *) <*> (x /) . (** pn)))) . snd)
   (x, x / n)
 where
   pn = pred n

TESTS --------------------------

main :: IO () main =

 putStrLn $
 fTable
   "Nth roots:"
   (\(a, b) -> show a ++ " `nthRoot` " ++ show b)
   show
   (uncurry nthRoot)
   [(2, 2), (5, 34), (10, 734 ^ 10), (0.5, 7)]

FORMAT OF RESULTS --------------------

fTable :: String -> (a -> String) -> (b -> String) -> (a -> b) -> [a] -> String fTable s xShow fxShow f xs =

 let w = maximum (length . xShow <$> xs)
     rjust n c = drop . length <*> (replicate n c ++)
 in unlines $
    s : fmap (((++) . rjust w ' ' . xShow) <*> (("  ->  " ++) . fxShow . f)) xs</lang>

Output:

Nth roots:
                  2.0 `nthRoot` 2.0  ->  1.414213562373095
                 5.0 `nthRoot` 34.0  ->  2.0243974584998847
10.0 `nthRoot` 4.539004352165717e28  ->  734.0
                  0.5 `nthRoot` 7.0  ->  49.0

HicEst

<lang HicEst>WRITE(Messagebox) NthRoot(5, 34) WRITE(Messagebox) NthRoot(10, 7131.5^10)

FUNCTION NthRoot(n, A)

  REAL :: prec = 0.001

  IF( (n > 0) * (A > 0) ) THEN
      NthRoot = A / n
      DO i = 1, 1/prec
        x = ((n-1)*NthRoot + A/(NthRoot^(n-1))) / n
        IF( ABS(x - NthRoot) <= prec ) THEN
            RETURN
        ENDIF
        NthRoot = x
      ENDDO
  ENDIF

  WRITE(Messagebox, Name) 'Cannot solve problem for:', prec, n, A

END</lang>

Icon and Unicon

All Icon/Unicon reals are double precision. <lang Icon>procedure main()

  showroot(125,3)
  showroot(27,3)
  showroot(1024,10)
  showroot(39.0625,4)
  showroot(7131.5^10,10)

end

procedure showroot(a,n)

  printf("%i-th root of %i = %i\n",n,a,root(a,n))

end

procedure root(a,n,p) #: finds the n-th root of the number a to precision p

  if n < 0 | type(n) !== "integer" then runerr(101,n)     
  if a < 0 then runerr(205,a)   
  /p := 1e-14                  # precision
  xn := a / real(n)            # initial guess
  while abs(a - xn^n) > p do       
     xn := ((n - 1) * (xi := xn) + a / (xi ^ (n-1))) / real(n)
  return xn

end

link printf</lang>

Output:

3-th root of 125 = 5.0
3-th root of 27 = 3.0
10-th root of 1024 = 2.0
4-th root of 39.0625 = 2.5
10-th root of 3.402584077894253e+038 = 7131.5

J

Translation of: E

J has a built in Nth root primitive, %:. For example, 7131.5 = 10 %: 7131.5^10. Also, the exponentiation primitive supports exponents < 1, e.g. 7131.5 = (7131.5^10)^(1%10).

But, since the talk page discourages using built-in facilities, here is a reimplementation, using the E algorithm:

<lang j> '`N X NP' =. (0 { [)`(1 { [)`(2 { [)

  iter      =.  N %~ (NP * ]) + X % ] ^ NP
  nth_root  =:  (, , _1+[) iter^:_ f. ]
  10 nth_root 7131.5^10

7131.5</lang>

Java

Translation of: Fortran

<lang java>public static double nthroot(int n, double A) { return nthroot(n, A, .001); } public static double nthroot(int n, double A, double p) { if(A < 0) { System.err.println("A < 0");// we handle only real positive numbers return -1; } else if(A == 0) { return 0; } double x_prev = A; double x = A / n; // starting "guessed" value... while(Math.abs(x - x_prev) > p) { x_prev = x; x = ((n - 1.0) * x + A / Math.pow(x, n - 1.0)) / n; } return x; }</lang>

Translation of: E

<lang java>public static double nthroot(int n, double x) {

 assert (n > 1 && x > 0);
 int np = n - 1;
 double g1 = x;
 double g2 = iter(g1, np, n, x);
 while (g1 != g2) {
   g1 = iter(g1, np, n, x);
   g2 = iter(iter(g2, np, n, x), np, n, x);
 }
 return g1;

}

private static double iter(double g, int np, int n, double x) {

 return (np * g + x / Math.pow(g, np)) / n;

}</lang>

JavaScript

Gives the n:nth root of num, with precision prec. (n defaults to 2 [e.g. sqrt], prec defaults to 12.)

<lang javascript>function nthRoot(num, nArg, precArg) {

 var n = nArg || 2;
 var prec = precArg || 12;
 
 var x = 1; // Initial guess.
 for (var i=0; i<prec; i++) {
   x = 1/n * ((n-1)*x + (num / Math.pow(x, n-1)));
 }
 
 return x;

}</lang>

jq

<lang jq># An iterative algorithm for finding: self ^ (1/n) to the given

absolute precision if "precision" > 0, or to within the precision
allowed by IEEE 754 64-bit numbers.

The following implementation handles underflow caused by poor estimates.

def iterative_nth_root(n; precision):

 def abs: if . < 0 then -. else . end;
 def sq: .*.;
 def pow(p): . as $in | reduce range(0;p) as $i (1; . * $in);
   def _iterate: # state: [A, x1, x2, prevdelta]
     .[0] as $A | .[1] as $x1 | .[2] as $x2 | .[3] as $prevdelta
     | ( $x2 | pow(n-1)) as $power
     | if $power <= 2.155094094640383e-309
       then  [$A, $x1, ($x1 + $x2)/2, n] | _iterate

else (((n-1)*$x2 + ($A/$power))/n) as $x1 | (($x1 - $x2)|abs) as $delta

       | if (precision == 0 and $delta == $prevdelta and $delta < 1e-15) 
            or (precision > 0 and $delta <= precision) or $delta == 0 then $x1
         else [$A, $x2, $x1, $delta] | _iterate
         end
       end
   ;
   if n == 1 then .
   elif . == 0 then 0
   elif . < 0 then error("iterative_nth_root: input \(.) < 0")
   elif n != (n|floor) then error("iterative_nth_root: argument \(n) is not an integer")
   elif n == 0 then error("iterative_nth_root(0): domain error")
   elif n < 0 then 1/iterative_nth_root(-n; precision)
   else [., ., (./n), n, 0]  | _iterate
   end

</lang>

Example: Compare the results of iterative_nth_root and nth_root implemented using builtins <lang jq>def demo(x):

 def nth_root(n): log / n | exp;
 def lpad(n): tostring | (n - length) * " " + .;
 . as $in
 | "\(x)^(1/\(lpad(5))): \(x|nth_root($in)|lpad(18)) vs \(x|iterative_nth_root($in; 1e-10)|lpad(18)) vs \(x|iterative_nth_root($in; 0))"

5^m for various values of n:

"5^(1/ n): builtin precision=1e-10 precision=0", ( (1,-5,-3,-1,1,3,5,1000,10000) | demo(5))</lang>

Output:

<lang sh>$ jq -n -r -f nth_root_machine_precision.jq 5^(1/ n): builtin precision=1e-10 precision=0 5^(1/ 1): 4.999999999999999 vs 5 vs 5 5^(1/ -5): 0.7247796636776955 vs 0.7247796636776956 vs 0.7247796636776955 5^(1/ -3): 0.5848035476425733 vs 0.5848035476425731 vs 0.5848035476425731 5^(1/ -1): 0.2 vs 0.2 vs 0.2 5^(1/ 1): 4.999999999999999 vs 5 vs 5 5^(1/ 3): 1.709975946676697 vs 1.709975946676697 vs 1.709975946676697 5^(1/ 5): 1.3797296614612147 vs 1.3797296614612147 vs 1.379729661461215 5^(1/ 1000): 1.0016107337527294 vs 1.0016107337527294 vs 1.0016107337527294 5^(1/10000): 1.0001609567433902 vs 1.0001609567433902 vs 1.0001609567433902 </lang>

Julia

Works with: Julia version 1.2

Julia has a built-in exponentiation function A^(1 / n), but the specification calls for us to use Newton's method (which we iterate until the limits of machine precision are reached):

<lang julia>function nthroot(n::Integer, r::Real)

   r < 0 || n == 0 && throw(DomainError())
   n < 0 && return 1 / nthroot(-n, r)
   r > 0 || return 0
   x = r / n
   prevdx = r
   while true
       y = x ^ (n - 1)
       dx = (r - y * x) / (n * y)
       abs(dx) ≥ abs(prevdx) && return x
       x += dx
       prevdx = dx
   end

end

@show nthroot.(-5:2:5, 5.0) @show nthroot.(-5:2:5, 5.0) - 5.0 .^ (1 ./ (-5:2:5))</lang>

Output:

nthroot.(-5:2:5, 5.0) = [0.7247796636776955, 0.5848035476425731, 0.2, 5.0, 1.709975946676697, 1.379729661461215]
nthroot.(-5:2:5, 5.0) - 5.0 .^ (1 ./ (-5:2:5)) = [0.0, -1.1102230246251565e-16, 0.0, 0.0, 0.0, 0.0]

Kotlin

Translation of: E

<lang scala>// version 1.0.6

fun nthRoot(x: Double, n: Int): Double {

   if (n < 2) throw IllegalArgumentException("n must be more than 1")
   if (x <= 0.0) throw IllegalArgumentException("x must be positive")
   val np = n - 1
   fun iter(g: Double) = (np * g + x / Math.pow(g, np.toDouble())) / n
   var g1 = x
   var g2 = iter(g1)
   while (g1 != g2) {
       g1 = iter(g1)
       g2 = iter(iter(g2))
   }
   return g1

}

fun main(args: Array<String>) {

  val numbers = arrayOf(1728.0 to 3, 1024.0 to 10, 2.0 to 2) 
  for (number in numbers)  
      println("${number.first} ^ 1/${number.second}\t = ${nthRoot(number.first, number.second)}")

}</lang>

Output:

1728.0 ^ 1/3     = 12.0
1024.0 ^ 1/10    = 2.0
2.0 ^ 1/2        = 1.414213562373095

langur

Langur has a root operator. Here, we show use of both the root operator and an nth root function.

Works with: langur version 0.8

Translation of: D

<lang langur>writeln "operator" writeln( (7131.5 ^ 10) ^/ 10 ) writeln 64 ^/ 6 writeln()

To make the example from the D language work, we set a low maximum for the number of digits after a decimal point in division.

mode divMaxScale = 7

val .nthroot = f(.n, .A, .p) {

   var .x = [.A, .A / .n]
   while abs(.x[2]-.x[1]) > .p {
       .x = [.x[2], ((.n-1) x .x[2] + .A / (.x[2] ^ (.n-1))) / .n]
   }
   simplify .x[2]

}

writeln "calculation" writeln .nthroot(10, 7131.5 ^ 10, 0.001) writeln .nthroot(6, 64, 0.001)</lang>

Output:

operator
7131.5
2

calculation
7131.5
2

Liberty BASIC

<lang lb> print "First estimate is: ", using( "#.###############", NthRoot( 125, 5642, 0.001 )); print " ... and better is: ", using( "#.###############", NthRoot( 125, 5642, 0.00001)) print "125'th root of 5642 by LB's exponentiation operator is "; using( "#.###############", 5642^(1 /125))

print "27^(1 / 3)", using( "#.###############", NthRoot( 3, 27, 0.00001)) print "2^(1 / 2)", using( "#.###############", NthRoot( 2, 2, 0.00001)) print "1024^(1 /10)", using( "#.###############", NthRoot( 10, 1024, 0.00001))

wait

function NthRoot( n, A, p)

 x( 0) =A
 x( 1) =A /n
 while abs( x( 1) -x( 0)) >p
   x( 0) =x( 1)
   x( 1) =( ( n -1.0) *x( 1) +A /x( 1)^( n -1.0)) /n
 wend
 NthRoot =x( 1)

end function

end

</lang>

First estimate is:          1.071559602191682    ... and better is:   1.071547591944771
125'th root of 5642 by LB's exponentiation operator is 1.071547591944767
27^(1 / 3)    3.000000000000002
2^(1 / 2)     1.414213562374690
1024^(1 /10)  2.000000000000000

Lingo

<lang lingo>on nthRoot (x, root)

 return power(x, 1.0/root)

end</lang> <lang lingo>the floatPrecision = 8 -- only about display/string cast of floats put nthRoot(4, 4) -- 1.41421356</lang>

Logo

<lang logo>to about :a :b

 output and [:a - :b < 1e-5] [:a - :b > -1e-5]

end

to root :n :a [:guess :a]

 localmake "next ((:n-1) * :guess + :a / power :guess (:n-1)) / n
 if about :guess :next [output :next]
 output (root :n :a :next)

end

show root 5 34 ; 2.02439745849989</lang>

Lua

<lang Lua> function nroot(root, num)

 return num^(1/root)

end </lang>

M2000 Interpreter

Using stack statements PUSH, READ, OVER, SHIFT, DROP, NUMBER, FLUSH

 
Flush empty stack
Over 2 copy 2nd as new top (so 2nd now is 3rd)
Over 2,2 repeat Over 2 two times.
Shift 2 send top to 2nd, and 2nd to top (1st) (there is a SHFITBACK to revesre action)
Drop drop top
Number get top if is number, else raise error
Read, read a variable form top. 
Functions parameters works with a read too
     Function Root {
             Read a, n%, d as double=1.e-4
      ......
      }
because we can send any type and number if function, interpreter can make conversions if we declare that,
or if it not possible (no conversion done to a numeric variable if a string is in top of stack) we get an error.
Also if we send less values, and we didn't initialize variable before, we get error too.  
Here we need to flush stack for other parameters if from an error anyone put more arguments. 
(interpreter never count before call a user function, except for calling events by using event object,
so there there is a signature to follow)


n% is double inside.

<lang M2000 Interpreter> Module Checkit {

     Function Root (a, n%, d as double=1.e-4) {
            if n%=0 then Error "Division by zero: 1/0" 
            if a<=0 then Error "Negative or zero number"
            if n%=1 then = a : exit
            Flush
            n2=1-1/n%:a/=n%:n%--:Push a
            {    Push 1: For i=1 to n% {Over 2 :Push Number*Number}
                 Over 2 : Push n2*Number + a/Number
                 Shift 2: Over 2, 2 :if Abs(Number-Number)>d Then loop
                 Drop
            }  Read a : = a
     }
     Print "square root single"
     Print root(1.3346767~, 2, 1.e-9)
     Print "square double"
     Print root(1.3346767, 2, 1.e-9)
     Print "square root decimal"
     Print root(1.3346767@, 2, 1.e-9)
     Print "internal square root, double"
     Print  1.3346767^(1/2)
     Print sqrt(1.3346767)

} Checkit </lang>

Maple

The root command performs this task. <lang Maple> root(1728, 3);

root(1024, 10);

root(2.0, 2); </lang>

Output:

                                     12

                                      2

                                 1.414213562

Mathematica

MATLAB

<lang MATLAB>function answer = nthRoot(number,root)

   format long

   answer = number / root;
   guess = number;
   
   while not(guess == answer)
      guess = answer;
      answer = (1/root)*( ((root - 1)*guess) + ( number/(guess^(root - 1)) ) ); 
   end

end</lang>

Sample Output: <lang MATLAB>>> nthRoot(2,2)

ans =

  1.414213562373095</lang>

Maxima

<lang maxima>nth_root(a, n) := block(

  [x, y, d, p: fpprec],
  fpprec: p + 10,
  x: bfloat(a),
  eps: 10.0b0^-p,
  y: do (
     d: bfloat((a / x^(n - 1) - x) / n),
     if abs(d) < eps * x then return(x),
     x: x + d
  ),
  fpprec: p,
  bfloat(y)

)$</lang>

Metafont

Metafont does not use IEEE floating point and we can't go beyond 0.0001 or it will loop forever. <lang metafont>vardef mnthroot(expr n, A) =

 x0 := A / n;
 m := n - 1;
 forever:
   x1 := (m*x0 + A/(x0 ** m)) / n;
   exitif abs(x1 - x0) < abs(x0 * 0.0001);
   x0 := x1;
 endfor;
 x1

enddef;

primarydef n nthroot A = mnthroot(n, A) enddef;

show 5 nthroot 34; % 2.0244 show 0.5 nthroot 7; % 49.00528

bye</lang>

МК-61/52

<lang>1/x <-> x^y С/П</lang> Instruction: number ^ degree В/О С/П

NetRexx

Translation of: REXX

<lang netrexx> /*NetRexx program to calculate the Nth root of X, with DIGS accuracy. */ class nth_root

 method main(args=String[]) static
   if args.length < 2 then
     do

say "at least 2 arguments expected" exit

     end
   x = args[0]
   root = args[1]
   if args.length > 2 then digs = args[2]

   if root== then root=2
   if digs = null, digs =  then digs=20
   numeric digits digs
   say '     x	= ' x
   say '  root	= ' root
   say 'digits	= ' digs
   say 'answer	= ' root(x,root,digs)
   
 method root(x,r,digs) static --procedure; parse arg x,R 1 oldR  /*assign 2nd arg-->r and rOrig.  */
   /*this subroutine will use the   */
   /*digits from the calling prog.  */
   /*The default digits is  9.      */
   R = r
   oldR = r
   if r=0 then do
     say
     say '*** error! ***'
     say "a root of zero can't be specified."
     say
     return '[n/a]'
   end
   
   R=R.abs()                              /*use absolute value of root.    */
   
   if x<0 & (R//2==0) then do
     say
     say '*** error! ***'
     say "an even root can't be calculated for a" -
     'negative number,'
     say 'the result would be complex.'
     say
     return '[n/a]'
   end
   
   if x=0 | r=1 then return x/1           /*handle couple of special cases.*/
   Rm1=R-1                                /*just a fast version of  ROOT-1 */
   oldDigs=digs                           /*get the current number of digs.*/
   dm=oldDigs+5                           /*we need a little guard room.   */
   ax=x.abs()                             /*the absolute value of  X.      */
   g=(ax+1)/r**r                          /*take a good stab at 1st guess. */
--   numeric fuzz 3                         /*fuzz digits for higher roots.  */
   d=5                                    /*start with only five digits.   */
   /*each calc doubles precision.   */
   
   loop forever
     
     d=d+d
     if d>dm then d = dm                        /*double the digits, but not>DM. */
     numeric digits d                     /*tell REXX to use   D   digits. */
     old=0                                /*assume some kind of old guess. */
     
     loop forever

_=(Rm1*g**R+ax)/R/g**rm1 /*this is the nitty-gritty stuff.*/ if _=g | _=old then leave /*computed close to this before? */ old=g /*now, keep calculation for OLD. */ g=_ /*set calculation to guesstimate.*/

     end
     
     if d==dm then leave                  /*found the root for DM digits ? */
   end

   _=g*x.sign()                           /*correct the sign (maybe).      */
   if oldR<0 then return _=1/_            /*root < 0 ?    Reciprocal it is.*/
   numeric digits oldDigs                 /*re-instate the original digits.*/
   return _/1                             /*normalize the number to digs.  */

</lang>

NewLISP

<lang NewLISP>(define (nth-root n a)

 (let ((x1 a)

(x2 (div a n)))

   (until (= x1 x2)
     (setq x1 x2

x2 (div (add (mul x1 (- n 1)) (div a (pow x1 (- n 1)))) n)))

   x2))</lang>

Nim

<lang nim>import math

proc nthRoot(a: float; n: int): float =

 var n = float(n)
 result = a
 var x = a / n
 while abs(result-x) > 1e-15:
   x = result
   result = (1/n) * (((n-1)*x) + (a / pow(x, n-1)))

echo nthRoot(34.0, 5) echo nthRoot(42.0, 10) echo nthRoot(5.0, 2)</lang> Output:

2.024397458499885
1.453198460282268
2.23606797749979

Objeck

Translation of: C

<lang objeck>class NthRoot {

 function : Main(args : String[]) ~ Nil {
   NthRoot(5, 34, .001)->PrintLine();
 }

 function : NthRoot(n : Int, A: Float, p : Float) ~ Float {
   x := Float->New[2];
   x[0] := A;
   x[1] := A / n;

   while((x[1] - x[0])->Abs() > p) {
     x[0] := x[1];
     x[1] := ((n - 1.0) * x[1] + A / x[1]->Power(n - 1.0)) / n;
   };

   return x[1];
 }

}</lang>

OCaml

Translation of: C

<lang ocaml>let nthroot ~n ~a ?(tol=0.001) () =

  let nf = float n in let nf1 = nf -. 1.0 in
  let rec iter x =
     let x' = (nf1 *. x +. a /. (x ** nf1)) /. nf in
     if tol > abs_float (x -. x') then x' else iter x' in
  iter 1.0

let () =

 Printf.printf "%g\n" (nthroot 10 (7131.5 ** 10.0) ());
 Printf.printf "%g\n" (nthroot 5 34.0 ());

</lang>

Octave

Octave has it's how nthroot function. <lang octave>

 r = A.^(1./n)

</lang>

Here it is another implementation (after Tcl)

Translation of: Tcl

<lang octave>function r = m_nthroot(n, A)

 x0 = A / n;
 m  = n - 1;
 while(1)
   x1 = (m*x0 + A./ x0 .^ m) / n;
   if ( abs(x1-x0) < abs(x0 * 1e-9) )
     r = x1;
     return
   endif
   x0 = x1;
 endwhile

endfunction</lang>

Here is an more elegant way by computing the successive differences in an explicit way: <lang octave>function r = m_nthroot(n, A)

 r = A / n;
 m = n - 1;
 do 
   d = (A ./ r .^ m - r) / n;
   r+= d;
 until (abs(d) < abs(r * 1e-9))

endfunction</lang>

Show its usage and the built-in nthroot function

<lang octave>m_nthroot(10, 7131.5 .^ 10) nthroot(7131.5 .^ 10, 10) m_nthroot(5, 34) nthroot(34, 5) m_nthroot(0.5, 7) nthroot(7, .5)</lang>

Oforth

<lang Oforth>Float method: nthroot(n)

  1.0 doWhile: [ self over n 1 - pow / over - n / tuck + swap 0.0 <> ] ;</lang>

Output:

734 10.0 powf nthroot(10) println
734

2.0 nthroot(2) println
1.41421356237309

34.0 nthroot(5) println
2.02439745849989

Oz

<lang oz>declare

 fun {NthRoot NInt A}
    N = {Int.toFloat NInt}

    fun {Next X}
       ( (N-1.0)*X + A / {Pow X N-1.0} ) / N
    end
 in
    {Until Value.'==' Next A/N}
 end

 fun {Until P F X}
    case {F X}
    of NX andthen {P NX X} then X
    [] NX then {Until P F NX}
    end
 end

in

 {Show {NthRoot 2 2.0}}</lang>

PARI/GP

Pascal

See Delphi

Perl

Translation of: Tcl

<lang perl>use strict;

sub nthroot ($$) {

   my ( $n, $A ) = @_;

   my $x0 = $A / $n;
   my $m = $n - 1.0;
   while(1) {

my $x1 = ($m * $x0 + $A / ($x0 ** $m)) / $n; return $x1 if abs($x1 - $x0) < abs($x0 * 1e-9); $x0 = $x1;

}</lang>

<lang perl>print nthroot(5, 34), "\n"; print nthroot(10, 7131.5 ** 10), "\n"; print nthroot(0.5, 7), "\n";</lang>

Phix

Translation of: AWK

(main loop)

Translation of: C

(use of pow_ instead of power)
<lang Phix>function pow_(atom x, integer e)

   atom r = 1
   for i=1 to e do
       r *= x
   end for
   return r

end function

function nth_root(atom y,n) atom eps = 1e-15 -- relative accuracy atom x = 1

   while 1 do

-- atom d = ( y / power(x,n-1) - x ) / n

       atom d = ( y / pow_(x,n-1) - x ) / n
       x += d
       atom e = eps*x   -- absolute accuracy       
       if d > -e and d < e then exit end if
   end while 
   return {y,n,x,power(y,1/n)}

end function

?nth_root(1024,10) ?nth_root(27,3) ?nth_root(2,2) ?nth_root(5642,125) --?nth_root(7,0.5) -- needs power(), not pow_() ?nth_root(4913,3) ?nth_root(8,3) ?nth_root(16,2) ?nth_root(16,4) ?nth_root(125,3) ?nth_root(1000000000,3) ?nth_root(1000000000,9)</lang>

Output:

Shows inputs and both the iterative and builtin results.

{1024,10,2,2}
{27,3,3,3}
{2,2,1.414213562,1.414213562}
{5642,125,1.071547592,1.071547592}
{4913,3,17,17.0}
{8,3,2,2}
{16,2,4,4}
{16,4,2,2}
{125,3,5,5}
{1000000000,3,1000,1000.0}
{1000000000,9,10,10.0}

Phixmonti

<lang Phixmonti>def nthroot var n var y 1e-15 var eps /# relative accuracy #/ 1 var x true while y x n 1 - power / x - n / var d x d + var x eps x * var e /# absolute accuracy #/ d 0 e - < d e > or endwhile x enddef

def printList len for get print endfor enddef

10 1024 3 27 2 2 125 5642 4 16 stklen tolist

len 1 swap 2 3 tolist for var i i get swap i 1 + get rot var e var b "The " e "th root of " b " is " b 1 e / power " (" b e nthroot ")" 9 tolist printList drop nl endfor</lang>

PHP

<lang PHP>function nthroot($number, $root, $p = P) {

   $x[0] = $number;
   $x[1] = $number/$root;
   while(abs($x[1]-$x[0]) > $p)
   {
       $x[0] = $x[1];
       $x[1] = (($root-1)*$x[1] + $number/pow($x[1], $root-1))/$root;
   }
   return $x[1];

}</lang>

PicoLisp

<lang PicoLisp>(load "@lib/math.l")

(de nthRoot (N A)

  (let (X1 A  X2 (*/ A N))
     (until (= X1 X2)
        (setq
           X1 X2
           X2 (*/
              (+
                 (* X1 (dec N))
                 (*/ A 1.0 (pow X1 (* (dec N) 1.0))) )
              N ) ) )
     X2 ) )

(prinl (format (nthRoot 2 2.0) *Scl)) (prinl (format (nthRoot 3 12.3) *Scl)) (prinl (format (nthRoot 4 45.6) *Scl))</lang> Output:

1.414214
2.308350
2.598611

PL/I

<lang PL/I>/* Finds the N-th root of the number A */ root: procedure (A, N) returns (float);

  declare A float, N fixed binary;
  declare (xi, xip1) float;

  xi = 1; /* An initial guess */
  do forever;
     xip1 = ((n-1)*xi + A/xi**(n-1) ) / n;
     if abs(xip1-xi) < 1e-5 then leave;
     xi = xip1;
  end;
  return (xi);

end root;</lang> Results:

The 2-th root of 4.00000E+0000 is  2.00000E+0000
The 5-th root of 3.20000E+0001 is  2.00000E+0000
The 3-th root of 2.70000E+0001 is  3.00000E+0000
The 2-th root of 2.00000E+0000 is  1.41422E+0000
The 3-th root of 1.00000E+0002 is  4.64159E+0000

PowerShell

This sample implementation does not use [System.Math] classes. <lang powershell>#NoTeS: This sample code does not validate inputs

Thus, if there are errors the 'scary' red-text
error messages will appear.
This code will not work properly in floating point values of n,
and negative values of A.
Supports negative values of n by reciprocating the root.

$epsilon=1E-10 #Sample Epsilon (Precision)

function power($x,$e){ #As I said in the comment $ret=1 for($i=1;$i -le $e;$i++){ $ret*=$x } return $ret } function root($y,$n){ #The main Function if (0+$n -lt 0){$tmp=-$n} else {$tmp=$n} #This checks if n is negative. $ans=1

do{ $d = ($y/(power $ans ($tmp-1)) - $ans)/$tmp $ans+=$d } while ($d -lt -$epsilon -or $d -gt $epsilon)

if (0+$n -lt 0){return 1/$ans} else {return $ans} }

Sample Inputs

root 625 2 root 2401 4 root 2 -2 root 1.23456789E-20 34 root 9.87654321E20 10 #Quite slow here, I admit...

((root 5 2)+1)/2 #Extra: Computes the golden ratio ((root 5 2)-1)/2</lang>

Output:

PS> .\NTH.PS1
25
7
0.707106781186548
0.259690655650288
125.736248016373
1.61803398874989
0.618033988749895
PS>

Prolog

Uses integer math, though via scaling, it can approximate non-integral roots to arbitrary precision. <lang Prolog> iroot(_, 0, 0) :- !. iroot(M, N, R) :-

   M > 1,
   (N > 0 ->
       irootpos(M, N, R)
   ;
       N /\ 1 =:= 1,
       NegN is -N, irootpos(M, NegN, R0), R is -R0).

irootpos(N, A, R) :-

   X0 is 1 << (msb(A) div N),  % initial guess is 2^(log2(A) / N)
   newton(N, A, X0, X1),
   iroot_loop(A, X1, N, A, R).

iroot_loop(X1, X2, _, _, X1) :- X1 =< X2, !. iroot_loop(_, X1, N, A, R) :-

   newton(N, A, X1, X2),
   iroot_loop(X1, X2, N, A, R).

newton(2, A, X0, X1) :- X1 is (X0 + A div X0) >> 1, !. % fast special case newton(N, A, X0, X1) :- X1 is ((N - 1)*X0 + A div X0**(N - 1)) div N. </lang>

Output:

?- iroot(3, 10000, X).
X = 21.

?- A is 2**(1/12).  % 12-root of 2 via built-in
A = 1.0594630943592953.

?- A is 2 * 10**(12 * 15), iroot(12, A, R), format("~15d", [R]). % 12-root of 2 via scaled int
1.059463094359295
A = 2000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000,
R = 1059463094359295.

?- iroot(2, 81, X).
X = 9.

?- iroot(4, -256, X).  % fails for negative even roots
false.

?- iroot(3, -27, X).  % succeeds for negative odd roots
X = -3.

PureBasic

<lang PureBasic>#Def_p=0.001

Procedure.d Nth_root(n.i, A.d, p.d=#Def_p)

 Protected Dim x.d(1) 
 x(0)=A: x(1)=A/n
 While Abs(x(1)-x(0))>p
   x(0)=x(1)
   x(1)=((n-1.0)*x(1)+A/Pow(x(1),n-1.0))/n
 Wend
 ProcedureReturn x(1)

EndProcedure

//////////////////////////////

Debug "125'th root of 5642 is" Debug Pow(5642,1/125) Debug "First estimate is:" Debug Nth_root(125,5642) Debug "And better:" Debug Nth_root(125,5642,0.00001)</lang> Outputs

125'th root of 5642 is
1.0715475919447675
First estimate is:
1.0715596021916822
And better:
1.0715475919447714

Python

<lang python>from decimal import Decimal, getcontext

def nthroot (n, A, precision):

   getcontext().prec = precision
   
   n = Decimal(n)
   x_0 = A / n #step 1: make a while guess.
   x_1 = 1     #need it to exist before step 2
   while True:
       #step 2:
       x_0, x_1 = x_1, (1 / n)*((n - 1)*x_0 + (A / (x_0 ** (n - 1))))
       if x_0 == x_1:
           return x_1</lang>

<lang python>print nthroot(5, 34, 10) print nthroot(10,42, 20) print nthroot(2, 5, 400)</lang>

Or, in terms of a general until function:

Works with: Python version 3.7

<lang python>Nth Root

from decimal import Decimal, getcontext from operator import eq

nthRoot :: Int -> Int -> Int -> Real

def nthRoot(precision):

   The nth root of x at the given precision.
   def go(n, x):
       getcontext().prec = precision
       dcn = Decimal(n)

       def same(ab):
           return eq(*ab)

       def step(ab):
           a, b = ab
           predn = pred(dcn)
           return (
               b,
               reciprocal(dcn) * (
                   predn * a + (
                       x / (a ** predn)
                   )
               )
           )
       return until(same)(step)(
           (x / dcn, 1)
       )[0]
   return lambda n: lambda x: go(n, x)

--------------------------TEST---------------------------

def main():

   Nth roots at various precisions

   def xShow(tpl):
       p, n, x = tpl
       return rootName(n) + (
           ' of ' + str(x) + ' at precision ' + str(p)
       )

   def f(tpl):
       p, n, x = tpl
       return nthRoot(p)(n)(x)

   print(
       fTable(main.__doc__ + ':\n')(xShow)(str)(f)(
           [(10, 5, 34), (20, 10, 42), (30, 2, 5)]
       )
   )

-------------------------DISPLAY-------------------------

fTable :: String -> (a -> String) ->
(b -> String) -> (a -> b) -> [a] -> String

def fTable(s):

   Heading -> x display function -> fx display function ->
      f -> xs -> tabular string.
   
   def go(xShow, fxShow, f, xs):
       ys = [xShow(x) for x in xs]
       w = max(map(len, ys))
       return s + '\n' + '\n'.join(map(
           lambda x, y: y.rjust(w, ' ') + ' -> ' + fxShow(f(x)),
           xs, ys
       ))
   return lambda xShow: lambda fxShow: lambda f: lambda xs: go(
       xShow, fxShow, f, xs
   )

-------------------------GENERIC-------------------------

rootName :: Int -> String

def rootName(n):

   English ordinal suffix.
   return ['identity', 'square root', 'cube root'][n - 1] if (
       4 > n or 1 > n
   ) else (str(n) + 'th root')

pred :: Enum a => a -> a

def pred(x):

   The predecessor of a value. For numeric types, (- 1).
   return x - 1

reciprocal :: Num -> Num

def reciprocal(x):

   Arithmetic reciprocal of x.
   return 1 / x

until :: (a -> Bool) -> (a -> a) -> a -> a

def until(p):

   The result of repeatedly applying f until p holds.
      The initial seed value is x.
   
   def go(f, x):
       v = x
       while not p(v):
           v = f(v)
       return v
   return lambda f: lambda x: go(f, x)

if __name__ == '__main__':

   main()</lang>

Output:

Nth roots at various precisions:

  5th root of 34 at precision 10 -> 2.024397458
 10th root of 42 at precision 20 -> 1.4531984602822678165
square root of 5 at precision 30 -> 2.23606797749978969640917366873

R

<lang R>nthroot <- function(A, n, tol=sqrt(.Machine$double.eps)) {

  ifelse(A < 1, x0 <- A * n, x0 <- A / n)
  repeat
  {
     x1 <- ((n-1)*x0 + A / x0^(n-1))/n
     if(abs(x1 - x0) > tol) x0 <- x1 else break
  }
  x1

} nthroot(7131.5^10, 10) # 7131.5 nthroot(7, 0.5) # 49</lang>

Racket

<lang Racket>#lang racket

(define (nth-root number root (tolerance 0.001))

 (define (acceptable? next current)
   (< (abs (- next current)) tolerance))
 
 (define (improve current)
   (/ (+ (* (- root 1) current) (/ number (expt current (- root 1)))) root))
 
 (define (loop current)
   (define next-guess (improve current))
   (if (acceptable? next-guess current)
       next-guess
       (loop next-guess)))
 (loop 1.0))</lang>

Raku

(formerly Perl 6)

<lang perl6>sub nth-root ($n, $A, $p=1e-9) {

   my $x0 = $A / $n;
   loop {
       my $x1 = (($n-1) * $x0 + $A / ($x0 ** ($n-1))) / $n;
       return $x1 if abs($x1-$x0) < abs($x0 * $p);
       $x0 = $x1;
   }

}

say nth-root(3,8);</lang>

RATFOR

<lang RATFOR> program nth

integer root real number, precision real temp0, temp1

1 format('Enter the base number: ') 2 format('Enter the desired root: ') 3 format('Enter the desired precision: ') 4 format(F12.6) 5 format(I6) write(6,1) read(5,4)number write(6,2) read(5,5)root write(6,3) read(5,4)precision

temp0 = number temp1 = number/root

while ( abs(temp0 - temp1) > precision )

  {
     temp0 = temp1
     temp1 = ((root - 1.0) * temp1 + number / temp1 ** (root - 1.0)) / root
  }

6 format(' number root precision') write(6,6) 7 format(f12.6,i6,f12.6) write (6,7)number,root,precision 8 format('The root is: ',F12.6) write (6,8)temp1

end </lang>


Results:

Enter the base number: 
25.0
Enter the desired root: 
2
Enter the desired precision: 
.0001
  number      root    precision
   25.000000     2    0.000100
The root is:     5.000000

Enter the base number: 
65536.0
Enter the desired root: 
16
Enter the desired precision: 
.0001
  number      root    precision
65536.000000    16    0.000100
The root is:     2.000000

REXX

<lang rexx>/*REXX program calculates the Nth root of X, with DIGS (decimal digits) accuracy. */ parse arg x root digs . /*obtain optional arguments from the CL*/ if x== | x=="," then x= 2 /*Not specified? Then use the default.*/ if root== | root=="," then root= 2 /* " " " " " " */ if digs== | digs=="," then digs=65 /* " " " " " " */ numeric digits digs /*set the decimal digits to DIGS. */ say ' x = ' x /*echo the value of X. */ say ' root = ' root /* " " " " ROOT. */ say ' digits = ' digs /* " " " " DIGS. */ say ' answer = ' root(x, root) /*show the value of ANSWER. */ exit /*stick a fork in it, we're all done. */ /*──────────────────────────────────────────────────────────────────────────────────────*/ root: procedure; parse arg x 1 Ox, r 1 Or /*arg1 ──► x & Ox, 2nd ──► r & Or*/

     if r==  then r=2                               /*Was root specified?  Assume √. */
     if r=0    then return '[n/a]'                    /*oops-ay!  Can't do zeroth root.*/
     complex= x<0 & R//2==0                           /*will the result be complex?    */
     oDigs=digits()                                   /*get the current number of digs.*/
     if x=0 | r=1  then return x/1                    /*handle couple of special cases.*/
     dm=oDigs+5                                       /*we need a little guard room.   */
     r=abs(r);   x=abs(x)                             /*the absolute values of R and X.*/
     rm=r-1                                           /*just a fast version of  ROOT -1*/
     numeric form                                     /*take a good guess at the root─┐*/
     parse value format(x,2,1,,0) 'E0' with ? 'E' _ . /* ◄────────────────────────────┘*/
     g= (? / r'E'_ % r)  +  (x>1)                     /*kinda uses a crude "logarithm".*/
     d=5                                              /*start with five decimal digits.*/
          do until d==dm;   d=min(d+d,dm)             /*each time,  precision doubles. */
          numeric digits d                            /*tell REXX to use   D   digits. */
          old=-1                                      /*assume some kind of old guess. */
                 do until old=g;   old=g              /*where da rubber meets da road─┐*/
                 g=format((rm*g**r+x)/r/g**rm,, d-2)  /* ◄────── the root computation─┘*/
                 end   /*until old=g*/                /*maybe until the cows come home.*/
          end          /*until d==dm*/                /*and wait for more cows to come.*/

     if g=0        then return 0                      /*in case the jillionth root = 0.*/
     if Or<0       then g=1/g                         /*root < 0 ?   Reciprocal it is! */
     if \complex   then g=g*sign(Ox)                  /*adjust the sign  (maybe).      */
     numeric digits oDigs                             /*reinstate the original digits. */
     return (g/1)  ||  left('j', complex)             /*normalize # to digs, append j ?*/</lang>

output when using the default inputs:

       x =  2
    root =  2
  digits =  65
  answer =  1.414213562373095048801688724209698078569671875376948073176679738

output when using for input: 10 3

       x =  10
    root =  3
  digits =  65
  answer =  2.1544346900318837217592935665193504952593449421921085824892355063

output when using for input: 625 -4

       x =  625
    root =  -4
  digits =  65
  answer =  0.2

output when using for input: 100.666 47

       x =  100.666
    root =  47
  digits =  65
  answer =  1.1030990940616109102886569991014966919115206420386192403152621652

output when using for input: -256 8

       x =  -256
    root =  8
  digits =  65
  answer =  2j

output when using for input: 12345678900098765432100.00987654321000123456789e333 19

       x =  12345678900098765432100.00987654321000123456789e333
    root =  19
  digits =  65
  answer =  4886828567991886455.3257854108687610458584138783288904955196401434

Ring

<lang ring> decimals(12) see "cube root of 5 is : " + root(3, 5, 0) + nl

func root n, a, d y = 0 x = a / n while fabs (x - y) > d

     y = ((n - 1)*x + a/pow(x,(n-1))) / n
     temp = x
     x = y
     y = temp

end return x </lang> Output:

cube root of 5 is : 1.709975946677

Ruby

<lang ruby>def nthroot(n, a, precision = 1e-5)

 x = Float(a)
 begin
   prev = x
   x = ((n - 1) * prev + a / (prev ** (n - 1))) / n
 end while (prev - x).abs > precision
 x

end

p nthroot(5,34) # => 2.02439745849989</lang>

Run BASIC

<lang runbasic>print "Root 125th Root of 5643 Precision .001 ";using( "#.###############", NthRoot( 125, 5642, 0.001 )) print "125th Root of 5643 Precision .001 ";using( "#.###############", NthRoot( 125, 5642, 0.001 )) print "125th Root of 5643 Precision .00001 ";using( "#.###############", NthRoot( 125, 5642, 0.00001)) print " 3rd Root of 27 Precision .00001 ";using( "#.###############", NthRoot( 3, 27, 0.00001)) print " 2nd Root of 2 Precision .00001 ";using( "#.###############", NthRoot( 2, 2, 0.00001)) print " 10th Root of 1024 Precision .00001 ";using( "#.###############", NthRoot( 10, 1024, 0.00001))

wait

function NthRoot( root, A, precision)

 x0 = A
 x1 = A /root
 while abs( x1 -x0) >precision
   x0 = x1
   x1 = x1 / 1.0                                ' force float
   x1 = (( root -1.0) *x1 +A /x1^( root -1.0)) /root
 wend
 NthRoot =x1

end function

end</lang>

125th Root of 5643 Precision .001   1.071559602456735
125th Root of 5643 Precision .00001 1.071547591944771
  3rd Root of   27 Precision .00001 3.000000000000001
  2nd Root of    2 Precision .00001 1.414213562374690
 10th Root of 1024 Precision .00001 2.000000000000000

Rust

Translation of: Raku

<lang rust>// 20210212 Rust programming solution

fn nthRoot(n: f64, A: f64) -> f64 {

  let      p  =  1e-9_f64 ;
  let mut x0  =     A / n ;

  loop {
     let mut x1 = ( (n-1.0) * x0 + A / f64::powf(x0, n-1.0) ) / n;
     if (x1-x0).abs() < (x0*p).abs() { return x1 };
     x0 = x1
  }

}

fn main() {

  println!("{}", nthRoot(3. , 8. ));

}</lang>

Sather

Translation of: Octave

<lang sather>class MATH is

 nthroot(n:INT, a:FLT):FLT
   pre n > 0
 is  
   x0 ::= a / n.flt;
   m  ::= n - 1;
   loop
     x1 ::= (m.flt * x0 + a/(x0^(m.flt))) / n.flt;
     if (x1 - x0).abs < (x0 * 1.0e-9).abs then
       return x1;
     end;
     x0 := x1;
   end;
 end;

end;</lang>

<lang sather>class MAIN is

 main is
   a:FLT := 2.5 ^ 10.0;
   #OUT + MATH::nthroot(10, a) + "\n";
 end;

end;</lang>

Scala

Using tail recursion: <lang Scala>def nroot(n: Int, a: Double): Double = {

 @tailrec
 def rec(x0: Double) : Double = {
   val x1 = ((n - 1) * x0 + a/math.pow(x0, n-1))/n
   if (x0 <= x1) x0 else rec(x1)
 }
 
 rec(a)

}</lang>

Alternatively, you can implement the iteration with an iterator like so: <lang scala>def fallPrefix(itr: Iterator[Double]): Iterator[Double] = itr.sliding(2).dropWhile(p => p(0) > p(1)).map(_.head) def nrootLazy(n: Int)(a: Double): Double = fallPrefix(Iterator.iterate(a){r => (((n - 1)*r) + (a/math.pow(r, n - 1)))/n}).next</lang>

Scheme

<lang scheme>(define (root number degree tolerance)

 (define (good-enough? next guess)
   (< (abs (- next guess)) tolerance))
 (define (improve guess)
   (/ (+ (* (- degree 1) guess) (/ number (expt guess (- degree 1)))) degree))
 (define (*root guess)
   (let ((next (improve guess)))
     (if (good-enough? next guess)
         guess
         (*root next))))
 (*root 1.0))

(display (root (expt 2 10) 10 0.1)) (newline) (display (root (expt 2 10) 10 0.01)) (newline) (display (root (expt 2 10) 10 0.001)) (newline)</lang> Output:

2.04732932236839
2.00463204835482
2.00004786858167

Seed7

The nth root of the number 'a' can be computed with the exponentiation operator: 'a ** (1 / n)'. An alternate function which uses Newton's method is:

<lang seed7>const func float: nthRoot (in integer: n, in float: a) is func

 result
   var float: x1 is 0.0;
 local
   var float: x0 is 0.0;
 begin 
   x0 := a;
   x1 := a / flt(n);
   while abs(x1 - x0) >= abs(x0 * 1.0E-9) do
     x0 := x1;
     x1 := (flt(pred(n)) * x0 + a / x0 ** pred(n)) / flt(n);
   end while;
 end func;</lang>

Original source: [1]

Sidef

Translation of: Ruby

<lang ruby>func nthroot(n, a, precision=1e-5) {

 var x    = 1.float
 var prev = 0.float
 while ((prev-x).abs > precision) {
   prev = x;
   x = (((n-1)*prev + a/(prev**(n-1))) / n)
 }
 return x

}

say nthroot(5, 34) # => 2.024397458501034082599817835297912829678314204</lang>

A minor optimization would be to calculate the successive int(n-1) square roots of a number, then raise the result to the power of 2**(int(n-1) / n).

<lang ruby>func nthroot_fast(n, a, precision=1e-5) {

 { a = nthroot(2, a, precision) } * int(n-1)
 a ** (2**int(n-1) / n)

}

say nthroot_fast(5, 34, 1e-64) # => 2.02439745849988504251081724554193741911462170107</lang>

Smalltalk

Works with: GNU Smalltalk

Translation of: Tcl

<lang smalltalk>Number extend [

   nthRoot: n [

|x0 m x1| x0 := (self / n) asFloatD. m := n - 1. [true] whileTrue: [ x1 := ( (m * x0) + (self/(x0 raisedTo: m))) / n. ((x1 - x0) abs) < ((x0 * 1e-9) abs) ifTrue: [ ^ x1 ]. x0 := x1 ]

].</lang>

<lang smalltalk>(34 nthRoot: 5) displayNl. ((7131.5 raisedTo: 10) nthRoot: 10) displayNl. (7 nthRoot: 0.5) displayNl.</lang>

SPL

<lang spl>nthr(n,r) <= n^(1/r)

nthroot(n,r)=

 a = n/r
 g = n
 > g!=a
   g = a
   a = (1/r)*(((r-1)*g)+(n/(g^(r-1))))
 <
 <= a

.

.output(nthr(2,2))
.output(nthroot(2,2))</lang>

Output:

1.4142135623731
1.41421356237309

Swift

<lang swift>extension FloatingPoint where Self: ExpressibleByFloatLiteral {

 @inlinable
 public func power(_ e: Int) -> Self {
   var res = Self(1)

   for _ in 0..<e {
     res *= self
   }

   return res
 }

 @inlinable
 public func root(n: Int, epsilon: Self = 2.220446049250313e-16) -> Self {
   var d = Self(0)
   var res = Self(1)

   guard self != 0 else {
     return 0
   }

   guard n >= 1 else {
     return .nan
   }

   repeat {
     d = (self / res.power(n - 1) - res) / Self(n)
     res += d
   } while d >= epsilon * 10 || d <= -epsilon * 10

   return res
 }

}

print(81.root(n: 4)) print(13.root(n: 5))</lang>

Output:

3.0
1.6702776523348104

Tcl

The easiest way is to just use the pow function (or exponentiation operator) like this: <lang tcl>proc nthroot {n A} {

   expr {pow($A, 1.0/$n)}

}</lang> However that's hardly tackling the problem itself. So here's how to do it using Newton-Raphson and a self-tuning termination test.

Works with: Tcl version 8.5

<lang tcl>proc nthroot {n A} {

   set x0 [expr {$A / double($n)}]
   set m [expr {$n - 1.0}]
   while 1 {
       set x1 [expr {($m*$x0 + $A/$x0**$m) / $n}]
       if {abs($x1 - $x0) < abs($x0 * 1e-9)} {
           return $x1
       }
       set x0 $x1
   }

}</lang> Demo: <lang tcl>puts [nthroot 2 2] puts [nthroot 5 34] puts [nthroot 5 [expr {34**5}]] puts [nthroot 10 [expr 7131.5**10]] puts [nthroot 0.5 7]; # Squaring!</lang> Output:

1.414213562373095
2.0243974584998847
34.0
7131.5
49.0

Ursala

The nthroot function defined below takes a natural number n to the function that returns the n-th root of its floating point argument. Error is on the order of machine precision because the stopping criterion is either a fixed point or a repeating cycle. <lang Ursala>#import nat

import flo

nthroot =

-+

  ("n","n-1"). "A". ("x". div\"n" plus/times("n-1","x") div("A",pow("x","n-1")))^== 1.,
  float^~/~& predecessor+-</lang>

This implementation is unnecessary in practice due to the availability of the library function pow, which performs exponentiation and allows fractional exponents. Here is a test program. <lang Ursala>#cast %eL

examples =

<

  nthroot2 2.,
  nthroot5 34.,
  nthroot5 pow(34.,5.),
  nthroot10 pow(7131.5,10.)></lang>

output:

<
   1.414214e+00,
   2.024397e+00,
   3.400000e+01,
   7.131500e+03>

VBA

Translation of: Phix

The internal power operator "^" is used in stead of an auxiliary pow_ function and the accuracy has been reduced. <lang vb>Private Function nth_root(y As Double, n As Double)

   Dim eps As Double: eps = 0.00000000000001 '-- relative accuracy
   Dim x As Variant: x = 1
   Do While True
       d = (y / x ^ (n - 1) - x) / n
       x = x + d
       e = eps * x '-- absolute accuracy
       If d > -e And d < e Then
           Exit Do
       End If
   Loop
   Debug.Print y; n; x; y ^ (1 / n)

End Function Public Sub main()

   nth_root 1024, 10
   nth_root 27, 3
   nth_root 2, 2
   nth_root 5642, 125
   nth_root 7, 0.5
   nth_root 4913, 3
   nth_root 8, 3
   nth_root 16, 2
   nth_root 16, 4
   nth_root 125, 3
   nth_root 1000000000, 3
   nth_root 1000000000, 9

End Sub</lang>

Output:

 1024  10  2  2 
 27  3  3  3 
 2  2  1,41421356237309  1,4142135623731 
 5642  125  1,07154759194477  1,07154759194477 
 7  0,5  49  49 
 4913  3  17  17 
 8  3  2  2 
 16  2  4  4 
 16  4  2  2 
 125  3  5  5 
 1000000000  3  1000  1000 
 1000000000  9  10  10

Wren

Translation of: E

<lang ecmascript>var nthRoot = Fn.new { |x, n|

   if (n < 2) Fiber.abort("n must be more than 1")
   if (x <= 0) Fiber.abort("x must be positive")
   var np = n - 1
   var iter = Fn.new { |g| (np*g + x/g.pow(np))/n }
   var g1 = x
   var g2 = iter.call(g1)
   while (g1 != g2) {
       g1 = iter.call(g1)
       g2 = iter.call(iter.call(g2))
   }
   return g1

}

var trios = [ [1728, 3, 2], [1024, 10, 1], [2, 2, 5] ] for (trio in trios) {

   System.print("%(trio[0]) ^ 1/%(trio[1])%(" "*trio[2]) = %(nthRoot.call(trio[0], trio[1]))")

}</lang>

Output:

1728 ^ 1/3   = 12
1024 ^ 1/10  = 2
2 ^ 1/2      = 1.4142135623731

XPL0

<lang XPL0>include c:\cxpl\stdlib;

func real NRoot(A, N); \Return the Nth root of A real A, N; real X, X0, Y; int I; [X:= 1.0; \initial guess repeat X0:= X;

       Y:= 1.0;
       for I:= 1 to fix(N)-1 do Y:= Y*X0;
       X:= ((N-1.0)*X0 + A/Y) / N;

until abs(X-X0) < 1.0E-15; \(until X=X0 doesn't always work) return X; ];

[Format(5, 15); RlOut(0, NRoot( 2., 2.)); CrLf(0); RlOut(0, Power( 2., 0.5)); CrLf(0); \for comparison RlOut(0, NRoot(27., 3.)); CrLf(0); RlOut(0, NRoot(1024.,10.)); CrLf(0); ]</lang>

Output:

    1.414213562373100
    1.414213562373100
    3.000000000000000
    2.000000000000000

Yabasic

Translation of: AWK

<lang Yabasic>data 10, 1024, 3, 27, 2, 2, 125, 5642, 4, 16, 0, 0

do

   read e, b
   if e = 0 break
   print "The ", e, "th root of ", b, " is ", b^(1/e), " (", nthroot(b, e), ")"

loop

sub nthroot(y, n)

   local eps, x, d, e
   
   eps = 1e-15     // relative accuracy
   x   = 1
   repeat
       d  = ( y / ( x^(n-1) ) - x ) / n
       x = x + d
       e = eps * x // absolute accuracy
       
   until(not(d < -e or d > e ))

   return x

end sub</lang>

zkl

Translation of: Ruby

<lang zkl>fcn nthroot(nth,a,precision=1.0e-5){

  x:=prev:=a=a.toFloat(); n1:=nth-1;
  do{
     prev=x;
     x=( prev*n1 + a/prev.pow(n1) ) / nth;
  }    
  while( not prev.closeTo(x,precision) );
  x

}

nthroot(5,34) : "%.20f".fmt(_).println() # => 2.02439745849988828041</lang>