Nth root

From Rosetta Code
Task
Nth root
You are encouraged to solve this task according to the task description, using any language you may know.
Task

Implement the algorithm to compute the principal   nth   root     of a positive real number   A,   as explained at the   Wikipedia page.



11l[edit]

Translation of: Nim
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))
Output:
2.0244
1.4532
2.23607

360 Assembly[edit]

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.

*        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
Output:
     1.414213

AArch64 Assembly[edit]

Works with: as version Raspberry Pi 3B version Buster 64 bits
/* 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"
Output:
Root= +2.000000000000000
Root= +1.414213562373095

Action![edit]

INCLUDE "H6:REALMATH.ACT"

PROC NthRoot(REAL POINTER a,n REAL POINTER res)
  REAL n1,eps,one,tmp1,tmp2,tmp3
  
  ValR("0.0001",eps)
  IntToReal(1,one)
  RealSub(n,one,n1)

  Sqrt(a,res)            ;res=sqrt(a)
  DO
    Power(res,n,tmp1)    ;tmp=res^n
    RealSub(a,tmp1,tmp2) ;tmp2=a-res^n
    RealAbs(tmp2,tmp1)       ;tmp1=abs(a-res^n)
    IF RealGreaterOrEqual(eps,tmp1) THEN
      RETURN
    FI

    Power(res,n1,tmp1)      ;tmp1=res^(n-1)
    RealDiv(a,tmp1,tmp2)    ;tmp2=a/(res^(n-1))
    RealMult(n1,res,tmp1)   ;tmp1=(n-1)*res
    RealAdd(tmp1,tmp2,tmp3) ;tmp3=((n-1)*res + a/(res^(n-1)))
    RealDiv(tmp3,n,res)     ;res=((n-1)*res + a/(res^(n-1)))/n
  OD
RETURN

PROC Test(CHAR ARRAY sa,sn)
  REAL a,n,res

  ValR(sa,a)
  ValR(sn,n)
  PrintR(n) Print(" root of ")
  PrintR(a) Print(" is ")
  NthRoot(a,n,res)
  PrintRE(res)
RETURN

PROC Main()
  Put(125) PutE() ;clear screen
  MathInit()
  Test("2","2")
  Test("81","4")
  Test("1024","10")
  Test("7","0.5")
  Test("12.34","56.78")
RETURN
Output:

Screenshot from Atari 8-bit computer

2 root of 2 is 1.41421355
4 root of 81 is 3.00000047
10 root of 1024 is 2.00000001
.5 root of 7 is 48.99975418
56.78 root of 12.34 is 1.04524972

Ada[edit]

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 xk+1xk.

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;

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[edit]

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
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))
)

Output:

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

ALGOL W[edit]

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.
Output:
    7131.500000
       2.000000

ARM Assembly[edit]

Works with: as version Raspberry Pi
/* 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

Arturo[edit]

Translation of: Nim
nthRoot: function [a,n][
    N: to :floating n
    result: a
    x: a / N
    while [0.000000000000001 < abs result-x][
        x: result
        result: (1//n) * add (n-1)*x a/pow x n-1
    ]
    return result
]

print nthRoot 34.0 5
print nthRoot 42.0 10
print nthRoot 5.0 2
Output:
2.024397458499885
1.453198460282268
2.23606797749979

AutoHotkey[edit]

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
}

Message box shows:

7131.500000
2.024397
1.414214
49.000000

AutoIt[edit]

;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

output :

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

AWK[edit]

#!/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
}

Sample output:

 2
 4 
 2
 5
 1.44225
 1.73205

BASIC[edit]

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.

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

Note that for the above to work in QBasic, the function definition needs to be changed like so:

FUNCTION RootX# (tBase AS DOUBLE, tExp AS DOUBLE, diffLimit AS DOUBLE)

The function is called like so:

PRINT "The "; e; "th root of "; b; " is "; RootX(b, e, .000001)

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[edit]

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

BASIC256[edit]

function nth_root(n, a)
    precision = 0.0001

    dim x(2)
    x[0] = a
    x[1] = a /n
    while abs(x[1] - x[0]) > precision
        x[0] = x[1]
        x[1] = ((n -1.0) * x[1] +a / x[1]^(n -1.0)) / n
    end while
    return x[1]
end function

print "  n    5643 ^ 1 / n        nth_root ^ n"
print " --------------------------------------"
for n = 3 to 11 step 2
    tmp = nth_root(n, 5643)
    print " "; n; "    "; tmp; chr(9); (tmp ^ n)
next n
print
for n = 25 to 125 step 25
    tmp = nth_root(n, 5643)
    print n; "    "; tmp; chr(9); (tmp ^ n)
next n

BBC BASIC[edit]

      *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

Output:

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

bc[edit]

/* 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)
}

BQN[edit]

There are two builtin methods to solve this problem( and ), both give a result at the highest precision possible.

Root2 is a translation of the Run BASIC implementation, which uses Newton's approximation method. It allows an optional argument for precision, and otherwise defaults to 10¯5.

_while_ is a BQNcrate idiom used for unbounded looping here.

_while_ ← {𝔽⍟𝔾∘𝔽_𝕣_𝔾∘𝔽⍟𝔾𝕩}
Root ← √
Root1 ← ⋆⟜÷˜
Root2 ← {
  n 𝕊 a‿prec:
  1⊑{
    p‿x:
    ⟨
      x
      ((p × n - 1) + a ÷ p ⋆ n - 1) ÷ n
    ⟩
  } _while_ {
    p‿x:
    prec ≤ | p - x
  } ⟨a, ⌊a÷n⟩;
  𝕨 𝕊 𝕩: 𝕨 𝕊 𝕩‿1E¯5
}

•Show 3 Root 5
•Show 3 Root1 5
•Show 3 Root2 5
•Show 3 Root2 5‿1E¯16
1.7099759466766968
1.7099759466766968
1.7099759641072136
1.709975946676697

Try It!

Bracmat[edit]

Bracmat does not have floating point numbers as primitive type. Instead we have to use rational numbers. This code is not fast!

( ( 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)
)

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[edit]

Implemented without using math library, because if we were to use pow(), the whole exercise wouldn't make sense.

#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;
}

C#[edit]

Almost exactly how C works.

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];
}

C++[edit]

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;
   };
double NthRoot(double value, double degree)
{
    return pow(value, (double)(1 / degree));
};

Clojure[edit]

(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

COBOL[edit]

       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.
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[edit]

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})"

output

> 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)

Common Lisp[edit]

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

(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))))

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

(let* ((r (nth-root 3 10))
       (rf (coerce r 'float)))
  (print (* r r r ))
  (print (* rf rf rf)))

produces the following output.

1176549099958810982335712173626176/117654909634627320192156007194483 
10.0

D[edit]

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));
}
Output:
7131.5
2

Delphi[edit]

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;

E[edit]

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.)

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
}

EasyLang[edit]

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
numfmt 0 4
print r

Elixir[edit]

Translation of: Erlang
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)
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[edit]

Done by finding the fixed point of a function, which aims to find a value of x for which f(x)=x:

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)).

The nth root function algorithm defined on the wikipedia page linked above can advantage of this:

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).

Excel[edit]

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;

=A1^(1/B1)
  • 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#[edit]

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
Compiled using fsc nthroot.fs example output:
nthroot 0.5 7
49.0

Factor[edit]

Translation of: Forth
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

Forth[edit]

: 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

Fortran[edit]

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

FreeBASIC[edit]

' 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
Next

Print
For n = 25 To 125 Step 25
    tmp = nth_root(n, 5643)
    Print Using " ###    ###.########    ####.########"; n; tmp; tmp ^ n
Next

' empty keyboard buffer
While Inkey <> "" : Wend
Print : Print "hit any key to end program"
Sleep
End
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[edit]

window 1

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

HandleEvents

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[edit]

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
}

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!

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
}

Groovy[edit]

Solution:

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)
}

Test:

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
}

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[edit]

Function exits when there's no difference between two successive values.

n `nthRoot` x = fst $ until (uncurry(==)) (\(_,x0) -> (x0,((n-1)*x0+x/x0**(n-1))/n)) (x,x/n)

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:

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
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[edit]

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

Icon and Unicon[edit]

All Icon/Unicon reals are double precision.

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
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[edit]

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:

   '`N X NP' =.  (0 { [)`(1 { [)`(2 { [)
   iter      =.  N %~ (NP * ]) + X % ] ^ NP
   nth_root  =:  (, , _1+[) iter^:_ f. ]
   10 nth_root 7131.5^10
7131.5

Java[edit]

Translation of: Fortran
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;
}
Translation of: E
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;
}

JavaScript[edit]

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

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;
}

jq[edit]

# 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
;

Example: Compare the results of iterative_nth_root and nth_root implemented using builtins

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))
Output:
$ 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

Julia[edit]

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):

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))
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[edit]

Translation of: E
// 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)}")
}
Output:
1728.0 ^ 1/3     = 12.0
1024.0 ^ 1/10    = 2.0
2.0 ^ 1/2        = 1.414213562373095

Lambdatalk[edit]

Translation of Scheme

{def root 
  {def good-enough? {lambda {next guess tol}
    {< {abs {- next guess}} tol} }}
  {def improve {lambda {guess num deg}
    {/ {+ {* {- deg 1} guess} 
             {/ num {pow guess {- deg 1}}}} deg} }}
  {def *root {lambda {guess num deg tol}
    {let { {guess guess} {num num} {deg deg} {tol tol}
           {next {improve guess num deg}}
         } {if {good-enough? next guess tol}
            then guess
            else {*root next num deg tol}} }}}
 {lambda {num deg tol}
  {*root 1.0 num deg tol} }}
-> root 

{root {pow 2 10} 10 0.1}
-> 2.0473293223683866
{root {pow 2 10} 10 0.01}
-> 2.004632048354822
{root {pow 2 10} 10 0.001}
-> 2.000047868581671

langur[edit]

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
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)
Output:
operator
7131.5
2

calculation
7131.5
2

Liberty BASIC[edit]

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
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[edit]

on nthRoot (x, root)
  return power(x, 1.0/root)
end
the floatPrecision = 8 -- only about display/string cast of floats
put nthRoot(4, 4)
-- 1.41421356

[edit]

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

Lua[edit]

function nroot(root, num)
  return num^(1/root)
end

M2000 Interpreter[edit]

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.


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

Maple[edit]

The root command performs this task.

root(1728, 3);

root(1024, 10);

root(2.0, 2);

Output:

                                     12

                                      2

                                 1.414213562

Mathematica/Wolfram Language[edit]

Root[A,n]

MATLAB[edit]

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

Sample Output:

>> nthRoot(2,2)

ans =

   1.414213562373095

Maxima[edit]

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)
)$

Metafont[edit]

Metafont does not use IEEE floating point and we can't go beyond 0.0001 or it will loop forever.

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

МК-61/52[edit]

1/x	<->	x^y	С/П

Instruction: number ^ degree В/О С/П

NetRexx[edit]

Translation of: REXX
/*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.  */

NewLISP[edit]

(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))

Nim[edit]

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)

Output:

2.024397458499885
1.453198460282268
2.23606797749979

Objeck[edit]

Translation of: C
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];
  }
}

OCaml[edit]

Translation of: C
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 ());
;;

Octave[edit]

Octave has it's how nthroot function.

  r = A.^(1./n)

Here it is another implementation (after Tcl)

Translation of: Tcl
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

Here is an more elegant way by computing the successive differences in an explicit way:

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

Show its usage and the built-in nthroot function

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)

Oforth[edit]

Float method: nthroot(n)
   1.0 doWhile: [ self over n 1 - pow / over - n / tuck + swap 0.0 <> ] ;
Output:
734 10.0 powf nthroot(10) println
734

2.0 nthroot(2) println
1.41421356237309

34.0 nthroot(5) println
2.02439745849989

Oz[edit]

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}}

PARI/GP[edit]

root(n,A)=A^(1/n);

Pascal[edit]

See Delphi

Perl[edit]

Translation of: Tcl
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;
    }
}
print nthroot(5, 34), "\n";
print nthroot(10, 7131.5 ** 10), "\n";
print nthroot(0.5, 7), "\n";

Phix[edit]

Main loop copied from AWK, and as per C uses pow_() instead of power() since using the latter would make the whole exercise somewhat pointless.

with javascript_semantics
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
         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 x
end function

procedure test(sequence yn)
    atom {y,n} = yn
    printf(1,"nth_root(%d,%d) = %.10g, builtin = %.10g\n",{y,n,nth_root(y,n),power(y,1/n)})
end procedure
papply({{1024,10},{27,3},{2,2},{5642,125},{4913,3},{8,3},{16,2},{16,4},{125,3},{1000000000,3},{1000000000,9}},test)

Note that a {7,0.5} test would need to use power() instead of pow_().

Output:
nth_root(1024,10) = 2, builtin = 2
nth_root(27,3) = 3, builtin = 3
nth_root(2,2) = 1.414213562, builtin = 1.414213562
nth_root(5642,125) = 1.071547592, builtin = 1.071547592
nth_root(4913,3) = 17, builtin = 17
nth_root(8,3) = 2, builtin = 2
nth_root(16,2) = 4, builtin = 4
nth_root(16,4) = 2, builtin = 2
nth_root(125,3) = 5, builtin = 5
nth_root(1000000000,3) = 1000, builtin = 1000
nth_root(1000000000,9) = 10, builtin = 10

Phixmonti[edit]

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

PHP[edit]

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];
}

Picat[edit]

go =>
  L = [[2,2],
       [34,5],
       [34**5,5],
       [7131.5**10],
       [7,0.5],
       [1024,10],
       [5642, 125]
  ],
  foreach([A,N] in L)
    R = nthroot(A,N),
    printf("nthroot(%8w,%8w) %20w (check: %w)\n",A,N,R,A**(1/N))
  end,
  nl.

%
% x^n = a
%
% Given a and n, find x (to Precision)
%
nthroot(A,N) = nthroot(A,N,0.000001).

nthroot(A,N,Precision) = X1 => 
  NF = N * 1.0, % float version of N
  X0 = A / NF,
  X1 = 1.0,
  do 
      X0 := X1,
      X1 := (1.0 / NF)*((NF - 1.0)*X0 + (A / (X0 ** (NF - 1))))
  while( abs(X0-X1) > Precision).
Output:
nthroot(       2,       2)    1.414213562373095 (check: 1.414213562373095)
nthroot(      34,       5)    2.024397458499885 (check: 2.024397458499885)
nthroot(45435424,       5)                 34.0 (check: 34.000000000000007)
nthroot(       7,     0.5)   48.999999999999993 (check: 49.0)
nthroot(    1024,      10)                  2.0 (check: 2.0)
nthroot(    5642,     125)    1.071547591944767 (check: 1.071547591944767)

PicoLisp[edit]

(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))

Output:

1.414214
2.308350
2.598611

PL/I[edit]

/* 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;

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[edit]

This sample implementation does not use [System.Math] classes.

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

Prolog[edit]

Uses integer math, though via scaling, it can approximate non-integral roots to arbitrary precision.

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.
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[edit]

#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)

Outputs

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

Python[edit]

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
print nthroot(5, 34, 10)
print nthroot(10,42, 20)
print nthroot(2, 5, 400)

Or, in terms of a general until function:

Works with: Python version 3.7
'''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()
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[edit]

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

Racket[edit]

#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))

Raku[edit]

(formerly Perl 6)

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);

RATFOR[edit]

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

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[edit]

/*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 ?*/

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[edit]

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

Output:

cube root of 5 is : 1.709975946677

Ruby[edit]

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

Run BASIC[edit]

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
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[edit]

Translation of: Raku
// 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. ));
}

Sather[edit]

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

S-BASIC[edit]

When single precision results are sufficient for the task at hand, resort to Newton's method seems unnecessarily cumbersome, given the ready availability of S-BASIC's built-in exp and natural log functions.

rem - return nth root of x
function nthroot(x, n = real) = real
end = exp((1.0 / n) * log(x))

rem - exercise the routine by finding successive roots of 144
var i = integer

print "Finding the nth root of x"
print "  x      n         root"
print "-----------------------"
for i = 1 to 8
   print using "###   ####    ###.####"; 144; i; nthroot(144, i)
next i

end

But if the six or seven digits supported by S-BASIC's single-precision REAL data type is insufficient, Newton's Method is the way to go, given that the built-in exp and natural log functions are only single-precision.

rem - return the nth root of real.double value x to stated precision
function nthroot(n, x, precision = real.double) = real.double
   var x0, x1 = real.double
   x0 = x
   x1 = x / n   rem - initial guess
   while abs(x1 - x0) > precision do
      begin
        x0 = x1
        x1 = ((n-1.0) * x1 + x / x1 ^ (n-1.0)) / n
      end
end = x1

rem -- exercise the routine

var i = integer
print "Finding the nth root of 144 to 6 decimal places"
print "  x      n        root"
print "------------------------"
for i = 1 to 8
   print using "###   ####    ###.######"; 144; i; nthroot(i, 144.0, 1E-7)
next i

end
Output:

From the second version of the program.

Finding the nth root of 144 to 6 decimal places
  x      n         root
-------------------------
144      1     144.000000
144      2      12.000000
144      3       5.241483
144      4       3.464102
144      5       2.701920
144      6       2.289428
144      7       2.033937
144      8       1.861210

Scala[edit]

Using tail recursion:

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)
}

Alternatively, you can implement the iteration with an iterator like so:

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

Scheme[edit]

(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)

Output:

2.04732932236839
2.00463204835482
2.00004786858167

Seed7[edit]

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:

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;

Original source: [1]

Sidef[edit]

Translation of: 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

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).

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

Smalltalk[edit]

Works with: GNU Smalltalk
Translation of: Tcl
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
	]
    ]
].
(34 nthRoot: 5) displayNl.
((7131.5 raisedTo: 10) nthRoot: 10) displayNl.
(7 nthRoot: 0.5) displayNl.

SPL[edit]

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))
Output:
1.4142135623731
1.41421356237309

Swift[edit]

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))
Output:
3.0
1.6702776523348104

Tcl[edit]

The easiest way is to just use the pow function (or exponentiation operator) like this:

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

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
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
    }
}

Demo:

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!

Output:

1.414213562373095
2.0243974584998847
34.0
7131.5
49.0

True BASIC[edit]

FUNCTION Nroot (n, a)
    LET precision = .00001

    LET x1 = a
    LET x2 = a / n
    DO WHILE ABS(x2 - x1) > precision
       LET x1 = x2
       LET x2 = ((n - 1) * x2 + a / x2 ^ (n - 1)) / n
    LOOP
    LET Nroot = x2
END FUNCTION

PRINT "   n    5643 ^ 1 / n     nth_root ^ n"
PRINT " ------------------------------------"
FOR n = 3 TO 11 STEP 2
    LET tmp = Nroot(n, 5643)
    PRINT USING "####": n;
    PRINT "    ";
    PRINT USING "###.########": tmp;
    PRINT "    ";
    PRINT USING "####.########": (tmp ^ n)
NEXT n
PRINT
FOR n = 25 TO 125 STEP 25
    LET tmp = Nroot(n, 5643)
    PRINT USING "####": n;
    PRINT "    ";
    PRINT USING "###.########": tmp;
    PRINT "    ";
    PRINT USING "####.########": (tmp ^ n)
NEXT n
END

Ursala[edit]

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.

#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+-

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.

#cast %eL

examples =

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

output:

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

VBA[edit]

Translation of: Phix

The internal power operator "^" is used in stead of an auxiliary pow_ function and the accuracy has been reduced.

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
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[edit]

Translation of: E
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]))")
}
Output:
1728 ^ 1/3   = 12
1024 ^ 1/10  = 2
2 ^ 1/2      = 1.4142135623731

XBS[edit]

func nthRoot(x,a){
	send x^(1/a);
}{a=2};
log(nthRoot(8,3));
Output:
2

XPL0[edit]

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);
]

Output:

    1.414213562373100
    1.414213562373100
    3.000000000000000
    2.000000000000000

Yabasic[edit]

Translation of: AWK
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

zkl[edit]

Translation of: Ruby
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