# Nth root

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

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

## 11l

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

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
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

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
/*******************************************/
/* UnInitialized data                      */
/*******************************************/
.bss
.align 4
/*******************************************/
/*  code section                           */
/*******************************************/
.text
.global main
main:                                   // entry of program

/* root 10ieme de 1024  */
ldr d0,[x0]                         // load number in registre d0
scvtf d0,d0                         // conversion in float
mov x0,#10                          // N
bl nthRoot
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
// 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

/******************************************************************/
/*     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
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                  //
cmp x2,x1                      // n -1 ?
blt 2b                         // no -> loop
fdiv d7,d0,d1                  // A / b pow (n-1)
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
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!

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
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:
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


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

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

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

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,[r0]
vmov s0,r0                          @
vcvt.f64.s32 d0, s0                 @conversion in float single précision (32 bits)
mov r0,#10                          @ N
bl nthRoot
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
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

/******************************************************************/
/*     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
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                  @
cmp r2,r1                          @ n -1 ?
blt 2b                             @ no -> loop
vdiv.f64 d7,d0,d1                  @ A / b pow (n-1)
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

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

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

;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

#!/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

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).

### Basic09

PROCEDURE nth
PARAM N : INTEGER; A, P : REAL
DIM   TEMP0, TEMP1 : REAL
TEMP0 := A
TEMP1 := A/N
WHILE ( abs(TEMP0 - TEMP1) > P) DO
TEMP0 := TEMP1
TEMP1 := (( N - 1.0) * TEMP1 + A / TEMP1 ^ (N - 1.0)) / N
ENDWHILE
PRINT "Root            Number          Precision"
PRINT N, A, P
PRINT "The Root is: ";TEMP1
END

### BASIC256

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

      *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


### Chipmunk Basic

Translation of: S-BASIC
10 rem Nth root
20 print "Finding the nth root of 144 to 6 decimal places"
30 print "  x      n        root"
40 print "------------------------"
50 for i = 1 to 8
60    print using "###   ";144;
70    print using "####    ";i;
80    print using "###.######";nthroot(i,144,1.000000E-07)
90 next i
100 end
1000 sub nthroot(n,x,precision)
1010   rem Returns the nth root of value x to stated precision
1020   x0 = x
1030   x1 = x/n ' - initial guess
1040   while abs(x1-x0) > precision
1050     x0 = x1
1060     x1 = ((n-1)*x1+x/x1^(n-1))/n
1070   wend
1080   nthroot = x1
1090 end sub

Output:
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


### Craft Basic

precision 6

let a = int(rnd * 5999) + 2

print "calculating nth root of ", a, "..."

for n = 1 to 10

gosub nroot
print n, " : ", y

next n

end

sub nroot

let p = .00001

let x = a
let y = a / n

do

if abs(x - y) > p then

let x = y
let y = ((n - 1) * y + a / y ^ (n - 1)) / n

endif

wait

loop abs(x - y) > p

return


### FreeBASIC

' version 14-01-2019
' compile with: fbc -s console

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

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

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

Return a1

End Function

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

Dim As UInteger n
Dim As Double tmp

Print
Print "   n    5643 ^ 1 / n     nth_root ^ n"
Print " ------------------------------------"
For n = 3 To 11 Step 2
tmp = nth_root(n, 5643)
Print Using " ###    ###.########    ####.########"; n; tmp; tmp ^ n
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

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


### Liberty BASIC

Works with: Just BASIC
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
Output:
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


### PureBasic

#Def_p=0.001

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

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

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


### Run BASIC

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

### S-BASIC

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


### True BASIC

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


### Yabasic

Translation of: AWK
data 10, 1024, 3, 27, 2, 2, 125, 5642, 4, 16, 0, 0

do
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

### VBA

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


## bc

/* Take the nth root of 'a' (a positive real number).
* 'n' must be an integer.
* Result will have 'd' digits after the decimal point.
*/
define r(a, n, d) {
auto e, o, x, y, z

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

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

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

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


## BQN

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


## Bracmat

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

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#

Almost exactly how C works.

static void Main(string[] args)
{
Console.WriteLine(NthRoot(81,2,.001));
Console.WriteLine(NthRoot(1000,3,.001));
}

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++

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

(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

       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.

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 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

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

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

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

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

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

proc power x n . r .
r = 1
for i = 1 to n
r *= x
.
.
proc 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 4 0
print r


## Elixir

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

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)).


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

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#

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

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

: 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

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


## Go

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


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 {
}

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

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

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 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 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 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 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 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 # 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:
$jq -n -r -f nth_root_machine_precision.jq 5^(1/ n): builtin precision=1e-10 precision=0 5^(1/ 1): 4.999999999999999 vs 5 vs 5 5^(1/ -5): 0.7247796636776955 vs 0.7247796636776956 vs 0.7247796636776955 5^(1/ -3): 0.5848035476425733 vs 0.5848035476425731 vs 0.5848035476425731 5^(1/ -1): 0.2 vs 0.2 vs 0.2 5^(1/ 1): 4.999999999999999 vs 5 vs 5 5^(1/ 3): 1.709975946676697 vs 1.709975946676697 vs 1.709975946676697 5^(1/ 5): 1.3797296614612147 vs 1.3797296614612147 vs 1.379729661461215 5^(1/ 1000): 1.0016107337527294 vs 1.0016107337527294 vs 1.0016107337527294 5^(1/10000): 1.0001609567433902 vs 1.0001609567433902 vs 1.0001609567433902  ## Julia Works with: Julia version 1.2 Julia has a built-in exponentiation function A^(1 / n), but the specification calls for us to use Newton's method (which we iterate until the limits of machine precision are reached): function nthroot(n::Integer, r::Real) r < 0 || n == 0 && throw(DomainError()) n < 0 && return 1 / nthroot(-n, r) r > 0 || return 0 x = r / n prevdx = r while true y = x ^ (n - 1) dx = (r - y * x) / (n * y) abs(dx) ≥ abs(prevdx) && return x x += dx prevdx = dx end end @show nthroot.(-5:2:5, 5.0) @show nthroot.(-5:2:5, 5.0) - 5.0 .^ (1 ./ (-5:2:5))  Output: nthroot.(-5:2:5, 5.0) = [0.7247796636776955, 0.5848035476425731, 0.2, 5.0, 1.709975946676697, 1.379729661461215] nthroot.(-5:2:5, 5.0) - 5.0 .^ (1 ./ (-5:2:5)) = [0.0, -1.1102230246251565e-16, 0.0, 0.0, 0.0, 0.0] ## Kotlin Translation of: E // version 1.0.6 fun nthRoot(x: Double, n: Int): Double { if (n < 2) throw IllegalArgumentException("n must be more than 1") if (x <= 0.0) throw IllegalArgumentException("x must be positive") val np = n - 1 fun iter(g: Double) = (np * g + x / Math.pow(g, np.toDouble())) / n var g1 = x var g2 = iter(g1) while (g1 != g2) { g1 = iter(g1) g2 = iter(iter(g2)) } return g1 } fun main(args: Array<String>) { val numbers = arrayOf(1728.0 to 3, 1024.0 to 10, 2.0 to 2) for (number in numbers) println("${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

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

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


## Lingo

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

## Logo

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

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

show root 5 34   ; 2.02439745849989

## Lua

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


## M2000 Interpreter

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


Flush empty stack
Over 2 copy 2nd as new top (so 2nd now is 3rd)
Over 2,2 repeat Over 2 two times.
Shift 2 send top to 2nd, and 2nd to top (1st) (there is a SHFITBACK to revesre action)
Drop drop top
Number get top if is number, else raise error
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

The root command performs this task.

root(1728, 3);

root(1024, 10);

root(2.0, 2);

Output:

                                     12

2

1.414213562


## Mathematica /Wolfram Language

Root[A,n]


## MATLAB

function answer = nthRoot(number,root)

format long

guess = number;

answer = (1/root)*( ((root - 1)*guess) + ( number/(guess^(root - 1)) ) );
end

end


Sample Output:

>> nthRoot(2,2)

ans =

1.414213562373095


## Maxima

nth_root(a, n) := block(
[x, y, d, p: fpprec],
fpprec: p + 10,
x: bfloat(a),
eps: 10.0b0^-p,
y: do (
d: bfloat((a / x^(n - 1) - x) / n),
if abs(d) < eps * x then return(x),
x: x + d
),
fpprec: p,
bfloat(y)
)$ ## Metafont 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 1/x <-> x^y С/П  Instruction: number ^ degree В/О С/П ## NetRexx 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 (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 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 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 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 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 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 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 root(n,A)=A^(1/n); ## Pascal See Delphi ## Perl 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 = $x1; } }  print nthroot(5, 34), "\n"; print nthroot(10, 7131.5 ** 10), "\n"; print nthroot(0.5, 7), "\n";  ## Phix Main loop copied from AWK, and as per C uses pow_() instead of power() since using the latter would make the whole exercise somewhat pointless. with javascript_semantics function pow_(atom x, integer e) atom r = 1 for i=1 to e do r *= x end for return r end function function nth_root(atom y, n) atom eps = 1e-15, -- relative accuracy x = 1 while 1 do -- atom d = ( y / power(x,n-1) - x ) / n atom d = ( y / pow_(x,n-1) - x ) / n x += d atom e = eps*x -- absolute accuracy if d > -e and d < e then exit end if end while return x end function procedure test(sequence yn) atom {y,n} = yn printf(1,"nth_root(%d,%d) = %.10g, builtin = %.10g\n",{y,n,nth_root(y,n),power(y,1/n)}) end procedure papply({{1024,10},{27,3},{2,2},{5642,125},{4913,3},{8,3},{16,2},{16,4},{125,3},{1000000000,3},{1000000000,9}},test)  Note that a {7,0.5} test would need to use power() instead of pow_(). Output: nth_root(1024,10) = 2, builtin = 2 nth_root(27,3) = 3, builtin = 3 nth_root(2,2) = 1.414213562, builtin = 1.414213562 nth_root(5642,125) = 1.071547592, builtin = 1.071547592 nth_root(4913,3) = 17, builtin = 17 nth_root(8,3) = 2, builtin = 2 nth_root(16,2) = 4, builtin = 4 nth_root(16,4) = 2, builtin = 2 nth_root(125,3) = 5, builtin = 5 nth_root(1000000000,3) = 1000, builtin = 1000 nth_root(1000000000,9) = 10, builtin = 10  ## Phixmonti def nthroot var n var y 1e-15 var eps /# relative accuracy #/ 1 var x true while y x n 1 - power / x - n / var d x d + var x eps x * var e /# absolute accuracy #/ d 0 e - < d e > or endwhile x enddef def printList len for get print endfor enddef 10 1024 3 27 2 2 125 5642 4 16 stklen tolist len 1 swap 2 3 tolist for var i i get swap i 1 + get rot var e var b "The " e "th root of " b " is " b 1 e / power " (" b e nthroot ")" 9 tolist printList drop nl endfor ## PHP function nthroot($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

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

(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

/* 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

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

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.


## Python

from decimal import Decimal, getcontext

def nthroot (n, A, precision):
getcontext().prec = precision

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

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

nthroot <- function(A, n, tol=sqrt(.Machine$double.eps)) { ifelse(A < 1, x0 <- A * n, x0 <- A / n) repeat { x1 <- ((n-1)*x0 + A / x0^(n-1))/n if(abs(x1 - x0) > tol) x0 <- x1 else break } x1 } nthroot(7131.5^10, 10) # 7131.5 nthroot(7, 0.5) # 49  ## Racket #lang racket (define (nth-root number root (tolerance 0.001)) (define (acceptable? next current) (< (abs (- next current)) tolerance)) (define (improve current) (/ (+ (* (- root 1) current) (/ number (expt current (- root 1)))) root)) (define (loop current) (define next-guess (improve current)) (if (acceptable? next-guess current) next-guess (loop next-guess))) (loop 1.0))  ## Raku (formerly Perl 6) sub nth-root ($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 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 /*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 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  ## RPL Works with: Halcyon Calc version 4.2.7 RPL code Comment ≪ → a n epsilon ≪ a n / DUP DO SWAP DROP a n / OVER INV n 1 - ^ * OVER n 1 - * n / + UNTIL DUP2 - ABS epsilon < END SWAP DROP ≫ ≫ 'NROOT' STO  ( a n error -- a^(1/n) ) Start with x0 = x1 = a/n both in stack Loop: Forget x(k-1) Calculate x(k+1) Exit loop when x(k+1) close to xk Forget xk  Input: 5 3 0.000000001 NROOT  Output: 1: 1.70997594668  ## Ruby def nthroot(n, a, precision = 1e-5) x = Float(a) begin prev = x x = ((n - 1) * prev + a / (prev ** (n - 1))) / n end while (prev - x).abs > precision x end p nthroot(5,34) # => 2.02439745849989  ## Rust 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 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; ## Scala 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 (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 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 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 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 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 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 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 $x1 } }  Demo: puts [nthroot 2 2] puts [nthroot 5 34] puts [nthroot 5 [expr {34**5}]] puts [nthroot 10 [expr 7131.5**10]] puts [nthroot 0.5 7]; # Squaring!  Output: 1.414213562373095 2.0243974584998847 34.0 7131.5 49.0 ## Ursala The nthroot function defined below takes a natural number n to the function that returns the n-th root of its floating point argument. Error is on the order of machine precision because the stopping criterion is either a fixed point or a repeating cycle. #import nat #import flo nthroot = -+ ("n","n-1"). "A". ("x". div\"n" plus/times("n-1","x") div("A",pow("x","n-1")))^== 1., float^~/~& predecessor+- This implementation is unnecessary in practice due to the availability of the library function pow, which performs exponentiation and allows fractional exponents. Here is a test program. #cast %eL examples = < nthroot2 2., nthroot5 34., nthroot5 pow(34.,5.), nthroot10 pow(7131.5,10.)> output: < 1.414214e+00, 2.024397e+00, 3.400000e+01, 7.131500e+03>  ## V (Vlang) import math fn main() { println("cube root of 5 is:${math.cbrt(5)}")
}
Output:
cube root of 5 is: 1.709975946676697


## Wren

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

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


## XPL0

include c:\cxpl\stdlib;

func real NRoot(A, N);  \Return the Nth root of A
real A, N;
real X, X0, Y;
int  I;
[X:= 1.0;               \initial guess
repeat  X0:= X;
Y:= 1.0;
for I:= 1 to fix(N)-1 do Y:= Y*X0;
X:= ((N-1.0)*X0 + A/Y) / N;
until   abs(X-X0) < 1.0E-15;            \(until X=X0 doesn't always work)
return X;
];

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

Output:

    1.414213562373100
1.414213562373100
3.000000000000000
2.000000000000000


## zkl

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`