# Euler's constant 0.5772...

Euler's constant 0.5772...
You are encouraged to solve this task according to the task description, using any language you may know.

Compute the Euler constant 0.5772...

Discovered by Leonhard Euler around 1730, it is the most ubiquitous mathematical constant after pi and e, but appears more arcane than these.

Denoted gamma (γ), it measures the amount by which the partial sums of the harmonic series (the simplest diverging series) differ from the logarithmic function (its approximating integral): lim n → ∞ (1 + 1/2 + 1/3 + … + 1/n − log(n)).

The definition of γ converges too slowly to be numerically useful, but in 1735 Euler himself applied his recently discovered summation formula to compute ‘the notable number’ accurate to 15 places. For a single-precision implementation this is still the most economic algorithm.

In 1961, the young Donald Knuth used Euler's method to evaluate γ to 1271 places. Knuth found that the computation of the Bernoulli numbers required in the Euler-Maclaurin formula was the most time-consuming part of the procedure.

The next year Dura Sweeney computed 3566 places, using a formula based on the expansion of the exponential integral which didn't need Bernoulli numbers. It's a bit-hungry method though: 2d digits of working precision obtain d correct places only.

This was remedied in 1988 by David Bailey; meanwhile Richard Brent and Ed McMillan had published an even more efficient algorithm based on Bessel function identities and found 30100 places in 20 hours time.

Nowadays the old records have far been exceeded: over 6·1011 decimal places are already known. These massive computations suggest that γ is neither rational nor algebraic, but this is yet to be proven.

References.

[1] Gourdon and Sebah, The Euler constant γ. (for all formulas)

[2] Euler's original journal article translated from the latin (p. 9)

## BASIC

### FreeBASIC

#### Single precision

```'**********************************************
'Subject: Comparing five methods for
'         computing Euler's constant 0.5772...
'tested : FreeBasic 1.08.1
'----------------------------------------------
const eps = 1e-6
dim as double a, b, h, n2, r, u, v
dim as integer k, k2, m, n

? "From the definition, err. 3e-10"

n = 400

h = 1
for k = 2 to n
h += 1 / k
next k
'faster convergence: Negoi, 1997
a = log(n +.5 + 1 / (24*n))

? "Hn   "; h
? "gamma"; h - a; !"\nk ="; n
?

? "Sweeney, 1963, err. idem"

n = 21

dim as double s(1) = {0, n}
r = n
k = 1
do
k += 1
r *= n / k
s(k and 1) += r / k
loop until r < eps

? "gamma"; s(1) - s(0) - log(n); !"\nk ="; k
?

? "Bailey, 1988"

n = 5

a = 1
h = 1
n2 = 2^n
r = 1
k = 1
do
k += 1
r *= n2 / k
h += 1 / k
b = a: a += r * h
loop until abs(b - a) < eps
a *= n2 / exp(n2)

? "gamma"; a - n * log(2); !"\nk ="; k
?

? "Brent-McMillan, 1980"

n = 13

a = -log(n)
b = 1
u = a
v = b
n2 = n * n
k2 = 0
k = 0
do
k2 += 2*k + 1
k += 1
a *= n2 / k
b *= n2 / k2
a = (a + b) / k
u += a
v += b
loop until abs(a) < eps

? "gamma"; u / v; !"\nk ="; k
?

? "How Euler did it in 1735"
'Bernoulli numbers with even indices
dim as double B2(9) = {1,1/6,-1/30,1/42,_
-1/30,5/66,-691/2730,7/6,-3617/510,43867/798}
m = 7
if m > 9 then end

n = 10

'n-th harmonic number
h = 1
for k = 2 to n
h += 1 / k
next k
? "Hn   "; h

h -= log(n)
? "  -ln"; h

'expansion C = -digamma(1)
a = -1 / (2*n)
n2 = n * n
r = 1
for k = 1 to m
r *= n2
a += B2(k) / (2*k * r)
next k

? "err  "; a; !"\ngamma"; h + a; !"\nk ="; n + m
?
? "C  =  0.57721566490153286..."
end```
output:
```From the definition, err. 3e-10
Hn    6.569929691176506
gamma 0.5772156645765731
k = 400

Sweeney, 1963, err. idem
gamma 0.5772156645636311
k = 68

Bailey, 1988
gamma 0.5772156649015341
k = 89

Brent-McMillan, 1980
gamma 0.5772156649015329
k = 40

How Euler did it in 1735
Hn    2.928968253968254
-ln 0.6263831609742079
err  -0.04916749607267539
gamma 0.5772156649015325
k = 17

C  =  0.57721566490153286...
```

#### Multi precision

From first principles

```'***************************************************
'Subject: Computation of Euler's constant 0.5772...
'         with the Brent-McMillan algorithm B1,
'         Math. Comp. 34 (1980), 305-312
'tested : FreeBasic 1.08.1 with gmp 6.2.0
'---------------------------------------------------
#include "gmp.bi"

'multi-precision float pointers
Dim as mpf_ptr a, b
Dim shared as mpf_ptr k2, u, v
'unsigned long integers
Dim as ulong k, n, n2, r, s, t
'precision parameters
Dim shared as ulong e10, e2
Dim shared e as clong
Dim shared f as double
Dim as double tim = TIMER
CLS

a = allocate(len(__mpf_struct))
b = allocate(len(__mpf_struct))
u = allocate(len(__mpf_struct))
v = allocate(len(__mpf_struct))
k2 = allocate(len(__mpf_struct))

'log(x/y) with the Taylor series for atanh(x-y/x+y)
Sub ln (byval s as mpf_ptr, byval x as ulong, byval y as ulong)
Dim as mpf_ptr d = u, q = v
Dim k as ulong
'Möbius transformation
k = x: x -= y: y += k

If x <> 1 Then
Print "ln: illegal argument x - y <> 1"
End
End If

's = 1 / (x + y)
mpf_set_ui (s, y)
mpf_ui_div (s, 1, s)
'k2 = s * s
mpf_mul (k2, s, s)
mpf_set (d, s)

k = 1
Do
k += 2
'd *= k2
mpf_mul (d, d, k2)
'q = d / k
mpf_div_ui (q, d, k)
's += q

f = mpf_get_d_2exp (@e, q)
Loop until abs(e) > e2

's *= 2
mpf_mul_2exp (s, s, 1)
End Sub

'Main

'n = 2^i * 3^j * 5^k

'log(n) = r * log(16/15) + s * log(25/24) + t * log(81/80)

'solve linear system for r, s, t
' 4 -3 -4| i
'-1 -1  4| j
'-1  2 -1| k

'examples
t = 1
select case t
case 1
n = 60
r = 41
s = 30
t = 18
'100 digits
case 2
n = 4800
r = 85
s = 62
t = 37
'8000 digits, 0.6 s
case 3
n = 9375
r = 91
s = 68
t = 40
'15625 digits, 2.5 s
case else
n = 18750
r = 98
s = 73
t = 43
'31250 digits, 12 s. @2.00GHz
end select

'decimal precision
e10 = n / .6
'binary precision
e2 = (1 + e10) / .30103

'initialize mpf's
mpf_set_default_prec (e2)
mpf_inits (a, b, u, v, k2, Cptr(mpf_ptr, 0))

'Compute log terms

ln b, 16, 15

'a = r * b
mpf_mul_ui (a, b, r)

ln b, 25, 24

'a += s * b
mpf_mul_ui (u, b, s)

ln b, 81, 80

'a += t * b
mpf_mul_ui (u, b, t)

''gmp_printf (!"log(%lu) %.*Ff\n", n, e10, a)

'B&M, algorithm B1

'a = -a, b = 1
mpf_neg (a, a)
mpf_set_ui (b, 1)
mpf_set (u, a)
mpf_set (v, b)

k = 0
n2 = n * n
'k2 = k * k
mpf_set_ui (k2, 0)
do
'k2 += 2k + 1
mpf_add_ui (k2, k2, (k shl 1) + 1)
k += 1

'b = b * n2 / k2
mpf_div (b, b, k2)
mpf_mul_ui (b, b, n2)
'a = (a * n2 / k + b) / k
mpf_div_ui (a, a, k)
mpf_mul_ui (a, a, n2)
mpf_div_ui (a, a, k)

'u += a, v += b

f = mpf_get_d_2exp (@e, a)
Loop until abs(e) > e2

mpf_div (u, u, v)
gmp_printf (!"gamma %.*Ff (maxerr. 1e-%lu)\n", e10, u, e10)

gmp_printf (!"k = %lu\n\n", k)

gmp_printf (!"time: %.7f s\n", TIMER - tim)
end```
output:
```gamma 0.5772156649015328606065120900824024310421593359399235988057672348848677267776646709369470632917467495 (maxerr. 1e-100)
k = 255
```

#### The easy way

```' ******************************************
'Subject: Euler's constant 0.5772...
'tested : FreeBasic 1.08.1 with mpfr 4.1.0
'-------------------------------------------
#include "gmp.bi"
#include "mpfr.bi"

dim as mpfr_ptr a = allocate(len(__mpfr_struct))
dim as ulong e2, e10
dim as double tim = TIMER

'decimal precision
e10 = 100

'binary precision
e2 = (1 + e10) / .30103
mpfr_init2 (a, e2)

mpfr_const_euler (a, MPFR_RNDN)
mpfr_printf (!"gamma %.*Rf\n\n", e10, a)

gmp_printf (!"time: %.7f s\n", TIMER - tim)
end```

## Burlesque

Subject to double rounding error, evaluates 33 partial terms in a series

`33ro{JroJJJL[\/JL[\/nr\/{-1}\/.*FLJL[ro?^?*\/JL[{2}\/.*FL\/?^?*\/{1.+}m[J{lg}m[\/?/?*++\/1 3.0./\/?^.*}m[++-2 3 2lg.*./.*2lg2./.+`
Output:
`0.5772156649015326`

## C

### Single precision

```/*********************************************
Subject: Comparing five methods for
computing Euler's constant 0.5772...
tested : tcc-0.9.27
--------------------------------------------*/
#include <math.h>
#include <stdio.h>

#define eps 1e-6

int main(void) {
double a, b, h, n2, r, u, v;
int k, k2, m, n;

printf("From the definition, err. 3e-10\n");

n = 400;

h = 1;
for (k = 2; k <= n; k++) {
h += 1.0 / k;
}
//faster convergence: Negoi, 1997
a = log(n +.5 + 1.0 / (24*n));

printf("Hn    %.16f\n", h);
printf("gamma %.16f\nk = %d\n\n", h - a, n);

printf("Sweeney, 1963, err. idem\n");

n = 21;

double s[] = {0, n};
r = n;
k = 1;
do {
k += 1;
r *= (double) n / k;
s[k & 1] += r / k;
} while (r > eps);

printf("gamma %.16f\nk = %d\n\n", s[1] - s[0] - log(n), k);

printf("Bailey, 1988\n");

n = 5;

a = 1;
h = 1;
n2 = pow(2,n);
r = 1;
k = 1;
do {
k += 1;
r *= n2 / k;
h += 1.0 / k;
b = a; a += r * h;
} while (fabs(b - a) > eps);
a *= n2 / exp(n2);

printf("gamma %.16f\nk = %d\n\n", a - n * log(2), k);

printf("Brent-McMillan, 1980\n");

n = 13;

a = -log(n);
b = 1;
u = a;
v = b;
n2 = n * n;
k2 = 0;
k = 0;
do {
k2 += 2*k + 1;
k += 1;
a *= n2 / k;
b *= n2 / k2;
a = (a + b) / k;
u += a;
v += b;
} while (fabs(a) > eps);

printf("gamma %.16f\nk = %d\n\n", u / v, k);

printf("How Euler did it in 1735\n");
//Bernoulli numbers with even indices
double B2[] = {1.0,1.0/6,-1.0/30,1.0/42,-1.0/30,\
5.0/66,-691.0/2730,7.0/6,-3617.0/510,43867.0/798};
m = 7;
if (m > 9) return(0);

n = 10;

//n-th harmonic number
h = 1;
for (k = 2; k <= n; k++) {
h += 1.0 / k;
}
printf("Hn    %.16f\n", h);

h -= log(n);
printf("  -ln %.16f\n", h);

//expansion C = -digamma(1)
a = -1.0 / (2*n);
n2 = n * n;
r = 1;
for (k = 1; k <= m; k++) {
r *= n2;
a += B2[k] / (2*k * r);
}

printf("err  %.16f\ngamma %.16f\nk = %d", a, h + a, n + m);

printf("\n\nC  =  0.57721566490153286...\n");
}
```
output:
```From the definition, err. 3e-10
Hn    6.5699296911765055
gamma 0.5772156645765731
k = 400

Sweeney, 1963, err. idem
gamma 0.5772156645636311
k = 68

Bailey, 1988
gamma 0.5772156649015341
k = 89

Brent-McMillan, 1980
gamma 0.5772156649015329
k = 40

How Euler did it in 1735
Hn    2.9289682539682538
-ln 0.6263831609742079
err  -0.0491674960726754
gamma 0.5772156649015325
k = 17

C  =  0.57721566490153286...
```

### Multi precision

From first principles

```/**************************************************
Subject: Computation of Euler's constant 0.5772...
with the Brent-McMillan algorithm B1,
Math. Comp. 34 (1980), 305-312
tested : tcc-0.9.27 with gmp 6.2.0
-------------------------------------------------*/
#include <gmp.h>
#include <stdio.h>
#include <stdlib.h>
#include <time.h>

//multi-precision float pointers
mpf_ptr u, v, k2;

//precision parameters
unsigned long e10, e2;
long e;
double f;

//log(x/y) with the Taylor series for atanh(x-y/x+y)
void ln (mpf_ptr s, unsigned long x, unsigned long y) {
mpf_ptr d = u, q = v;
unsigned long k;
//Möbius transformation
k = x; x -= y; y += k;

if (x != 1) {
printf ("ln: illegal argument x - y != 1");
exit;
}

//s = 1 / (x + y)
mpf_set_ui (s, y);
mpf_ui_div (s, 1, s);
//k2 = s * s
mpf_mul (k2, s, s);
mpf_set (d, s);

k = 1;
do {
k += 2;
//d *= k2
mpf_mul (d, d, k2);
//q = d / k
mpf_div_ui (q, d, k);
//s += q

f = mpf_get_d_2exp (&e, q);
} while (abs(e) < e2);

//s *= 2
mpf_mul_2exp (s, s, 1);
}

int main (void) {
mpf_ptr a = malloc(sizeof(__mpf_struct));
mpf_ptr b = malloc(sizeof(__mpf_struct));
u = malloc(sizeof(__mpf_struct));
v = malloc(sizeof(__mpf_struct));
k2 = malloc(sizeof(__mpf_struct));
//unsigned long integers
unsigned long k, n, n2, r, s, t;

clock_t tim = clock();

// n = 2^i * 3^j * 5^k

// log(n) = r * log(16/15) + s * log(25/24) + t * log(81/80)

// solve linear system for r, s, t
//  4 -3 -4| i
// -1 -1  4| j
// -1  2 -1| k

//examples
t = 1;
switch (t) {
case 1 :
n = 60;
r = 41;
s = 30;
t = 18;
//100 digits
break;
case 2 :
n = 4800;
r = 85;
s = 62;
t = 37;
//8000 digits, 0.6 s
break;
case 3 :
n = 9375;
r = 91;
s = 68;
t = 40;
//15625 digits, 2.5 s
break;
default :
n = 18750;
r = 98;
s = 73;
t = 43;
//31250 digits, 12 s. @2.00GHz
}

//decimal precision
e10 = n / .6;
//binary precision
e2 = (1 + e10) / .30103;

//initialize mpf's
mpf_set_default_prec (e2);
mpf_inits (a, b, u, v, k2, (mpf_ptr)0);

//Compute log terms

ln (b, 16, 15);

//a = r * b
mpf_mul_ui (a, b, r);

ln (b, 25, 24);

//a += s * b
mpf_mul_ui (u, b, s);

ln (b, 81, 80);

//a += t * b
mpf_mul_ui (u, b, t);

//gmp_printf ("log(%lu) %.*Ff\n", n, e10, a);

//B&M, algorithm B1

//a = -a, b = 1
mpf_neg (a, a);
mpf_set_ui (b, 1);
mpf_set (u, a);
mpf_set (v, b);

k = 0;
n2 = n * n;
//k2 = k * k
mpf_set_ui (k2, 0);
do {
//k2 += 2k + 1
mpf_add_ui (k2, k2, (k << 1) + 1);
k += 1;

//b = b * n2 / k2
mpf_div (b, b, k2);
mpf_mul_ui (b, b, n2);
//a = (a * n2 / k + b) / k
mpf_div_ui (a, a, k);
mpf_mul_ui (a, a, n2);
mpf_div_ui (a, a, k);

//u += a, v += b

f = mpf_get_d_2exp (&e, a);
} while (abs(e) < e2);

mpf_div (u, u, v);
gmp_printf ("gamma %.*Ff (maxerr. 1e-%lu)\n", e10, u, e10);

gmp_printf ("k = %lu\n\n", k);

tim = clock() - tim;
printf("time: %.7f s\n",((double)tim)/CLOCKS_PER_SEC);
}
```
output:
```gamma 0.5772156649015328606065120900824024310421593359399235988057672348848677267776646709369470632917467495 (maxerr. 1e-100)
k = 255
```

### The easy way

```/*******************************************
Subject: Euler's constant 0.5772...
tested : tcc-0.9.27 with mpfr 4.1.0
------------------------------------------*/
#include <gmp.h>
#include <mpfr.h>
#include <stdio.h>
#include <stdlib.h>
#include <time.h>

int main (void) {
mpfr_ptr a = malloc(sizeof(__mpfr_struct));
unsigned long e2, e10;
clock_t tim = clock();

//decimal precision
e10 = 100;

//binary precision
e2 = (1 + e10) / .30103;
mpfr_init2 (a, e2);

mpfr_const_euler (a, MPFR_RNDN);
mpfr_printf ("gamma %.*Rf\n\n", e10, a);

tim = clock() - tim;
gmp_printf ("time: %.7f s\n",((double)tim)/CLOCKS_PER_SEC);
}
```

## C++

```#include <array>
#include <cmath>
#include <iomanip>
#include <iostream>

double ByVaccaSeries(int numTerms)
{
// this method is simple but converges slowly
// calculate gamma by:
// 1 * (1/2 - 1/3) +
// 2 * (1/4 - 1/5 + 1/6 - 1/7) +
// 3 * (1/8 - 1/9 + 1/10 - 1/11 + 1/12 - 1/13 + 1/14 - 1/15) +
// 4 * ( . . . ) +
// . . .
double gamma = 0;
size_t next = 4;

for(double numerator = 1; numerator < numTerms; ++numerator)
{
double delta = 0;
for(size_t denominator = next/2; denominator < next; denominator+=2)
{
// calculate terms two at a time
delta += 1.0/denominator - 1.0/(denominator + 1);
}

gamma += numerator * delta;
next *= 2;
}
return gamma;
}

// based on the C entry
double ByEulersMethod()
{
//Bernoulli numbers with even indices
const std::array<double, 8> B2 {1.0, 1.0/6, -1.0/30, 1.0/42, -1.0/30,
5.0/66, -691.0/2730, 7.0/6};

const int n = 10;

//n-th harmonic number
const double h = [] // immediately invoked lambda
{
double sum = 1;
for (int k = 2; k <= n; k++) { sum += 1.0 / k; }
return sum - log(n);
}();

//expansion C = -digamma(1)
double a = -1.0 / (2*n);
double r = 1;
for (int k = 1; k < ssize(B2); k++)
{
r *= n * n;
a += B2[k] / (2*k * r);
}

return h + a;
}

int main()
{
std::cout << std::setprecision(16) << "Vacca series:  " << ByVaccaSeries(32);
std::cout << std::setprecision(16) << "\nEulers method: " << ByEulersMethod();
}
```
Output:
```Vacca series:  0.5772156610598274
Eulers method: 0.5772156649015325
```

## Clojure

```(let [n 1e10]
(loop [i 1
out (- (Math/log n))]
(if (<= i n)
(recur (inc i) (+ out (/ 1.0 i)))
out)))
```
Output:
```Gives:                      0.5772156649484047
Compared to the true value: 0.5772156649015328...
```

## Delphi

Works with: Delphi version 6.0

This programs demonstrates the basic method of calculating the Euler number. It illustrates how the number of iterations increases the accuracy.

```function ComputeEuler(N: int64): double;
{Compute Eurler number with N-number of iterations}
var I: integer;
var A: double;
begin
Result:=0;
for I:=1 to N-1 do
Result:=Result + 1 / I;
A:=Ln(N + 0.5 + 1/(24.0*N));
Result:=Result-A;
end;

procedure ShowEulersNumber(Memo: TMemo);
{Show Euler numbers at various levels of precision}
var Euler,G,A,Error: double;
var N: integer;
const Correct =0.57721566490153286060651209008240243104215933593992;

procedure ShowEulerError(N: int64);
{Show Euler number and Error}
begin
Euler:=ComputeEuler(N);
Error:=Correct-Euler;
end;

begin
{Compute Euler number with iterations ranging 10 to 10^9}
for N:=1 to 9 do
begin
ShowEulerError(Trunc(Power(10,N)));
end;

end;
```
Output:
```N =   10
Euler=0.477196250122325250
Error=0.100019414779207616

N =   100
Euler=0.567215644212103243
Error=0.010000020689429618

N =   1,000
Euler=0.576215664880711742
Error=0.001000000020821119

N =   10,000
Euler=0.577115664901475256
Error=0.000100000000057605

N =   100,000
Euler=0.577205664901386584
Error=0.000010000000146277

N =   1,000,000
Euler=0.577214664900591146
Error=0.000001000000941715

N =   10,000,000
Euler=0.577215564898391875
Error=0.000000100003140985

N =   100,000,000
Euler=0.577215654895336883
Error=0.000000010006195978

N =   1,000,000,000
Euler=0.577215664060567235
Error=0.000000000840965626

Elapsed Time: 4.716 Sec.
```

## EasyLang

```fastfunc gethn n .
i = 1
while i <= n
hn += 1 / i
i += 1
.
return hn
.
e = 2.718281828459045235
n = 10e8
numfmt 9 0
print gethn n - log10 n / log10 e```

## FutureBasic

```void local fn Euler
long   n = 10000000, k
double a, h = 1.0

for k = 2 to n
h += 1 / k
next
a = log( n + 0.5 + 1 / ( 24 * n ) )

printf @"From the definition, err. 3e-10"
printf @"Hn    %.15f", h
printf @"gamma %.15f", h - a
printf @"k = %ld\n", n
printf @"C     %.15f", 0.5772156649015328
end fn

CFTimeInterval t
t = fn CACurrentMediaTime
fn Euler
printf @"\vCompute time: %.3f ms",(fn CACurrentMediaTime-t)*1000

HandleEvents```
Output:
```From the definition, err. 3e-10
Hn    16.695311365857272
gamma 0.577215664898954
k = 10000000

C     0.577215664901533
Compute time: 67.300 ms
```

## J

We can approximate euler's constant using the difference between the reciprocal and the gamma function of a small number:

```   (% - !@<:) 2^_27
0.577216
```

For higher accuracy we can use a variation on the C implementation which was attributed to Richard P. Brent and Edwin M. McMillan:

```Euler=: {{
A=.B=. ^.1r13 1x1
r=. j=. 0
whilst. (r=.%/B)~:!.0(r) do.
B=. B+A=. (j,1)%~+/\.A*169%(1,j)*(j=.j+1)
end.
r
}}0
```

With J configured for 16 digits of display precision (`9!:11(16)`) we can see that this gave us:

```   Euler
0.5772156649015329
```

## Java

```/**
* Using a simple formula derived from Hurwitz zeta function,
* as described on https://en.wikipedia.org/wiki/Euler%27s_constant,
* gives a result accurate to 12 decimal places.
*/
public class EulerConstant {

public static void main(String[] args) {
System.out.println(gamma(1_000_000));
}

private static double gamma(int N) {
double gamma = 0.0;

for ( int n = 1; n <= N; n++ ) {
gamma += 1.0 / n;
}

gamma -= Math.log(N) + 1.0 / ( 2 * N );

return gamma;
}

}
```
Output:
`0.5772156649007147`

## jq

Translated from C (Bailey, 1988)

Works with: jq

Works with gojq, the Go implementation of jq

```# Bailey, 1988
def bailey(\$n; \$eps):
pow(2; \$n) as \$n2
| {a :1, b: 0, h: 1, r: 1, k: 1}
| until( (.b - .a)|fabs <= \$eps;
.k += 1
| .r *= (\$n2 / .k)
| .h += (1.0 / .k)
| .b = .a
| .a += (.r * .h) )
| (.a * \$n2 / (\$n2|exp) ) - (\$n * (2|log)) ;```
Output:
```bailey(5; 1E-6)
0.5772156649015341
```

## Julia

Translation of: PARI/GP
```display(MathConstants.γ)  # γ = 0.5772156649015...
```

## Lambdatalk

Following the definition with an improvment from Negoi.

```{def negoi
{lambda {:n}
{let { {:n :n}
{:h {+ {S.map {lambda {:k} {/ 1 :k}} {S.serie 1 :n}}} }
{:a {log {+ :n 0.5 {/ 1 {* 24 :n}}}}}   // Negoi, 1997
} {div}-> Hn :h
{div}gamma {- :h :a}
{div}k :n
}}}
-> negoi

{negoi 400}
-> Hn 6.5699296911765055
gamma 0.5772156645765731 with k = 400
(0.57721566457657 target)
```

Following Sweeney

```{def sweeney
{def sweeney.set!
{lambda {:s :r :k :i}
{A.set! :i {+ {A.get :i :s} {/ :r :k}} :s}
}}
{def sweeney.loop
{lambda {:n :s :r :k}
{if {<= :r 1.e-10}
then gamma = {- {A.get 1 :s} {A.get 0 :s} {log :n}} with k=:k
else {sweeney.loop :n
{sweeney.set! :s {* :r {/ :n :k}} :k {% :k 2}}
{* :r {/ :n :k}}
{+ :k 1} }
}}}
{lambda {:n}
{sweeney.loop :n {A.new 0 :n} :n 2} }}
-> sweeney

{sweeney 21}
-> gamma = 0.577215664563631 with k=76
(0.57721566456363 target)
```

## Lua

A value using 100,000,000 iterations of the harmonic series computes in less than a second and is correct to eight decimal places.

```function computeGamma (iterations, decimalPlaces)
local Hn = 1
for i = 2, iterations do
Hn = Hn + (1/i)
end
local gamma = tostring(Hn - math.log(iterations))
end

print(computeGamma(10^8, 8))
```
Output:
`0.57721566`

## Mathematica/Wolfram Language

```N[EulerGamma, 1000]
```
Output:
```0.57721566490153286060651209008240243104215933593992359880576723488486772677766467093694706329174674
9514631447249807082480960504014486542836224173997644923536253500333742937337737673942792595258247094
9160087352039481656708532331517766115286211995015079847937450857057400299213547861466940296043254215
1905877553526733139925401296742051375413954911168510280798423487758720503843109399736137255306088933
1267600172479537836759271351577226102734929139407984301034177717780881549570661075010161916633401522
7893586796549725203621287922655595366962817638879272680132431010476505963703947394957638906572967929
6010090151251959509222435014093498712282479497471956469763185066761290638110518241974448678363808617
4945516989279230187739107294578155431600500218284409605377243420328547836701517739439870030237033951
8328690001558193988042707411542227819716523011073565833967348717650491941812300040654693142999297779
569303100503086303418569803231083691640025892970890985486825777364288253954925873629596133298574739302```

## Maxima

```%gamma,numer;
```
Output:
```0.5772156649015329
```

## Nim

```import std/math

const n = 1e6
var result = 1.0

for i in 2..int(n):
result += 1/i

echo result - ln(n)
```
Output:
```0.5772161649007153
```

## PARI/GP

built-in:

```\l "euler_const.log"
\p 100
print("gamma ", Euler);
\q```

## Perl

```#!/usr/bin/perl

use strict; # https://en.wikipedia.org/wiki/Euler%27s_constant
use warnings;
use List::Util qw( sum );

print sum( map 1 / \$_, 1 .. 1e6) - log 1e6, "\n";
```
Output:
```0.577216164900715
```

## Phix

### part 1 (max 12dp)

Translation of Perl, with the same accuracy limitation

```-- demo\rosetta\Eulers_constant.exw
with javascript_semantics
constant C = sum(sq_div(1,tagset(1e6)))-log(1e6)
printf(1,"gamma %.12f  (max 12d.p. of accuracy)\n",C)
```
Output:
```gamma 0.577216164901  (max 12d.p. of accuracy)
```

### part 2 (first principles)

Translation of C, from first principles.

```without js  -- no mpfr_get_d_2exp() in mpfr.js as yet
include mpfr.e
mpfr u, v, k2;
integer e, e10, e2
atom f

//log(x/y) with the Taylor series for atanh(x-y/x+y)
procedure ln(mpfr s, integer x, y)
mpfr d = u, q = v;
assert((x-y)==1)
mpfr_set_si(s, x+y)
mpfr_si_div(s, 1, s)            // s = 1 / (x + y)
mpfr_mul(k2, s, s)              // k2 = s * s
mpfr_set(d, s)
integer k = 1
while true do
k += 2;
mpfr_mul(d, d, k2)          // d *= k2
mpfr_div_si(q, d, k)        // q = d / k
mpfr_add(s, s, q)           // s += q
{f,e} = mpfr_get_d_2exp(q)
if abs(e)>=e2 then exit end if
end while
mpfr_mul_si(s, s, 2)            //s *= 2
end procedure

mpfr a, b
integer k,
n = 60,     -- (required precision in decimal dp *6/10)
n2,
r = 41,
s = 30,
t = 18;

// n = 2^i * 3^j * 5^k

// log(n) = r * log(16/15) + s * log(25/24) + t * log(81/80)

// solve linear system for r, s, t
//  4 -3 -4| i
// -1 -1  4| j
// -1  2 -1| k

//decimal precision
e10 = floor(n/0.6)
//binary precision
e2 = floor((1 + e10) / 0.30103)

mpfr_set_default_precision(e2)
{a, b, u, v, k2} = mpfr_inits(5)

//Compute log terms
ln(b, 16, 15)  mpfr_mul_si(a, b, r)     // a = r * b
ln(b, 25, 24)  mpfr_addmul_si(a, b, s)  // a += s * b
ln(b, 81, 80)  mpfr_addmul_si(a, b, t)  // a += t * b

mpfr_neg(a, a)          // a = -a
mpfr_set_si(b, 1)       // b = 1
mpfr_set (u, a)
mpfr_set (v, b)

k = 0;
n2 = n * n;
mpfr_set_si(k2, 0)      // k2 = k * k (as below)
while true do
mpfr_add_si(k2, k2, k*2+1)      // k2 += 2k + 1
k += 1;

mpfr_div(b, b, k2)
mpfr_mul_si(b, b, n2)           // b = b * n2 / k2

mpfr_div_si(a, a, k)
mpfr_mul_si(a, a, n2)
mpfr_div_si(a, a, k)            // a = (a * n2 / k + b) / k

mpfr_add(u, u, a)               // u += a
mpfr_add(v, v, b)               // v += b

{f,e} = mpfr_get_d_2exp (a)
if abs(e)>=e2 then exit end if
end while

mpfr_div(u, u, v)
string su = mpfr_get_fixed(u,e10)
printf(1,"gamma %s (maxerr. 1e-%d)\n", {su, e10})
```
Output:
```gamma 0.5772156649015328606065120900824024310421593359399235988057672348848677267776646709369470632917467494 (maxerr. 1e-100)
```

### part 3 (the easy way)

```without js  -- no mpfr_const_euler() in mpfr.js as yet
requires("1.0.1") -- mpfr_const_euler()
include mpfr.e
mpfr gamma = mpfr_init(0,-100)
mpfr_const_euler(gamma)
printf(1,"gamma %s (mpfr_const_euler)\n",{mpfr_get_fixed(gamma,100)})
```

(matches part 2 except for the 100th decimal place)

Output:
```gamma 0.5772156649015328606065120900824024310421593359399235988057672348848677267776646709369470632917467495 (mpfr_const_euler)
```

## Picat

### List comprehension

```main =>
Gamma = 0.57721566490153286060651209008240,
println(Gamma),
foreach(N in 1..8)
G = e(10**N),
println([n=N,g=G,diff=G-Gamma])
end.

e(N) = [1.0/I : I in 1..N].sum-log(N).```
Output:
```0.577215664901533
[n = 1,g = 0.626383160974208,diff = 0.049167496072675]
[n = 2,g = 0.582207331651529,diff = 0.004991666749996]
[n = 3,g = 0.577715581568206,diff = 0.000499916666674]
[n = 4,g = 0.577265664068165,diff = 0.000049999166632]
[n = 5,g = 0.577220664893106,diff = 0.000004999991574]
[n = 6,g = 0.577216164900715,diff = 0.000000499999182]
[n = 7,g = 0.577215714898951,diff = 0.000000049997419]
[n = 8,g = 0.577215669900188,diff = 0.000000004998655]```

### Loop

```e2(N) = E-log(N) =>
E = 1,
foreach(I in 2..N)
E := E + 1/I
end.```

Translation of: Rust
```main =>
Gamma = 0.577215664901532860606512090082402,
println(gamma=Gamma),
member(N, 1..23),
G = gamma(N),
println([n=N,g=G,diff=G-Gamma]),
fail,
nl.

gamma(N) = Gamma =>
Gamma = 1/2 - 1/3,
foreach(I in 2..N)
Power = 2**I,
Sign = -1,
Term = 0,
foreach(Denominator in Power..(2*Power-1))
Sign := Sign * -1,
Term := Term + Sign / Denominator
end,
Gamma := Gamma + I*Term
end.```
Output:
```[n = 1,g = 0.166666666666667,diff = -0.410548998234866]
[n = 2,g = 0.314285714285714,diff = -0.262929950615819]
[n = 3,g = 0.416741591741592,diff = -0.160474073159941]
[n = 4,g = 0.482164184398886,diff = -0.095051480502647]
[n = 5,g = 0.522141654090275,diff = -0.055074010811257]
[n = 6,g = 0.545853770405349,diff = -0.031361894496184]
[n = 7,g = 0.559605750992416,diff = -0.017609913909116]
[n = 8,g = 0.567441138957738,diff = -0.009774525943794]
[n = 9,g = 0.571842107494027,diff = -0.005373557407506]
[n = 10,g = 0.574285301882304,diff = -0.002930363019229]
[n = 11,g = 0.57562856705805,diff = -0.001587097843483]
[n = 12,g = 0.576361123043496,diff = -0.000854541858037]
[n = 13,g = 0.576757887880701,diff = -0.000457777020832]
[n = 14,g = 0.576971520706463,diff = -0.00024414419507]
[n = 15,g = 0.577085964243776,diff = -0.000129700657756]
[n = 16,g = 0.577147000098518,diff = -0.000068664803014]
[n = 17,g = 0.577179425210813,diff = -0.00003623969072]
[n = 18,g = 0.577196591397621,diff = -0.000019073503912]
[n = 19,g = 0.577205651316587,diff = -0.000010013584946]
[n = 20,g = 0.57721041969158,diff = -0.000005245209953]
[n = 21,g = 0.577212923087556,diff = -0.000002741813977]
[n = 22,g = 0.577214234389975,diff = -0.000001430511558]
[n = 23,g = 0.577214919843452,diff = -0.000000745058081]```

## Processing

Translation of: C
```/*********************************************
Subject: Comparing five methods for
computing Euler's constant 0.5772...
// https://rosettacode.org/wiki/Euler%27s_constant_0.5772...
--------------------------------------------*/
double a, b, h, n2, r, u, v;
float floatA, floatB, floatN2;
int k, k2, m, n;
double eps = 1e-6;

void setup() {
size(100, 100);
noLoop();
}

void draw() {
println("From the definition, err. 3e-10\n");

n = 400;

h = 1;

for (int k = 2; k <= n; k++) {
h += 1.0 / k;
}
//faster convergence: Negoi, 1997
a = log(n +.5 + 1.0 / (24*n));

println("Hn    ", h);
println("gamma ", h - a);
println("k = ", n);
println("");

println("Sweeney, 1963, err. idem");
n = 21;

double s[] = {0, n};
r = n;
k = 1;
while (r > eps) {
k ++;
r *= (double) n / k;
s[k & 1] = s[k & 1] + r / k;
}

// println("gamma %.16f\nk = %d\n\n", s[1] - s[0] - log(n), k);

println("Hn    ", h);
println("gamma ", s[1] - s[0] - log(n));
println("k = ", k);
println("");

println("Bailey, 1988");
n = 5;
floatA = 1;
h = 1;
floatN2 = pow(2, n);
r = 1;
k = 1;
while (abs(floatB - floatA) > eps) {
k += 1;
r *= floatN2 / k;
h += 1.0 / k;
floatB = floatA;
floatA += r * h;
}
floatA *= floatN2 / exp(floatN2);

println("gamma ", floatA - n * log(2));
println("k = ", k);
println("");

println("Brent-McMillan, 1980");

n = 13;

floatA = -log(n);
floatB = 1;
u = a;
v = b;
n2 = n * n;
k2 = 0;
k = 0;

while (abs(floatA) > eps) {
k2 += 2*k + 1;
k += 1;
floatA *= n2 / k;
floatB *= n2 / k2;
floatA = (floatA + floatB) / k;
u += floatA;
v += floatB;
}
println("gamma  ", u / v);
println("k      ", k);

println("How Euler did it in 1735\n");
//Bernoulli numbers with even indices

double[] B2 = new double[11];

B2[1] = 1.0;
B2[2] = 1.0/6;
B2[3] = -1.0/30;
B2[4] = 1.0/42;
B2[5] = -1.0/30;
B2[6] = 5.0/66;
B2[7] = -691.0/2730;
B2[8] = 7.0/6;
B2[9] = -3617.0/510;
B2[10]= 43867.0/798;

m = 7;
n = 10;

//n-th harmonic number
h = 1;
for (k = 2; k <= n; k++) {
h += 1.0 / k;
}
println("Hn    ", h);

h -= log(n);
println("  -ln ", h);

//expansion C = -digamma(1)
a = -1.0 / (2*n);
n2 = n * n;
r = 1;
for (k = 1; k <= m; k++) {
r *= n2;
a += B2[k] / (2*k * r);
}

println("");
println("err  ", a);
println("gamma ", h + a );
println("k = ", n + m);
println("");
println("C  =  0.57721566490153286...\n");
}```
Output:
```From the definition, err. 3e-10

Hn     6.569929751800373
gamma  0.577215823577717
k =  400

Sweeney, 1963, err. idem
Hn     6.569929751800373
gamma  0.5772155784070492
k =  68

Bailey, 1988
gamma  0.5772176
k =  67

Brent-McMillan, 1980
gamma   0.5772157269353247
k       40

How Euler did it in 1735

Hn     2.9289682805538177
-ln  0.6263831555843353
err   -0.044995839604388355
gamma  0.581387315979947
k =  17

C  =  0.57721566490153286...
```

## Python

```# /**************************************************
# Subject: Computation of Euler's constant 0.5772...
#          with Euler's Zeta Series.
# tested : Python 3.11
# -------------------------------------------------*/

from scipy import special as s

def eulers_constant(n):
k = 2
euler = 0
while k <= n:
euler += (s.zeta(k) - 1)/k
k += 1
return 1 - euler

print(eulers_constant(47))
```
Output:
```0.577215664901533
```

## Racket

```#lang racket/base

(require math/number-theory
math/base)

gamma.0

;; if you want to work it out the hard way...
(define (H n)
(for/sum ((i n)) (/ (add1 i))))

(define (g #:n (n 10) #:k (k 7))
(+ (- (H n)
(log n)
(/ (* n 2)))
(for/sum ((2k (in-range 2 (* 2 (add1 k)) 2)))
(/ (bernoulli-number 2k) (* (expt n 2k) 2k)))))

(g)
```
Output:
```0.5772156649015329
0.5772156649015324```

## Raku

```# 20211124 Raku programming solution

sub gamma (\N where N > 1) { # Vacca series https://w.wiki/4ybp
# convert terms to FatRat for arbitrary precision
return  (1/2 - 1/3) + [+] (2..N).race.map: -> \n {

my (\$power, \$sign, \$term) = 2**n, -1;

for (\$power..^2*\$power) { \$term += (\$sign = -\$sign) / \$_ }

n*\$term
}
}

say gamma 23 ;
```
Output:
```0.5772149198434515
```

## RPL

Translation of: FreeBASIC

From the definition, with Negoi's convergence improvement

```≪ → n
≪ 1
2 n FOR k k INV + NEXT
n .5 + 24 n * INV + LN -
≫ ≫ ‘GAMMA1’ STO
```

Sweeney method

```≪ 0 OVER R→C 1E-6 → n s eps
≪ n 1
DO
1 + SWAP
OVER / n * SWAP
DUP2 /
IF OVER 2 MOD THEN (0,1) * END 's' STO+
UNTIL OVER eps < END
DROP2
s C→R SWAP - n LN -
≫ ≫ ‘GAMMA2’ STO
```
```400 GAMMA1
21 GAMMA2
```
Output:
```2: .57721566456
1: .57717756228
```

## Ruby

```n = 1e6
p (1..n).sum{ 1.0/_1 } - Math.log(n)
```
Output:
```0.5772161649014507 #12 digits accurate
```

## Rust

Translation of: Raku
```// 20220322 Rust programming solution

fn gamma(N: u32) -> f64 { // Vacca series https://w.wiki/4ybp

return 1f64 / 2f64 - 1f64 / 3f64
+ ((2..=N).map(|n| {
let power: u32 = 2u32.pow(n);
let mut sign: f64 = -1f64;
let mut term: f64 = 0f64;

for denominator in power..=(2 * power - 1) {
sign *= -1f64;
term += sign / f64::from(denominator);
}

return f64::from(n) * term;
}))
.sum::<f64>();
}

fn main() {
println!("{}", gamma(23));
}
```
Output:
`0.5772149198434514`

## Scala

```/**
* Using a simple formula derived from Hurwitz zeta function,
* as described on https://en.wikipedia.org/wiki/Euler%27s_constant,
* gives a result accurate to 11 decimal places: 0.57721566490...
*/

object EulerConstant extends App {

println(gamma(1_000_000))

private def gamma(N: Int): Double = {
val sumOverN = (1 to N).map(1.0 / _).sum
sumOverN - Math.log(N) - 1.0 / (2 * N)
}

}
```
Output:
```0.5772156649007153
```

## Scheme

Works with: Chez Scheme
```; Procedure to compute factorial.
(define fact
(lambda (n)
(if (<= n 0)
1
(* n (fact (1- n))))))

; Compute Euler's gamma constant as the difference of log(n) from a sum.
; See section 2.3 of <http://numbers.computation.free.fr/Constants/Gamma/gamma.html>.
(define gamma
(lambda (n)
(let ((sum 0))
(do ((k 1 (1+ k)))
((> k (* 3.5911 n)) (- sum (log n)))
(set! sum (+ sum (/ (* (expt -1 (1- k)) (expt n k)) (* k (fact k)))))))))

; Show Euler's gamma constant computed at log(100).
(printf "Euler's gamma constant: ~a~%" (gamma 100))
```
Output:
```Euler's gamma constant: 0.5772156649015328
```

## Sidef

Built-in:

```# 100 decimals of precision
local Num!PREC = 4*100
say Num.EulerGamma
```
Output:
```0.5772156649015328606065120900824024310421593359399235988057672348848677267776646709369470632917467495
```

Several formulas:

```const n = (ARGV ? Num(ARGV[0]) : 50)       # number of iterations

define ℯ = Num.e
define π = Num.pi
define γ = Num.EulerGamma

func display(r, t) {
say "#{r}\terror: #{ '%.0g' % abs(r - t) }"
}

# Original definition of the Euler-Mascheroni constant, due to Euler (1731)
display(sum(1..n, {|n| 1/n }) - log(n), γ)

# Formula due to Euler (best convergence)
display(harmfrac(n) - log(n) - 1/(2*n) - sum(1..n, {|k|
-bernoulli(2*k) / (2*k) / n**(2*k)
}), γ)

# Formula derived from the above formula of Euler,
# using approximations of Bernoulli numbers.
display(harmfrac(n) - log(n) - 1/(2*n) - sum(1..n, {|k|
(-1)**k * 4 * sqrt(π*k) * (π * ℯ)**(-2*k) * k**(2*k) / (2*k) / n**(2*k)
}), γ)

# Euler-Mascheroni constant, involving zeta(n)
display(1 - sum(2..(n+1), {|n|
(zeta(n) - 1) / n
}), γ)

# Limit_{n->Infinity} zeta((n+1)/n) - n} = gamma
display(zeta((n+1)/n) - n, γ)

# Series due to Euler (1731).
display(sum(2..(n+1), {|n|
(-1)**n * zeta(n) / n
}),  γ)

# Formula due to Euler in terms of log(2) and the odd zeta values
display(3/4 - log(2)/2 + sum(1..n, {|n|
(1 - 1/(2*n + 1)) * (zeta(2*n + 1) - 1)
}), γ)

# Formula due to Euler in terms of log(2) and the odd zeta values (VII)
display(log(2) - sum(1..n, {|n|
zeta(2*n + 1) / (2*n + 1) / 2**(2*n)
}), γ)

# Formula due to Vacca (1910)
display(sum(1..n, {|n|
(-1)**n * floor(log2(n)) / n
}), γ)
```
Output:
```0.58718233290127899894172100505421724484389898107	error: 0.01
0.57721566490153286060651209008240243104215933594	error: 2e-58
0.577201775274185974649592917416402750312879661543	error: 1e-05
0.577215664901532868991217643213156967011395887777	error: 8e-18
0.57867004101560321761330253540741574969540128177	error: 0.001
0.567507841748305076772446986005418728718501189232	error: 0.01
0.577215664901532860606512090082272222693164663104	error: 1e-31
0.57721566490153286060651209008240496797924366087	error: 3e-33
0.611009392556929160631547597803236563977259982583	error: 0.03
```

## V (Vlang)

Translation of: C
```import math
const eps = 1e-6
fn main() {
//.5772
println("From the definition, err. 3e-10")
mut n := 400
mut h := 1.0
for k in 2..n+1 {
h += 1.0/f64(k)
}
//faster convergence: Negoi, 1997
mut a := math.log(f64(n) + 0.5 + 1.0/f64(24*n))

println("Hn    \${h:0.16f}")
println("gamma \${h-a:0.16f}\nk = \$n\n")

println("Sweeney, 1963, err. idem")
n = 21
mut s := [0.0, f64(n)]
mut r := f64(n)
mut k := 1
for {
k++
r *= f64(n) / f64(k)
s[k & 1] += r/f64(k)
if r <= eps {
break
}
}
println("gamma \${s[1] - s[0] - math.log(n):0.16f}\nk = \$k\n")

println("Bailey, 1988")
n = 5
a = 1.0
h = 1.0
mut n2 := math.pow(f64(2),f64(n))
r = 1
k = 1
for {
k++
r *= n2 / f64(k)
h += 1/f64(k)
b := a
a += r * h
if math.abs(b-a) <= eps {
break
}
}
a *= n2 / math.exp(n2)
println("gamma \${a - n * math.log(2):.16f}\nk = \$k\n")

println("Brent-McMillan, 1980")
n = 13
a = -math.log(n)
mut b := 1.0
mut u := a
mut v := b
n2 = n * n
mut k2 := 0
k = 0
for {
k2 += 2*k + 1
k++
a *= n2 / f64(k)
b *= n2 / f64(k2)
a = (a + b)/f64(k)
u += a
v += b
if math.abs(a) <= eps {
break
}
}
println("gamma \${u/v:0.16f}\nk = \$k\n")

println("How Euler did it in 1735")
// Bernoulli numbers with even indices
b2 := [1.0, 1.0/6, -1.0/30, 1.0/42, -1.0/30, 5.0/66, -691.0/2730, 7.0/6, -3617.0/510, 43867.0/798]
m := 7
n = 10
// n'th harmonic number
h = 1.0
for kz in 2..n+1 {
h += 1.0/f64(kz)
}
println("Hn    \${h:0.16f}")
h -= math.log(n)
println("  -ln \${h:0.16f}")
// expansion C = -digamma(1)
a = -1.0 / (2.0*f64(n))
n2 = f64(n * n)
r = 1
for kq in 1..m+1 {
r *= n2
a += b2[kq] / (2.0*f64(kq)*r)
}
println("err  \${a:0.16f}\ngamma \${h+a:0.16f}\nk = \${n+m}")
println("\nC = 0.57721566490153286...")
}```
Output:
`Same as C entry`

## Wren

Translation of: C

### Single precision (Cli)

Library: Wren-fmt

Note that, whilst internally double arithmetic is carried out to the same precision as C (Wren is written in C), printing doubles is effectively limited to a maximum of 14 decimal places.

```import "./fmt" for Fmt

var eps = 1e-6

System.print("From the definition, err. 3e-10")
var n = 400
var h = 1
for (k in 2..n) h = h + 1/k
//faster convergence: Negoi, 1997
var a = (n + 0.5 + 1/(24*n)).log

Fmt.print("Hn    \$0.14f", h)
Fmt.print("gamma \$0.14f\nk = \$d\n", h - a, n)

System.print("Sweeney, 1963, err. idem")
n = 21
var s = [0, n]
var r = n
var k = 1
while (true) {
k = k + 1
r = r * n / k
s[k & 1] = s[k & 1] + r/k
if (r <= eps) break
}
Fmt.print("gamma \$0.14f\nk = \$d\n", s[1] - s[0] - n.log, k)

System.print("Bailey, 1988")
n = 5
a = 1
h = 1
var n2 = 2.pow(n)
r = 1
k = 1
while (true) {
k = k + 1
r = r * n2 / k
h = h + 1/k
var b = a
a = a + r * h
if ((b-a).abs <= eps) break
}
a = a * n2 / n2.exp
Fmt.print("gamma \$0.14f\nk = \$d\n", a - n * 2.log, k)

System.print("Brent-McMillan, 1980")
n = 13
a = -n.log
var b = 1
var u = a
var v = b
n2 = n * n
var k2 = 0
k = 0
while (true) {
k2 = k2 + 2*k + 1
k = k + 1
a = a * n2 / k
b = b * n2 / k2
a = (a + b)/k
u = u + a
v = v + b
if (a.abs <= eps) break
}
Fmt.print("gamma \$0.14f\nk = \$d\n", u / v, k)

System.print("How Euler did it in 1735")
// Bernoulli numbers with even indices
var b2 = [1, 1/6, -1/30, 1/42, -1/30, 5/66, -691/2730, 7/6, -3617/510, 43867/798]
var m = 7
n = 10
// n'th harmonic number
h = 1
for (k in 2..n) h = h + 1/k
Fmt.print("Hn    \$0.14f", h)
h = h - n.log
Fmt.print("  -ln \$0.14f", h)
// expansion C = -digamma(1)
a = -1 / (2*n)
n2 = n * n
r = 1
for (k in 1..m) {
r = r * n2
a = a + b2[k] / (2*k*r)
}
Fmt.print("err  \$0.14f\ngamma \$0.14f\nk = \$d", a, h + a, n + m)
System.print("\nC = 0.57721566490153286...")
```
Output:
```From the definition, err. 3e-10
Hn    6.56992969117651
gamma 0.57721566457657
k = 400

Sweeney, 1963, err. idem
gamma 0.57721566456363
k = 68

Bailey, 1988
gamma 0.57721566490154
k = 89

Brent-McMillan, 1980
gamma 0.57721566490153
k = 40

How Euler did it in 1735
Hn    2.92896825396825
-ln 0.62638316097421
err  -0.04916749607268
gamma 0.57721566490153
k = 17

C = 0.57721566490153286...
```

### Multi precision (Embedded)

Library: Wren-gmp

The display is limited to 100 digits for all four examples as I couldn't see much point in showing them all.

```import "./gmp" for Mpf

var euler = Fn.new { |n, r, s, t|
// decimal precision
var e10 = (n/0.6).floor

// binary precision
var e2 = ((1 + n/0.6)/0.30103).round
Mpf.defaultPrec = e2

var b = Mpf.new().log(Mpf.from(16).div(15))
var a = b.mul(r)
b = Mpf.new().log(Mpf.from(25).div(24))
b = Mpf.new().log(Mpf.from(81).div(80))
var u = b * t
b.set(1)
u.set(a)
var v = Mpf.from(b)
var k = 0
var n2 = n * n
var k2 = Mpf.zero
while (true) {
k = k + 1
b.mul(n2).div(k2)
var e = Mpf.frexp(a)[1]
if (e.abs >= e2) break
}
u.div(v)
System.print("gamma %(u.toString(10, 100)) (maxerr. 1e-%(e10))")
System.print("k = %(k)")
}

var start = System.clock
euler.call(60, 41, 30, 18)
euler.call(4800, 85, 62, 37)
euler.call(9375, 91, 68, 40)
euler.call(18750, 98, 73, 43)
System.print("\nTook %(System.clock - start) seconds.")
```
Output:
```gamma 0.5772156649015328606065120900824024310421593359399235988057672348848677267776646709369470632917467495 (maxerr. 1e-100)
k = 255
gamma 0.5772156649015328606065120900824024310421593359399235988057672348848677267776646709369470632917467495 (maxerr. 1e-8000)
k = 20462
gamma 0.5772156649015328606065120900824024310421593359399235988057672348848677267776646709369470632917467495 (maxerr. 1e-15625)
k = 39967
gamma 0.5772156649015328606065120900824024310421593359399235988057672348848677267776646709369470632917467495 (maxerr. 1e-31250)
k = 79936

Took 2.330538 seconds.
```

### The easy way (Embedded)

```import "./gmp" for Mpf

var prec = (101/0.30103).round
var gamma = Mpf.euler(prec)
System.print(gamma.toString(10, 100))
```
Output:
```0.5772156649015328606065120900824024310421593359399235988057672348848677267776646709369470632917467495
```

## XPL0

Translation of: C
```\*********************************************
\Subject: Comparing five methods for
\         computing Euler's constant 0.5772...
\---------------------------------------------

include xpllib; \for Print
define Epsilon = 1e-6;

real A, B, H, N2, R, U, V, S(2), B2;
int  K, K2, M, N;
[Print("From the definition, error 3e-10\n");
N:= 400;  H:= 1.;
for K:= 2 to N do
H:= H + 1.0/float(K);
\Faster convergence: Negoi, 1997
A:= Ln(float(N) + 0.5 + 1.0/(24.*float(N)));
Print("Hn    %1.16f\n", H);
Print("gamma %1.16f\nK = %d\n\n", H-A, N);

Print("Sweeney, 1963, error 3e-10\n");
N:= 21;  S(0):= 0.;  S(1):= float(N);
R:= float(N);  K:= 1;
repeat
K:= K+1;
R:= R * float(N) / float(K);
S(K&1):= S(K&1) + R/float(K);
until R <= Epsilon;
Print("gamma %1.16f\nK = %d\n\n", S(1)-S(0)-Ln(float(N)), K);

Print("Bailey, 1988\n");
N:= 5;  A:= 1.;  H:= 1.;
N2:= Pow(2., float(N));
R:= 1.;  K:= 1;
repeat
K:= K+1;
R:= R * N2 / float(K);
H:= H + 1.0/float(K);
B:= A;  A:= A + R*H;
until abs(B-A) <= Epsilon;
A:= A * N2 / Exp(N2);
Print("gamma %1.16f\nK = %d\n\n", A-float(N)*Ln(2.), K);

Print("Brent-McMillan, 1980\n");
N:= 13;  A:= -Ln(float(N));
B:= 1.;  U:= A;  V:= B;
N2:= float(N*N);  K2:= 0;  K:= 0;
repeat
K2:= K2 + 2*K + 1;
K:= K+1;
A:= A * N2 / float(K);
B:= B * N2 / float(K2);
A:= (A + B) / float(K);
U:= U + A;
V:= V + B;
until abs(A) <= Epsilon;
Print("gamma %1.16f\nK = %d\n\n", U/V, K);

Print("How Euler did it in 1735\n");
\Bernoulli numbers with even indices
B2:= [1.0, 1.0/6., -1.0/30., 1.0/42., -1.0/30.,
5.0/66., -691.0/2730., 7.0/6., -3617.0/510., 43867.0/798.];
M:= 7;  N:= 10;
\Nth harmonic number
H:= 1.;
for K:= 2 to N do
H:= H + 1.0/float(K);
Print("Hn    %1.16f\n", H);
H:= H - Ln(float(N));
Print("  -ln %1.16f\n", H);
\Expansion C:= -digamma(1)
A:= -1.0 / (2.*float(N));
N2:= float(N*N);
R:= 1.;
for K:= 1 to M do [
R:= R * N2;
A:= A + B2(K)/(2.*float(K)*R);
];
Print("err  %1.16f\ngamma %1.16f\nK = %d", A, H+A, N+M);

Print("\n\nC  =  0.57721566490153286...\n");
]```
Output:
```From the definition, error 3e-10
Hn    6.5699296911765100
gamma 0.5772156645765730
K = 400

Sweeney, 1963, error 3e-10
gamma 0.5772156645636310
K = 68

Bailey, 1988
gamma 0.5772156649015350
K = 89

Brent-McMillan, 1980
gamma 0.5772156649015330
K = 40

How Euler did it in 1735
Hn    2.9289682539682500
-ln 0.6263831609742080
err  -0.0491674960726750
gamma 0.5772156649015330
K = 17

C  =  0.57721566490153286...
```