Compute the first N digits of e

Question

Challenge

Write a program to compute the the first N (<= 10^3) digits of e.

Your program should take an integer N as input.

Input:

Output:

2.718281828459045235360287471352662497757247093699959574966967627724076630353547594571382178525166427

Input:

Output:

2.718282

Input:

Output:

2.718281828459045235360287471352662497757247093699959574966967627724076630353547594571382178525166427427466391932003059921817413596629043572900334295260595630738132328627943490763233829880753195251019011573834187930702154089149934884167509244761460668082264800168477411853742345442437107539077744992069551702761838606261331384583000752044933826560297606737113200709328709127443747047230696977209310141692836819025515108657463772111252389784425056953696770785449969967946864454905987931636889230098793127736178215424999229576351482208269895193668033182528869398496465105820939239829488793320362509443117301238197068416140397019837679320683282376464804295311802328782509819455815301756717361332069811250996181881593041690351598888519345807273866738589422879228499892086805825749279610484198444363463244968487560233624827041978623209002160990235304369941849146314093431738143640546253152096183690888707016768396424378140592714563549061303107208510383750510115747704171898610687396965521267154688957035035

Shortest solution wins!

Quixotic · Accepted Answer · 2011-03-13 01:00:03Z

9

Mathematica (12 bytes)

N[E,Input[]]

answered Mar 13, 2011 at 1:00

Quixotic

2,37915 silver badges27 bronze badges

\$\begingroup\$ You can try it online: there is a e-digits command on yubnub.org e.g., yubnub.org/parser/parse?command=e-digits+1000 that uses wolframalpha wolframalpha.com/input/… \$\endgroup\$
– jfs
Mar 13, 2011 at 10:09
1

\$\begingroup\$ -1 because you do not implement the calculation by your self. \$\endgroup\$
– FUZxxl
Mar 14, 2011 at 14:13
2

\$\begingroup\$ @FUZxxl:The author didn't asked for an implementation,only shortest ones is preferred. \$\endgroup\$
– Quixotic
Mar 22, 2011 at 0:34
8

\$\begingroup\$ @Quixotic The question does say "compute" not just output. Would you have argued the same thing if the question had said "calculate" instead of "compute"? \$\endgroup\$
– nitro2k01
Feb 2, 2014 at 20:44

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hallvabo · Accepted Answer · 2011-03-13 17:52:59Z

9

Python, 69

e=f=n=1;N=input()+2;exec"e+=10**N/f;f*=n;n+=1;"*N;print'2.'+`e`[1:-4]

Computes N+2 iterations of the standard power series for e.

edited Mar 13, 2011 at 17:52

answered Mar 13, 2011 at 13:21

hallvabo

1,67014 silver badges18 bronze badges

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jfs · Accepted Answer · 2011-03-13 09:50:20Z

7

Python, 67

import decimal as d
d.getcontext().prec=input()
print d._One.exp()

answered Mar 13, 2011 at 9:50

jfs

2291 silver badge6 bronze badges

\$\begingroup\$ Save 3 chars? from decimal import* and remove both d. \$\endgroup\$
– Timtech
Feb 2, 2014 at 22:16
2

\$\begingroup\$ @Timtech import * won't import _One due to the leading underscore. \$\endgroup\$
– jfs
Apr 24, 2016 at 10:24
\$\begingroup\$ Okay, didn't know that, sorry/ \$\endgroup\$
– Timtech
Apr 24, 2016 at 13:44

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Jesse Millikan · Accepted Answer · 2011-03-13 17:32:35Z

6

J, 20...ish.

Computational, but... Very inefficient. With n defined (20):

(0 j.<:n)":+/%!i.x:n+9

As a monadic function (32):

ge =. 3 : '(0 j.<:y)":+/%!i.x:y+9'

As a tacit function (33, fixed):

(0 j.<:)":(+/)&:(%&!&i.&x:&(9&+))

edited Mar 13, 2011 at 17:32

answered Mar 13, 2011 at 4:18

Jesse Millikan

1,4588 silver badges11 bronze badges

\$\begingroup\$ This converges very fast (for 100 digits you need "just" 70 terms) so you can drop that +9. \$\endgroup\$
– Eelvex
Mar 13, 2011 at 10:05
\$\begingroup\$ For 1000 digits this: (0 j.<:1000)":+/%!i.x:450 takes just a few seconds. \$\endgroup\$
– Eelvex
Mar 13, 2011 at 10:09
\$\begingroup\$ Would you care to explain your code? \$\endgroup\$
– aaaaaaaaaaaa
Mar 13, 2011 at 14:49
\$\begingroup\$ @Eelvex: But I need extra terms up until N=30 or so. (Any hints on reducing that mess of composes, or is that about right?) \$\endgroup\$
– Jesse Millikan
Mar 13, 2011 at 18:04
\$\begingroup\$ Ah, you're right, that's unfortunate. (That "mess" is shorter than anything else I could come up with - unless, off course, you use a fixed number for i. like: (0 j.<:n)":+/%!i.999x; then tacitly:(+/%!i.999x)":~0 j.<: ) \$\endgroup\$
– Eelvex
Mar 13, 2011 at 18:20

| Show 2 more comments

FlipTack · Accepted Answer · 2017-12-28 20:16:28Z

3

05AB1E, 4 3 2 bytes

_{Crossed out 4 is still regular 4 ;(}

Thanks to @Adnan for a byte.

žt

Uses CP-1252 encoding.

Explanation:

žt - Push input. Pop a, push e to a places (up to 10000).

Update:

Remove I, as žt takes input anyway if none is on the stack.

edited Dec 28, 2017 at 20:16

FlipTack

14.7k3 gold badges53 silver badges101 bronze badges

answered May 2, 2016 at 7:13

user53406

1

\$\begingroup\$ Welcome to Programming Puzzles and Code Golf! The ž character is 1 byte in the CP-1252 encoding, so this has a total score of 2 bytes instead of 3 :p. \$\endgroup\$
– Adnan
May 2, 2016 at 17:46
\$\begingroup\$ @Adnan Thanks! Nice language, btw. \$\endgroup\$
– user53406
May 2, 2016 at 17:47
\$\begingroup\$ @GeorgeGibson Thank you! :) \$\endgroup\$
– Adnan
May 2, 2016 at 17:47

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Community · Accepted Answer · 2020-06-17 09:04:33Z

1

Ruby, 68

require 'bigdecimal/math';include BigMath;puts E(gets.to_i).to_s 'F'

edited Jun 17, 2020 at 9:04

CommunityBot

1

answered Mar 13, 2011 at 1:08

david4dev

6933 silver badges10 bronze badges

\$\begingroup\$ it doesn't work on ruby1.8. It produces invalid results on ruby1.9 \$\endgroup\$
– jfs
Mar 13, 2011 at 9:57
\$\begingroup\$ You're right, but it should work in Ruby 1.9. There seems to be something wrong with precision in the E function. \$\endgroup\$
– david4dev
Mar 13, 2011 at 10:48
\$\begingroup\$ Looks fine here, it just adds some imprecise digits. Mayby like this? require'bigdecimal/math';puts BigMath::E(a=gets.to_i).to_s(?F)[0,a+1] \$\endgroup\$
– steenslag
Mar 13, 2011 at 22:48

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aaaaaaaaaaaa · Accepted Answer · 2011-03-13 16:02:39Z

GolfScript 43 41

~10\?1:b 461,{)b*:b@+\}460*;*b/`);(;'2.'\

Edit: I might as well replace the 0 with the leftover 1 from initializing b, the resulting difference is far too small to make it to the output.

I left the old version as that is what I have documented below.

~10\?1:b;0 461,{)b*:b@+\}460*;*b/`);(;'2.'\

~10\? Take input and calculate 10^input, leave the result on the stack.
1:b; Store 1 in b.
0 461, Put 0 on the stack, put the array [0 1 ... 459 460] on the stack.
{ }460* Execute function 460 times.
)b*:b Take the last element of the array, multiply it by b, store result in b and leave the result on the stack.
@+\ Switch the 0 (which is only a zero at the first iteration) to the top of the stack, add it to the leftover b value, and switch the result back again.
; Remove the rest of the array (only [0] is left).
The number that was initialized to 0 now hold the value e*460! and b hold 460!
* Multiply 10^input by e*460! (they are at this point the only 2 elements left on the stack).
b/ Divide the result by b.
The stack now hold the value e*10^input which when converted to a string will hold all the decimals, but not the dot.
`);(;'2.'\ A bunch of string operations to fit in the dot.

e*460! is calculated as 1 + 460 + 460*459 + 460*459*458 etc.

Community · Accepted Answer · 2020-06-17 09:04:33Z

1

bc, 17 chars

scale=read()
e(1)

edited Jun 17, 2020 at 9:04

CommunityBot

1

answered Feb 2, 2014 at 20:36

Tomas

2,54915 silver badges19 bronze badges

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pacman256 · Accepted Answer · 2022-11-06 17:33:14Z

1

Vyxal, 2 bytes

∆Ė

Try it Online!

Built in solution.

answered Nov 6, 2022 at 17:33

pacman256

3,7601 gold badge7 silver badges30 bronze badges

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barbermot · Accepted Answer · 2014-02-02 22:43:58Z

0

J, 17

(":&(x:^1)@*&0j1)

Example:

(":&(x:^1)@*&0j1) 50
2.71828182845891281655718620537435347047040502245993

Uses built in exponential verb - so, "compute" is on shaky grounds. Basically:

^1 - computes e**1
x: - does extended precision
0jy ": - formats the number to y digit

answered Feb 2, 2014 at 22:43

barbermot

1413 bronze badges

\$\begingroup\$ *&0j1 is simply j. \$\endgroup\$
– FrownyFrog
Oct 27, 2017 at 0:29

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Stuart Bruff · Accepted Answer · 2016-05-02 13:08:53Z

0

Mathcad, [tbd] bytes

Uses a spigot algorithm - http://comjnl.oxfordjournals.org/content/11/2/229.full.pdf+html.

Byte count not given as Mathcad byte count equivalence is yet to be determined. However, using a symbol equivalence it is roughly 121 bytes ... and not going to win any brevity prizes no matter how the equivalence is determined.

answered May 2, 2016 at 13:08

Stuart Bruff

5412 silver badges6 bronze badges

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RGS · Accepted Answer · 2020-01-23 18:15:03Z

0

Mathematica, 7 bytes

N[E,#]&

My answer is a further golfing of this answer.

Try it online!

For some reason, the output over at TIO looks funny but I tested it on my machine and the code works just fine.

answered Jan 23, 2020 at 18:15

RGS

14.1k1 gold badge56 silver badges116 bronze badges

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Timtech · Accepted Answer · 2014-02-02 22:57:54Z

-1

GTB, 14

eS?`AS;_,1,A+1

Explanation

e - Put e as the last calculated value

S? - Convert e to string _

`A - Input A

S;_,1,A+1 Display the first A digits of e

answered Feb 2, 2014 at 22:57

Timtech

12.6k2 gold badges44 silver badges61 bronze badges

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Xcali · Accepted Answer · 2020-01-23 22:18:37Z

-1

Perl 5 `-Mbignum=bexp -p`, 12 bytes

$_=bexp 1,$_

Try it online!

answered Jan 23, 2020 at 22:18

Xcali

14.9k2 gold badges14 silver badges38 bronze badges

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Stack Exchange Network

Compute the first N digits of e

Challenge

14 Answers 14

Mathematica (12 bytes)

Python, 69

Python, 67

J, 20...ish.

05AB1E, 4 3 2 bytes

Ruby, 68

GolfScript 43 41

bc, 17 chars

Vyxal, 2 bytes

Mathcad, [tbd] bytes

Mathematica, 7 bytes

GTB, 14

Perl 5 `-Mbignum=bexp -p`, 12 bytes

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Compute the first N digits of e

Challenge

14 Answers 14

Mathematica (12 bytes)

Python, 69

Python, 67

J, 20...ish.

05AB1E, 4 3 2 bytes

Ruby, 68

GolfScript 43 41

bc, 17 chars

Vyxal, 2 bytes

Mathcad, [tbd] bytes

Mathematica, 7 bytes

GTB, 14

Perl 5 -Mbignum=bexp -p, 12 bytes

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Perl 5 `-Mbignum=bexp -p`, 12 bytes

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