# Digits of e (#226)

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## Digits of e (#226)

Wayumbe Rubyists,

The mathematical constant e is the unique real number such that the
value of the derivative (slope of the tangent line) of the function
f(x) = e^x at the point x = 0 is exactly 1. The function e^x so
defined is called the exponential function, and its inverse is the
natural logarithm, or logarithm to base e.1

e is one of the most important numbers in mathematics, alongside the
additive and multiplicative identities 0 and 1, the constant ð, and
the imaginary unit i. These are the five constants appearing in one
formulation of Euler’s identity.

This week¢s quiz is to write a Ruby program that can compute the first
100,000 digits of e.

Have fun!

Daniel M. wrote:

This week¢s quiz is to write a Ruby program that can compute the first
100,000 digits of e.

Have fun!

I used the fast converging continued fraction series from the first
Wikipedia article:

# compute e*10**digits as a rounded integer

def calc_e(digits)
limit = 10**((digits+3)/2)
p = [2, 3]
q = [2, 1]
n = 1
begin
if p.length.even?
a = 4*(4n-1)
else
a = 4
n+1
n += 1
end
p << ap[-1] + p[-2]
q << a
q[-1] + q[-2]
end while p.last <= limit
(p.last*10**(digits+3)/q.last+500)/1000
end

if FILE == $0 p calc_e(if ARGV.length.zero? then 100_000 else Integer(ARGV[0]) end) end Hello, 2010/1/8 Daniel M. [email protected]: This week’s quiz is to write a Ruby program that can compute the first 100,000 digits of e. Have fun! Well I had :o) I got inspired from Quiz Digits of Pi (#202) and tried to estimate how long the standard library call could take (I didn’t want to waste too much CPU on this): require ‘bigdecimal’ require ‘bigdecimal/math’ include Math include BigMath d = [Time.now] 1000.step(20000,1000) do |i| E(i) d << Time.now puts i.to_s + " needing #{d[-1] - d[-2]} s" end which gives 1000 needing 0.147869 s 2000 needing 0.956326 s 3000 needing 2.993516 s 4000 needing 6.738271 s 5000 needing 12.78967 s 6000 needing 21.584826 s 7000 needing 34.154156 s 8000 needing 49.223259 s 9000 needing 69.57902 s 10000 needing 94.507271 s 11000 needing 123.327777 s 12000 needing 158.839553 s 13000 needing 198.976605 s 14000 needing 246.62736 s 15000 needing 300.471167 s 16000 needing 364.013845 s ^C almost scaling as the cube of the digit number you ask for. So that it should take around 10**5 s to compute E(100_000). I then tried my own implementation using the definition exp(x) = \Sum{n=0}{\infty} x^n/n! applied in x=1 and checking from E(number) that it was correct. It’s converging pretty fast (in term of iteration number, only 34_000 to get 100_000 digits) but it’s quite exactly 10 times slower than the standard method (and I didn’t want to wait 10**6 s to get all the 100_000 digits :o). The needed number of iterations is guessed using Stirling formula and I discovered that each step does not take the same time (it gets slower as time goes) whereas it does not depend on the initial number of digits asked for (that is the 1000 first iterations take approximately the same time when you ask for 10_000 digits or for 100_000 digits) which I couldn’t really understand. I first thought that the slowing-down was due to the fact that for more digits you need bigger integers to play with but the latter observations seems to invalidate this argument… If anybody have an explanation, I will be happy to hear it. require ‘rational’ require ‘enumerator’ require ‘bigdecimal’ require ‘bigdecimal/math’ include Math include BigMath d1 = Time.now precision = (ARGV[0] || 1000).to_i iterations = precision /(log10(precision)) + precision/(log10(precision))*log10(log10(precision)) puts 'Approximate number of iterations needed: ’ + iterations.to_i.to_s accuracy = 10**precision.to_i fact = 1 final = accuracy other = false d = [d1] 1.upto(iterations) do |i| if i%100 == 0 d << Time.now puts i.to_s + " needing #{d[-1] - d[-2]} s" end fact *= i final_old = final final += Rational(accuracy,fact) end d2 = Time.now puts "Time elapsed in computation: " + (d2 - d1).to_s + ’ s’ puts “Computation terminated: computing E(#{precision}) now.” puts "Delta * 10**(#{precision}): " + (final.round - E(precision)*accuracy).to_s d3 = Time.now puts "Time elapsed calling E(#{precision}): " + (d3 - d2).to_s + ’ s’ Last but not least, the same method could be called as one line in irb using inject and getting the result in the form of a Rational: fact = 1 ; (1…34_000).inject(1) {|sum,i| fact *= i ; sum + Rational(1,fact)} but you will rather want to try with a smaller number of steps (say 500 to get 1000 digits, remember: 34_000 steps will take around 10**6 s…) Cheers, Hello, based on the spigot algorithm of Rabonitz and Wagon [1]: #!/usr/bin/ruby def spigot n e = [] arr = [2] + Array.new(n+1,1) (n-1).times do |i| (n+1).downto 0 do |j| arr[j] *= 10 end q = 0 (n+1).downto 0 do |j| arr[j] += q q = arr[j]/(j+1) arr[j] %= j+1 end e << q end e[0]/= 10.0 puts e.join(‘’) end if$0 == FILE
spigot ARGV[0].to_i
end

[1] http://www.mathpropress.com/stan/bibliography/spigot.pdf : citing
the easier case of e

Am 08.01.2010 06:26, schrieb Daniel M.:

Sorry, Rabonitz meant Rabinowitz, ugly typo.

Am 11.01.2010 14:57, schrieb Thorsten H.:

After recovering some vaguely remembered math, and trying not to look to
closely at any code
on the net, here is my solution.

The reference data for testing, from Project Gutenberg, has been
truncated for
this posting. The program will report “Incorrect” for more than a couple
of hundred digits.

Bill Rutiser
wrutiser AT gmail DOT com

e.

generated.

used

i.

actually

nn,

# previous term, one digit is extracted.

def digits_int( n_terms )

nn = 0
dd = 1
mm = 1

q1 = -1

z = “2.\n”
digits = 0

2.upto(n_terms) do |i|

dd *= i
nn *= i
nn += mm

q2, r2 = (nn * 100).divmod( dd )

if( q1 == q2 )
q, r = (nn * 10).divmod( dd )
z << q.to_s

digits += 1
z << " " if 0 == digits % 10
z << "\n" if 0 == digits % 50

mm *= 10
nn = r
q1 = -1
else
q1 = q2
end


end

puts( “\ndigits: #{digits} n_terms: #{n_terms}”
" terms/digit: #{Float(n_terms) /digits}" )
puts( “dd.size: #{dd.size} nn.size: #{nn.size}” )
puts z

if compare?( $gutenberg_digits, z ) puts “Correct!!” else puts “Not correct.” end end def compare?( ssx, ssy ) sx = ssx.delete( " \n" ) sy = ssy.delete( " \n" ) sx.start_with?( sy ) end # Digits from http://www.gutenberg.org/files/127/127.txt http://www.gutenberg.org/files/127/127.txt # Computed by Robert Nemiroff and Jerry Bonnell. # See the file itself for details, license, copyrights, etc.$gutenberg_digits = <<HERE
2.
7182818284590452353602874713526624977572470936999595749669676277240766303535
47594571382178525166427427466391932003059921817413596629043572900334295260595630
73813232862794349076323382988075319525101901157383418793070215408914993488416750

#### thousands of digits removed from email posting

HERE

$gutenberg_digits =$gutenberg_digits.delete( " \n" )
puts “gutenberg_digits.length: #{$gutenberg_digits.length}” Hello, 2010/1/8 Daniel M. [email protected] Suggestions?: http://rubyquiz.strd6.com/suggestions defined is called the exponential function, and its inverse is the Have fun! -Daniel http://rubyquiz.strd6.com I got some fun playing with computing e. I didn’t get any formal solution (it definitely takes too long …) So, e.rb contains 8 basic methods to compute e, using basic Float of ruby(so don’t exepect to get a ot of precision). Anyway, we can see the continuous fraction is good, and it’s the only one who worked easily with BigDecimal(I got only 113 digits, shame on me …). ## Here is the output of e.rb 2.718281828459045 lim(n->âˆž) (1+1/n)**n with n = 100000000 2.7182817983473577 3.011168736577474e-08 lim(n->0) (1+n)**(1/n) with n = 1.0e-08 2.7182817983473577 3.011168736577474e-08 Î£(n=0,âˆž) 1/n! with n = 17 2.7182818284590455 -4.440892098500626e-16 lim(n->âˆž) n/(âˆš(n,n!)) with n : 170 2.663087878748024 0.055193949711020984 [[2;1,2,1,1,4,1,1,6,1,1,8,1,1,…,2n,1,1,…]] with 23 numbers 2.718281828459045 0.0 [[1,0,1,1,2,1,1,4,1,1,6,1,1,8,1,1,…]] with 25 numbers 2.718281828459045 0.0 [[1,0.5,12,5,28,9,44,13,60,17,…,4(4n-1),4n+1,…]] with 10 numbers 2.718281828459045 0.0 ## Global maximum of f(x) = âˆš(x,x) with p = 16 2.7182818284590446 4.440892098500626e-16 And I got also a strange thing. I saw somewhere it’s possible to compute e with lim(n->inf) ((2n+1)/(2n-1))**n, but changing the limit to 0 gives Ï€ (hum, the complex part of the result, divided by n), awesome (or this is surprising me at least ) include Math p E # 2.718281828459045 puts formula = → n { ((2n+1)/(2n-1).to_f)**n } puts “lim(n->inf) ((2n+1)/(2n-1))**n with n = #{n = 100_000}” puts “e” p e = formula[n] # 2.718281828493031 p e - E # 3.398570314061544e-11 puts “lim(n->0) ((2n+1)/(2n-1))**n with n = #{n = 1.0/100_000_000}” puts “Ï€” p pi = formula[n] # (1.0+3.141592653589794e-08i) p pi = pi.imaginary / n # 3.1415926535897936 p pi - PI # 4.440892098500626e-16 Cheers, B.D. ## Digits of e (#226) This weekï¿½s quiz is to write a Ruby program that can compute the first 100,000 digits of e.$ cat e.rb
digits = 100000
fudge = 10
unity = 10**(digits + fudge)
e = unity
n = unity
i = 0
while (n>0)
i += 1
n /= i
e += n
end
e /= 10**fudge
p e
\$ time ruby e.rb > /dev/null

real 0m4.023s
user 0m3.940s
sys 0m0.087s

This uses the taylor series definition of e which actually converges
quite fast.

-----Jay

Math::E**Math::PI-Math::PI ###
=> 19.9990999791895

## I decided to work on an old quiz.

This week’s quiz is to write a Ruby program that can compute the first
100,000 digits of e.

## Here is what I came up with:

PLACES = 100_000

euler = 0
big_one = 10**(PLACES+5)
nxt_one = big_one
n = 1

while (nxt_one != 0)
n += 1
nxt_one /= n
euler += nxt_one
end

## puts “2.”+euler.to_s[0…PLACES-1]

It ended up looking a lot like Jay’s solution.

This quiz harkens back to another mathematically themed computation
competition Digits of Pi (#202)[1]. Many of the same techniques that can
be
applied to the computation of pi can be applied to the computation of e,
and
there are some surprising results.

Jack Rouse used the fast converging continued fraction series from
Wikipedia[2] to create a short program that quickly calculates e to
100_000
digits. Jack’s solution is quite quick, yielding the output within
several
seconds.

Jean-Julien F. started with a benchmark of the the standard library
call
and estimated that it would take 10**5 seconds to compute the first
100_000
digits. Jean-Julien provides a handy one liner that can be used in irb:

fact = 1 ; (1..34_000).inject(1) {|sum,i| fact *= i ; sum +


Rational(1,fact)}

This computation makes use of the Taylor series definition. The results
use
number
of iterations, possibly 500 or so.

Thorsten H. created a program based on the spigot algorithm from
Rabonitz
and Wagon[3]. From the paper:

> This algorithm is a "spigot" algorithm: it pumps out digits one
> at a time and does not use the digits after they are computed


[…];
> the entire algorithm uses only ordinary integer arithmetic on
> relatively small integers.

The paper is actually about computing the digits of pi with the same
method,
but in doing so covers the case of e, which is simpler. An interesting
approach, and worth looking at if you are into mathematics.

Benoit D. explored eight different methods of computing e with a
focus
on breadth rather than depth. See them all in the attached solutions
supplement. Benoit also came across this interesting bit of information:

lim(n->inf) ((2n+1)/(2n-1))**n    => e
lim(n->0)   ((2n+1)/(2n-1))**n    => Ï€


Jay A. and and David S. used the Taylor series definition of
e,
which is quite fast at converging and delivering 100_000 digits. These
solutions are much faster that other Taylor series implementations
because
they use only integer arithmatic and a very simple loop.

Here is Jay A.'s solution:

digits = 100000
fudge = 10
unity = 10**(digits + fudge)

e = unity
n = unity
i = 0

while (n>0)
i += 1
n /= i
e += n
end

e /= 10**fudge
p e


Setting unity to be 10^digits shifts everything into the integers so
that
floating point math won’t be required. The extra fudge factor ensures
that
there aren’t rounding errors near the end. Each step of the loop
requires
only an increment, an integer division, and an addition. Notice how the
division in the loop accumulates the factorial, because the result of
1/2 is
stored when it is divided by 3 it is set equal to 1/3!. The loop
terminates
when the series term becomes 0, i.e. too small to add anything that will
matter within the chosen number of digits.

Thanks everyone for your solutions to the quiz!

[Digits of e (#226) - Solutions][4]

P.S. Brian C. linked to an xkcd comic involving e and Ï€[5], not all
contributions need to be code.