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Montegasppα Cacilhας
Montegasppα Cacilhας

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How To Compute Arbitrary Precision Transcendental Numbers

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Transcendental numbers are probably the most useful tools in Mathematics. They are irrational numbers that cannot be expressed using a finite formula.

For example, the Euler’s identity is considered the most beautiful Mathematical formula (perhaps I’m gonna explore it in the future):

𝑒𝑖π + 1 = 0

It exposes a relation between the two most important transcendental numbers, the complex numbers, the unit and the null / zero. It’s used mostly in rotation over real 𝑣𝑠 imaginary axes, on complex exponential, roots, and other useful operations.

Yet, it’s useless if one doesn’t know how to get 𝑒 and π in the required precision.

Euler’s constant

The Euler’s constant or Euler’s number, 𝑒 for short, is the ratio describing any constant growth. It’s defined as:

𝑒 = lim_(n→+∞)[1+(1/n)]^n

It’s about 2.71828…. 𝑒 is kinda magical number, poping up in a lot of Mathematical problems, offering good and easy solutions, since complex number operations to logarithm, exponential, and other kinds of growth behaviour.

But how to get to the desired precision?

One can take this formula and compute greater and greater 𝑛 values, until it reaches the required precision. Nevertheless, the exponent can be quite intimidating, leading to an undesired weak performance – similar or worst than the π computation below. Yet, it’s hard to say how long it must go to getta the desired precision.

Fortunately another Euler’s formula gives us a better solution:

𝑒 = Σ(1/n!)

This formula is very convenient, ’cause it increases the precision every step in an easly predictable way:

1/0! = 1/1 = 1
1 + 1/1! = 1 + 1/1 = 2
2 + 1/2! = 2 + 1/2 = 2.5
2.5 + 1/3! = 2.5 + 1/6 2.6667
2.6667 + 1/4! 2.6666 + 1/24 2.7083
2.7083 + 1/5! 2.7083 + 1/120 2.7167
2.7167 + 1/6! 2.7177 + 1/720 2.7180
prev. + 1/∞! = 𝑒

It’s easly predictable ’cause the precision equals to log₁₀(n!):

n = 0 → log₁₀(0!) = log₁₀(1) = 0
n = 1 → log₁₀(1!) = log₁₀(1) = 0
n = 2 → log₁₀(2!) = log₁₀(2) 0.3010
n = 3 → log₁₀(3!) = log₁₀(6) 0.7782
n = 4 → log₁₀(4!) = log₁₀(24) 1.3802
n = 5 → log₁₀(5!) = log₁₀(120) 2.0792
n = 1000 → log₁₀(1000!) 2’567.6046

It gets big very quickly.

More everyday case

Image one needs to calculate 𝑒 to the 80-bit precision. That doesn’t differ from before. Bits are binary digits, i.e, 2-base numbers. So, instead of log₁₀, one must use log₂:

  • log₂(n!) ≈ 80
  • 2log₂(n!) ≈ 280
  • n! ≈ 2⁸⁰
  • 25! < 2⁸⁰ < 26!
  • 25! < n! < 26!
  • 25 < n < 26

So, 26 steps are good enough.

Let’s implement it using Python:

from typing import Callable

# Lazy factorial implementation
fact: Callable[[int], int] = lambda n: prod(range(1, n+1))

# An LRU cached version, if you prefer:
# from functools import lru_cache
# fact: Callable[[int], int] = lru_cache(lambda n:
#     1 if n <= 0 else (n * fact(n - 1))
# )

def steps(prec: int) -> int:
    res = 1
    while fact(res) >> prec == 0:
        res += 1
    return res
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The right shift (>>) returns zero as long as the value‘s less bit wide than the precision.

Now let’s compute 𝑒 itself:

compute_e: Callable[[int], int] = lambda prec: sum(
    1./fact(x) for x in range(prec)
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The solutions coming out from it is:

>>> compute_e(steps(64))  # Python uses 64-bit floats
>>> math.e
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Not bad at all. In fact, we might reach the 64-bit precision with only 18 iterations.


We’re not worried about performance here, it’s not this post’s scope. We’re just showing how it works and how you can implement and use it.

What about π?

π is the ratio of any circle’s perimeter (circumference) to its diameter. That’s π very definition.

However, it’s a transcendental number and needs an infinity serie to be computed. Leibniz gave us a neat solution:

π = 4Σ(1/(4n+1) - 1/(4n+3))

The process is quite the same used for 𝑒, take the formula:

π = 4Σ(1/(4n+1) - 1/(4n+3))

Then get the precision:

compute_pi: Callable[[int], int] = lambda prec: 4. * sum(
    (1./(4*n+1) - 1./(4*n+3)) for n in range(prec)
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Tip: the precision is log₂(4n), so n = 2prec-2, which one can get by left shifting. Note that we got exactly the inverse of the previous calculation: here the necessary steps amount grows very fast with small gain.

So you may get disappointed when running:

>>> compute_pi(1 << (64 - 2))
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This computation is way more inefficient than the previous one – Python isn’t that good in dealing with floating-point operation and it’s necessary a humongous amount of steps to getta the target. I recomend 5M steps:

>>> compute_pi(5000000)
>>> math.pi
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To this task, one’s gonna need a tougher tool, as C and multithreading. Again performance is not this post’s scope. Probably it requires using some C cast spells in order to optimise the computation.

Or you can go down through the Wolfram MathWorld’s or Wolfram Alpha’s π page references in search of better formulæ. I did it, and I can ensure they are, including serie representations.


>> is good, << is bad (in a very generic and shallow fashion).

Top comments (6)

miguelmj profile image

Very interesting! Honestly It's refreshing to feel that you really learn something here. Great post!

cacilhas profile image
Montegasppα Cacilhας

In fact, I follow a lotta Mathematician YouTube channels, and I read a lotta scientific papers.

I just go to Kodumaro and come here to share what I’ve got (mostly after months or years of learning). 😬

If anyone wants it, I can share some of the channels I follow.

miguelmj profile image

I've seen your blog before and I think that's the kind of scientific dissemination we need. For me it's similar with Computer Science. I started my degree to learn it and I follow experts on the matter, but reaching a general audience is harder. Getting better at it will take a lot of learning, as you say.
There are lots of people writing in tech that contribute nothing. But people like you, that have an educational spirit and care about producing quality content, are the ones that deserve more attention!
Also, if you want to share some of those channels it would be much appreciated 😄

Thread Thread
cacilhas profile image
Montegasppα Cacilhας • Edited

Okay, @miguelmj ,

Let’s go into my YouTube Math&Tech channels list:


I put a mark in my favorities.


cacilhas profile image
Montegasppα Cacilhας • Edited

I recently updated this post, ’cause I’m used to write my posts for Kodumaro, and then I just copy the Markdown code and paste here in

However the interface deals with Markdown in a different way than Kodumaro does, driving the resultant presentation to be a little odd.

I updated it fixing the differences.

cacilhas profile image
Montegasppα Cacilhας • Edited

In time: I use Online LaTeX Equation Editor in order to generate the formulæ.