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submitted 2 months ago by [email protected] to c/[email protected]

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[–] [email protected] 31 points 2 months ago (1 children)

The approximation is only around 4.2810^-101 off from pi.
https://www.wolframalpha.com/input?i=integral+from+-a+to+a+of+%28integral+from+-a+to+a+of+e%5E-%28x%5E2%2By%5E2%29dy%29dx
https://www.wolframalpha.com/input?i=%281-erf%28e%5Ee%29%5E2%29pi

[–] [email protected] 41 points 2 months ago (3 children)

And because it always bears repeating;

According to JPL’s Chief Engineer for Mission Operations and Science, Marc Rayman-

Let's go to the largest size there is: the known universe. The radius of the universe is about 46 billion light years. Now let me ask (and answer!) a different question: How many digits of pi would we need to calculate the circumference of a circle with a radius of 46 billion light years to an accuracy equal to the diameter of a hydrogen atom, the simplest atom? It turns out that 37 decimal places (38 digits, including the number 3 to the left of the decimal point) would be quite sufficient.

[–] [email protected] 1 points 2 months ago* (last edited 2 months ago)

You can quote with the “greater than” sign (>). Backticks mark text as source code.

> quote

[–] [email protected] 7 points 2 months ago (2 children)

So the number 3 should be close enough for home use. Good to know. Thanks!

[–] [email protected] 0 points 2 months ago (1 children)

My maths exam asked me to consider pi=5.

[–] [email protected] 1 points 2 months ago

"I will... consider it."

[–] [email protected] 4 points 2 months ago

As an engineer, I approve this message!

[–] [email protected] 14 points 2 months ago* (last edited 2 months ago)

Technically you need another 20 digits if you want to get down to a Planck length. (57 digits in total)