Kind of intentionally obtuse since they used eₑ as a variable and eₑₑ as another variable, and used (e-e) as an exponent a few times, which is basically the equivalent of multiplying by 1 in a fancy way. The first and last term also perfectly cancel out.
The same integral written in a saner form is:
integral from -e^e to e^e of (integral from -e^e to e^e of e^-(x^2+y^2)dy)dx
Wait… that’s not an approximation at all! That equals exactly pi. If I understand the math correctly, it’s effectively a formula for the area of a unit circle.
That should be an approximation. To get exactly pi the range of both integrals should be from minus infinity to infinity like this. It’s the integral of the 2D Gaussian, which is fairly known.
Ah, you’re right. I was thrown off by WolframAlpha saying the integral = π ≈ 3.1416 Both of those should be ≈
(x^2 + y^2)=1 is the equation for a unit circle, so it’s definitely related. Just not quite how I thought.
The approximation is only around 4.28*10^-101 off from pi.
https://www.wolframalpha.com/input?i=integral+from+-a+to+a+of+(integral+from+-a+to+a+of+e^-(x^2%2By^2)dy)dx
https://www.wolframalpha.com/input?i=(1-erf(e^e)^2)*piAnd 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.So the number 3 should be close enough for home use. Good to know. Thanks!
My maths exam asked me to consider pi=5.
“I will… consider it.”
Technically you need another 20 digits if you want to get down to a Planck length. (57 digits in total)
