[VA] SRC #009 - Pi Day 2021 Special

[robve]

(03-16-2021 07:17 PM)Albert Chan Wrote:  

(03-14-2021 07:00 PM)Valentin Albillo Wrote:   [*]d.  Conversely, the volume enclosed by the n-dimensional sphere of radius R is given by:



Go on and evaluate the \(\pi\)-th root of the summation for even dimensions from 0 to \(\infty\) of the volumes enclosed by the respective n-dimensional unit spheres (R = 1).

sum = 1 + pi/1! + pi²/2! + pi³/3! + ... = e^pi
sum ^ (1/pi) = e

Comment: formula give 1 for volume of 0-dimensional sphere, which seems weird.

Yes, it is kind of weird. But this is connecting two seemingly unrelated formulae, which is nice.

1. Taylor series:

$$ e^x = 1+\frac{x}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}+\cdots $$

2. The volume of an n-ball with radius R:

$$ V_n = \frac{\pi^{\frac{n}{2}}}{\Gamma(\frac{n}{2}+1)}R^n $$

The latter simplifies to

$$ \frac{\pi^k}{k!} $$

for k=2n and R=1 (the conditions stated in the challenge). Therefore, the answer is e as the π root of the sum:

$$ \sum_{k=0}^\infty \frac{\pi^k}{k!} = \mathrm{e}^\pi $$

I recognized the Taylor series after simplifying the sum's terms. Whenever you see a factorial in a denominator in a term of a series sum, there may be a Taylor series lurking beneath.

Nice piece of natural pie...

- Rob