there is any relation between $\pi$, $\sqrt{2}$ or a generic polygon?

Question

I'm a programmer, I'm always looking for new formulas and new way of computing things, to satisfy my curiosity I would like to know if there are any formulas, or I should say equalities, that make use of both $\pi$ and $\sqrt{2}$ .

I would also like to know if it's possible to generalize this relatively to any n-sided polygon ( even a 3D figure ), $\sqrt{2}$ that usually appears in quadrilaterals only.

Of course I would like to know about any possible domain, but since we should start from something, I would say that the domain of polygons and polyhedron triggers my interest in the first place.

Answer 1

$$\exp(i (\sqrt{2})^2 \pi)=1$$

Answer 2

Stirling's approximation: $$ n! \sim \sqrt{2 \pi n}\left(\frac{n}{e}\right)^n $$

Answer 3

$$ \pi = 2\sqrt 2 \cdot \frac{2}{\sqrt{2+\sqrt2}} \cdot \frac{2}{\sqrt{2+\sqrt{2+\sqrt2}}} \cdots. $$

Answer 4

$$t_n=\bigg(-\dfrac14\bigg)^n\cdot\dfrac{\displaystyle{2n\choose n}}{1-2n}\qquad=>\qquad\sum_{n=0}^\infty t_n=\sqrt2~,\qquad\qquad\sum_{n=0}^\infty t_n^2=\dfrac4\pi$$

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This is based on the fact that $~\displaystyle\sum_{n=0}^N{N\choose n}=2^N,~$ and $\quad\displaystyle\sum_{n=0}^N{N\choose n}^2={2N\choose N}.~$ Now let $N=\dfrac12$

and use the fact that $\Big(\tfrac12\Big)!=\dfrac{\sqrt\pi}2$ . See binomial series, Vandermonde's identity, and $\Gamma$ function

for more details.

Answer 5

For a programmer the Ramanujan formula(e) might be quite useful.

Answer 6

Here are some formulas relating $\pi$ and $\sqrt{2}$. Given the Riemann zeta function,

$$\zeta(s) = \sum_{k=1}^\infty \frac{1}{k^s}$$

Let $q=e^{\pi\sqrt{2}}$. Then,

$$\begin{aligned}\frac{3}{2}\,\zeta(3) &= \frac{\pi^3}{24}\sqrt{2}-2\sum_{k=1}^\infty \frac{1}{k^3(q^{k}-1)}-\sum_{k=1}^\infty\frac{1}{k^3(q^{2k}-1)}\\ \frac{3}{2}\,\zeta(5) &= \frac{\pi^5}{270}\sqrt{2}-4\sum_{k=1}^\infty \frac{1}{k^5(q^{k}-1)}+\sum_{k=1}^\infty \frac{1}{k^5(q^{2k}-1)}\\ \frac{9}{2}\,\zeta(7) &= \frac{41\pi^7}{37800}\sqrt{2}-8\sum_{k=1}^\infty\frac{1}{k^7(q^{k}-1)}-\sum_{k=1}^\infty\frac{1}{k^7(q^{2k}-1)} \end{aligned}$$

Answer 7

Pretty trivial, but for the sake of completeness:

$$\sin\frac\pi4=\cos\frac\pi4=\frac1{\sqrt2}$$

Here I'd say the $\frac\pi4$ relates to a regular octagon, or the canonical eighth root of unity, so yes, it generalizes to other regular $n$-gons, resulting in other algebraic numbers for the sine and cosine of the corresponding angles resp. the real and imaginary part of the corresponding root of unity.

If you want to go 3D, you'd probably start using Euler angles and quaternions, but I'll not go into this.