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.
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$\begingroup$$$ \pi = 2\sqrt 2 \cdot \frac{2}{\sqrt{2+\sqrt2}} \cdot \frac{2}{\sqrt{2+\sqrt{2+\sqrt2}}} \cdots. $$
$\endgroup$ 1 $\begingroup$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.
$\endgroup$ $\begingroup$Stirling's approximation: $$ n! \sim \sqrt{2 \pi n}\left(\frac{n}{e}\right)^n $$
$\endgroup$ 1 $\begingroup$I had fun doing that back in the days; lots of paper and the smallest mistake would end in chaos. I felt I had discovered something in the end :) (Note: bug in the code snippet: k=3 is skipped; the line "if k>3 then" should be "if k>2 then".)
I was $17$ years old when I took on that task, alone at home with a lot of scratch paper. I remember the wow effect when the pattern emerged. It was about expressing PI in function of the square root of $2$. Focus recommended! In the last $35$ years, I mentioned it $3$ or $4$ times and it never raised much interest, probably because expressing an irrational number in function of another no more rational one does not make sense :). In iteration $3$, the function starts to look like the precedent one; then it's obvious between $3$ and $5$ and it goes on, making it easy to program. It will get closer to PI before breaking due to the variable's limitations. Here a code snippet (too) quickly put together for demo purposes.
Original post:
$$t_n=\bigg(-\dfrac14\bigg)^n\cdot\dfrac{\displaystyle{2n\choose n}}{1-2n}\qquad=>\qquad\sum_{n=0}^\infty t_n=\sqrt2~,\qquad\qquad\sum_{n=0}^\infty t_n^2=\dfrac4\pi$$
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.
$\endgroup$ 4 $\begingroup$You can use the Taylor series of $\sin(x)$ to calculate square roots using $\pi$.
For example, you can calculate the square root of 2 as follows:
$$\sin{(45^\circ)}=\sin{(\frac\pi4)}=\frac{\sqrt2}{2}$$
Therefore:
$$\sqrt2=2\cdot\sum_{x=1}^n \frac{(\frac{\pi}{4})^{4x-3}}{(4x-3)!}-\frac{(\frac{\pi}{4})^{4x-1}}{(4x-1)!}$$
I haven't tried personally, but I imagine you could relate a lot of square roots back to $\pi$ like this. Simply find the angle, in radians of a triangle with a side length of $\sqrt x$ and then expand by expressing the angle in terms of $\pi$ using the Taylor series.
$\endgroup$ $\begingroup$$$\exp(i (\sqrt{2})^2 \pi)=1$$
$\endgroup$ 8 $\begingroup$For a programmer the Ramanujan formula(e) might be quite useful.
$\endgroup$ $\begingroup$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,
$$\begin{aligned}\frac{3}{2}\,\zeta(3) &= \frac{\pi^3}{24}\sqrt{2}-2\sum_{k=1}^\infty \frac{1}{k^3(q^{k}-1)}-\sum_{k=1}^\infty\frac{1}{k^3(q^{2k}-1)}\\ \frac{3}{2}\,\zeta(5) &= \frac{\pi^5}{270}\sqrt{2}-4\sum_{k=1}^\infty \frac{1}{k^5(q^{k}-1)}+\sum_{k=1}^\infty \frac{1}{k^5(q^{2k}-1)}\\ \frac{9}{2}\,\zeta(7) &= \frac{41\pi^7}{37800}\sqrt{2}-8\sum_{k=1}^\infty\frac{1}{k^7(q^{k}-1)}-\sum_{k=1}^\infty\frac{1}{k^7(q^{2k}-1)} \end{aligned}$$
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