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Something similar to this has probably been posted, but since I can't find any at the moment I will post it here.

There are many numerical expressions to do with $\pi$, $e$ and $163$ (Wikipedia has many of these). The following are some of the approximations I have discovered when trying out different operations using the three numbers on my calculator:

$$e^\pi - \pi^{1-e} \approx 23$$ $$\sqrt[e]{\pi} \approx \dfrac{\pi+1}e$$ $$\sqrt{\pi+e+163} \approx 13$$ $$\sqrt[3]{163}-\sqrt[3]{\pi} \approx 4 $$ $$\sqrt{163}-\sqrt{\pi}\approx11$$ $$\dfrac{\sqrt{163}}{\sqrt[3]e} \approx 6+\pi $$ $$\dfrac{\pi}{2e} \approx \dfrac1{\sqrt3}$$ $$\sqrt[3]{\dfrac{\pi^3}{\sqrt[3]e}+\dfrac{e^3}{\sqrt[3]{\pi}}}\approx 3.3 \,\text{(my favourite)}$$ $$ e^\pi-2(4\pi-1)\approx0$$ $$ \dfrac{\pi}e\left(e^{\sqrt[3]{\pi}}\right)\approx5$$

EDIT: Inspired by @Raffaele's approximation I find that if $$x=\frac{163}{e}+\frac{e}{163}+\frac{\pi}{163}-e^{\pi}$$ then $\sin x \approx 0.6$, $\cos x \approx 0.8$ and $\tan x \approx 0.75$.

Do you have any others?

Martin Sleziak
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ə̷̶̸͇̘̜́̍͗̂̄︣͟
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6 Answers6

10

enter image description here

The diagram shows $n$ circles and $n-1$ equilateral triangles (and $6n-5$ cells).

We have the following elegant result:

There exists a constant $C$ such that, if the total area is $2^nC$ then
{product of the areas of the cells} $=2C$.

Numerical investigation shows that $C=2.71858...$

So $C=e$, right?

Wrong!

$C=\dfrac{3\left(2^{1/6}\right)}{\sqrt{\dfrac{15\sqrt3}{\pi}-\dfrac{27}{\pi^2}-4}}\approx 2.718586969\approx 1.000112e$.

A mathematical imposter.

Calculation

$r_k=\{\text{radius of $k$th circle from outside}\}=\sqrt{\dfrac{2^nC}{\pi}}\left(\frac{1}{2}\right)^{k-1}$

$A_k=\{\text{area of $k$th circle from outside}\}=\pi{r_k}^2$

$B_k=\{\text{area of $k$th triangle from outside}\}=\frac{3\sqrt3}{4}{r_k}^2$

$\text{product areas of cells}=A_n\prod\limits_{k=1}^{n-1}\left(\dfrac{A_k-B_k}{3}\right)^3\left(\dfrac{B_k-A_{k+1}}{3}\right)^3=2C$

This leads to $C=\dfrac{3\left(2^{1/6}\right)}{\sqrt{\dfrac{15\sqrt3}{\pi}-\dfrac{27}{\pi^2}-4}}$

Dan
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  • A bit late to the party, but do you have a source for the result? And how do you count the enclosed regions? Are there $n+(n-1)$ or $3(n-1)+1$? – Benjamin Wang Oct 26 '23 at 17:56
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    @BenjaminWang I found this result by myself, so I can't quote a source. I have edited my answer to address your question about the enclosed regions. – Dan Oct 27 '23 at 06:36
  • Change the triangles to squares, and we have: If the total area is ${\sqrt2}^nC$ then the product of the areas of the cells is $\sqrt2 C$, where $C=\dfrac{2^{17/16}}{\sqrt{\frac{3}{\pi}-\frac{4}{\pi^2}-\frac12}}\approx9.37361$. – Dan Oct 27 '23 at 11:12
5

$$\frac{163}{e}+\frac{e}{163}+\frac{\pi }{163}\approx 60$$

It's mine :)

Hope you like it

EDIT

$163 (\pi -e)\approx 69$

Raffaele
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3

The number $e^{\pi\sqrt{163}}$ is very close to the integer $262537412640768744$ the difference is about $7.5\times 10^{-13}$

Ben P.
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  • How did you find this coincidence? – Gribouillis Dec 02 '17 at 12:19
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    It's fairly well known, that number is sometimes called the "Ramanujan constant". – Ben P. Dec 02 '17 at 12:45
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    This is not a coincidence. – MJD Dec 02 '17 at 14:26
  • @MJD Ah, you got me... still it'll look like a coincidence for "most people". – Ben P. Dec 02 '17 at 15:04
  • $e^{\pi\sqrt{163}}=2.62537412640768743999999999999250\times 10^{17}$ – Claude Leibovici Dec 03 '17 at 08:31
  • We know that $\mathbb{Z}[\frac{1+\sqrt{-163}}{2}]$ is a Dedekind domain with class number one, thus (class field theory of quadratic imaginary fields) $j(\frac{1+\sqrt{-163}}{2}) $ is an integer. But we also have the Fourier expansion : $j(z) =e^{-2i \pi z}+ \sum_{n=1}^\infty c_n e^{2i \pi n z}$, thus $j(\frac{1+\sqrt{-163}}{2}) = e^{2i \pi \frac{1+\sqrt{-163}}{2}} + \mathcal{O}(e^{-\pi\sqrt{163}})$ and $e^{2i \pi \frac{1+\sqrt{-163}}{2}} = -e^{-\pi \sqrt{163}}$ is very close to an integer. – reuns Dec 14 '17 at 17:26
1

This is not an approximation but an equality,

$$\big(12\sqrt2\big)^2 \sum_{k=1}^\infty\frac1{k^6\,\binom{2k}{k}}+\big(12\sqrt3\big)^2\sum_{k=1}^\infty\frac1{k^2\,\binom{2k}{k}}\sum_{n=1}^{k-1}\frac1{n^4}=\color{blue}{163}\,\zeta(6)$$

where $\displaystyle\zeta(6) = \frac{\pi^6}{945}$, but the appearance of $163$ here is just a coincidence, and not related to its being a Heegner number.

Tito Piezas III
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1

Just sharing one of mine. [Sorry: It is only for $e$ and $\pi$]

$$(1+9^{-4^{6\times 7}})^{3^{2^{85}}}\approx e$$

The left-hand side is equal to first 18,457,734,525,36,901,453,873,570 (18 trillion trillion) digits of $e$ after decimal.

See that the left-hand side consists of all the numbers from $1$ to $9$.

A similar thing can be formed for $\pi$ also, $$\pi\approx 2^{5^{.4}}-.6-(\frac{.3^9}{7})^{.8^{.1}}$$

Correct up to 10 digits of $\pi$.

Video_Link

Sujit Bhattacharyya
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-1

$$ \frac{1}{\pi} + \frac{4}{\pi^2 - 4} \approx 1 $$

Nilotpal Sinha
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