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While study Numerics and playing with famous constants ($e$, $\pi$, Golden ratio) I came across the following relation

$$ \color{blue}{1.6^2+2.7^2 = 9.85\approx 3.14^2}$$

This is nothing special but patently, by first oder approximation one glimpses that,

$$ e \approx 2.7,~~~~~~\phi\approx1.6~~~~~~\text{and}~~~~~\pi\approx3.14$$ where $\phi =\frac{1+\sqrt5}{2}$ is the well known Golden ratio. Putting this in the previous relation we get

$$ \color{blue}{\phi^2+e^2 \approx \pi^2}.$$

Although it is hopeless to get the perfect relation $ \color{blue}{\phi^2+e^2 =\pi^2},$ which could be an amazing Pythagorean relation between famous constants $\color{red}{e,\phi} $ and $\color{red}{\pi}$. I believe there is a chance that there exists a well know constant $\color{red}{\delta}$ (of this kind) such that we have the following approximation.

$$ \color{blue}{\phi^2+e^2 +\delta^2\approx \pi^2}.\tag{I}\label{Eq}$$ Which is also a Pythagorean relation in three dimension.

But since $ \color{blue}{\phi^2+e^2 >\pi^2}$, the best idea to find such constant is obviously to consider $\delta $ satisfying the relation.

$$\color{red}{\pm i\delta= \sqrt{\phi^2+e^2-\pi^2}\approx 0.37079062365 }$$

My Question: Which Well known constant could be suitable in other to improve this Pythagorean relation in $\eqref{Eq}$?

I am thankful to all your propositions. Please I would like if it is possible some very close approximation like here, Proving that: $9.9998\lt \frac{\pi^9}{e^8}\lt 10$? or here How to prove that: $19.999<e^\pi-\pi<20$?

Michael Hardy
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Guy Fsone
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    Why did you take $e$ and $\phi$ to two significant digits but $\pi$ to three ? –  Oct 27 '17 at 18:41
  • Why should we assume there's anything special about $\delta$? We can easily come up with many relations like $\varphi^2+e^2\approx \pi^2$. I don't see any reason to assume that there's anything very deep here apart from that the numbers happen to be quite close together. Furthermore, what are you really asking? For us to find whether $\delta$ appears anywhere else or whether it's got a different closed for? I think either case is quite unlikely. – Jam Oct 27 '17 at 21:35
  • If you'd like to find constants close to $\delta\approx0.37$, you could try this list of mathematical constants (http://bit.ly/2yPS2Db) and pick one such as Artin's constant (http://bit.ly/2iG683B) or $\frac1e$ or the median of the Gumbel distribution, $-\ln\ln2$ (http://bit.ly/2iG683B), which are all equal to $0.37\pm0.01$. – Jam Oct 27 '17 at 21:42
  • Please note that after all the goal is to improve the Pythagorean relation in (I). choosing a suitable delta which for sure a pure complex number . – Guy Fsone Oct 27 '17 at 22:29
  • https://math.stackexchange.com/questions/108510/approximation-of-e-using-pi-and-phi?rq=1 – Guy Fsone Jan 28 '18 at 08:12

3 Answers3

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$\delta = \sqrt{e^2 + \phi^2 - \pi^2}$ is presumably a transcendental number (though AFAIK there's no proof that it isn't rational). A good rational approximation of it is $33/89$. I don't know if you'd call that a "well known constant".

Robert Israel
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With a similar approach to the other question,

  1. $\phi^2=\phi+1 = [2;1,1,1,\ldots]$ has simple convergents;
  2. $e^2=[7; 2, 1, 1, 3, 18, 5, 1, 1, 6, 30, 8, 1, 1,\ldots]$ has simple convergents too;
  3. $\pi^2$ is related to the solution of the Basel problem through creative telescoping: $$\pi^2 = \sum_{n\geq 1}\frac{18}{n^2 \binom{2n}{n}} $$ and the last series is pretty fast-convergent.

  4. It follows that $$ \phi^2=\frac{1597}{610}\pm 2\cdot 10^{-6} $$ $$ e^2 = \frac{12288}{1663}\pm 2\cdot 10^{-7}$$ $$ \pi^2 = \frac{16929464521}{1715313600}\pm 2\cdot 10^{-6}$$ hence $\phi^2+e^2$ is very close to $10$, which on its turn is not very far from $\pi^2$ ($\pi\approx\sqrt{10}$ was already known to Babylonians). According to the inverse symbolic calculator we have

    $$ \phi^2+ e^2 \approx \pi^2+\delta^2 $$ with $$ \delta = G\left[\zeta\left(\frac{1}{2}\right)+\Gamma\left(\frac{11}{12}\right)\right] $$ and $G$ being Catalan's constant.

Jack D'Aurizio
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  • For instant somebody proposed to tried $\delta =e^{-1}$ which is a good approximation so far. If you have anything in this direction it will be great. – Guy Fsone Oct 27 '17 at 20:57
  • @GuyFsone: a quick look at the inverse symbolic calculator (https://isc.carma.newcastle.edu.au/standardCalc) returns $$ \phi^2+ e^2 \approx \pi^2+\delta^2 $$ with $$ \delta = G\left[\zeta\left(\frac{1}{2}\right)+\Gamma\left(\frac{11}{12}\right)\right] $$ and $G$ being Catalan's constant. – Jack D'Aurizio Oct 27 '17 at 21:02
  • It seems I cannot access that link could you tries with this equation $\phi^2+ e^2+\delta^2 \approx \pi^2$ which exactly the approximation I want. The other way was my attempt. I have tried your link but Cannot access to it from here. thanks very much – Guy Fsone Oct 27 '17 at 21:37
  • @GuyFsone: as already remarked by other users, there is no way an approximate identity like $\phi^2+e^2+\delta^2\approx \pi^2$ might hold, since $\phi^2+e^2\approx 10$ is already greater than $\pi^2$. What you really want is $\phi^2+e^2\approx \pi^2+\delta^2$. – Jack D'Aurizio Oct 27 '17 at 21:43
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I also once played with these famous constants- I came across to this following relation :

$e+\pi \cong 3\phi+1$ or $\phi\approx\frac{e+\pi-1}{3}$

Approximation error:$|3\phi-(e+\pi-1)|\approx0,00577 < \frac{3}{500}$

Being in Pythagorean relations- let's check for $e+\pi \cong 3\phi+1$ squaring the sides :

Left side:$(e+\pi)^2 =e^2+2(e+\pi)+\pi^2 $

Right side:$(3\phi+1)^2 =9\phi^2+2(3\phi+1)+1 $

Hense:

$e^2+\pi^2+2(e+\pi)\cong9\phi^2+2(3\phi+1)+1 $

$e^2+\pi^2\cong9\phi^2+1 $

The approximation error of: $\sqrt{|e^2+\pi^2-(9\phi^2+1)|} \approx e$

Krzysztof Myśliwiec
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