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Saw it in the news: $$(\pi^4 + \pi^5)^{\Large\frac16} \approx 2.71828180861$$

Is this just pigeon-hole?


DISCUSSION: counterfeit $e$ using $\pi$'s

Given enough integers and $\pi$'s we can approximate just about any number. In formal mathematical language we say this set is dense in the real numbers:

$$ \overline{ \mathbb{Z}[\pi]} = \mathbb{R}$$

This is only part of the story since it doesn't tell us how big our integers have to be in order to approximate the constant of our choosing? Maybe we can quantify this with a notion of density?

$$ \mu_N([a,b]) = \frac{\# |\{ m + n \pi: -N \leq m,n \leq N \}\cap[a,b]|}{N^2} $$

The example above works because of the constants 4, 5 and 6.
We can focus on a particular constant and ask how much effort it takes to approximate a given constant:

$$ \big\{ (m,n)\in \mathbb{Z}^2: \big| m + n \pi - \alpha \big|< \epsilon \big\} $$

In our case we need to incorporate for square roots, cube roots and higher.


Generalization How closely can we approximate $e$ using powers of $\pi$ and $n$-th roots?

$$\displaystyle ( a + b\pi )^{1/p} \approx e $$

Here $0 \leq |a|,|b|,p \leq 10$

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    $\begingroup$ A good approximation of order approx. $\;10^{-7}\;$ ....that's all. There are literally thousands of different such approximations to all kinds of numbers. $\endgroup$ – DonAntonio Jun 14 '14 at 11:07
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    $\begingroup$ This is most likely a coincidence, as the approximation is explained well by a long-known astonishing coincidence $\pi \approx e^\pi - 20$. For example, $\log(\pi^4 + \pi^5) = 4\log(\pi)+\log(1+\pi)$ and applying the above with the crude estimation $\exp(\pi) \approx 23$ in mind, one has $\log(\pi) \approx \log(\exp(\pi)-20) \approx \pi + \log(1-20/\exp(\pi)) \approx \pi + \log(3/23)$ and similarly $\log(1+\pi) \approx \log(\exp(\pi) - 19) \approx \pi + \log(4/23)$. Adding up, we have $5\pi + \log(3^4/23^4\cdot 4/23) \approx 5.811$ which is much close to $6$. $\endgroup$ – Balarka Sen Jun 14 '14 at 12:23
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    $\begingroup$ Is there a question here? $\endgroup$ – Alexander Gruber Jun 14 '14 at 12:48
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    $\begingroup$ @johnmangual I see a question mark but the words behind it do not make sense. Please clarify your meaning of "is this pigeonhole" as requested. $\endgroup$ – Alexander Gruber Jun 14 '14 at 16:10
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    $\begingroup$ $\pi\approx\dfrac{\ln(640320^3+744)}{\sqrt{163}}$ , with a precision of $30$ exact decimals. Discovered independently by both Charles Hermite and Srinivasa Ramanujan. Based on the fact that $e^{\pi\sqrt H}$ is an almost integer when H is a Heegner number. $\endgroup$ – Lucian Jun 14 '14 at 17:23
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Well known approximations for $\pi$, $\pi^2$ and $\pi^3$ can be related to the question.

$$e^6 \approx 403 = 13·31 = (3+10)·31 \approx \left(\pi+\pi^2\right)\pi^3= \pi^4+\pi^5$$

The approximations $\pi \approx 3$ and $\pi^2 \approx 10$ have similar absolute errors with opposite sign so the combination $\pi+\pi^2 \approx 13$ is more precise. The largest root of the polynomial $x^2+x-13$ is $\frac{\sqrt{53}-1}{2}\approx 3.140$, which approximates $\pi$ with an accuracy between that of $\sqrt{10}$ (one digit) and $31^\frac{1}{3}$ (three digits).

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  • $\begingroup$ this is the right idea. i think the best way is to use Poisson summation and truncate. usually one side or the other converges quickly $\endgroup$ – cactus314 Apr 15 '17 at 14:27
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    $\begingroup$ Do you mean that there is a series $$\left(\frac{e^3}{\pi^2}\right)^2 = \sum_{n=0}^{\infty} k_n\pi^n$$ with $k_0=k_1=1$ and the other $k_n$ very small? $\endgroup$ – Jaume Oliver Lafont Apr 15 '17 at 15:30

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