Is the solution to $\frac{e^n}{n} = e^2$ related to $\pi$?

Question

I learned recently that the solution to $\frac{e^n}{n}=e^2$ is 3.146.., which is very near $\pi$. Is this just a coincidence, or is there something to the expression that leads to it being so close to $\pi$?

I feel like my question might be related to this, but I'm not too sure how to relate the two: Curious relation between $e$ and $\pi$ that produces almost integers

Answer 1

(A proof and explanation based on the comments)

Let $\bbox[5px,border:2px solid red]{p>0 \land p\neq1}$, $q$ be 'know' and $x$ -unknow: $$ \frac{p^x}{x}=p^q\implies x\neq0 $$ $$ p^{x-q}=x $$ From: $a^c=b \iff log_ab=c$

We get $$ log_px=x-q \implies \bbox[5px,border:2px solid green]{q=x-log_px }\leftarrow comments $$ Because of the logarithm $log_px \Rightarrow\bbox[5px,border:2px solid yellow]{x>0}$

Conclusion:

$$ \bbox[5px,border:2px solid blue]{log_px= x-q}\iff\bbox[5px,border:2px solid blue]{q+log_px=x}\iff \bbox[5px,border:2px solid cyan]{p^{x-q}=x}$$$$ x;p>0 \land p\neq1$$

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EDIT: We can also say:$q+log_px=p^{x-q}$

When $q=0$ $\implies \exists!x:log_px=p^x$

I) For $ a\in(0;1) $ $\implies \exists!x:q+log_px=p^{x-q}$

We just transform $a^x\rightarrow p^{x-q}$ $\land$ $log_ax\rightarrow q+log_px$

II) For $a>1$:

• $q>0$ $\rightarrow$ $2$ solutions
• $q<0$ $\rightarrow$ $0$ solutions

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I think $x$ can be found at least via numerical analysis

or via a graph-sample pair functions: $$f_1(x)=log_px\land g_1(x)=x-q$$ $$f_2(x)=p^{x-q}\land g_2(x)=x$$

in the points of intersection.

For the $(p;q)$ pairs:

1)$(e;2)\mapsto x\approx\pi$ @Peter

2)$(5;2)\mapsto x\cong2.59$ @Pedro

I am gonna find the other solutions:

I. $x=e^{x-2}$ $x_1\approx\frac{1}{2\pi+\frac{e}{122.39}}$ and $x_2\approx\pi$

II. $x=5^{x-2}$ $x_1\approx \frac{1}{0.973+10\sqrt{5}}$ and $x_2\approx2.591708$

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Properties:

I.

For the pair $(p;1)$ at $x=1\rightarrow p$ can be everything from the domain.

Proof: Let's put $x=q$ (to see what happens): $$p^{q-q}=q $$ $$p^{0}=q\implies q=1\iff x=1 $$

II.

For the pair $(p;p-1)$ at $x=p\rightarrow p$ can be everything from the domain.

Proof: Let's put $x=p$ (to see what happens): $$p^{p-q}=p $$ $$p^{p-q}=p^1\implies p-q=1\implies q=p-1 $$

Note that base on the domain:$p\neq1$ that $q\in\mathbb{R}$ still is true.

Answer 2

$3.146$ is not very near $\pi$. There are million expressions as close. $\sqrt2+\sqrt3$, $\dfrac{22}7$, or $\sqrt{31}$ for example. This means nothing. We would be more surprised by a match on the first twenty decimals.

Try this: http://isc.carma.newcastle.edu.au/standard

In any case, close values do not mean that there is a deeper meaning. I find it more intriguing to see $\pi$ appear a the constant in Stirling's asymptotic formula for the factorial ($\sqrt{2\pi}$), though the numerical value is different.