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How to prove $\lfloor{\pi}\rfloor+\lfloor{\pi^e}\rfloor=\lfloor{e}\rfloor+\lfloor{e^\pi}\rfloor$?(without using the calculator)

$e$ is Euler's number. I tried to solve this question. But I could only get to the following inequality: $e^\pi >\pi^e$. I got this inequality by deriving from the function $f(x)=e^x-x^e$. so $f'(x) \geq 0$ for $x > e$. This means that $f$ is an increasing function, and $f(e) = 0$.Thus, $f(\pi) > 0$, so $e^{\pi} - \pi^{e} > 0$. This implies that $e^{\pi}$ is greater than $\pi^{e}$.

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    $\begingroup$ How exactly did you get the inequality? Please share (with an edit). That would prevent further down- or close votes. $\endgroup$
    – Shaun ♦
    Commented Nov 19, 2021 at 19:16
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    $\begingroup$ You have the equation backwards. It should be $$\lfloor \pi \rfloor + \left\lfloor \pi^e \right\rfloor = \lfloor e\rfloor + \left\lfloor e^\pi \right\rfloor.$$ This follows since $\displaystyle 22 < \pi^e < 23 < e^\pi < 24.$ $\endgroup$
    – user2661923
    Commented Nov 19, 2021 at 19:24
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    $\begingroup$ Personally, if I was going to try to attack this problem, I would try to focus on the distinction between $[e \times \log \pi]$ and $\pi.$ However, I (also) see no way of attacking this problem. Normally, I would upvote your question, but the issue of upvoting to reverse perceived undeserved downvotes has been identified on meta math SE as bad. $\endgroup$
    – user2661923
    Commented Nov 19, 2021 at 19:28
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    $\begingroup$ As @user2661923 has commented, the assertion is wrong and there is a difference of $2$. But even with the correct version, there is little hope without a calculator: $\lfloor x \rfloor + \left\lfloor x^e \right\rfloor = \lfloor e\rfloor + \left\lfloor e^x \right\rfloor$ changes from correct to incorrect and back again $8$ times between $e$ and $\pi$ $\endgroup$
    – Henry
    Commented Nov 19, 2021 at 19:34
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    $\begingroup$ Can you define "without using a calculator"? IE There are ways to show that $ 22 < \pi^e < 23 < e^{\pi} $ by taking approximation and logarithms. Does that count? $\endgroup$
    – Calvin Lin
    Commented Nov 19, 2021 at 19:57

1 Answer 1

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This is not a full answer, however I do explain one approach to establishing the value of the RHS, and at the end I will give an observation that hopefully someone can build off of to prove the LHS.

For a start, I think it is pretty well established that $\lfloor e\rfloor=2$ and $\lfloor\pi\rfloor=3$. These follow immediately from continued-fraction expansions and other well-known formulas.

To find a lower-bound on $e^\pi$, we can use the exponential taylor series combined with a well-known approximation $\pi>\frac{339}{108}$. So, we find that $$e^\pi>\sum_{n=0}^\infty\frac{1}{n!}\left(\frac{339}{108}\right)^n.$$

Although tedious, truncating after $n=9$ terms, we thus find $$e^\pi>\frac{849221956019853023669}{36854077075834798080},$$ which is not hard to show is greater than $23$.

Furthermore, we can establish an upper bound by noting that $\pi<\frac{256}{81}$, and thus

$$\sum_{n=0}^\infty\frac{1}{n!}\left(\frac{256}{81}\right)^n>e^\pi,$$

which can be shown to be greater than 23 (by truncating after $n=7$), but the remaining terms can never account for a difference that brings the full sum past $24$ (which is difficult to explain in brevity, but very doable). Thus, $$23<e^\pi<24\implies\lfloor e^\pi\rfloor=23\implies\lfloor e\rfloor+\lfloor e^\pi \rfloor=25$$

Establishing bounds for $\pi^e$ is more difficult however. As pointed out, it may be easier to work with $\ln\pi$. One idea is to make use of the fact that

$$\zeta'(0)=-\ln\sqrt{2\pi},$$ $$\implies \ln\pi = -\ln 2-2\zeta'(0).$$

From there, one could use the many series and approximations for $\ln(2)$ and $\zeta'$, respectively.

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