# How to prove $\lfloor{\pi}\rfloor+\lfloor{\pi^e}\rfloor=\lfloor{e}\rfloor+\lfloor{e^\pi}\rfloor$?

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 $$\pi^{e}$$.

• How exactly did you get the inequality? Please share (with an edit). That would prevent further down- or close votes.
– Shaun
Commented Nov 19, 2021 at 19:16
• 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 $\displaystyle 22 < \pi^e < 23 < e^\pi < 24.$ Commented Nov 19, 2021 at 19:24
• 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. Commented Nov 19, 2021 at 19:28
• 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 $\pi$ Commented Nov 19, 2021 at 19:34
• 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? Commented Nov 19, 2021 at 19:57

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

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.