1

I have already know how to prove $\pi^{e} < e^{\pi}$(solution), but I cannot figure out how to prove $\pi^{e} + 1 > e^{\pi}$. You can use approximation to prove it but calculators are prohibited. Any help is appreciated, thanks!

I have tried to convert $\pi^{e}$ to $e^{e\ln\pi}$ but I couldn't think of what to do next.

Seanmamasde
  • 35
  • 7
    Note that $\pi^e<e^\pi$ only hinges on properties of $e$, since $\pi$ can be replaced with anything else (other than $e$). However, what you are trying to show is clearly more challenging since it does not enjoy this generality: it must depend on properties of both $e$ and $\pi$. – Golden_Ratio Dec 29 '21 at 11:11
  • I didn't follow through with it completely, but I recommend looking at the behaviour of $f(x) = x/\log(x)$. Note that $e^{\pi} > \pi^{e}$ follows trivially from the observation that $e$ is a local minimum of $f$. I tried approximating near $x=e$ to second order, and got $$e^{\pi} - \pi^{e} < \pi^{e} \frac{(\pi-e)^2}{2e}\log \pi$$ which I can't prove is less than 1 (but it is). – preferred_anon Dec 29 '21 at 12:15

  • Note that $e^{\pi} > \pi^{e}$ follows trivially from theobservation that e is a local minimum of f.

    @preferred_anon Thank you for your input. I can understand your comment till the quoted part, but may I ask how do you jump to the inquality? My brain can't process that fast:(

     – Seanmamasde Dec 29 '21 at 13:19

1 Answers1

0

Just some numerical thoughts, take this as a long comment. I will add something later!

If we think about it as the following inequality

$$x^e > e^x -1$$ Then all we need is to find the range of $x$ for which that inequality is true. This reduces to study

$$\dfrac{\ln(e^x-1)}{\ln(x)} - e < 0$$

If $\pi$ lies in that range, we are done! One interval for which this holds is surely $x\in(0,1)$. The other one can be found numerically and it's very approximatively $x \in (1.87, 3.22) \ni \pi$

Clearly this is not a number theory proof, nor an analytical proof. Nor a rigorous one, but it might be a starting point. Will add more later!

Enrico M.
  • 26,114
  • It is often useful to consider your inequality as an analytic function and differentiate it as many times as necessary to find turning points and analyse their convexity/concavity to identify the positive and negative regions. The continuity of the first and second derivatives is important there – FShrike Dec 29 '21 at 12:58
  • @FShrike Absolutely Right! I wanted to add something about indeed. But I did not have time (and I am still a bit busy), so I will write more later. Thank you! – Enrico M. Dec 29 '21 at 13:49
  • Apparently, a downvoter ghost is walking around us... The same brat who doesn't even have the courage to give a motivation. – Enrico M. Dec 29 '21 at 13:52
  • @EllipticCurve The downvote probably comes from the fact that the question is an obvious duplicate. – TheSilverDoe Dec 29 '21 at 14:37
  • @TheSilverDoe I do not really know the rules for the downvotes, but I wonder if a duplicate make it "legit" to downvote. People here, myself too, were not aware of this question being a duplicate, and we don't waste our times in searching for duplicates. We prefer to invest the time in thinking and answering, and this effort must at least receive a null vote or some upvotes (if the answer is correct or helpful at least), not a downvote. Anyway, this has always been a deep problem on this site so be it... – Enrico M. Dec 29 '21 at 14:40
  • @EllipticCurve "we don't waste our times in searching for duplicates" : this is precisely the problem. Please read this. – TheSilverDoe Dec 29 '21 at 14:44
  • @TheSilverDoe Well, as I said I was not aware of... You know, sometimes there are questions that are rather good, and you just want to reason over them without searching for duplicates. It may be wrong but assuming there is no old similar question makes you to think and have fun. If you find the duplicate the game is over. Anyway I got it, I will try next time :) – Enrico M. Dec 29 '21 at 14:47