5

When playing in a Python console I observed $\,\pi^\pi\approx 36.46\,$ and $\,e^e\approx 15.15$, and that their ratio is close to $\dfrac{12}{5} =$ a fraction with small numerator and denominator, hence $5\pi^\pi\approx 12\,e^e$.

Do you see a way to solve $$5\,\pi^\pi\;\stackrel{?}{\lessgtr}\; 12\,e^e$$ without calculator use ?

I would estimate that many Math.stackexchangers do not appreciate this kind of "without calculator" questions.
It is asked here in the hope that some elegant solution is found/presented, possibly in the spirit of
this awesome answer to $\pi^e\stackrel{?}{<} e^\pi$ as of ten years ago.

I have none to offer.
My approach was to consider the natural logarithm of the quotient $$\log\frac{\pi^\pi}{e^e} = \pi\log\pi - e\log e$$ and to relate it somehow to the definite integral of the logarithm $$\int_1^t\log x\:dx \;=\; t\,(\log t - 1) +1\,,$$ but without seeing a tangible result.

J. W. Tanner
  • 63,683
  • 4
  • 43
  • 88
Hanno
  • 6,776
  • Using your integral idea, let $A = \int_e^\pi{\ln x} dx = \pi \ln \pi - \pi$. Comparing $A$ to the trapezoidal area that approximates it leads to a nice equality, $\ln \pi \ge \frac{3\pi-e}{\pi+e}$. This could be a useful intermediate step. – Théophile Feb 08 '23 at 20:31
  • If that helps, $$\int_e^\pi (1+\log x),dx=\pi\log\pi-e\log e.$$

    By convexity of the curve, this area exceeds that of the trapezoid $\frac12(\pi-e)(3+\log\pi)>\log(\frac{12}5)$.

     –  Sep 02 '24 at 14:19

1 Answers1

-1

Here is my (unfinished) take. Hope this helps!

$5\pi^{\pi} \approx 12e^e$

Therefore, $\frac{5}{12}\pi^{\pi} \approx e^e$.

But, $e^x > 1+x$, $x \neq 0$.

Hence, $\frac{5}{12}\pi^{\pi} > 1 + e$.

Rearranging and manipulating, we get $(\frac{5}{12}\pi^{\pi} - 1)^e > e^e$.

(Alternatively, you could have used $e^x > 1 + x$, $x\neq 0$ from the beginning and shown that $5\pi^{\pi} > 12(1 + e)$, and hence that ($5\pi^{\pi} - 12)^e > 12e^e$.)

Hopefully, I will finish my take on this sometime soon.

Edit 1: I've made no progress, so my take will remain unfinished.

Yajat Shamji
  • 570