2020年10月14日 · {e^ {\pi}=e^e (e^ {\frac {\pi} {e}-1})^e> e^e\left (\frac {\pi} {e}-1+1\right)^e=\pi^e.} 对于正数a和b，如果我们要说明 a^b>b^a，则须先说明 b\ln a>a\ln b 即 {\ln a\over a}> {\ln b\over b} 。 因此我们将通过研究函数 f (x)= {\ln x\over x} 的 单调性 来比大小，对其求导可得： f' (x)= {1-\ln x\over x^2} 所以当 x>e 时 f' (x)<0 ，因此根据 e<\pi 可得 {\ln e\over e}> {\ln\pi\over\pi} ，于是 e^\pi>\pi^e. 怎么比较？

E and Pi are both mathematical constants that are irrational and transcendental numbers. While Pi represents the ratio of a circle's circumference to its diameter, E represents the base of the natural logarithm. Both constants have infinite decimal representations and are used in various mathematical and scientific calculations.

Note that $\ln(e^\pi) = \pi$ and $\ln(\pi^e) = e\ln(\pi)$, and we have that $\pi > e\ln(\pi)$. Hence $\;e^\pi > \pi^e$. But admittedly, the inequality isn't immediately obvious!

In mathematics, a transcendental number is a real or complex number that is not algebraic: that is, not the root of a non-zero polynomial with integer (or, equivalently, rational) coefficients. The best-known transcendental numbers are π and e. [1][2] The quality of a number being transcendental is called transcendence.

3 天之前 · At t=0, we have Vs(1-e 0), which is 0. At t=0.5, the voltage should be about 7.86V; at t=1, it should be up to 10.57V. ... (Title image showing e living in pi’s shadow adapted from “Pi” by ...

当 x\\ne0 时 e^x>x+1,\\\\ 将 x 替换为 \\pi/e-1, 两边取 e 次方即 e^\\pi>\\pi^e.\\\\ 事实上 \\pi 换成任何不等于 e 的正实数都成立.

2013年8月5日 · Answer to e^pi versus pi^e. The answer is e π is larger. There are several ways you can solve this problem. I’ve listed a few proofs below. I feel the first proofs are the most intuitive methods to approach the problem, but they are relatively involved proofs.

What is e? Top ln(e^10) reasons why e is better than pi. 10) e is easier to spell than pi. 9) pi ~= 3.14 while e ~=2.718281828459045. 8) The character for e can be found on a keyboard, but pi sure can't. 7) Everybody fights for their piece of the pie. 6) …

2010年10月26日 · As a hint, it's easier to consider the more general problem: for which positive x is ex> xe? Alternatively, we can compare e1 / e and π1 / π. Let f(x) = x1 / x. Then f ′ (x) = x1 / x(1 − log(x)) / x2. Since log(x)> 1 for x> e, we see that f ′ (x) <0 for e <x <π. We conclude that π1 / π <e1 / e, and so πe <eπ.

2016年6月30日 · We are going to compare e^pi vs pi^e to see which result is larger? This is a very classic calculus problem where we compare a^b vs b^a. We will be using calculus to find out, not...