Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

How can I calculate without calculator or something like this

the values of $\pi^{e}$ and $e^{\pi}$

in order to compare them ?

share|improve this question

7 Answers 7

up vote 77 down vote accepted

Another proof uses the fact that $\displaystyle \pi \ne e$ and that $e^x > 1 + x$ for $x \ne 0$.

We have $$e^{\pi/e -1} > \pi/e,$$

and so

$$e^{\pi/e} > \pi.$$

Thus,

$$e^{\pi} > \pi^e.$$

Note: This proof is not specific to $\pi$.

share|improve this answer
10  
This is from The Book! Impeccable proof. –  Prism Sep 14 '13 at 22:20
    
@Prism: Thanks! –  Aryabhata Sep 17 '13 at 3:27
    
if x is negative than would your first equation which is e^x > 1 + x be true ? –  Murtuza Vadharia Jan 15 at 17:36
1  
@MurtuzaVadharia: Yes, it is true. Consider $f(x) = e^x -1 -x$. It's derivative is $e^x -1$ which is $\lt 0$ for $x \lt 0$ and $\gt 0$ for $x \gt 0$, so $f$ decreases from $-\infty$ to $0$, and increases from $0$ to $\infty$. Since $f(0) = 0$... –  Aryabhata Jan 21 at 8:11
1  
I think it's worth mentioning that this works for all positive numbers which aren't $e$. The proof has nothing to do with $\pi$. –  NikolajK Feb 7 at 20:48

Let f (x) = x^(1/x) Find value of x such that function gets maximum value For this functions for x=e function will get the maximum value so e^1/e is greater than pi^1/pi so e^pi is greater than pi ^e

share|improve this answer

Hint:

Prove that the function $f(x)=\frac{e^x}{x^e}, x\geq e$ is strictly increasing on the interval $x\in \left [ e,\pi \right ]$. What is $f(e)$ and $f(\pi)$?

share|improve this answer

From Proofs without Words.

alt text

share|improve this answer
2  
+1 because pictures are always nice to have –  Guess who it is. Nov 8 '10 at 15:50
    
Which software did you use to plot this figure? –  metacompactness Aug 5 '13 at 16:29

Alternatively, we can compare $e^{1/e}$ and $\pi^{1/\pi}$.

Let $f(x) = x^{1/x}$. Then $f'(x) = x^{1/x} (1 - \log(x))/x^2$. Since $\log(x) > 1$ for $x > e$, we see that $f'(x) < 0$ for $e < x < \pi$. We conclude that $\pi^{1/\pi} < e^{1/e}$, and so $\pi^e < e^\pi$.

The same calculation shows that $f(x)$ reaches its maximum at $e^{1/e}$, and so in general $x^e < e^x$.

share|improve this answer
    
Correction: $\text{log}(x)<1$ for $x<e$. –  Robert Smith Oct 26 '10 at 15:39
    
Your argument appears to need $x>e$, but it is easy to also include $x<e$ as well i the $x^e<e^x$ inequality. –  Thomas Andrews May 22 '13 at 12:43
    
It doesn't really need $\log x > 1$. All you need is that the only solution of $\log x = 1$ is $x = e$. –  Yuval Filmus May 22 '13 at 14:29

Elaborating Robin's answer take $f(x) = \log{x} - \frac{x}{e}$. We have $$f'(x)= \frac{e-x}{xe}$$ Thus $f'(x)>0$ for $0 < x < e$ and $f'(x) <0$ if $x > e$. Consequently, we have $f(x) < f(e)$ if $x \neq e$.

Exercise: Try to prove this using the same methods: $2^{\sqrt{2}} < e$.

share|improve this answer

This is an old chestnut. As a hint, it's easier to consider the more general problem: for which positive $x$ is $e^x>x^e$?

share|improve this answer
1  
With the new font it looks like vareps instead of e... Can someone reverse the font?? :/ –  AD. Oct 26 '10 at 12:09
    
What are vareps? –  Mariano Suárez-Alvarez Oct 26 '10 at 13:54
    
\vareps = $\vareps$ ? –  Guess who it is. Oct 26 '10 at 14:32
1  
@Mariano, @J.M. I think AD. means \varepsilon = $\varepsilon$ which is an alternate form of $\epsilon$. –  Rahul Oct 26 '10 at 16:32
1  
@Rahul Narain: Right, sorry it is my personal abbreviation. –  AD. Oct 26 '10 at 20:36

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.