Which is greater: e^π or π^e?

**Jamin2112** · December 5th 2009, 10:15 PM

My Math 324 professor said we should try this for fun. Can someone explain how to come up with the answer? I thought maybe do a Taylor approximation since e^x is ∑ x^n / n!, but then I got something divergent.

Please contribute.

**VonNemo19** · December 5th 2009, 10:23 PM

Originally Posted by Jamin2112

My Math 324 professor said we should try this for fun. Can someone explain how to come up with the answer? I thought maybe do a Taylor approximation since e^x is ∑ x^n / n!, but then I got something divergent.

Please contribute.

Could you write the instructions exactly as they were given please? I'm having trouble understanding what the problem is.

And BTW, the taylor series for e^x converges for all x.

**RockHard** · December 5th 2009, 10:28 PM

This is very interesting, do you have to provide some sort of proof?

**Bruno J.** · December 5th 2009, 10:33 PM

$\log(e^n)=n\log(e)$

$\log(n^e)=e\log(n)$

What do you think?

**simplependulum** · December 5th 2009, 10:59 PM

the answer is $e^x$ is greater

let $f(x) = x - e\ln(x)$

we have $f(x) \to +\infty$ , when $x \to 0$

and

$\lim_{x\to+\infty} x - e\ln(x) = \lim_{x\to +\infty} x ( 1 - \frac{e\ln(x)}{x} ) = +\infty$

consider

$f'(x) = 1 - \frac{e}{x}$

we can see when $x = e$ , $f'(x) = 0$

its second derivative gives $f''(e) = \frac{1}{e} > 0$

means $f(x)$ has only one local min.

come back to $f(e) = e - e\ln(e) = e-e = 0$

from the above calculations .

we can see $f(x) \geq 0$

$\implies x \geq e \ln(x) \implies e^x \geq x^e$

equality holds when $x = e$ (obviously )

ps . add a constraint $x > 0$

**Krizalid** · December 6th 2009, 11:31 AM

consider the function $f(x)=\frac{\ln x}x,$ which is strictly decreasing for $x>e.$

since $\pi>e,$ we get $\frac{\ln\pi}\pi<\frac{\ln e}e\implies e\ln\pi<\pi\ln e,$ and that yields $\ln\pi^e<\ln e^\pi,$ and finally since $g(x)=e^x$ is a strictly increasing function, then $e^{\ln \pi^e}<e^{\ln e^\pi}\implies \pi^e<e^\pi,$ and we're done.

**Jamin2112** · December 6th 2009, 08:27 PM

Originally Posted by Krizalid

consider the function $f(x)=\frac{\ln x}x,$ which is strictly decreasing for $x>e.$

since $\pi>e,$ we get $\frac{\ln\pi}\pi<\frac{\ln e}e\implies e\ln\pi<\pi\ln e,$ and that yields $\ln\pi^e<\ln e^\pi,$ and finally since $g(x)=e^x$ is a strictly increasing function, then $e^{\ln \pi^e}<e^{\ln e^\pi}\implies \pi^e<e^\pi,$ and we're done.

Thanks, dude! I went through all your steps and now I understand it!

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