Which is greater: e^π or π^e?

• Dec 5th 2009, 10:15 PM
Jamin2112
Which is greater: e^π or π^e?
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.

• Dec 5th 2009, 10:23 PM
VonNemo19
Quote:

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.

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.
• Dec 5th 2009, 10:28 PM
RockHard
This is very interesting, do you have to provide some sort of proof?
• Dec 5th 2009, 10:33 PM
Bruno J.
$\displaystyle \log(e^n)=n\log(e)$

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

What do you think?
• Dec 5th 2009, 10:59 PM
simplependulum
the answer is $\displaystyle e^x$ is greater

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

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

and

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

consider

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

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

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

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

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

from the above calculations .

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

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

equality holds when $\displaystyle x = e$ (obviously )

ps . add a constraint $\displaystyle x > 0$
• Dec 6th 2009, 11:31 AM
Krizalid
consider the function $\displaystyle f(x)=\frac{\ln x}x,$ which is strictly decreasing for $\displaystyle x>e.$

since $\displaystyle \pi>e,$ we get $\displaystyle \frac{\ln\pi}\pi<\frac{\ln e}e\implies e\ln\pi<\pi\ln e,$ and that yields $\displaystyle \ln\pi^e<\ln e^\pi,$ and finally since $\displaystyle g(x)=e^x$ is a strictly increasing function, then $\displaystyle e^{\ln \pi^e}<e^{\ln e^\pi}\implies \pi^e<e^\pi,$ and we're done.
• Dec 6th 2009, 08:27 PM
Jamin2112
Quote:

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

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

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