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.

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- Dec 5th 2009, 10:15 PMJamin2112Which 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.

Please contribute. - Dec 5th 2009, 10:23 PMVonNemo19
- Dec 5th 2009, 10:28 PMRockHard
This is very interesting, do you have to provide some sort of proof?

- Dec 5th 2009, 10:33 PMBruno J.
$\displaystyle \log(e^n)=n\log(e)$

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

What do you think? - Dec 5th 2009, 10:59 PMsimplependulum
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 AMKrizalid
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 PMJamin2112