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.
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.
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 $
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.