# Prove pi^e < e^pi

• Aug 29th 2009, 03:01 AM
Khonics89
Prove pi^e < e^pi
Consider the function f(x)= ln(x)/x which is defined for all x>0.

Show that this functions is strictly increasing on the interval (0,e) [ This part is done just find f'(x) and state its increasing for all x between 0 and e]

strictly decreasing on the interval (e,infinity) [ same as the above part..]

and thus has a global maximum at x=e ... [ DONE, proving its station point is at that point].

HENCE SHOW THAT F(X) <= 1/e FOR ALL x>0.

USE THIS RESULT WITH x=pi TO SHOW THAT pi^e < e^pi.....

just that last part i dont even know where to start.....
• Aug 29th 2009, 03:06 AM
ynj
Quote:

Originally Posted by Khonics89
Consider the function f(x)= ln(x)/x which is defined for all x>0.

Show that this functions is strictly increasing on the interval (0,e) [ This part is done just find f'(x) and state its increasing for all x between 0 and e]

strictly decreasing on the interval (e,infinity) [ same as the above part..]

and thus has a global maximum at x=e ... [ DONE, proving its station point is at that point].

HENCE SHOW THAT F(X) <= 1/e FOR ALL x>0.

USE THIS RESULT WITH x=pi TO SHOW THAT pi^e < e^pi.....

just that last part i dont even know where to start.....

$a^b\frac{\ln a}{a}$
• Aug 29th 2009, 03:19 AM
Khonics89
Hmmm ^^^^^^^^

where do I go ??
• Aug 29th 2009, 03:32 AM
skeeter
Quote:

Originally Posted by Khonics89
Consider the function f(x)= ln(x)/x which is defined for all x>0.

Show that this functions is strictly increasing on the interval (0,e) [ This part is done just find f'(x) and state its increasing for all x between 0 and e]

strictly decreasing on the interval (e,infinity) [ same as the above part..]

and thus has a global maximum at x=e ... [ DONE, proving its station point is at that point].

HENCE SHOW THAT F(X) <= 1/e FOR ALL x>0.

USE THIS RESULT WITH x=pi TO SHOW THAT pi^e < e^pi.....

just that last part i dont even know where to start.....

$f(\pi) = \frac{\ln(\pi)}{\pi} < \frac{1}{e}$

$\ln(\pi) < \frac{\pi}{e}$

$e\ln(\pi) < \pi$

$\ln(\pi^e) < \pi$

$e^{\ln(\pi^e)} < e^{\pi}$

$\pi^e < e^{\pi}$