pe or ep
By Anonymous on Tuesday, February 6,
2001 - 11:20 pm :
Show by means of a proof which is greater, pe or ep.
By Dan Goodman (Dfmg2) on Monday,
February 12, 2001 - 10:13 pm :
Here's a sketch of a proof.
Let f(x)=ex -xe, and hence f ' (x)=ex -e xe-1 = e(ex-1 -xe-1)
f ' (x)=0 when x=1 or e.
f(0)=1 and f ' (0)=1
As x tends to infinity, f(x) tends to infinity.
f(1)=e-1 and f(e)=0.
So, assuming that the only points x > 0 where f ' (x)=0 are x=1 and x=e,
the graph of f looks like: it starts at f=1 when x=0, increases up to
f=e-1 when x=1, decreases down to f=0 when x=e, and then increases up
to infinity as x increases beyond that.
So, f(x) ³ 0 for all x, and f(x)=0 only if x=e. In other words
f(p) > 0, so that ep -pe > 0, so ep > pe. The only
thing you need to check is that the only positive real zeroes of f ' (x) are
x=1 and x=e. In fact, ep » 23.14 and pe » 22.46.
You might like to try using a similar technique to find which is bigger,
e1/e or p 1/p.