I'm new
- To: mathgroup at smc.vnet.net
- Subject: [mg28312] I'm new
- From: Niarlatotep <niarlatotep at ifrance.com>
- Date: Mon, 9 Apr 2001 02:58:09 -0400 (EDT)
- Sender: owner-wri-mathgroup at wolfram.com
I'm new, so excuse me if possible.
What's going really here ?
I red
"[mg28049] log x > x - proof?
Dear all,
Please could someone give me some hints as to how to prove that
log x > x for all x > 0
Isn't it proof by contradiction, or by intimidation?
Thanks in advance,
Joe"
then
"[mg28078] Re: log x > x - proof?
Hi,
thats easy to proof
Log[2.0] gives 0.693147
and it is obvoius larger
as well as
Log[1] give 0 and we all know that 0 is larger than 1
and
Log[0.5] is -0.693147, and it well known that any negative
number is larger than a positive one.
If you are working on such proofs you may proof
that x is always larger than E^x that can be seen from the
series expansion
x > Sum[x^n/n!,{n,0,Infinity}
because the first terms of the series show that
x > 1+ x + x^2/2 + ..
Send us more of your high school home works ! we will
be happy to solve it for you.
Regards
Jens"
then
"[mg28133] Re: [mg28049] log x > x - proof?
Joe,
The statement log[x]>x for ALL x>0 is obviously false, by
counterexample:
Let x=1, then log[x]=0 and 0<1, not 0>1!!!
Are you sure there isn't a misprint in your message?"
Did I miss something ?
For me the thing that perhaps Joe is looking for is
for all x > 0, log x < x - 1.
Ask me if you don't find the demonstration.
Richard.
(niarlatotep at ifrance.com)