Re: log x > x - proof?
- To: mathgroup at smc.vnet.net
- Subject: [mg28078] Re: log x > x - proof?
- From: Jens-Peer Kuska <kuska at informatik.uni-leipzig.de>
- Date: Fri, 30 Mar 2001 04:12:28 -0500 (EST)
- Organization: Universitaet Leipzig
- References: <99us89$60p@smc.vnet.net>
- Sender: owner-wri-mathgroup at wolfram.com
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
Joe wrote:
>
> 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