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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


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