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)