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MathGroup Archive 2001

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  • 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)


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