Property of Natural Log- Inequality equation

  • #1
chetanladha
59
0
Hi.

I just saw on wikipedia that natural logarithm has such a property:
[x/(1+x)] < ln (1 + x) < x

(http://en.wikipedia.org/wiki/Natural_logarithm)

Can anyone pls tell me how to prove this?

Proving [x/(1+x)] and ln (1 + x) less than 'x' is easy.. But how abt [x/(1+x)] < ln (1 + x) ??

Thanks in advance..
 

Answers and Replies

  • #2
dextercioby
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It's wrong. It's smaller or equal.

Consider the function

[tex] f:(-1,+\infty)\longrightarrow \mathbb{R} [/tex]

[tex] f(x) = \frac{x}{x+1} - \ln (1+x) [/tex]

Plot it.
 
  • #3
BloodyFrozen
353
1
Following what dextercioby said, the following inequality does not always hold true.

[x/(1+x)] < ln (1 + x) < x

For example, let x = 0. We the have:

0/(1+0) < ln(1+0) < 0
0 < 0 < 0

This makes no sense.
 
  • #4
chetanladha
59
0
Thanks for response.

Yes its smaller or equal (Sorry couldn't rite it..)..

Is plotting the function f(x) = [x/(1+x)] - ln (1 + x) best way 2 prove it??
 
  • #5
dextercioby
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You needn't plot it, just make the standard analysis of 1st and 2nd derivatives of f to be able to draw the desired conclusion.
 
  • #6
chetanladha
59
0
Thanks a lot..!!
 

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