4

Prove $\log(1+x) \lt x$ for $x\gt0$

my attempt:

I show $e^{x}\gt 1+x$ for $x\gt0$ since

$e^{x}=1+x+\frac{n(n-1)}{2}\frac{x^2}{n^2}+...$

so if $x\gt0$ then all terms are positive

so $e^{x}\gt 1+x$ for $x\gt0$

now given $e^{x}\gt 1+x$ for $x\gt0$, can I take $\log$ on both sides and show

$\log(1+x)\lt x$ for $x\gt0$

or do I have to prove firstly that $\exp(x)=e^x$ for $x\gt 0$ then I can take $\log$..

Zhanxiong
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ketan
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  • But $\exp(x)$ is defined as $e^{x}$, so I'm not sure what you mean by prove it? – Thomas Russell Aug 14 '15 at 07:45
  • @Shaktal actually that depends on the situation. You can define $$\exp(x)=\sum\limits_{k=0}^\infty \frac{x^k}{k!}$$ and show that this series converges absolutely for all $x\in\mathbb R$, but this doesn't give you $\exp(x)=e^x$. This is a different result one has to prove first. – Hirshy Aug 14 '15 at 07:47
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    similar http://math.stackexchange.com/questions/741600/prove-that-logx-x-for-x-0-x-in-mathbbn – Taylor Ted Aug 14 '15 at 07:47
  • @Shaktal exp is defined as expx=$(1+\frac{x}{n})^n$ without limit. only while $e^x= lim_{n \to \infty}(1+\frac{x}{n})^n)$ isnt it? – ketan Aug 14 '15 at 07:53
  • @ketan Why should exp be defined like that? That is completely wrong. – Hirshy Aug 14 '15 at 07:56
  • @Hirshy then what is exp and when is it e^x? – ketan Aug 14 '15 at 07:59
  • @ketan See my comment above, it depends on your approach. You can independently show: $$\lim\limits_{n\to\infty} \left(1+\frac{1}{n}\right)^n=e=2.7182\dots$$ and $$\exp(x):=\sum\limits_{k=0}^\infty \frac{x^k}{k!}$$ converges absolutely for all $x\in\mathbb R$; thus $$\exp:\mathbb R\rightarrow\mathbb R,~x\mapsto\exp(x)$$ is a well-defined function. But this does not directly imply $e^x=\exp(x)$, especially we don't even know if $$e^x\stackrel{?}{=}\lim\limits_{n\to\infty}\left(1+\frac{x}{n}\right)^n$$ exists or what it means. This is a result one has to prove before using it. – Hirshy Aug 14 '15 at 08:08
  • @ketan I answered the question of $e^x=\exp(x)$ in http://math.stackexchange.com/questions/1397018/is-ex-expx-and-why so you might want to check that out. – Hirshy Aug 14 '15 at 12:18
  • Possible duplicate of https://math.stackexchange.com/questions/380963/prove-that-log-x-x-for-all-x-0 – Frank Aug 29 '19 at 00:39

5 Answers5

9

Let $$f(x) = \ln(1+x)-x\;,$$ Where $x>0$

Now $$\displaystyle f'(x) = \frac{1}{1+x}-1 = -\frac{x}{1+x}<0$$ for all $x>0$

So function $f(x)$ is Strictly Decreasing function

So Here $$x>0\Rightarrow f(x)<f(0)\;,$$ bcz function $f(x)$ is Strictly Decreasing function.

So $$\ln(1+x)-x<0\Rightarrow \ln(1+x)<x\;,$$ for $x>0$

juantheron
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1

Let $f: x \mapsto x - \log (1+x)$ on $]0, +\infty[$. Since $f'(x) = 1 - \frac{1}{1+x} > 0$ for all $x > 0$, so $f$ is strictly increasing on $]0, +\infty[$. Can you finish it?

Yes
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1

Your derivation is fine. In particular note that when $x > 0$ then we have $$ \forall x > 0 : e^x > 1, \ 1 + x > 1$$ Now when we take the logarithm of both sides the inequality sign does not turn around.

Kayle of the Creeks
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1

The function $\ln (1+x)$ is strictly concave on $(-1,\infty).$ Thus this function is below any of its tangent lines in this domain (except for the point of tangency). Since $y=x$ is the line tangent to $y=\ln (1+x)$ at $(0,0),$ we have the desired result (and more).

zhw.
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0

Your inequality can be improved. The $\exp(x)$ is strictly convex. (Because the second derivative is positive).

If you no nothing about convexity, you can take the argument that if Hessian is a positive definite of the function then you can use Taylor (1685-1731) series and use a reminder in a Lagrange form and lower bound value of quadratic form from the Hessian by zero.

In any case you should come in some way to notation: $f(x) > f(x_0)+\langle \nabla f(x),x-x_0\rangle$, or in case when function image is $\mathbb{R}$ you will have: $f(x)>f(x_0) + f'(x_0)(x-x_0)$. Now you use exponent as $f$ and $x_0=0$ and so:

$exp(x)>1+1(x-0)=1+x \implies \forall x>-1,\ln(1+x)<x$

#!/usr/bin/env python

You can visualize the functions using a python interpreter


import numpy as np
import matplotlib.pyplot as plt


x = np.arange(-1, 10, 0.01) 
plt.plot (x, x) 
plt.plot (x, np.log(1+x)) 
plt.grid ( True ) 
plt.show ( )
Konstantin Burlachenko
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