Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Without using differentiation, logarithmic function, rigorously, prove that $$e^x\ge x+1$$ for all real values of $x$.

share|improve this question
    
Please avoid using display math (double \$\$) in titles. –  Jonathan Christensen Dec 6 '12 at 20:28
    
Are you alowed to use MVT? –  user127.0.0.1 Dec 6 '12 at 20:30
6  
How do you define $e$ and/or $e^x$? What kind of answer is expected depends on the definition you are using. –  Andres Caicedo Dec 6 '12 at 20:31
    
Yes,I can use MVT, I want to prove it using only lim(1+x/n)n as n goest to inf –  p.s Dec 6 '12 at 20:35
    
For positive x's I am done,but negative,I need help, have been trying it for hours... –  p.s Dec 6 '12 at 20:37

8 Answers 8

Hint: Break it into two cases, $x \leq 0$ and $x > 0$. One of them is trivial. For the other case, consider the Taylor Series expansion.

share|improve this answer
2  
Taylor expansion without knowledge about differentiation? –  user127.0.0.1 Dec 6 '12 at 20:29
    
Knowing a Taylor Series expansion is not the same as using differentiation. Maybe my math education was weird, but I learned Taylor Series in 10th grade, long before I learned differentiation. –  Jonathan Christensen Dec 6 '12 at 20:33
    
I want to prove it rigourously,Taylor expansion is not allowed –  p.s Dec 6 '12 at 20:36
1  
I think the Taylor series expansion is plenty rigorous, but fair enough. –  Jonathan Christensen Dec 6 '12 at 20:40

Hint:

Prove for rational $x=\frac{a}{b}$ ($b > 0$) that $e^{\frac{a}{b}} \ge \frac{a}{b}+1$. You can do this by showing that $$e^a \ge \left(\frac{a}{b}+1\right)^b.$$ Finally, argue by continuity.

share|improve this answer

Bernoulli's Inequality: for any $\,n\in\Bbb N\,$

$$1+x\leq\left(1+\frac{x}{n}\right)^n\xrightarrow [n\to\infty]{} e^x$$

The inequality above is true for $\,x\geq -1\,$ , and since the wanted inequality is trivial for $\,x<-1\,$ we're done.

share|improve this answer
    
Showing Bernoulli's inequality for $-1 < x < 0$ does require some effort though... –  Zarrax Dec 6 '12 at 21:06
    
@DonAntonio: Nice. Might be clearer if said inequality is trivial for $x\lt -1$. –  André Nicolas Dec 6 '12 at 21:11
    
@AndréNicolas, that was the intention yet some of my fingers believe they have free will...Thanks. –  DonAntonio Dec 6 '12 at 21:53
    
@Zarrax, a very minimal effort: the inequality follows by induction in the general case. –  DonAntonio Dec 6 '12 at 21:53
    
It is quite easy to demonstrate that Bernoull’s inequality is strict for $x\neq 0$ and $n>1$, but the limit introduces it again. Can we extend this to show that the (question) inequality is strict iff $x\neq 0$? –  Steve Powell Sep 25 '13 at 14:26

If $ x\ge 0 $ as $ \exp $ is increase we have $ e^{x} \ge 1 $. Then by mean value theorem there exist $ c \in (0,x) $ such that \begin{equation} e^x - 1 = e^c x. \end{equation} Thus \begin{equation} e^x \ge x +1. \end{equation} If $ x \le -1 $, we have \begin{equation} e^x \ge 0 > 1+x \end{equation} Now if $ -1 < x < 0 $ we have \begin{equation} e^c < 1 \Rightarrow -e^c > -1 \Rightarrow -e^c x < -x \end{equation} Thus \begin{equation} 1 - e^x = e^c (-x) < -x \end{equation} and \begin{equation} e^x > 1+x \end{equation}

share|improve this answer
    
What if $x \in (-1, 0)$? –  Andrew Uzzell Dec 6 '12 at 21:05
    
I included this case now, Ok? –  user29999 Dec 6 '12 at 21:08
    
How is the negative number $-e^c$ larger than $1$? –  Andres Caicedo Dec 6 '12 at 21:10
    
Thank you, you are right. I corrected this. –  user29999 Dec 6 '12 at 21:15
    
Your 3rd equation, $ e^x \ge 0 > 1+x $ is incorrect. $ 1 + x $ is zero when $ x = -1 $ so the second inequality needs to be $ \ge $ –  Alexis Wilke Sep 25 '13 at 21:52

Assuming $x > 0$ and regardless of your definition of $e$, you can show that $e^x = \displaystyle\lim_{n \to \infty} \left( 1 + \frac{x}{n} \right)^n$. Then:

\begin{align} \left( 1 + \frac{x}{n} \right)^n &= \sum_{k=0}^n {n \choose k}\frac{x^k}{n^k} \\ &= 1 + x + \sum_{k=2}^n {n \choose k} \frac{x^k}{n^k} \end{align}

Since each term ${n \choose k} (x/n)^k$ is positive, it follows that

$$ \left( 1 + \frac{x}{n} \right)^n - x - 1 > 0. $$ Hence,

$$ \lim_{n \to \infty} \left( 1 + \frac{x}{n} \right)^n - x - 1 \geq 0. $$ from which the desired conclusion follows for $x > 0$.

share|improve this answer
    
Why the downvote? –  JavaMan Dec 7 '12 at 3:18

Write out the expansion of $(1 + {x \over n})^n$ as $$(1 + {x \over n})^n = 1 + n {x \over n} + {n(n-1) \over 2}{x^2 \over n^2} + ...$$ If $x \geq 0$, then all terms are nonnegative, so the sum is at least the sum of the first two terms, namely $1 + x$. If $x \leq - 1$, then for even $n$, the expression $(1 + {x \over n})^n $ is nonnegative, so the limit must be at least zero, which is greater than or equal to $1 + x$.

So it remains to look at the case where $-1 < x < 0$. In this case write $x = -y$ and you have $$(1 - {y \over n})^n = 1 - n {y \over n} + {n(n-1) \over 2}{y^2 \over n^2} - ...$$ This is an (finite) alternating series, and the absolute value of the ratio of two consecutive terms of this series is of the form ${n - k \over k+ 1} {y \over n}$, which is less than $1$ since $0 < y < 1$. So the terms decrease in absolute value. Hence the overall sum is at least what you get if you truncate after a negative term. So truncating after two terms you get $$(1 - {y \over n})^n \geq 1 - y$$ Equivalently, $$(1 + {x \over n})^n \geq 1 + x$$ Taking limits as $n$ goes to infinity gives what you're looking for.

share|improve this answer

Suppose $x \le 0$. From $t^n - 1 = (t-1)(1 + t + \ldots + t^{n-1})$ (with $t = 1 + x/n$) we get $$(1+x/n)^n - 1 = \frac{x}{n} \sum_{j=0}^{n-1} (1+x/n)^j$$ If $n$ is large enough that $1+x/n \ge 0$ we have $(1+x/n)^j \le 1$, so $$\sum_{j=0}^{n-1} (1+x/n)^j \le \sum_{j=0}^{n-1} 1 = n $$ and since $x/n \le 0$ $$ \frac{x}{n} \sum_{j=0}^{n-1} (1+x/n)^j \ge \frac{x}{n} n = x $$ Thus $$(1+x/n)^n \ge 1 + x$$ Now take the limit as $n \to \infty$.

share|improve this answer

I don't know if it's cheating, but you didn't say that integration is forbidden. Since $\exp$ is increasing we know that $e^{-x} \leq 1$ for all $x \geq 0$. Integrating, we get $$ \forall x\geq 0,\qquad 0 = \int_0^x 0\,dt \leq \int_0^x (1-e^{-t})\,dt = x + e^{-x} - 1. $$

The inequality $\forall x \geq 0, \;e^x \geq 1 + x$ follows in the same lines as the former.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.