Without using differentiation, logarithmic function, rigorously, prove that $$e^x\ge x+1$$ for all real values of $x$.
Prove that $e^x\ge x+1$ for all real $x$
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0Please avoid using display math (double \$\$) in titles. – 2012-12-06
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0Are you alowed to use MVT? – 2012-12-06
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6How do you define $e$ and/or $e^x$? What kind of answer is expected depends on the definition you are using. – 2012-12-06
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0Yes,I can use MVT, I want to prove it using only lim(1+x/n)n as n goest to inf – 2012-12-06
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0For positive x's I am done,but negative,I need help, have been trying it for hours... – 2012-12-06
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0Try using Bernoulli's inequality, see here http://en.wikipedia.org/wiki/Bernoulli%27s_inequality#Proof_For_Rational_Case – 2012-12-06
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0This is wrongly marked as duplicate since it from Dec 2012 and the other post is from 2013 – 2017-11-10
8 Answers
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.
Bernoulli inequality $$1+ny\leq\left(1+y\right)^n$$ using y=x/n we get $$1+x\leq\left(1+\frac{x}{n}\right)^n$$
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1Showing Bernoulli's inequality for $-1 < x < 0$ does require some effort though... – 2012-12-06
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0@DonAntonio: Nice. Might be clearer if said inequality is trivial for $x\lt -1$. – 2012-12-06
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1@AndréNicolas, that was the intention yet some of my fingers believe they have free will...Thanks. – 2012-12-06
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1@Zarrax, a very minimal effort: the inequality follows by induction in the general case. – 2012-12-06
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0It 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 $x\neq 0$? – 2013-09-25
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0it's wrong. This is not Bernoulli inequality – 2018-02-10
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0@frhack Of ourse it is. If you were more careful you would check the above thoroughly before downvoting a correct answer. – 2018-02-10
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0Bernoulli's Inequality is different (1+nx)<=(1+x)^n and not directly useful to demostrate the target inequality. So if you have the arguments you have to show them given that the question ask to prove it. – 2018-02-11
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1@frhack That's **exactly what is written there** with $\;\cfrac xn\;$ instead of $\;x\;$ alone. Do you **Really** have problems to see this, or you just wanted to downvote something that is correct? – 2018-02-11
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0ok damn me, you are right ! sorry ! – 2018-02-11
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1@frhack Don't worry, it happens to us all some time. It is advisable, though, to wait a little more for downvoting **even** if there's an outstanding mistake in the answer. Sometimes typos go under the radar or someone gets confused. Better, to comment and wait for the answerer to address the doubt. – 2018-02-11
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0@DonAntonio i waited a few hours before downvoting. :( Peraphs i was a little bit confused by this demostration. https://jeremykun.com/2015/11/23/the-inequality/ – 2018-02-11
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.
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.
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}
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0What if $x \in (-1, 0)$? – 2012-12-06
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0I included this case now, Ok? – 2012-12-06
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0How is the negative number $-e^c$ larger than $1$? – 2012-12-06
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0Your 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 $ – 2013-09-25
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0What's the problem? $e^x > 0$ for all $x \in \mathbb{R}^n$. – 2013-10-08
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.
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 $x > 0$.
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0Why the downvote? – 2012-12-07
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 $n \to \infty$.
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.
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2Taylor expansion without knowledge about differentiation? – 2012-12-06
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0Knowing 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. – 2012-12-06
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0I want to prove it rigourously,Taylor expansion is not allowed – 2012-12-06
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1I think the Taylor series expansion is plenty rigorous, but fair enough. – 2012-12-06