Prove inequality $e^{x}>1+x$

Question

Prove that $e^{x}>1+x$ when $x\neq0$. I can't make for all $x$. But as $e^{x}>0$ for all $x$, it's obvious that $e^{x}>1+x$ when $x\le-1$. It's where I got got stuck

Depends on how you've defined $e^x$. – Thomas Andrews Dec 24 '16 at 15:45 — Thomas Andrews, Dec 24 '16 at 15:45

score 7 · Accepted Answer · answered Dec 24 '16 at 15:38

7

Take $f(x) = e^x -(1+x)$. Since $f'(x) = e^x -1 >0$ iff $x>0$, it means that $f$ is increasing on $(0,+\infty)$ and decreasing on $(-\infty,0)$. Moreover $f(0)=0$. Then it follows that $f>0$ if $x \neq 0$.

answered Dec 24 '16 at 15:38

Harnak

1,587
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score 5 · Answer 2 · answered Dec 24 '16 at 15:36

5

The graph of $e^x$ is concave up everywhere, and thus lies above any tangent line to that graph. Since $y = 1+x$ is the tangent line at $x = 0$, the result follows.

answered Dec 24 '16 at 15:36

hunter

29,847

score 3 · Answer 3 · answered Dec 24 '16 at 15:35

3

Hint:$$e^x=\sum_{n=0}^{\infty}\frac{x^n}{n!}=1+x+\sum_{n=2}^{\infty}\frac{x^n}{n!}$$

answered Dec 24 '16 at 15:35

Jimmy R.

35,868

Can we use this expansion if $x$ doesn't tend to 0? – ioleg19029700 Dec 24 '16 at 15:36
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Needs a bit more to prove when $x<0$, but it is pretty clear here. – Simply Beautiful Art Dec 24 '16 at 15:38
Yes, you can use this expansion for any $x\in \mathbb R$, since the convergence radius of this series is equal to $\infty$. But, you should pay attention for $-1<x<0$. For these values of $x$ you need one more step to reach the conclusion (Hint2: alternating series, group consecutive terms appropriately) – Jimmy R. Dec 24 '16 at 15:38

score 3 · Answer 4 · answered Dec 24 '16 at 15:40

3

$f(x)=e^x$ is a convex function because $f''(x)>0$, thus $$f'(x)>f(x_1)+f'(x_1)(x-x_1)$$ set $x_1=0$

answered Dec 24 '16 at 15:40

Behrouz Maleki

11,182

Arnaldo · Answer 5 · 2016-12-24T16:08:48.770

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$f(x)=e^x-(1+x) \Rightarrow f'(x)=e^x-1 \Rightarrow f''(x)=e^x>0$

So the minimum point is given by $f'(x)=0 \Rightarrow x=0$, then $f(x)>0$ if $x \ne 0$

edited Dec 24 '16 at 16:08

answered Dec 24 '16 at 15:47

Arnaldo

21,342

score 1 · Answer 6 · edited Apr 13 '17 at 12:19

1

In THIS ANSWER, I showed that the sequence $\displaystyle e_n(x)=\left(1+\frac xn\right)^n$ monotonically increases for $x>-n$.

Therefore, for $n>1$ and $x>-n$

$$e_n(x)\ge e_1(x)=1+x$$

Letting $n\to \infty$, we find the coveted inequality

$$e^x=\lim_{n\to \infty}\left(1+\frac xn\right)^n\ge 1+x$$

for all $x$. And we are done!

edited Apr 13 '17 at 12:19

Community

1

answered Dec 24 '16 at 15:58

Mark Viola

179,405

score 0 · Answer 7 · answered Dec 24 '16 at 15:52

These sorts of questions are inherently problematic because you have not specified the definition of $e^x$ you're working with. I usually define $\log$ as an integral. It's easy to show that it's a strictly increasing $C^{\infty}$ bijection $\mathbb{R_+} \to \mathbb{R}$. We define $e^x$ as its inverse. It is also $C^{\infty}$, strictly increasing and bijective $\mathbb{R} \to \mathbb{R_+}$.

After that, it's not too hard to prove the standard properties of the function, such as $(e^x)' = e^x$.

Note

$$e^x= 1 + \int_{0}^x e^t dt \ \ (1) $$

We've specified $e^t$ is positive, so $e^x \geq 1$ since the integral in $(1)$ must be non-negative. Referring to $(1)$ in a recursive manner, we can see that since $e^x \geq 1$, $\int_{0}^x e^t dt \geq \int_{0}^x \ dt = x$ so that $e^x \geq x+1$.

This argument can be generalized to show that $$e^x \geq \sum_{i=0}^n \frac{x^i}{i!}$$ for all $n \in \mathbb{N}$.

Prove inequality $e^{x}>1+x$

7 Answers7