Bernoulli inequality and Exponential Bound

Question

I'm asking a complete proof of the Exponential Bound using the Bernoulli Inequality.

Exponential bound:

$$1+x\leq e^x$$

Bernoulli inequality

$$1+nx\leq\left(1+x\right)^n$$ for all $$x>= -1$$ and $$n=1,2,...$$

Here it's a partial (x>=0) proof but lacks the most interesting part, when x<0.

@chess4ever when x<1 the inequality is obvious: x<-1 => 1+x<0<=e^x — frhack, Nov 25 '20 at 23:54

score 3 · Answer 1 · answered Feb 12 '18 at 11:29

3

Note that

$$e\ge \left(1+\frac1x\right)^x \implies e^x\ge \left(1+\frac1x\right)^{x^2}\implies e^x\ge 1+x^2\cdot\frac1x=1+x$$

answered Feb 12 '18 at 11:29

user

score 2 · Accepted Answer · answered Feb 12 '18 at 11:30

2

Use the Bernoulli inequality with $\;\frac xn\;$ instead of $\;x\,$ , and $\;x>-1\;$ (otherwise it is very simple...) :

$$1+x\stackrel{\text{Bernoulli}}\le\left(1+\frac xn\right)^n\le e^x$$

because the central expression is a monotonic ascendent sequence whose limit is $\;e^x\;$ .

answered Feb 12 '18 at 11:30

DonAntonio

1

this is such a beatiful proof ! And to prove that it's monotonic ascendent we can use Bernoulli again ! https://math.stackexchange.com/questions/167843/i-have-to-show-1-frac1nn-is-monotonically-increasing-sequence – frhack Feb 12 '18 at 11:53

score 0 · Answer 3 · answered Feb 12 '18 at 11:23

0

1.) The exponential bound is trivial for $x<-1$ so it keeps to show it for $x \ge -1$

2.) Consider that in the proof of the given link you can choose $z \ge -1$ instead of $z\ge 0$ and the proof is still valid...

And you are done.

answered Feb 12 '18 at 11:23

Gono

3 Answers3