Bernoulli's inequality

An illustration of Bernoulli's inequality. Here the two functions (red graph) and (blue graph) with the specific value . The red graph is always above the blue graph. ${\displaystyle f(x)=(1+x)^{n}}$ ${\displaystyle g(x)=1+n\cdot x}$ ${\displaystyle n=3}$ ${\displaystyle x\geq -1}$

In mathematics , the Bernoullian inequality is a simple but important inequality that can be used to estimate a power function downwards.

For every real number and every integer holds ${\displaystyle x\geq -1}$ ${\displaystyle n\geq 0}$

${\displaystyle (1+x)^{n}\geq 1+nx}$.

The inequality is named after the Swiss mathematician Jakob I Bernoulli .

history

Jakob Bernoulli first published this inequality in his work Positiones Arithmeticae de Seriebus Infinitis (Basel, 1689), in which he frequently used this inequality.

According to Joseph E. Hofmann, the inequality goes back to the mathematician Sluse, who is said to have published it in 1668 in his work Mesolabum.

proof

Proof of complete induction

The Bernoulli inequality can be proved with complete induction . The induction start is fulfilled: ${\displaystyle n=0}$

${\displaystyle (1+x)^{0}=1\geq 1=1+0x}$.

As induction hypothesis now applies for , and . Then because of and the induction hypothesis follows ${\displaystyle (1+x)^{n}\geq 1+nx}$ ${\displaystyle n\in \mathbb {N} _{0}}$ ${\displaystyle x\in \mathbb {R} }$ ${\displaystyle x\geq -1}$ ${\displaystyle 1+x\geq 0}$

${\displaystyle {\begin{array}{lcl}(1+x)^{n+1}&=&(1+x)^{n}\cdot (1+x)\\&{\stackrel {\mathrm {I.\,V.} }{\geq }}&(1+nx)\cdot (1+x)\\&=&1+x+nx+nx^{2}\\&\geq &1+x+nx\\&=&1+(n+1)x.\\\end{array}}}$

According to the principle of induction, the claim applies to everyone . ${\displaystyle n\in \mathbb {N} _{0}}$

Alternative proof for non-negative x

For , the Bernoulli inequality can also be proven using the binomial theorem . It applies here ${\displaystyle x\geq 0}$

${\displaystyle {\begin{aligned}(1+x)^{n}&=\sum _{k=0}^{n}{\binom {n}{k}}x^{k}\\&=1+n\cdot x+\underbrace {{\frac {n(n-1)}{2}}x^{2}+\ldots } _{\geq \ 0{\text{ wegen }}x\geq 0}\\&\geq 1+n\cdot x\end{aligned}}}$

example

Claim:

${\displaystyle \lim _{n\to \infty }{\sqrt[{n}]{a}}=1}$

for all real ones . ${\displaystyle a\geq 1}$

Proof: Let first be defined by ${\displaystyle x_{n}\geq 0}$

${\displaystyle {\sqrt[{n}]{a}}=1+x_{n}}$.

Then the Bernoulli inequality applies

${\displaystyle a=(1+x_{n})^{n}\geq 1+nx_{n}}$,

so

${\displaystyle {\frac {a-1}{n}}\geq x_{n}\geq 0}$.

But it is

${\displaystyle \lim _{n\to \infty }{\frac {a-1}{n}}=0}$.

That’s then too

${\displaystyle \lim _{n\to \infty }x_{n}=0}$

and ultimately

${\displaystyle \lim _{n\to \infty }{\sqrt[{n}]{a}}=1+\lim _{n\to \infty }x_{n}=1+0=1.}$

Related inequalities

Strict inequality

The following inequality is also referred to as Bernoulli's inequality, which uses a "strictly greater than" instead of "greater than or equal to":

For all real numbers , and all natural numbers applies ${\displaystyle x>-1}$ ${\displaystyle x\neq 0}$ ${\displaystyle n\geq 2}$

${\displaystyle (1+x)^{n}>1+nx}$.

The proof can also be carried out with induction according to the same pattern as the proof for the formulation “greater than or equal”.

Real exponents

For real exponents , the following generalizations by comparing the derivatives show: For all true ${\displaystyle x>-1}$

${\displaystyle (1+x)^{r}\geq 1+rx}$,

if , and ${\displaystyle r\geq 1}$

${\displaystyle (1+x)^{r}\leq 1+rx}$,

if . ${\displaystyle 0\leq r\leq 1}$

Variable factors

If one does not consider a potency, but a product of different factors, the following generalization can be shown by means of complete induction:

${\displaystyle \prod _{i=1}^{n}(1+x_{i})>1+\sum _{i=1}^{n}x_{i}}$

if for all or if for all and . ${\displaystyle -1<x_{i}<0\;}$ ${\displaystyle x_{i}\;}$ ${\displaystyle x_{i}>0\;}$ ${\displaystyle x_{i}\;}$ ${\displaystyle n}$ ${\displaystyle {\!}^{\geq }}$ ${\displaystyle 2}$

If one sets and considers the special case , that is , one obtains the so-called Weierstrass product inequality ${\displaystyle u_{i}:=-x_{i}\;}$ ${\displaystyle -1\leq x_{i}\leq 0}$ ${\displaystyle 0\leq u_{i}\leq 1}$

${\displaystyle \prod _{i=1}^{n}(1-u_{i})\geq 1-\sum _{i=1}^{n}u_{i}.}$

Applications

Exponential function

The Bernoullian inequality is useful for many estimates. It is fixed, then it is for big enough . With the Bernoullian inequality we therefore have ${\displaystyle x\in \mathbb {R} }$ ${\displaystyle {\frac {x}{n}}\geq -1}$ ${\displaystyle n}$

${\displaystyle \left(1+{\frac {x}{n}}\right)^{n}\geq 1+n\cdot {\frac {x}{n}}=1+x}$for big enough . ${\displaystyle n}$

Because of

${\displaystyle e^{x}=\lim _{n\to \infty }\left(1+{\frac {x}{n}}\right)^{n}}$

is thus the inequality

${\displaystyle 1+x\leq e^{x}}$ for all ${\displaystyle x\in \mathbb {R} }$

proven.

Proof of inequalities with powers

In order to prove the convergence for real numbers with, one has to find, among other things, a so that is given for an arbitrarily . The Bernoulli inequality can be used for this. First, one reshapes the target inequality through equivalence transformations: ${\displaystyle \lim _{n\rightarrow \infty }q^{n}=0}$ ${\displaystyle q}$ ${\displaystyle 0<q<1}$ ${\displaystyle N\in \mathbb {N} }$ ${\displaystyle q^{N}<\epsilon }$ ${\displaystyle \epsilon >0}$ ${\displaystyle q^{N}<\epsilon }$

${\displaystyle {\begin{array}{lrl}&q^{N}&<\epsilon \\\iff \ &\left({\tfrac {1}{q}}\right)^{N}&>{\tfrac {1}{\epsilon }}\end{array}}}$

Because is . Let us put is and also according to the Bernoulli inequality ${\displaystyle 0<q<1}$ ${\displaystyle {\tfrac {1}{q}}>1}$ ${\displaystyle 1+x={\tfrac {1}{q}}}$ ${\displaystyle x>0}$

${\displaystyle \left({\tfrac {1}{q}}\right)^{N}=(1+x)^{N}\geq 1+N\cdot x}$

Alternatively, one can also be found so that is. Is because then it follows from the above inequality that is also automatically . The existence of is guaranteed by the Archimedean axiom . ${\displaystyle N\in \mathbb {N} }$ ${\displaystyle 1+N\cdot x>{\tfrac {1}{\epsilon }}}$ ${\displaystyle 1+N\cdot x>{\tfrac {1}{\epsilon }}}$ ${\displaystyle \left({\tfrac {1}{q}}\right)^{N}\geq 1+N\cdot x}$ ${\displaystyle \left({\tfrac {1}{q}}\right)^{N}>\epsilon }$ ${\displaystyle N}$

The advantage of the above procedure is that the logarithm does not have to be used in the proof, which is usually not yet available at the beginning of an analysis lecture.

Inequality of the arithmetic and geometric mean

→ Main article : Proof of the inequality of the arithmetic and geometric mean

Using an estimate with Bernoulli's inequality, the inequality of the arithmetic and geometric mean can be proven via complete induction . In fact, Bernoulli's inequality is equivalent to the inequality of the arithmetic and geometric mean.

Web links

Wikibooks: Math for Non-Freaks: Bernoulli Inequality  - Learning and Teaching Materials

Commons : Bernoulli's inequality  - collection of images, videos and audio files

• Yuan-Chuan Li, Cheh-Chih Yeh: Some Equivalent Forms of Bernoulli's Inequality: A Survey. In: Applied Mathematics. 04, 2013, p. 1070, doi : 10.4236 / am.2013.47146

Sources and Notes

1. ↑ In fact, the inequality even holds for and odd , but this can no longer be so directly with complete induction. B. show by comparing the derivatives . One shows that for negative derivative and thus has no extrema, while the value for and is positive. In this case has a local maximum in . For even , the inequality even applies to all real ones , since here the left side of the inequality always remains positive, while the right side is certainly negative. ${\displaystyle x\geq -2}$ ${\displaystyle n\geq 3}$ ${\displaystyle f(x):=(1+x)^{n}-(1+nx)}$ ${\displaystyle -2<x<-1}$ ${\displaystyle x=-2}$ ${\displaystyle x=-1}$ ${\displaystyle f}$ ${\displaystyle x=-2}$ ${\displaystyle n\geq 2}$ ${\displaystyle x}$ ${\displaystyle x<-1}$
2. ↑ ^{a b} In the event and must be agreed. ${\displaystyle x=-1}$ ${\displaystyle n=0}$ ${\displaystyle 0^{0}=1}$
3. ↑ ^{a b c} Harro Heuser, Textbook of Analysis, Part 1. , BG Teubner Stuttgart, 1984, ISBN 3-519-22221-3 , p. 61, chapter 7.9 and p. 68, exercise 7.17
4. ↑ ^{a b} History of Science and Mathematics.
5. ^ About the Exercitatio Geometrica by MA Ricci. (1963), p. 177.
6. ↑ http://mo.mathematik.uni-stuttgart.de/inhalt/erlaeuterung/erlaeuterung39/
7. ↑ http://planetmath.org/encyclopedia/WeierstrassProductInequality.html
8. ^ Adam Kertesz and Eric Weisstein: Weierstrass Product Inequality . In: MathWorld (English).
9. ↑ http://www.cut-the-knot.org/Generalization/wineq.shtml
10. ^ Yuan-Chuan Li, Cheh-Chih Yeh: Some Equivalent Forms of Bernoulli's Inequality: A Survey. In: Applied Mathematics. 04, 2013, p. 1070, doi : 10.4236 / am.2013.47146 .