Bernoulli's inequality: Difference between revisions
en>Kasirbot m r2.7.1) (Robot: Adding fa:نابرابری برنولی |
→Alternate form: Corrected "alternate" to "alternative". |
||
Line 1: | Line 1: | ||
{{no footnotes|date=March 2013}} | |||
[[File:Bernoulli inequality.svg|right|thumb|An illustration of Bernoulli's inequality, with the graphs of <math>y=(1 + x)^r</math> and <math>y=1 + rx</math> shown in red and blue respectively. Here, <math>r=3.</math>]] | |||
In [[real analysis]], '''Bernoulli's inequality''' (named after [[Jacob Bernoulli]]) is an [[inequality (mathematics)|inequality]] that approximates [[exponentiation]]s of 1 + ''x''. | |||
The inequality states that | |||
:<math>(1 + x)^r \geq 1 + rx\!</math> | |||
for every [[integer]] ''r'' ≥ 0 and every [[real number]] ''x'' ≥ −1. If the exponent ''r'' is [[even number|even]], then the inequality is valid for ''all'' real numbers ''x''. The strict version of the inequality reads | |||
:<math>(1 + x)^r > 1 + rx\!</math> | |||
for every integer ''r'' ≥ 2 and every real number ''x'' ≥ −1 with ''x'' ≠ 0. | |||
Bernoulli's inequality is often used as the crucial step in the [[proof (math)|proof]] of other inequalities. It can itself be proved using [[mathematical induction]], as shown below. | |||
==Proof of the inequality== | |||
For ''r'' = 0, | |||
:<math>(1+x)^0 \ge 1+0x \, </math> | |||
is equivalent to 1 ≥ 1 which is true as required. | |||
Now suppose the statement is true for ''r'' = ''k'': | |||
:<math>(1+x)^k \ge 1+kx. \, </math> | |||
Then it follows that | |||
:<math> | |||
\begin{align} | |||
& {} \qquad (1+x)(1+x)^k \ge (1+x)(1+kx)\quad\text{(by hypothesis, since }(1+x)\ge 0) \\ | |||
& \iff (1+x)^{k+1} \ge 1+kx+x+kx^2, \\ | |||
& \iff (1+x)^{k+1} \ge 1+(k+1)x+kx^2. | |||
\end{align} | |||
</math> | |||
However, as 1 + (''k'' + 1)''x'' + ''kx''<sup>2</sup> ≥ 1 + (''k'' + 1)''x'' (since ''kx''<sup>2</sup> ≥ 0), it follows that (1 + ''x'')<sup>''k'' + 1</sup> ≥ 1 + (''k'' + 1)''x'', which means the statement is true for ''r'' = ''k'' + 1 as required. | |||
By induction we conclude the statement is true for all ''r'' ≥ 0. | |||
==Generalization== | |||
The exponent ''r'' can be generalized to an arbitrary real number as follows: if ''x'' > −1, then | |||
:<math>(1 + x)^r \geq 1 + rx\!</math> | |||
for ''r'' ≤ 0 or ''r'' ≥ 1, and | |||
:<math>(1 + x)^r \leq 1 + rx\!</math> | |||
for 0 ≤ ''r'' ≤ 1. | |||
This generalization can be proved by comparing [[derivative]]s. | |||
Again, the strict versions of these inequalities require ''x'' ≠ 0 and ''r'' ≠ 0, 1. | |||
== Related inequalities == | |||
The following inequality estimates the ''r''-th power of 1 + ''x'' from the other side. For any real numbers ''x'', ''r'' > 0, one has | |||
:<math>(1 + x)^r \le e^{rx},\!</math> | |||
where ''e'' = [[e (number)|2.718...]]. This may be proved using the inequality (1 + 1/''k'')<sup>''k''</sup> < ''e''. | |||
==Alternative form== | |||
An alternative form of Bernoulli's inequality for <math> t\geq 1 </math> and <math> 0\le x\le 1 </math> is: | |||
:<math> (1-x)^t \ge 1-xt. </math> | |||
This can be proved (for integer t) by using the formula for [[geometric series]]: (using y=1-x) | |||
:<math>t=1+1+\dots+1 \ge 1+y+y^2+\ldots+y^{t-1}=\frac{1-y^t}{1-y}</math> | |||
or equivalently <math>xt \ge 1-(1-x)^t. </math> | |||
==Proof for rational case== | |||
An "elementary" proof can be given using the fact that [[Inequality of arithmetic and geometric means|geometric mean of positive numbers is less than arithmetic mean]] | |||
First assume <math>t=\frac{a}{b}\leq 1</math> | |||
By comparing [[Arithmetic mean|Arithmetic]] and [[Geometric mean]] of <math>b</math> numbers | |||
(<math> (1+x) </math> occurs <math> a </math> times): | |||
<math>1,1, \ldots, (1+x),(1+x),\ldots (1+x) </math> | |||
we get | |||
<math> (1+x)^{a/b} \leq \left(1+\frac{a}{b}x\right)</math> | |||
or equivalently | |||
<math> (1+x)^{t} \leq \left(1+t x\right)</math> | |||
This proves inequality for <math> t \leq 1</math> case. | |||
For <math> s \geq 1</math> case, | |||
let <math> z=\frac{a}{b}x</math> As <math>x\geq -1, z\geq -\frac{a}{b}</math> | |||
we get with <math>s=\frac{1}{t}=\frac{b}{a}\geq 1</math>, | |||
<math> \left(1+ z \right)^{s}\geq 1+sz</math> | |||
This proves inequality for <math> s \geq 1</math> case. | |||
As these inequalities are true for all rational numbers <math> t \leq 1</math> and <math> s \geq 1</math>, | |||
they are also true for all real numbers. this is because, any real number can be | |||
approximated by rational numbers to arbitrary precision (this formally follows from the [[Construction of the real numbers|Cauchy construction of real numbers]]). | |||
==References== | |||
* {{cite book |last=Carothers |first=N. |title=Real Analysis |year=2000 |publisher=Cambridge University Press |location=Cambridge |isbn=978-0-521-49756-5 |page=9 }} | |||
* {{cite book |last=Bullen |first=P.S. |title=Handbook of Means and Their Inequalities |year=1987 |publisher=Springer |location=Berlin |isbn=978-1-4020-1522-9 |page=4 }} | |||
* {{cite book |last=Zaidman |first=Samuel |title=Advanced Calculus |year=1997 |publisher=World Scientific Publishing Company |location=City |isbn=978-981-02-2704-3 |page=32 }} | |||
== External links == | |||
* {{MathWorld |title= Bernoulli Inequality |urlname= BernoulliInequality}} | |||
* [http://demonstrations.wolfram.com/BernoulliInequality/ Bernoulli Inequality] by Chris Boucher, [[Wolfram Demonstrations Project]]. | |||
* {{cite web|title=Introduction to Inequalities|url=http://www.mediafire.com/?1mw1tkgozzu |author=Arthur Lohwater|year=1982|publisher=Online e-book in PDF format}} | |||
* Sanjeev Saxena, [http://www.cse.iitk.ac.in/users/ssax/bernoulii-inequality.pdf "A Simple Proof of Bernoulli's Inequality"], [http://vixra.org/abs/1205.0068 viXra:1205.0068], May 2012 | |||
{{DEFAULTSORT:Bernoulli's Inequality}} | |||
[[Category:Inequalities]] |
Revision as of 12:03, 23 January 2014
In real analysis, Bernoulli's inequality (named after Jacob Bernoulli) is an inequality that approximates exponentiations of 1 + x.
The inequality states that
for every integer r ≥ 0 and every real number x ≥ −1. If the exponent r is even, then the inequality is valid for all real numbers x. The strict version of the inequality reads
for every integer r ≥ 2 and every real number x ≥ −1 with x ≠ 0.
Bernoulli's inequality is often used as the crucial step in the proof of other inequalities. It can itself be proved using mathematical induction, as shown below.
Proof of the inequality
For r = 0,
is equivalent to 1 ≥ 1 which is true as required.
Now suppose the statement is true for r = k:
Then it follows that
However, as 1 + (k + 1)x + kx2 ≥ 1 + (k + 1)x (since kx2 ≥ 0), it follows that (1 + x)k + 1 ≥ 1 + (k + 1)x, which means the statement is true for r = k + 1 as required.
By induction we conclude the statement is true for all r ≥ 0.
Generalization
The exponent r can be generalized to an arbitrary real number as follows: if x > −1, then
for r ≤ 0 or r ≥ 1, and
for 0 ≤ r ≤ 1.
This generalization can be proved by comparing derivatives. Again, the strict versions of these inequalities require x ≠ 0 and r ≠ 0, 1.
Related inequalities
The following inequality estimates the r-th power of 1 + x from the other side. For any real numbers x, r > 0, one has
where e = 2.718.... This may be proved using the inequality (1 + 1/k)k < e.
Alternative form
An alternative form of Bernoulli's inequality for and is:
This can be proved (for integer t) by using the formula for geometric series: (using y=1-x)
Proof for rational case
An "elementary" proof can be given using the fact that geometric mean of positive numbers is less than arithmetic mean
By comparing Arithmetic and Geometric mean of numbers ( occurs times):
we get
or equivalently
This proves inequality for case.
For case, let As we get with ,
This proves inequality for case.
As these inequalities are true for all rational numbers and , they are also true for all real numbers. this is because, any real number can be approximated by rational numbers to arbitrary precision (this formally follows from the Cauchy construction of real numbers).
References
- 20 year-old Real Estate Agent Rusty from Saint-Paul, has hobbies and interests which includes monopoly, property developers in singapore and poker. Will soon undertake a contiki trip that may include going to the Lower Valley of the Omo.
My blog: http://www.primaboinca.com/view_profile.php?userid=5889534 - 20 year-old Real Estate Agent Rusty from Saint-Paul, has hobbies and interests which includes monopoly, property developers in singapore and poker. Will soon undertake a contiki trip that may include going to the Lower Valley of the Omo.
My blog: http://www.primaboinca.com/view_profile.php?userid=5889534 - 20 year-old Real Estate Agent Rusty from Saint-Paul, has hobbies and interests which includes monopoly, property developers in singapore and poker. Will soon undertake a contiki trip that may include going to the Lower Valley of the Omo.
My blog: http://www.primaboinca.com/view_profile.php?userid=5889534
External links
I had like 17 domains hosted on single account, and never had any special troubles. If you are not happy with the service you will get your money back with in 45 days, that's guaranteed. But the Search Engine utility inside the Hostgator account furnished an instant score for my launched website. Fantastico is unable to install WordPress in a directory which already have any file i.e to install WordPress using Fantastico the destination directory must be empty and it should not have any previous installation files. When you share great information, others will take note. Once your hosting is purchased, you will need to setup your domain name to point to your hosting. Money Back: All accounts of Hostgator come with a 45 day money back guarantee. If you have any queries relating to where by and how to use Hostgator Discount Coupon, you can make contact with us at our site. If you are starting up a website or don't have too much website traffic coming your way, a shared plan is more than enough. Condition you want to take advantage of the worldwide web you prerequisite a HostGator web page, -1 of the most trusted and unfailing web suppliers on the world wide web today. Since, single server is shared by 700 to 800 websites, you cannot expect much speed.
Hostgator tutorials on how to install Wordpress need not be complicated, especially when you will be dealing with a web hosting service that is friendly for novice webmasters and a blogging platform that is as intuitive as riding a bike. After that you can get Hostgator to host your domain and use the wordpress to do the blogging. Once you start site flipping, trust me you will not be able to stop. I cut my webmaster teeth on Control Panel many years ago, but since had left for other hosting companies with more commercial (cough, cough) interfaces. If you don't like it, you can chalk it up to experience and go on. First, find a good starter template design. When I signed up, I did a search for current "HostGator codes" on the web, which enabled me to receive a one-word entry for a discount. Your posts, comments, and pictures will all be imported into your new WordPress blog.- Bernoulli Inequality by Chris Boucher, Wolfram Demonstrations Project.
- Template:Cite web
- Sanjeev Saxena, "A Simple Proof of Bernoulli's Inequality", viXra:1205.0068, May 2012