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"graphs of y = (1+x)^r and y = 1+rx shown in red and blue respectively. Here, r=3. In mathematics, Bernoulli's inequality (named after Jacob Bernoulli) is an inequality that approximates exponentiations of 1 + x. It is often employed in real analysis. The inequality states that :(1 + x)^r \geq 1 + rx 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 :(1 + x)^r > 1 + rx for every integer r ≥ 2 and every real number x ≥ −1 with x ≠ 0\. There is also a generalized version that says for every real number r ≥ 1 and real number x ≥ −1, :(1 + x)^r \geq 1 + rx, while for 0 ≤ r ≤ 1 and real number x ≥ −1, :(1 + x)^r \leq 1 + rx. 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. History Jacob Bernoulli first published the inequality in his treatise “Positiones Arithmeticae de Seriebus Infinitis” (Basel, 1689), where he used the inequality often.mathematics - First use of Bernoulli's inequality and its name - History of Science and Mathematics Stack Exchange According to Joseph E. Hofmann, Über die Exercitatio Geometrica des M. A. Ricci (1963), p. 177, the inequality is actually due to Sluse in his Mesolabum (1668 edition), Chapter IV "De maximis & minimis". Proof of the inequality We proceed with mathematical induction in the following form: * we prove the inequality for r\in\\{0,1\\}, * from validity for some r we deduce validity for r + 2\. For r = 0, :(1+x)^0 \ge 1+0x is equivalent to 1 ≥ 1 which is true. Similarly, for r = 1 we have :(1+x)^r=1+x\ge 1+x=1+rx. Now suppose the statement is true for r = k: :(1+x)^k \ge 1+kx. Then it follows that : \begin{align} (1+x)^{k+2} &= (1+x)^k(1+x)^2 \\\ &\ge (1+kx)\left(1+2x+x^2\right) \qquad\qquad\qquad\text{ by hypothesis and }(1+x)^2\ge 0 \\\ &=1+2x+x^2+kx+2kx^2+kx^3 \\\ &=1+(k+2)x+kx^2(x+2)+x^2 \\\ &\ge 1+(k+2)x \end{align} since x^2\ge 0 as well as x+2\ge0. By the modified induction we conclude the statement is true for every non-negative integer r. Generalizations= Generalization of exponent = The exponent r can be generalized to an arbitrary real number as follows: if x > −1, then :(1 + x)^r \geq 1 + rx for r ≤ 0 or r ≥ 1, and :(1 + x)^r \leq 1 + rx 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\. = Generalization of base = Instead of (1+x)^n the inequality holds also in the form (1+x_1)(1+x_2)\dots(1+x_r) \geq 1+x_1+x_2 + \dots + x_r where x_1, x_2, \dots , x_r are real numbers, all greater than -1, all with the same sign. The Bernoulli's inequality is special case when x_1 = x_2 = \dots = x_r = x. This generalized inequality can be proved by mathematical induction. Proof In the first step we take n=1. In this case inequality 1+x_1 \geq 1 + x_1 is obviously true. In the second step we assume validity of inequality for r numbers and deduce validity for r+1 numbers. We assume that(1+x_1)(1+x_2)\dots(1+x_r) \geq 1+x_1+x_2 + \dots + x_ris valid. After multiplying both sides with a positive number (x_{r+1} + 1) we get: \begin{alignat}{2} (1+x_1)(1+x_2)\dots(1+x_r)(1+x_{r+1}) \geq & (1+x_1+x_2 + \dots + x_r)(1+x_{r+1}) \\\ \geq & (1+x_1+x_2+ \dots + x_r) \cdot 1 + (1+x_1+x_2+ \dots + x_r) \cdot x_{r+1} \\\ \geq & (1+x_1+x_2+ \dots + x_r) + x_{r+1} + x_1 x_{r+1} + x_2 x_{r+1} + \dots + x_r x_{r+1} \\\ \end{alignat} As x_1, x_2, \dots x_r, x_{r+1} have all equal sign, the products x_1 x_{r+1}, x_2 x_{r+1}, \dots x_r x_{r+1} are all positive numbers. So the quantity on the right-hand side can be bounded as follows:(1+x_1+x_2+ \dots + x_r) + x_{r+1} + x_1 x_{r+1} + x_2 x_{r+1} + \dots + x_r x_{r+1} \geq 1+x_1+x_2+ \dots + x_r + x_{r+1},what was to be shown. Related inequalities The following inequality estimates the r-th power of 1 + x from the other side. For any real numbers x, r with r > 0, one has :(1 + x)^r \le e^{rx}, 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 t\geq 1 and 0\le x\le 1 is: :(1-x)^t \ge 1-xt. This can be proved (for any integer t) by using the formula for geometric series: (using y = 1 − x) :t=1+1+\dots+1 \ge 1+y+y^2+\ldots+y^{t-1} = \frac{1-y^t}{1-y}, or equivalently xt \ge 1-(1-x)^t. Alternative proof Using AM-GM An elementary proof for 0\le r\le 1 and x ≥ -1 can be given using weighted AM-GM. Let \lambda_1, \lambda_2 be two non-negative real constants. By weighted AM-GM on 1,1+x with weights \lambda_1, \lambda_2 respectively, we get :\dfrac{\lambda_1\cdot 1 + \lambda_2\cdot (1+x)}{\lambda_1+\lambda_2}\ge \sqrt[\lambda_1+\lambda_2]{(1+x)^{\lambda_2}}. Note that :\dfrac{\lambda_1\cdot 1 + \lambda_2\cdot (1+x)}{\lambda_1+\lambda_2}=\dfrac{\lambda_1+\lambda_2+\lambda_2x}{\lambda_1+\lambda_2}=1+\dfrac{\lambda_2}{\lambda_1+\lambda_2}x and :\sqrt[\lambda_1+\lambda_2]{(1+x)^{\lambda_2}} = (1+x)^{\frac{\lambda_2}{\lambda_1+\lambda_2}}, so our inequality is equivalent to :1 + \dfrac{\lambda_2}{\lambda_1+\lambda_2}x \ge (1+x)^{\frac{\lambda_2}{\lambda_1+\lambda_2}}. After substituting r = \dfrac{\lambda_2}{\lambda_1+\lambda_2} (bearing in mind that this implies 0\le r\le 1) our inequality turns into :1+rx \ge (1+x)^r which is Bernoulli's inequality. Using the formula for geometric series Bernoulli's inequality is equivalent to and by the formula for geometric series (using y = 1 + x) we get which leads to Now if x \ge 0 then by monotony of the powers each summand (1+x)^k - 1 = (1+x)^k - 1^k \ge 0, and therefore their sum is greater 0 and hence the product on the LHS of (). If 0 \ge x\ge -2 then by the same arguments 1\ge(1+x)^k and thus all addends (1+x)^k-1 are non-positive and hence so is their sum. Since the product of two non-positive numbers is non- negative, we get again (). Using the binomial theorem One can prove Bernoulli's inequality for x ≥ 0 using the binomial theorem. It is true trivially for r = 0, so suppose r is a positive integer. Then (1+x)^r = 1 + rx + \tbinom r2 x^2 + ... + \tbinom rr x^r. Clearly \tbinom r2 x^2 + ... + \tbinom rr x^r \ge 0, and hence (1+x)^r \ge 1+rx as required. NotesReferences External links Bernoulli Inequality by Chris Boucher, Wolfram Demonstrations Project. * Inequalities "
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"The Bastard Operator From Hell (BOFH) is a fictional rogue computer operator who takes out his anger on users and others who pester him with their computer problems, uses his expertise against his enemies and manipulates his employer. Several other people have written stories about BOFHs, but those by Simon Travaglia are considered canonical. The BOFH stories were originally posted in 1992 to Usenet by Travaglia, with some being reprinted in Datamation. They were published weekly from 1995 to 1999 in Network Week.Network Week, London APT Data Services, 1995–1999 Since 2000 they have been published regularly in The Register (UK).Simon Travaglia, The Register, accessed 18 January 2013 Several collections of the stories have been published as books. By extension, the term is also used to refer to any system administrator who displays the qualities of the original. The early accounts of the BOFH took place in a university; later the scenes were set in an office workplace. In 2000 (BOFH 2k), the BOFH and his pimply-faced youth (PFY) assistant moved to a new company. Other characters * The PFY (Pimply-Faced Youth, the assistant to the BOFH. Real name is Stephen) Possesses a temperament similar to the BOFH, and often either teams up with or plots against him. * The Boss (often portrayed as having no IT knowledge but believing otherwise; identity changes as successive bosses are sacked, leave, are committed, or have nasty "accidents") * CEO of the company — The PFY's uncle Brian from 1996 until 2000, when the BOFH and PFY moved to a new company. * The Head of IT, almost as disposable as the Boss. * The help desk operators, referred to as the "Helldesk" and often scolded for giving out the BOFH's personal number. * The Boss's secretary, Sharon. * The security department * George, the cleaner (an invaluable source of information to the BOFH and PFY) Books * References in other media The main protagonist in Charles Stross's The Laundry Files series of novels named himself Bob Oliver Francis Howard in reference to the BOFH. As Bob Howard is a self-chosen pseudonym, and Bob is a network manager when not working as a computational demonologist, the name is all too appropriate. In the novella 'Pimpf' he acquires a pimply-faced young assistant by the name of Peter-Fred Young. References External links * Travaglia, Simon. http://www.bofharchive.com The author's official archive and blog. Further reading * Computer humor Internet culture Internet slang System administration Fictional characters introduced in 1992 Fictional people in information technology "