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| Line 1: |
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| In [[mathematics]], and more precisely in [[mathematical analysis|analysis]],
| | What improved tool to help you you stay on track this year than a sharp looking organizer. Hermes purses and scarves are famous, but simple examples of a Brand which has been built from scratch, painstakingly after a while and by being extremely protective of distribution channels for his or her limited production, hand crafted products. Sweet arm candy is dangling from every corner of summer 2011 fashion. [http://imgur.com/hot?q=Tory+Burch Tory Burch] is one in the most fashionable ladies fashion brand in America. because we've got the right to do lots of issues,along with all the most thrilling is we are in a position to obtain lots of our favorite items. <br><br><br><br> |
| the '''Wallis' integrals''' constitute a family of [[integral]]s introduced by [[John Wallis]]. | |
|
| |
|
| == Definition, basic properties ==
| | tory burch wallet saleWear start using machine to try the particular the washing. To get one of the most from heels, it's recommended to practice walking included before putting them on out. Consignment stores really are a great way to be stylish without spending a ton of money. I have experienced my best love before me, I did right treasure it. Vitamin C is really a nutrient which has been demonstrated to use a substantial influence on human health. <br><br>tory burch handbags saleWe have in mind the Tory Burch Handbags are fit to the young people particularly the young ladies and also the teenagers. Associated most with bringing back the charm bracelet, their particular metal and gemstone beads also allow women to customize necklaces, rings, earrings and bracelets. Jimmy Choo Outlet They provide a wide range a fabulous bouquets together with bunches you may choose and maybe they aren't necessarily pricey in anyway. Harcombe allows that vegetables can be a great addition towards the diet ' if served in butter to offer the [http://www.thefreedictionary.com/fat-soluble+vitamins fat-soluble vitamins] they contain ' but. Some type of candy like M&M's, mints, Hershey's Kisses (again you'll be able to match your candy color for a theme). <br><br><br><br>tory burch walletscom and Better Homes and Garden have cute crafts that you could make using your children or grandchildren. And although this function has being retained through the modern designers, some improvements are actually made to make garment more fashionable and appealing to the modern woman. Tory Burch, the famed New York dressmaker, will bring a downtown preppy vibe to South - Park. The stylish sandal, raised sac [http://www.pcs-systems.co.uk/Images/celinebag.aspx Cheap Celine Bags] high with a rustic-looking cork wedge, is held by rounded straps of various leathers. Bi-fold wallets, for the other hand, only folds in half. <br><br>tory burch saleOlivia has recentl tweaked clearl to make sure ou Secretar about Immunit Walternate (Kim Chevalier) experiments of our own couple of ears best that this There Olivia,Celine Bags Louis Vuitton Shoes alias Bolivia (furthermore Torv). Still, it's best to come to be discreet along with all the embellishments, rather than just produce over-decorating all of the running sneakers. Tory Burch males Leopard City Zip Tote Camel Leopard Print in 2009 are loved by countless layout males also it will last for 2010 despite belonging inside sac celine pas cher the direction in the easy actuality that within of belonging in the direction of the layout world. It could also aid in posture, forcing the wearer to handle the body more upright and provides them a much more confident stride. The two products that have grown to be the trademark of Burberry are their wide variety of perfumes in addition to their check bags. <br><br>tory burch pursesStill, watch as a possible accessory is demand especially for men who don't prefer to use bracelets. The enterprise might be grown at a pace that is certainly more easily handled by thinly resourced entrepreneurs. A quick search on line for "poems about mom" will arrive an endless selection of selections which you are able to then cut and paste into your individual do-it-yourself book; just be sure to credit the authors. Lucky Brand name Fly from the Night time Leather Hobo: Drawstring bags day back for the 30s whenever they ended up used to handle wine bottles. Sighing, I groped around my Moleskin notebook, my keys, a couple packs of Trident gum and my Mac lipgloss. <br><br>tory burch handbags outletas well as stealing content Web e-mail lists, and looking for Web addresses. This summer months sees the wedge shoe return with stunning colors and sensational towering heights. An peanut butter let alone jam pendant is additionall a interesting equipment to get lovers. You can Lv Monogram sac celine Speedy 20 intended for $199 or Louis Vuitton Immediate 25 for $100 tiny websites. Almost any set out to end up getting periodic glimpses involved with Lagoon Athabaska spanning not known wetlands together with crushed stone bars. |
| | |
| The ''Wallis' integrals'' are the terms of the sequence
| |
| <math>(W_n)_{\,n\, \in\, \mathbb{N}\,}</math> defined by:
| |
| :<math> W_n = \int_0^{\frac{\pi}{2}} \sin^n(x)\,dx, </math>
| |
| or equivalently (through a substitution: <math>x = \frac{\pi}{2} - t</math>):
| |
| :<math> W_n = \int_0^{\frac{\pi}{2}} \cos^n(x)\,dx</math>
| |
| In particular, the first two terms of this sequence are:
| |
| :<math>\quad W_0=\frac{\pi}{2}\qquad \text{ and }\quad W_1=1\,</math>
| |
| | |
| The sequence <math>\ (W_n)</math> is decreasing and has strictly positive terms.
| |
| In fact, for all <math>n \in\, \mathbb{N}</math> :
| |
| *<math>\ W_n > 0</math>, because it is an integral of a non-negative continuous function which is not all zero in the integration interval
| |
| *<math>W_{n} - W_{n + 1}= \int_0^{\frac{\pi}{2}} \sin^{n}(x)\,dx - \int_0^{\frac{\pi}{2}} \sin^{n + 1}(x)\,dx = \int_0^{\frac{\pi}{2}} \sin^{n}(x)\, [1 - \sin(x)]\,dx \geqslant 0</math>
| |
| :(by the linearity of integration and because the last integral is an integral of a non-negative function within the integration interval)
| |
| Note: Since the sequence <math>\ (W_n)</math> is decreasing and bounded below by 0, it converges to a non-negative limit. Indeed, the limit is non-zero (see below).
| |
| | |
| == Recurrence relation, evaluating the Wallis' integrals ==
| |
| | |
| By means of [[integration by parts]],
| |
| an interesting [[recurrence relation]] can be obtained:
| |
| : Noting that for all real <math>x</math>, <math>\quad \sin^2(x) = 1-\cos^2(x)</math>, we have, for all natural numbers <math>n \geqslant 2</math>,
| |
| | |
| :<math>\int_0^{\frac{\pi}{2}} \sin^{n}(x)\,dx = \int_0^{\frac{\pi}{2}} \sin^{n-2}(x) \left[1-\cos^2(x)\right]\,dx</math>
| |
| | |
| :<math>\int_0^{\frac{\pi}{2}} \sin^{n}(x)\,dx = \int_0^{\frac{\pi}{2}} \sin^{n-2}(x)\,dx - \int_0^{\frac{\pi}{2}} \sin^{n-2}(x) \cos^2(x)\,dx</math> (equation <math>\mathbf{(1)}</math>)
| |
| | |
| Integrating the second integral by parts, with:
| |
| :* <math>u'(x)=\cos (x) \sin^{n-2}(x)</math>, whose [[anti-derivative]] is <math>u(x) = \frac{1}{n-1} \sin^{n-1}(x)</math>
| |
| :* <math>v(x)=\cos (x)</math>, whose [[derivative]] is <math> v'(x) = - \sin(x)</math>
| |
| we have:
| |
| :<math>\int_0^{\frac{\pi}{2}} \sin^{n-2}(x) \cos^2(x)\,dx = \left[ \frac{1}{n-1} \sin^{n-1}(x) \cos(x)\right]_0^{\frac{\pi}{2}} + \int_0^{\frac{\pi}{2}} \ \frac{1}{n-1} \sin^{n-1}(x) \sin(x)\,dx = 0 + {1\over {n-1}}\,W_{n} </math>
| |
| | |
| Substituting this result into <math>\mathbf{(1)}</math> gives:
| |
| :<math>W_n=W_{n-2} - {1\over {n-1}}\,W_{n}</math>
| |
| and thus | |
| :<math> \qquad \left(1+ \frac{1}{n-1}\right)W_n=W_{n-2}</math> (equation <math>\mathbf{(2)}</math>)
| |
| | |
| This gives the well-known identity:
| |
| | |
| :<math>n\,W_n = (n-1)\,W_{n-2}\qquad \,</math>, valid for all <math>n \geqslant 2\qquad \,</math>.
| |
| | |
| This is a recurrence relation giving <math>W_n</math> in terms of <math>W_{n-2}</math>.
| |
| This, together with the values of <math>W_0</math> and <math>W_1</math>,
| |
| give us two sets of formulae for the terms in the sequence <math>\ (W_n)</math>,
| |
| depending on whether <math>n</math> is odd or even.
| |
| | |
| * for <math>\quad n=2\,p</math>, <math>\quad W_{2\,p}=\frac{2\,p-1}{2\,p}\times\frac{2\,p-3}{2\,p-2}\times\cdots\times\frac{1}{2}\,W_0=\frac{2\,p}{2\,p}\times\frac{2\,p-1}{2\,p}\times\frac{2\,p-2}{2\,p-2}\times\frac{2\,p-3}{2\,p-2}\times\cdots\times\frac{2}{2}\times\frac{1}{2}\,W_0 = \frac{(2\,p)!}{2^{2\,p}\, (p!)^2} \frac{\pi}{2}</math>
| |
| * for <math>\quad n=2\,p+1</math>, <math>\quad W_{2\,p+1}=\frac{2\,p}{2\,p+1}\,\frac{2\,p-2}{2\,p-1}\cdots\frac{2}{3}\,W_1=\frac{2^{2\,p}\, (p!)^2}{(2\,p +1)!}~</math>
| |
| Note that all the even terms are irrational,
| |
| whereas the odd terms are all rational.
| |
| | |
| == Equivalence ==
| |
| | |
| * From the recurrence formula above <math>\mathbf{(2)}</math>, we can deduce that
| |
| :<math>\ W_{n + 1} \sim W_n</math> (equivalence of two sequences).
| |
| | |
| :Indeed, for all <math>n \in\, \mathbb{N}</math> :
| |
| :<math>\ W_{n + 2} \leqslant W_{n + 1} \leqslant W_n</math> (since the sequence is decreasing)
| |
| :<math>\frac{W_{n + 2}}{W_n} \leqslant \frac{W_{n + 1}}{W_n} \leqslant 1</math> (since <math>\ W_n > 0</math>)
| |
| :<math>\frac{n + 1}{n + 2} \leqslant \frac{W_{n + 1}}{W_n} \leqslant 1</math> (by equation <math>\mathbf{(2)}</math>).
| |
| :By the [[sandwich theorem]], we conclude that <math>\frac{W_{n + 1}}{W_n} \to 1</math>, and hence <math>\ W_{n + 1} \sim W_n</math>.
| |
| | |
| *By examining <math>W_nW_{n+1}</math>, one obtains the following equivalence:
| |
| | |
| :<math>W_n \sim \sqrt{\frac{\pi}{2\, n}}\quad</math> ( and consequently <math>\quad\lim_{n \rightarrow \infty} \sqrt n\,W_n=\sqrt{\pi /2}\quad </math> ).
| |
| {{Hidden|Proof|
| |
| For all <math>n \in\, \mathbb{N}</math>, let <math>u_n = (n + 1)\, W_n\, W_{n + 1}</math>.
| |
| | |
| It turns out that, <math>\forall n\in \N,\, u_{n + 1} = u_n</math> because of equation <math>\mathbf{(2)}</math>.
| |
| In other words <math>\ (u_n)</math> is a constant.
| |
| | |
| It follows that for all <math>n \in\, \mathbb{N}</math>,
| |
| <math>u_n = u_0 = W_0\, W_1 = \frac{\pi}{2}</math>.
| |
| | |
| Now, since <math>\ n + 1 \sim n</math> and <math>\ W_{n + 1} \sim W_n</math>, we have, by the product rules of equivalents, <math>\ u_n \sim n\, W_n^2</math>.
| |
| | |
| Thus, <math>\ n\, W_n^2 \sim \frac{\pi}{2}</math>,
| |
| from which the desired result follows
| |
| (noting that <math>\ W_n > 0</math>).
| |
| }}
| |
| | |
| == Deducing Stirling's formula ==
| |
| | |
| Suppose that we have the following equivalence
| |
| (known as [[Stirling's formula]])
| |
| :<math>\ n\,! \sim C\, \sqrt{n}\left(\frac{n}{\mathrm{e}}\right)^n</math>, where <math>\ C \in \R^*</math>.
| |
| We now want to determine the value of this constant <math>\ C</math>
| |
| using the formula for <math>W_{2\, p}</math>.
| |
| *From above, we know that:
| |
| :<math>W_{2\, p} \sim \sqrt{\frac{\pi}{4\, p}} = \frac{\sqrt{\pi}}{2}\, \frac{1}{\sqrt{p}}</math> (equation <math>\mathbf{(3)}</math>)
| |
| *Expanding <math>W_{2\,p}</math> and using the formula above for the factorials, we get:
| |
| :<math>W_{2\,p}=\frac{(2\,p)!}{2^{2\,p}\, (p\,!)^2}\, \frac{\pi}{2} \sim \frac{C\, \left(\frac{2\, p}{\mathrm{e}}\right)^{2p}\, \sqrt{2\, p}}{2^{2p}\, C^2\, \left(\frac{p}{\mathrm{e}}\right)^{2p}\, \left(\sqrt{p}\right)^2}\, \frac{\pi}{2} </math> and hence:
| |
| :<math>W_{2\,p} \sim \frac{\pi}{C\, \sqrt{2}}\, \frac{1}{\sqrt{p}}</math> (equation <math>\mathbf{(4)}</math>)
| |
| :From <math>\mathbf{(3)}</math> and <math>\mathbf{(4)}</math>, we obtain, by transitivity,
| |
| :<math>\frac{\pi}{C\, \sqrt{2}}\, \frac{1}{\sqrt{p}} \sim \frac{\sqrt{\pi}}{2}\, \frac{1}{\sqrt{p}}</math>, which gives :
| |
| :<math>\frac{\pi}{C\, \sqrt{2}} = \frac{\sqrt{\pi}}{2}</math>, and hence <math>C = \sqrt{2\, \pi}</math>.
| |
| | |
| :We have thus proved Stirling's formula:
| |
| :<math>\ n\,! \sim \sqrt{2\, \pi\, n}\, \left(\frac{n}{\mathrm{e}}\right)^n</math>.
| |
| | |
| == Evaluating the Gaussian Integral ==
| |
| | |
| The [[Gaussian integral]] can be evaluated through the use of Wallis' integrals.
| |
| | |
| We first prove the following inequalities:
| |
| *<math>\forall n\in \mathbb N^* \quad \forall u\in\mathbb R_+ \quad u\leqslant n\quad\Rightarrow\quad (1-u/n)^n\leqslant e^{-u}</math>
| |
| *<math>\forall n\in \mathbb N^* \quad \forall u \in\mathbb R_+ \qquad e^{-u} \leqslant (1+u/n)^{-n} </math>
| |
| In fact, letting <math>\quad u/n=t</math>,
| |
| the first inequality (in which <math>t \in [0,1]</math>) is | |
| equivalent to <math>1-t\leqslant e^{-t}</math>;
| |
| whereas the second inequality reduces to
| |
| <math>e^{-t}\leqslant (1+t)^{-1}</math>,
| |
| which becomes <math>e^t\geqslant 1+t </math>.
| |
| These 2 latter inequalities follow from the convexity of the
| |
| exponential function
| |
| (or from an analysis of the function <math>t \mapsto e^t -1 -t</math>).
| |
| | |
| Letting <math>u=x^2</math> and
| |
| making use of the basic properties of improper integrals
| |
| (the convergence of the integrals is obvious),
| |
| we obtain the inequalities:
| |
| | |
| <math> \int_0^{\sqrt n}(1-x^2/n)^n dx \leqslant \int_0^{\sqrt n} e^{-x^2} dx \leqslant \int_0^{+\infty} e^{-x^2} dx \leqslant \int_0^{+\infty} (1+x^2/n)^{-n} dx</math>
| |
| for use with the [[sandwich theorem]] (as <math>n \to \infty</math>).
| |
| | |
| The first and last integrals can be evaluated easily using | |
| Wallis' integrals.
| |
| For the first one, let <math> x=\sqrt n\, \sin\,t </math>
| |
| (t varying from 0 to <math>\pi /2</math>).
| |
| Then, the integral becomes <math>\sqrt n \,W_{2n+1}</math>.
| |
| For the last integral, let <math>x=\sqrt n\, \tan\, t</math>
| |
| (t varying from <math>0</math> to <math>\pi /2</math>).
| |
| Then, it becomes <math>\sqrt n \,W_{2n-2}</math>.
| |
| | |
| As we have shown before,
| |
| <math> \lim_{n\rightarrow +\infty} \sqrt n\;W_n=\sqrt{\pi /2}</math>. So, it follows that
| |
| <math>\int_0^{+\infty} e^{-x^2} dx = \sqrt{\pi} /2</math>.
| |
| | |
| Remark: There are other methods of evaluating the Gaussian integral.
| |
| Some of them are [[Gaussian integral#Computation|more direct]].
| |
| | |
| == Relation with the Beta and Gamma functions ==
| |
| | |
| One of the definitions of the [[Beta function]] reads:
| |
| :<math>
| |
| \Beta(x,y) =
| |
| 2\int_0^{\pi/2}(\sin\theta)^{2x-1}(\cos\theta)^{2y-1}\,d\theta,
| |
| \qquad \mathrm{Re}(x)>0,\ \mathrm{Re}(y)>0 \!
| |
| </math>
| |
| Putting <math>x = \frac{n+1}{2}</math>, <math>y = \frac{1}{2}</math> into this equation gives us an expression of the Wallis' integrals in terms of the Beta function:
| |
| :<math>
| |
| \Beta \left( \frac{n+1}{2},\frac{1}{2} \right) =
| |
| 2\int_0^{\pi/2}(\sin\theta)^{n}(\cos\theta)^{0}\,d\theta
| |
| = 2\int_0^{\pi/2}(\sin\theta)^{n}\,d\theta
| |
| = 2 W_n
| |
| </math>
| |
| or equivalently,
| |
| :<math>
| |
| W_n = \frac{1}{2} \Beta \left( \frac{n+1}{2},\frac{1}{2} \right)
| |
| </math>.
| |
| | |
| Exploiting the identity relating the Beta function to [[Gamma function]]:
| |
| :<math>
| |
| \Beta(x,y)=
| |
| \dfrac{\Gamma(x)\,\Gamma(y)}{\Gamma(x+y)}
| |
| </math>
| |
| We can rewrite the above in terms of the Gamma function:
| |
| :<math>
| |
| W_n
| |
| = \frac{1}{2} \frac{\Gamma \left( \frac{n+1}{2} \right)
| |
| \Gamma \left( \frac{1}{2} \right)
| |
| }{
| |
| \Gamma \left( \frac{n+1}{2} + \frac{1}{2} \right)
| |
| }
| |
| = \frac{\Gamma \left( \frac{n+1}{2} \right)
| |
| \Gamma \left( \frac{1}{2} \right)
| |
| }{
| |
| 2 \, \Gamma \left( \frac{n+2}{2} \right)
| |
| }
| |
| </math> | |
| So, for odd <math>n</math>, writing <math>n = 2p+1</math>, we have:
| |
| :<math>
| |
| W_{2p+1}
| |
| = \frac{\Gamma \left( p+1 \right)
| |
| \Gamma \left( \frac{1}{2} \right)
| |
| }{
| |
| 2 \, \Gamma \left( p+1 + \frac{1}{2} \right)
| |
| }
| |
| = \frac{p!
| |
| \Gamma \left( \frac{1}{2} \right)
| |
| }{
| |
| (2p+1) \, \Gamma \left( p + \frac{1}{2} \right)
| |
| }
| |
| = \frac{2^p \; p!
| |
| }{
| |
| (2p+1)!!
| |
| }
| |
| = \frac{4^p \; (p!)^2
| |
| }{
| |
| (2p+1)!
| |
| }
| |
| </math>
| |
| whereas for even <math>n</math>, writing <math>n = 2p</math>, we get:
| |
| :<math>
| |
| W_{2p}
| |
| = \frac{\Gamma \left( p + \frac{1}{2} \right)
| |
| \Gamma \left( \frac{1}{2} \right)
| |
| }{
| |
| 2 \, \Gamma \left( p+1 \right)
| |
| }
| |
| = \frac{(2p-1)!! \; \pi
| |
| }{
| |
| 2^{p+1} \; p!
| |
| }
| |
| = \frac{(2p)!
| |
| }{
| |
| 4^p \; (p!)^2
| |
| }
| |
| \cdot
| |
| \frac{\pi}{2}
| |
| </math>
| |
| | |
| == Note ==
| |
| | |
| The same properties lead to [[Wallis product]],
| |
| which expresses <math>\frac{\pi}{2}\,</math>
| |
| (see [[Pi|<math>\pi</math>]])
| |
| in the form of an [[infinite product]].
| |
| | |
| == External links ==
| |
| * Pascal Sebah and Xavier Gourdon. ''Introduction to the Gamma Function''. In [http://numbers.computation.free.fr/Constants/Miscellaneous/gammaFunction.ps PostScript] and [http://numbers.computation.free.fr/Constants/Miscellaneous/gammaFunction.html HTML] formats.
| |
| | |
| {{Analysis-footer}}
| |
| | |
| {{DEFAULTSORT:Wallis' Integrals}}
| |
| [[Category:Integrals]]
| |
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