9.7 Pascal s Formula and the Binomial Theorem

Transcription

1 592 Chapte 9 Coutig ad Pobability Example 971 Values of 97 Pascal s Fomula ad the Biomial Theoem I m vey well acquaited, too, with mattes mathematical, I udestad equatios both the simple ad quadatical About biomial theoem I am teamig with a lot of ews, With may cheeful facts about the squae of the hypoteuse William S Gilbet, The Piates of Pezace, 1880 I this sectio we deive seveal fomulas fo values of The most impotat is Pascal s fomula, which is the basis fo Pascal s tiagle ad is a cucial compoet of oe of the poofs of the biomial theoem We offe two distict poofs fo both Pascal s fomula ad the biomial theoem Oe of them is called algebaic because it elies to a geat extet o algebaic maipulatio, ad the othe is called combiatoial, because it is based o the id of coutig agumets we have bee discussig i this chapte,, 1 2 Thi of Theoem 951 as a geeal template: Regadless of what oegative umbes ae placed i the boxes, if the umbe i the lowe box is o geate tha the umbe i the top box, the!!! Use Theoem 951 to show that fo all iteges 0, 1 971, if , if Solutio 1 2!!! 1 0! 1!!!!! +! 1! 2! 2! 2! 2!2! sice 0! 1 by defiitio 2 Note that the esult deived algebaically above, that equals 1, agees with the fact that a set with elemets has just oe subset of size, amely itself Similaly, execise 1 at the ed of the sectio ass you to show algebaically that 1, which agees with the fact that a set with elemets has oe subset, the empty set, of size 0 I execise 2 you ae also ased to show algebaically that This esult agees with the fact that thee ae subsets of size 1 that ca be chose fom a set with elemets, amely the subsets cosistig of each elemet tae aloe

2 Example Pascal s Fomula ad the Biomial Theoem 593 I execise 5 at the ed of the sectio you ae ased to veify algebaically that fo all oegative iteges ad with A alteative way to deduce this fomula is to itepet it as sayig that a set A with elemets has exactly as may subsets of size as it has subsets of size Deive the fomula usig this easoig Solutio Obseve that ay subset of size ca be specified eithe by sayig which elemets lie i the subset o by sayig which elemets lie outside the subset A, A Set with Elemets B, a subset with elemets A B, a subset with elemets Ay subset B with elemets completely detemies a subset, A B, with elemets Suppose A has subsets of size : B 1, B 2,,B The each B i ca be paied up with exactly oe set of size, amely its complemet A B i as show below Subsets of Size Subsets of Size B 1 B 2 A B 1 A B 2 B A B All subsets of size ae listed i the left-had colum, ad all subsets of size ae listed i the ight-had colum The umbe of subsets of size equals the umbe of subsets of size, ad so The type of easoig used i this example is called combiatoial, because it is obtaied by coutig thigs that ae combied i diffeet ways A umbe of theoems have both combiatoial poofs ad poofs that ae puely algebaic Blaise Pascal Hulto-Deutch Collectio/CORBIS Pascal s Fomula Pascal s fomula, amed afte the seveteeth-cetuy Fech mathematicia ad philosophe Blaise Pascal, is oe of the most famous ad useful i combiatoics which is the fomal tem fo +1 the study of coutig ad listig poblems It elates the value of to the values of ad Specifically, it says that wheeve ad ae positive iteges with This fomula maes it easy to compute highe combiatios i tems of lowe oes: If all the values of ae ow, the the +1 values of ca be computed fo all such that 0 <

3 594 Chapte 9 Coutig ad Pobability Pascal s tiagle, show i Table 971, is a geometic vesio of Pascal s fomula Sometimes it is simply called the aithmetic tiagle because it was used cetuies befoe Pascal by Chiese ad Pesia mathematicias But Pascal discoveed it idepedetly, ad eve sice 1654, whe he published a teatise that exploed may of its featues, it has geeally bee ow as Pascal s tiagle Table 971 Pascal s Tiagle fo Values of Each ety i the tiagle is a value of Pascal s fomula taslates ito the fact that the ety i ow + 1, colum equals the sum of the ety i ow, colum 1plus the ety i ow, colum That is, the ety i a give iteio positio equals the sum of the two eties diectly above ad to the above left The left-most ad ight-most eties i each ow ae 1 because 1 by Example 971 ad 1 by execise 1 at the ed of this sectio Example 973 Calculatig Usig Pascal s Tiagle Use Pascal s tiagle to compute the values of 6 6 ad 2 3 Solutio By costuctio, the value i ow, colum of Pascal s tiagle is the value of +1, fo evey pai of positive iteges ad with By Pascal s fomula, ca be computed by addig togethe ad, which ae located diectly above ad +1 above left of Thus, ad

4 97 Pascal s Fomula ad the Biomial Theoem 595 Pascal s fomula ca be deived by two etiely diffeet agumets Oe is algebaic; it uses the fomula fo the umbe of -combiatios obtaied i Theoem 951 The othe is combiatoial; it uses the defiitio of the umbe of -combiatios as the umbe of subsets of size tae fom a set with a cetai umbe of elemets We give both poofs sice both appoaches have applicatios i may othe situatios Theoem 971 Pascal s Fomula Let ad be positive iteges ad suppose The Poof algebaic vesio: Let ad be positive iteges with By Theoem 951,! + 1!! +!!!!! +! +!!! To add these factios, a commo deomiato is eeded, so multiply the umeato ad deomiato of the left-had factio by ad multiply the umeato ad deomiato of the ight-had factio by + The! + 1! +! +! +!! +!!! +! + +!! +!!! +!! +!! + +!! +! +!! !! Poof combiatoial vesio: Let ad be positive iteges with Suppose S is a set with + 1 elemets The umbe of subsets of S of size ca be calculated by thiig of S as cosistig of two pieces: oe with elemets {x 1, x 2,,x } ad the othe with oe elemet {x +1 } Ay subset of S with elemets eithe cotais x +1 o it does ot If it cotais x +1, the it cotais 1 elemets fom the set {x 1, x 2,,x } If it does ot cotai x +1, the it cotais elemets fom the set {x 1, x 2,,x } cotiued o page 596

5 596 Chapte 9 Coutig ad Pobability Subsets of Size of {x 1, x 2,, x +1 } subsets of size that cosist etiely of elemets fom {x 1, x 2,, x } subsets of size that cotai x +1 ad 1 elemets fom {x 1, x 2,, x } Thee ae of these Thee ae of these By the additio ule, umbe of subsets of umbe of subsets of umbe of subsets of {x 1, x 2,,x, x +1 } {x 1, x 2,,x } + {x 1, x 2,,x } of size of size 1 of size By Theoem 951, the set {x 1, x 2,,x, x +1 } has + subsets of size, theset {x 1, x 2,,x } has 1 subsets of size 1, ad the set {x1, x 2,,x } has subsets of size Thus + 1 +, 1 aswastobeshow Example 974 Deivig New Fomulas fom Pascal s Fomula +2 Use Pascal s fomula to deive a fomula fo i tems of values of,, ad 2 Assume ad ae oegative iteges ad 2 Solutio By Pascal s fomula, Now apply Pascal s fomula to 1 ad ad substitute ito the above to obtai [ ] [ ] Combiig the two middle tems gives fo all oegative iteges ad such that 2 The Biomial Theoem I algeba a sum of two tems, such as a + b, is called a biomial The biomial theoem gives a expessio fo the powes of a biomial a + b, fo each positive itege ad all eal umbes a ad b

6 97 Pascal s Fomula ad the Biomial Theoem 597 Coside what happes whe you calculate the fist few powes of a + b Accodig to the distibutive law of algeba, you tae the sum of the poducts of all combiatios of idividual tems: a + b 2 a + ba + b aa + ab + ba + bb, a + b 3 a + ba + ba + b aaa + aab + aba + abb + baa + bab + bba + bbb, a + b 4 a + b a + b a + b a + b }{{}}{{}}{{}}{{} 1st 2d 3d 4th facto facto facto facto aaaa + aaab + aaba + aabb + abaa + abab + abba + abbb + baaa + baab + baba + babb + bbaa + bbab + bbba + bbbb Now focus o the expasio of a + b 4 It is cocete, ad yet it has all the featues of the geeal case A typical tem of this expasio is obtaied by multiplyig oe of the two tems fom the fist facto times oe of the two tems fom the secod facto times oe of the two tems fom the thid facto times oe of the two tems fom the fouth facto Fo example, the tem abab is obtaied by multiplyig the a s ad b s maed with aows below a + ba + ba + ba + b Sice thee ae two possible values a o b fo each tem selected fom oe of the fou factos, thee ae tems i the expasio of a + b 4 Now some tems i the expasio ae lie tems ad ca be combied Coside all possible odeigs of thee a s ad oe b, fo example By the techiques of Sectio 95, 4 thee ae 4 of them Ad each of the fou occus as a tem i the expasio of a + b 4 : aaab aaba abaa baaa By the commutative ad associative laws of algeba, each such tem equals a 3 b,soall fou ae lie tems Whe the lie tems ae combied, theefoe, the coefficiet of a 3 b 4 equals Similaly, the expasio of a + b 4 cotais the diffeet odeigs of two a s ad two b s, aabb abab abba baab baba bbaa, all of which equal a 2 b 2, so the coefficiet of a 2 b 2 4 equals 2 By a simila aalysis, the coefficiet of ab 3 4 equals 3 Also, sice thee is oly oe way to ode fou a s, the coefficiet of a 4 4 is 1 which equals, ad sice thee is oly oe way to ode fou b s, the coefficiet of b 4 4 is 1 which equals 4 Thus, whe all of the lie tems ae combied, a + b 4 4 a a 3 4 b + a 2 2 b ab b 4 4 a 4 + 4a 3 b + 6a 2 b 2 + 4ab 3 + b 4 The biomial theoem geealizes this fomula to a abitay oegative itege

7 598 Chapte 9 Coutig ad Pobability Theoem 972 Biomial Theoem Give ay eal umbes a ad b ad ay oegative itege, a + b a b 0 a + a 1 b 1 + a 2 2 b a 1 b 1 + b 1 Note that the secod expessio equals the fist because 1 ad 1, fo all oegative iteges, povided that b 0 1 ad a 1 It is istuctive to see two poofs of the biomial theoem: a algebaic poof ad a combiatoial poof Both equie a pecise defiitio of itege powe Defiitio Note This is the defiitio of O 0 give by Doald E Kuth i The AtofCompute Pogammig, Volume 1: Fudametal Algoithms, Thid Editio Readig, Mass: Addiso-Wesley, 1997, p 57 Fo ay eal umbe a ad ay oegative itege, theoegative itege powes of a ae defied as follows: { a 1 if 0 a a 1 if > 0 I some mathematical cotexts, 0 0 is left udefied Defiig it to be 1, as is doe hee, maes it possible to wite geeal fomulas such as x i 1 without havig to 1 x i0 exclude values of the vaiables that esult i the expessio 0 0 The algebaic vesio of the biomial theoem uses mathematical iductio ad calls upo Pascal s fomula at a cucial poit Poof of the Biomial Theoem algebaic vesio: Suppose a ad b ae eal umbes We use mathematical iductio ad let the popety P be the equatio a + b a b P 0 Show that P is tue: Whe 0, the biomial theoem states that: 0 a + b 0 0 a 0 b P 0 But the left-had side is a + b 0 1 [by defiitio of powe], ad the ight-had side is 0 0 a 0 b 0 a 0 0 b 0 0 0! 0! 0! also [sice 0! 1, a 0 1, ad b 0 1] Hece P is tue

8 97 Pascal s Fomula ad the Biomial Theoem 599 Show that fo all iteges m 0, ifpmistuethepm+ istue:let a itege m 0 be give, ad suppose Pm is tue That is, suppose a + b m 0 m a m b We eed to show that Pm + is tue: 0 Pm iductive hypothesis m+1 a + b m+1 m + 1 a m+ b Pm + Now, by defiitio of the m + st powe, a + b m+1 a + b a + b m, so by substitutio fom the iductive hypothesis, a + b m+1 a + b a m a m b m a m b + b m a m+1 b m a m b m a m b +1 by the geealized distibutive law ad the facts that a a m a 1+m a m+1 ad b b b 1+ b +1 We tasfom the secod summatio o the ight-had side by maig the chage of vaiable j + 1 Whe 0, the j 1 Whe m, the j m + 1 Ad sice j 1, the geeal tem is m a m b +1 m j 1 a m j b j Hece the secod summatio o the ight-had side above is m+1 m a m+1 j b j j 1 j1 m a m+1 j b j j 1 But the j i this summatio is a dummy vaiable; it ca be eplaced by the lette, as log as the eplacemet is made eveywhee the j occus: m+1 m a m+1 j b j j 1 j1 0 m+1 1 m a m+1 b 1 Substitutig bac, we get m+1 a + b m+1 m a m+1 b m + a m+1 b 1 [The easo fo the above maeuves was to mae the powes of a ad b agee so that we ca add the summatios togethe tem by tem, except fo the fist ad the last tems, which we must wite sepaately] 1 cotiued o page 600

9 600 Chapte 9 Coutig ad Pobability Thus a + b m+1 But Hece m 0 a m+1 0 b 0 + a m+1 + a + b m+1 a m+1 + m [ ] m m + a m+1 b 1 1 m + a m+1 m+ b m+1 m + 1 [ ] m m + a m+1 b + b m+1 1 [ ] m m m + 1 m + 1 m + 1 a m+ b + b m+1 a m+ b which is what we eeded to show because m sice a 0 b 0 1ad m m 1 0 m by Pascal s fomula m m + 1 It is istuctive to wite out the poduct a + b a + b m without usig the summatio otatio but usig the iductive hypothesis about a + b m : [ a + b m+1 a + b a m m + a m 1 m b + + a m b ] m + a m b m + + ab m 1 + b m m 1 You will see that the fist ad last coefficiets ae clealy 1 ad that the tem cotaiig a m+1 b is obtaied fom multiplyig a m b by a ad a m b 1 by b [because m + 1 m ] Hece the coefficiet of a m+1 b m equals the sum of ad m This is the cux of the algebaic poof If ad ae oegative iteges ad, the is called a biomial coefficiet because it is oe of the coefficiets i the expasio of the biomial expessio a + b The combiatoial poof of the biomial theoem follows Poof of Biomial Theoem combiatoial vesio: [The combiatoial agumet used hee to pove the biomial theoem wos oly fo 1 If we wee givig oly this combiatoial poof, we would have to pove the case 0 sepaately Sice we have aleady give a complete algebaic poof that icludes the case 0, we do ot pove it agai hee] Let a ad b be eal umbes ad a itege that is at least 1 The expessio a + b ca be expaded ito poducts of lettes, whee each lette is eithe a o b

10 97 Pascal s Fomula ad the Biomial Theoem 601 Fo each 0, 1, 2,,, the poduct a b a } a a {{ a } b b b }{{ b } factos factos occus as a tem i the sum the same umbe of times as thee ae odeigs of a s ad b s But this umbe is, the umbe of ways to choose positios ito which to place the b s [The othe positios will be filled by a s] Hece, whe lie tems ae combied, the coefficiet of a b i the sum is Thus This is what was to be poved a + b 0 a b Example 975 Substitutig ito the Biomial Theoem Expad the followig expessios usig the biomial theoem: a a + b 5 b x 4y 4 Solutio a a + b a a 5 b 5 a 5 1 b a b a b a b 4 + b 5 a 5 + 5a 4 b + 10a 3 b a 2 b 3 + 5ab 4 + b 5 b Obseve that x 4y 4 x + 4y 4 Soleta x ad b 4y, ad substitute ito the biomial theoem 4 x 4y 4 4 x 4 4y 0 x x 4 1 4y x y x y 3 + 4y 4 x 4 + 4x 3 4y + 6x 2 16y 2 + 4x 1 64y y 4 x 4 16x 3 y + 96x 2 y 2 256xy y 4 Example 976 Deivig Aothe Combiatoial Idetity fom the Biomial Theoem Use the biomial theoem to show that fo all iteges 0 Solutio Sice , Apply the biomial theoem to this expessio by lettig a 1 ad b 1 The

11 602 Chapte 9 Coutig ad Pobability sice 1 1 ad 1 1 Cosequetly, Example 977 Usig a Combiatoial Agumet to Deive the Idetity Accodig to Theoem 631, a set with elemets has 2 subsets Apply this fact to give a combiatoial agumet to justify the idetity Solutio Suppose S is a set with elemets The evey subset of S has some umbe of elemets, whee is betwee 0 ad It follows that the total umbe of subsets of S, NPS, ca be expessed as the followig sum: umbe of umbe of umbe of umbe of subsets subsets of + subsets of + + subsets of of S size 0 size 1 size Now the umbe of subsets of size of a set with elemets is Hece the umbe of subsets of S But by Theoem 631, S has 2 subsets Hece Example 978 Usig the Biomial Theoem to Simplify a Sum Expess the followig sum i closed fom without usig a summatio symbol ad without usig a ellipsis : 9 Solutio 0 0 Whe the umbe 1 is aised to ay powe, the esult is still 1 Thus by the biomial theoem with a 1adb 9 10 Test Youself 1 If ad ae oegative iteges with, the the elatio betwee ad is 2 Pascal s fomula says that if ad ae positive iteges with, the 3 The cux of the algebaic poof of Pascal s fomula is that to add two factios you eed to expess both of them with a 4 The cux of the combiatoial poof of Pascal s fomula is that the set of subsets of size of a set {x 1, x 2,,x +1 } ca be patitioed ito the set of subsets of size that cotai ad those that 5 The biomial theoem says that give ay eal umbes a ad b ad ay oegative itege, 6 The cux of the algebaic poof of the biomial theoem is that, afte maig a chage of vaiable so that two

12 97 Pascal s Fomula ad the Biomial Theoem 603 summatios have the same lowe ad uppe limits ad the expoets of a ad b ae the same, you use the fact that m m + 1 Execise Set 97 I 1 4, use Theoem 951 to compute the values of the idicated quatities Assume is a itege 1,fo 0 2,fo ,fo 2 4,fo 3 2 3, 5 Use Theoem 951 to pove algebaically that fo iteges ad with 0 This ca be doe by diect calculatio; it is ot ecessay to use mathematical iductio Justify the equatios i 6 9 eithe by deivig them fom fomulas i Example 971 o by diect computatio fom Theoem 951 Assume m,,, ad ae iteges m + 6 m +,fom + 1 m ,fo 1 2 1, fo 0 fo 0 10 a Use Pascal s tiagle give i Table 971 to compute the values of,,,ad b Use the esult of pat a ad Pascal s fomula to compute 7 7 3,,ad c Complete the ow of Pascal s tiagle that coespods to 7 11 The ow of Pascal s tiagle that coespods to 8isas follows: What is the ow that coespods to 9? 12 Use Pascal s fomula epeatedly to deive a fomula fo +3 i tems of values of with Assume ad ae iteges with 3 13 Use Pascal s fomula to pove by mathematical iductio that if is a itege ad 1, the +1 i 2 i The cux of the combiatoial poof of the biomial theoem is that the umbe of ways to aage b s ad a s i ode is H 14 Pove that if is a itege ad 1, the i Pove the followig geealizatio of execise 13: Let be a fixed oegative itege Fo all iteges with, i Thi of a set with m + elemets as composed of two pats, oe with m elemets ad the othe with elemets Give a combiatoial agumet to show that m + m m m + + +, whee m ad ae positive iteges ad is a itege that is less tha o equal to both m ad This idetity gives ise to may useful additioal idetities ivolvig the quatities Because Alexade Vade- mode published a ifluetial aticle about it i 1772, it is geeally called the Vademode covolutio Howeve, it was ow at least i the 1300s i Chia by Chu Shih-chieh H 17 Pove that fo all iteges 0, Let m be ay oegative itege Use mathematical iductio ad Pascal s fomula to pove that fo all iteges 0, m m + 1 m + m Use the biomial theoem to expad the expessios i x 7 20 p + q x 6 22 u v 5 23 p 2q 4 24 u 2 3v 4 25 x x a a 5 27 x x 28 I Example 975 it was show that a + b 5 a 5 + 5a 4 b + 10a 3 b a 2 b 3 + 5ab 4 + b 35 Evaluate a + b 6 by substitutig the expessio above ito the equatio a + b 6 a + ba + b 5 ad the multiplyig out ad combiig lie tems

13 604 Chapte 9 Coutig ad Pobability I 29 34, fid the coefficiet of the give tem whe the expessio is expaded by the biomial theoem 29 x 6 y 3 i x + y 9 30 x 7 i 2x a 5 b 7 i a 2b u 16 v 4 i u 2 v p 16 q 7 i 3p 2 2q x 9 y 10 i 2x 3y As i the poof of the biomial theoem, tasfom the summatio m a m b+1 0 by maig the chage of vaiable j + 1 Use the biomial theoem to pove each statemet i Fo all iteges 1, Hit: Use the fact that H 37 Fo all iteges 0, m 38 Fo all iteges m 0, i 2 m i 1 i 39 Fo all iteges 0, i0 i 3 i 2 i i0 40 Fo all iteges 0 ad fo all oegative eal umbes x, 1 + x 1 + x H 41 Fo all iteges 1, if is eve if is odd Use mathematical iductio to pove that fo all iteges 1, if S is a set with elemets, the S has the same umbe of subsets with a eve umbe of elemets as with a odd umbe of elemets Use this fact to give a combiatoial agumet to justify the idetity of execise 36 Expess each of the sums i i closed fom without usig a summatio symbol ad without usig a ellipsis m i i 0 i0 m 45 x i 46 2 m x i i0 2 j0 i0 j 2 j x j x 2 m 1 p m i q 2i 50 i 2 0 m 1 i i 2 i 0 i 5 i 2 i i i0 i0 Fo studets who have studied calculus a Explai how the equatio below follows fom the biomial theoem: 1 + x 0 0 x b Wite the fomula obtaied by taig the deivative of both sides of the equatio i pat a with espect to x c Use the esult of pat b to deive the fomulas below i [ ] ii 0 1 d Expess 3 i closed fom without usig a 1 summatio sig o ellipsis Aswes fo Test Youself a + b a b 6 0 m commo deomiato 4 x +1 ; do ot cotai x +1