Math 475, Problem Set #6: Solutions


 Quentin Clyde Spencer
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1 Math 475, Problem Set #6: Solutios A (a) For each poit (a, b) with a, b oegative itegers satisfyig ab 8, cout the paths from (0,0) to (a, b) where the legal steps from (i, j) are to (i 2, j), (i, j 2), ad (i, j ) I ll do this i the first quadrat usig i ad j as x ad ycoordiates (though you could use the i ad j as rowidex ad columidex; you d get differet pictures but the same umbers) (b) Compute the coefficiets of (x 2 xy y 2 ) for 0,, 2, 3, 4 (x 2 xy y 2 ) 0, (x 2 xy y 2 ) x 2 xy y 2, (x 2 xy y 2 ) 2 x 4 2x 3 y 3x 2 y 2 2xy 3 y 4, (x 2 xy y 2 ) 3 x 6 3x 5 y 6x 4 y 2 7x 3 y 3 6x 2 y 4 3xy 5 y 6, ad (x 2 xy y 2 ) 4 x 8 4x 7 y 0x 6 y 2 6x 5 y 3 9x 4 y 4 6x 3 y 5 0x 2 y 6 4xy 7 y 8 (c) Based o parts (a) ad (b), formulate a precise cojecture of the form for all oegative itegers a ad b, the umber of paths from (0, 0) to (a, b) is equal to the coefficiet of i the polyomial The umber of paths from (0, 0) to (a, b) is equal to the coefficiet of x a y b i (x 2 xy y 2 ) (ab)/2 (This assumes that a b is eve; if a b is odd, the umber of paths from (0, 0) to (a, b) is zero) You ca thik of the three terms of x 2 xy y 2 as represetig the three possible steps you re allowed to take: x 2 x 2 y 0 correspods to movig 2 steps to the right, xy x y correspods to movig step to the right ad step up, ad y 2 x 0 y 2 correspods to movig 2 steps upward The terms i the expasio of (x 2 xy y 2 ) m correspod to all the places (a, b)
2 you ca get to usig m steps of the allowed kids, ad the coefficiet of x a y b i this expasio is the umber of ways to get there B Chapter 5, problem 2 Algebraic proof: Rewritig the expressio as ( ), we see that it is the coefficiet of x i the product of x x x 2 ( ) 0 2 ad x 0 x x 2 2 But by applyig the biomial theorem to the two factors, we see that this polyomial is equal to (x ) (x ) (x 2 ) All the terms i this sum are of eve degree, so if is odd, the coefficiet of x i the polyomial vaishes O the other had, if is eve (say 2k), the, settig y x 2, we see that the coefficiet of x i (x 2 ) is the same as the coefficiet of y /2 i (y ), which is ( ) /2 /2 (Remark: Eve if you did t fid this, you could have still gotte partial credit if you oticed that, whe is odd, the terms of the alteratig ( sum of the squares of the biomial coefficiets cacel i pairs: 2 0) cacels ( ), 2 cacels, etc) Combiatorial proof: Imagie me ad wome, from whom we wish to choose idividuals to form a committee Call a committee eve if it cotais a eve umber of wome, ad odd otherwise The is the umber of eve committees, ad 3 3
3 is the umber of odd committees We wish to show that the umber of eve commitees mius the umber of odd commitees is 0 whe is odd ad ( ) /2 /2 whe is eve We will do this by pairig up eve committees with odd committees i such a way that whe is odd, there are o upaired committees, while if is eve, there are exactly /2 upaired committees, all of which are eve if /2 is eve ad odd if /2 is odd To accomplish this, first marry off the me ad wome, ad umber the resultig couples from to If a committee cosists completely of married couples, we do t pair it with aother committee; otherwise, we pair it with the committee obtaied by replacig the lowestumbered committtee member whose spouse is ot o the committee by that committeemember s spouse (Here lowestumbered could mea with the lowest social security umber, or aythig else that lets us break ties) This pairig clearly pairs odd committees with eve committees If is odd, there are o upaired committees If is eve, the there are /2 upaired committees, each of which is either eve or odd accordig to the parity of /2 C Solve Brualdi, Chapter 5, problem 8 i two differet ways: oce usig problem 6 as a model, ad oce usig problem 7 as a model First method: Itegratig k0 x k k ( x) from x 0, we get ) k0 k xk( k (( x) ), ad settig x, we get ) k0 ( ) k( k k (0 ) Multiplyig by, we get ) k0 ( ) k( k k Secod method: ( ) [ ] ( ) 2 3 0
4 D What is the coefficiet of x 3 x 3 2x 3 x 2 4 i the expasio of (x x 2 2x 3 2x 4 ) 9? The relevat term of the multiomial theorem is equal to the product of 9 9! !3!!2! 5040 ad (x ) 3 ( x 2 ) 3 (2x 3 ) ( 2x 4 ) 2 8x 2 x 3 2x 3x 2 4; the total coefficiet is thus (5040) ( 8) E Brualdi, Chapter 5, problem 46 Retai all terms that are greater tha 0 3 ; discard the rest Let x 2 ad y 8 Sice 0 2 8, Theorem 56 applies: (2 8) /3 / /3 /3 2 8 /3 / /3 2 / /3 3 ) /3 4 ()(2)(/3)(2 )( (/3)( 2/3) )(2 3 )( (/3)( 2/3)( 5/3) 2! ( /3 4 ( (/3)( 2/3)( 5/3)( 8/3) 4! )(2 7 ) 2/6 /725/2592 5/5552 These terms decrease i absolute value ad alterate i sig (leavig aside the first two terms), ad the fifth term is less tha 0 3, so we may approximate the ifiite sum to withi 0 3 by takig just the first four terms: 2 /6 /72 5/ / (The true value is 2544 ) F Fix positive itegers, k 3 Cosider a covex go with vertices labelled through Call a covex kgo, whose vertices are a subset of the vertices of the go, a iteral kgo if all of its sides are diagoals of the go (a) How may iteral kgos are there cotaiig the vertex labelled? The iteral kgos cotaiig vertex are i  correspodece with the ktuples (x, x 2,, x k ) satisfyig x x 2 x k k, where each x i is a positive iteger (Thik of x i as the umber of vertices of the go that lie betwee successive vertices of the k go) Lettig y i x i, we see that these i tur correspod to k tuples (y, y 2,, x k ) satisfyig y y 2 y k 2k where each y i is a oegative iteger Applyig oeadstars i the usual 3! )(2 5 )
5 way we see that the umber of such ktuples is ( 2k)(k ) k ) ( k k (b) How may iteral kgos are there all together? (Hit: What do you kow ahead of time about the ratio betwee the aswer to (a) ad the aswer to (b)?) Claim (a) is true about ay particular vertex, ad ot just the vertex labelled So if we multiply k k by (the umber of vertices) ad divide by k (to take ito accout the fact that each kgo has k differet vertices), we get the correct aswer, amely k k k ( k )! k!( 2k)! Alterative aalysis: Let s cout, i two differet ways, the umber of ways to draw a iteral kgo ad pick a vertex of that kgo ad color it blue O the oe had, the aswer is N b times k, where N b is the aswer to problem (b), sice there are N b possible iteral kgos, each of which has k vertices that are available to be colored blue O the other had, you could choose the blue vertex first There are vertices of the origial go that could be colored blue For each choice of the blue vertex, there are N a iteral kgos that cotai that vertex, where N a is the aswer to part (a) (We proved this i the case where the special vertex is vertex, but there s othig special about the vertex that we happeed to call vertex ; the aswer is N a for ay particular vertex) Hece N b k N a, from which it follows that N b (/k)n a
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