5. Recurrences. The recursive denition of the Fibonacci numbers is well-known: if F n is the n th Fibonacci number, then

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1 5. Recurreces The recursive deitio of the Fiboacci umbers is well-kow: if F is the th Fiboacci umber, the F 0 = 0, F 1 = 1, F + = F +1 + F, if 0. We are iterested i a explicit form of the umbers F for all atural umbers. Actually, the problem is to solve a equatio where the ukow is give recursively, i which case the equatio is called a recurrece equatio. The solutio ca be cosidered as a fuctio over atural umbers, because F is deed for all. Such recurrece equatios are also kow as differece equatios, but could be amed as discrete differetial equatios for their similarities to differetial equatios. Deitio 5.1 A k th th order recurrece equatio, (k 1) is a equatio of the form where x must be give i a explicit form. f (x, x +1,...,, x +k ) = 0, 0, (5.1) For a uique determiatio of x, k iitial values must be give. Usually these values are x 0, x 1,..., x k 1. These ca be cosidered as iitial coditios. I the case of the equatio for Fiboacci-umbers, which is of secod order, two iitial values must be give. The sequece x = g() satisfyig equatio (5.1) ad the correspodig iitial coditios is called a particular solutio. If all particular solutios of equatio (5.1) ca be obtaied from the sequece x = h(, C 1, C,..., C k ), by adequately choosig of the costats C 1, C,..., C k, the this sequece x is a geeral solutio. Solvig recurrece equatios is ot a easy task. I the chapter we will discuss methods which ca be used i special cases. For simplicity of writig we will use the otatio x istead of x() as it appears i several books (sequeces ca be cosidered as fuctios over atural umbers). The chapter is divided ito three sectios. I sectio 5.1 we deal with solvig liear recurrece equatios, i sectio 5. with geeratig fuctios ad their use i solvig recurrece equatios ad i sectio 5.3 we focus our attetio o umerical solutio of recurrece equatios.

2 5.1. Liear recurrece equatios 07 If the recurrece equatio is of the form 5.1. Liear recurrece equatios f 0 ()x + f 1 ()x f k ()x +k = f (), 0, where f, f 0, f 1,..., f k are fuctios deed over atural umbers, f 0, f k 0, ad x must be give explicitly, the the recurrece equatio is liear. If f is the zero fuctio, the the equatio is homogeeous, otherwise ohomogeeous. If all the fuctios f 0, f 1,..., f k are costat, the equatio is called a liear recurrece equatio with costat coefficiets Liear homogeeous equatios with costat coefficiets Let the equatio be a 0 x + a 1 x a k x +k = 0, k, (5.) where a 0, a 1,..., a k are real costats, a 0, a k 0, k 1. If k iitial coditios are give (usually x 0, x 1,..., x k 1 ), the the geeral solutio of this equatio ca be uiquely give. To solve the equatio let us cosider its characteristic equatio a 0 + a 1 r + + a k 1 r k 1 + a k r k = 0, (5.3) a polyomial equatio with real coefficiets. This equatio has k roots i the eld of complex umbers. It ca easily be see after a simple substitutio that if r 0 is a real solutio of the characteristic equatio, the C 0 r 0 is a solutio of (5.), for arbitrary C 0. The geeral solutio of equatio (5.) is x = C 1 x (1) + C x () + + C k x (k), where x (i) (i = 1,,..., k) are the liearly idepedet solutios of equatio (5.). The costats C 1, C,..., C k ca be determied from the iitial coditios by solvig a system of k equatios. The liearly idepedet solutios are supplied by the roots of the characteristic equatio by the followig way. A fudametal solutio of equatio (5.) ca be associated with each root of the characteristic equatio. Let us cosider the followig cases. Distict real roots Let r 1, r,..., r p be distict real roots of the characteristic equatio. The are solutios of equatio (5.), ad r 1, r,..., r p C 1 r 1 + C r + + C pr p (5.4) is also a solutio, for arbitrary costats C 1, C,..., C p. If p = k, the (5.4) is the geeral solutio of the recurrece equatio.

3 08 5. Recurreces Example 5.1 Solve the recurrece equatio The correspodig characteristic equatio is with the solutios x + = x +1 + x, x 0 = 0, x 1 = 1. r r 1 = 0, r 1 = 1 + 5, r = 1 5. These are distict real solutios, so the geeral solutio of the equatio is x = C C 1 5. The costats C 1 ad C ca be determied usig the iitial coditios. From x 0 = 0, x 1 = 1 the followig system of equatios ca be obtaied. C C 1 + C = 0, C = 1. The solutio of this system of equatios is C 1 = 1/ 5, C = 1/ 5. Therefore the geeral solutio is x = , 5 5 which is the th Fiboacci umber F. Multiple real roots Let r be a real root of the characteristic equatio with multiplicity p. The r, r, r,..., p 1 r are solutios of equatio (5.) (fudametal solutios correspodig to r), ad ( C0 + C 1 + C + + C p 1 p 1) r (5.5) is also a solutio, for ay costats C 0, C 1,..., C p 1. If the characteristic equatio has o other solutios, the (5.5) is a geeral solutio of the recurrece equatio. Example 5. Solve the recurrece equatio The characteristic equatio is with r = a solutio with multiplicity. The is a geeral solutio of the recurrece equatio. x + = 4x +1 4x, x 0 = 1, x 1 = 3. r 4r + 4 = 0, x = (C 0 + C 1 )

4 5.1. Liear recurrece equatios 09 From the iitial coditios we have C 0 = 1, C 0 + C 1 = 3. From this system of equatios C 0 = 1, C 1 = 1/, so the geeral solutio is x = (1 + 1 ) or x = ( + ) 1. Distict complex roots If the complex umber a(cos b + i si b), writte i trigoometric form, is a root of the characteristic equatio, the its cojugate a(cos b i si b) is also a root, because the coefficiets of the characteristic equatio are real umbers. The are solutios of equatio (5.) ad a cos b ad a si b C 1 a cos b + C a si b (5.6) is also a solutio, for ay costats C 1 ad C. If these are the oly solutios of the characteristic equatio, the (5.6) is a geeral solutio. Example 5.3 Solve the recurrece equatio The correspodig characteristic equatio is x + = x +1 x, x 0 = 0, x 1 = 1. r r + = 0, with roots 1 + i ad 1 i. These ca be writte i trigoometric form as (cos(π/4) + i si(π/4)) ad (cos(π/4) i si(π/4)). Therefore x = C 1 ( ) cos π 4 + C ( ) si π 4 is a geeral solutio of the recurrece equatio. From the iitial coditios C 1 = 0, π C 1 cos 4 + C π si = 1. 4 Therefore C 1 = 0, C = 1. Hece the geeral solutio is x = ( ) si π 4.

5 10 5. Recurreces Multiple complex roots If the complex umber writte i trigoometric form as a(cos b + i si b) is a root of the characteristic equatio with multiplicity p, the its cojugate a(cos b i si b) is also a root with multiplicity p. The a cos b, a cos b,..., p 1 a cos b ad a si b, a si b,..., p 1 a si b are solutios of the recurrece equatio (5.). The (C 0 + C C p 1 p 1 )a cos b + (D 0 + D D p 1 p 1 )a si b is also a solutio, where C 0, C 1,..., C p 1, D 0, D 1,..., D p 1 are arbitrary costats, which ca be determied from the iitial coditios. This solutio is geeral if the characteristic equatio has o other roots. Example 5.4 Solve the recurrece equatio The characteristic equatio is x +4 + x + + x = 0, x 0 = 0, x 1 = 1, x =, x 3 = 3. r 4 + r + 1 = 0, which ca be writte as (r + 1) = 0. The complex umbers i ad i are double roots. The trigoometric form of these are i = cos π + i si π, ad i = cos π i si π respectively. Therefore the geeral solutio is x = (C 0 + C 1 ) cos π + (D 0 + D 1 ) si π. From the iitial coditios we obtai C 0 = 0, (C 0 + C 1 ) cos π + (D 0 + D 1 ) si π = 1, (C 0 + C 1 ) cos π + (D 0 + D 1 ) si π =, (C 0 + 3C 1 ) cos 3π + (D 0 + 3D 1 ) si 3π = 3, that is C 0 = 0, D 0 + D 1 = 1, C 1 =, D 0 3D 1 = 3. Solvig this system of equatios C 0 = 0, C 1 = 1, D 0 = 3 ad D 1 =. Thus the geeral solutio is x = (3 ) si π cos π.

6 5.1. Liear recurrece equatios 11 Usig these four cases all liear homogeeous equatios with costat coefficiets ca be solved, if we ca solve their characteristic equatios. Example 5.5 Solve the recurrece equatio The characteristic equatio is x +3 = 4x + 6x x, x 0 = 0, x 1 = 1, x = 1. r 3 4r + 6r 4 = 0, with roots, 1 + i ad 1 i. Therefore the geeral solutio is After determiig the costats we obtai ( ) x = 1 + x = C 1 + C ( ) cos π 4 + C 3( ) si π 4. ( cos π si π ). 4 The geeral solutio The characteristic equatio of the kth order liear homogeeous equatio (5.) has k roots i th eld of complex umbers, which are ot ecessarily distict. Let these roots be the followig: r 1 real, with multiplicity p 1 (p 1 1), r real, with multiplicity p (p 1),... r t real, with multiplicity p t (p t 1), s 1 = a 1 (cos b 1 + i si b 1 ) complex, with multiplicity q 1 (q 1 1), s = a (cos b + i si b ) complex, with multiplicity q (q 1),... s m = a m (cos b m + i si b m ) complex, with multiplicity q m (q m 1). Sice the equatio has k roots, p 1 + p + + p t + (q 1 + q + + q m ) = k. I this case the geeral solutio of equatio (5.) is where C ( j) 0 D (l) j), C( 1 x = + +,..., C( j) p j 1 t ( C ( j) 0 + C ( j) ) j) C( p j 1 p j 1 r j j=1 m j=1 ( D ( j) 0 j) + D( ) j) D( q j 1 q j 1 a j cos b j m ( E ( j) 0 + E ( j) ) j) E( q j 1 q j 1 a j si b j, (5.7) j=1, j = 1,,..., t,, l = 1,,..., m are costats, which ca be 0, E(l) 0, D(l) 1, E(l) 1,..., D(l) p l 1, E(l) p l 1 determied from the iitial coditios. The above statemets ca be summarised i the followig theorem.

7 1 5. Recurreces Theorem 5.1 Let k 1 be a iteger ad a 0, a 1,..., a k real umbers with a 0, a k 0. The geeral solutio of the liear recurrece equatio (5.) ca be obtaied as a liear combiatio of the terms j r i, where r i are the roots of the characteristic equatio (5.3) with multiplicity p i (0 j < p i ) ad the coefficiets of the liear combiatio deped o the iitial coditios. The proof of the theorem is left to the reader (see exercise ). The algorithm for the geeral solutio is the followig. LINEAR-HOMOGENEOUS 1 determie the characteristic equatio of the recurrece equatio d all roots of the characteristic equatio with their multiplicities 3 d the geeral solutio (5.7) based o the roots 4 determie the costats of (5.7) usig the iitial coditios, if these exists Liear ohomogeeous recurrece equatios with costat coefficiets Cosider the liear ohomogeeous recurrece equatio with costat coefficiets a 0 x + a 1 x a k x +k = f (), (5.8) where a 0, a 1,..., a k are real costats, a 0, a k 0, k 1, ad f is ot the zero fuctio. The correspodig liear homogeeous equatio (5.) ca be solved usig Theorem 5.1. If a particular solutio of equatio (5.8) is kow, the equatio (5.8) ca be solved. Theorem 5. Let k 1 be a iteger, a 0, a 1,..., a k real umbers, a 0, a k 0. If x (1) is a particular solutio of the liear ohomogeeous equatio (5.8) ad x (0) is a geeral solutio of the liear homogeeous equatio (5.), the is a geeral solutio of the equatio (5.8). x = x (0) + x (1) The proof of the theorem is left to the reader (see exercise ). Example 5.6 Solve the recurrece equatio First we solve the homogeeous equatio ad obtai the geeral solutio x + + x +1 x =, x 0 = 0, x 1 = 1. x + + x +1 x = 0, x (0) = C 1 ( ) + C, sice the roots of the characteristic equatio are ad 1. It is easy to see that x = C 1 ( ) + C +

8 5.1. Liear recurrece equatios 13 f () x (1) p a (C 0 + C C p p )a a p si b (C 0 + C C p p )a si b + (D 0 + D D p p )a cos b a p cos b (C 0 + C C p p )a si b + (D 0 + D D p p )a cos b Figure 5.1. The form of particular solutios. is a solutio of the ohomogeeous equatio. Therefore the geeral solutio is x = 1 4 ( ) + or x = ( ), 4 The costats C 1 ad C ca be determied usig the iitial coditios. Thus, that is { 0, if is eve, x = 1, if is odd. A particular solutio ca be obtaied usig the method of variatio of costats. However, there are cases whe there is a easier way of dig a particular solutio. I gure 5.1 we ca see types of fuctios f (), for which a particular solutio x (1) ca be obtaied i the give form i the table. The costats ca be obtaied by substitutios. I the previous example f () =, so the rst case ca be used with a = ad p = 0. Therefore we try to d a particular solutio of the form C 0. After substitutio we obtai C 0 = 1/4, thus the particular solutio is x (1) =. Exercises Solve the recurrece equatio H = H 1 + 1, ha 1, és H 0 = 0. (Here H is the optimal umber of moves i the problem of Towers of Haoi.) 5.1- Aalyse the problem of Towers of Haoi if discs have to be moved from stick A to stick C i such a way that o disc ca be moved directly from A to C ad vice versa. Hit. Show that if the optimal umber of moves is deoted by M, ad 1, the M = 3M 1 +.

9 14 5. Recurreces Solve the recurrece equatio ( + 1)R = ( 1)R 1, ha 1, és R 0 = Solve the liear ohomogeeous recurrece equatio x = + x 1, ha, és x 1 = 0. Hit. Try to d a particular solutio of the form C 1 + C Prove Theorem Prove Theorem Geeratig fuctios ad recurrece equatios Geeratig fuctios ca be used, amog others, to solve recurrece equatios, cout objects (e.g. biary trees), prove idetities ad solve partitio problems. Coutig the umber of objects ca be doe by statig ad solvig recurrece equatios. These equatios are usually ot liear, ad geeratig fuctios ca help us i solvig them Defiitio ad operatios Associate a series with the iite sequece (a ) = a 0, a 1, a,..., a,... the followig way A(z) = a 0 + a 1 z + a z + + a z + = a z. This is called the geeratig fuctio of the sequece (a ). For example, i the case of the Fiboacci umbers this geeratig fuctio is F(z) = F z = z + z + z 3 + 3z 4 + 5z 5 + 8z z 7 +. Multiplyig both sides of the equatio by z, the by z, we obtai F(z) = F 0 + F 1 z + F z + F 3 z F z +, zf(z) = F 0 z + F 1 z + F z F 1 z +, z F(z) = F 0 z + F 1 z F z +. If we subtract the secod ad the third equatio from the rst oe term by term, the use the deig formula of the Fiboacci umbers, we get F(z)(1 z z ) = z, that is F(z) = z 1 z z. (5.9)

10 5.. Geeratig fuctios ad recurrece equatios 15 The correctess of these operatios ca be proved mathematically, but here we do ot wat to go ito details. The formulae obtaied usig geeratig fuctios ca usually also be proved usig other methods. Let us cosider the followig geeratig fuctios A(z) = a z ad B(z) = b z. The geeratig fuctios A(z) ad B(z) are equal, if ad oly if a = b for all atural umbers. Now we dee the followig operatios with the geeratig fuctios: additio, multiplicatio by real umber, shift, multiplicatio, derivatio ad itegratio. Additio ad multiplicatio by real umber αa(z) + βb(z) = (αa + βb )z. Shift The geeratig fuctio z k A(z) = a z +k = a k z represets the sequece < 0, 0,..., 0, a } {{ } 0, a 1,... >, while the geeratig fuctio k 1 z (A(z) a k 0 a 1 z a z a k 1 z k 1 ) = a z k = a k+ z represets the sequece < a k, a k+1, a k+,... >. Example 5.7 Let A(z) = 1 + z + z +. The 1 ( ) 1 A(z) 1 = A(z) ad A(z) = z 1 z. k k Multiplicatio If A(z) ad B(z) are geeratig fuctios, the where s = A(z)B(z) = (a 0 + a 1 z + + a z + )(b 0 + b 1 z + + b z + ) a k b k. k=0 = a 0 b 0 + (a 0 b 1 + a 1 b 0 )z + (a 0 b + a 1 b 1 + a b 0 )z + = s z, Special case. If b = 1 for all atural umbers, the 1 A(z) 1 z = a k z. (5.10) k=0

11 16 5. Recurreces If, i additio, a = 1 for all, the 1 (1 z) = ( + 1)z. (5.11) Derivatio A (z) = a 1 + a z + 3a 3 z + = ( + 1)a +1 z. Example 5.8 After differetiatig the both sides of the geeratig fuctio we obtai A(z) = z = 1 1 z, 1 A (z) = z 1 1 = (1 z). Itegratio z 0 A(t)dt = a 0 z + 1 a 1z a z = a 1z. 1 Example 5.9 Let After itegratig both sides we get l 1 1 z = 1 + z + z + z z = z + 1 z z3 + = z. 1 Multiplyig the above geeratig fuctios we obtai 1 1 z l 1 1 z = H z, 1 where H = (H 0 = 0, H 1 = 1) are the so-called harmoic umbers. Chagig the argumets Let A(z) = a z represet the sequece < a 0, a 1, a,... >, the A(cz) = c a z represets the sequece < a 0, ca 1, c a,... c a,... >. The followig statemets holds 1( ) A(z) + A( z) = a0 + a z + + a z +, 1( ) A(z) A( z) = a1 z + a 3 z a 1 z 1 +.

12 5.. Geeratig fuctios ad recurrece equatios 17 Example 5.10 Let A(z) = 1 + z + z + z 3 + = 1 1 z. The 1 + z + z 4 + = 1 ( ) 1 A(z) + A( z) = ( 1 1 z z ) = 1 1 z, which ca also be obtaied by substitutig z with z i A(z). We ca obtai the sum of the odd power terms the same way, z + z 3 + z 5 + = 1 ( ( ) 1 1 A(z) A( z) = 1 z 1 ) z = 1 + z 1 z. Usig geeratig fuctios we ca obtai iterestig formulae. For example, let A(z) = 1/(1 z) = 1 + z + z + z 3 +. The za ( z(1 + z) ) = F(z), which is the geeratig fuctio of the Fiboacci umbers. From this za ( z(1 + z) ) = z + z (1 + z) + z 3 (1 + z) + z 4 (1 + z) 3 +. The coefficiet of z +1 o the left-had side is F +1, that is the ( + 1)th Fiboacci umber, while the coefficiet of z +1 o the right-had side is ( ) k, k k 0 after usig the biomial formula i each term. Hece ( ) k +1 ( ) k F +1 = =. (5.1) k k k 0 Remember that the biomial formula ca be geeralised for all real r, amely ( ) r (1 + z) r = z, ( ) r which is the geeratig fuctio of the biomial coefficiets. Here is a geeralisatio of the combiatios for ay real umber r, that is ( ) r(r 1)(r )... (r + 1), if > 0, r ( 1)... 1 = 1, = 0, 0, if < 0. We ca obtai useful formulae usig this geeralisatio for egative r. Let 1 ( ) m (1 z) m = (1 z) m = ( z) k. k Sice, by a simple computatio, we get ( ) ( ) m m + k 1 = ( 1) k, k k k=0 k 0

13 18 5. Recurreces the followig formula ca be obtaied The ad z m (1 z) = m+1 where m is a atural umber. k 0 1 (1 z) = m+1 ( m + k k k 0 k 0 ) z m+k = ( m + k k ) z k. ( m + k m k 0 ) z m+k = k 0 ( ) k z k, m ( ) k z k z m =, (5.13) m (1 z) m Solvig recurrece equatios with geeratig fuctios If the geeratig fuctio of the geeral solutio of a recurrece equatio to be solved ca be expaded i such a way that the coefficiets are i closed form, the this method is successful. Let the recurrece equatio be To solve it, let us cosider the geeratig fuctio X(z) = x z. F(x, x 1,..., x k ) = 0. (5.14) If (5.14) ca be writte as G ( X(z) ) = 0 ad ca be solved for X(z), the X(z) ca be expaded ito series i such a way that x ca be writte i closed form, equatio (5.14) ca be solved. Now we give a geeral method for solvig liear ohomogeeous recurrece equatios. After this we give three examples for the oliear case. I the rst two examples the umber of elemets i some sets of biary trees, while i the third example the umber of leaves of biary trees is computed. The correspodig recurrece equatios (5.15), (5.17) ad (5.18) will be solved usig geeratig fuctios. Liear ohomogeeous recurrece equatios with costat coefficiets Multiply both sides of equatio (5.8) by z. The a 0 x z + a 1 x +1 z + + a k x +k z = f ()z. Summig up both sides of the equatio term by term we get a 0 x z + a 1 x +1 z + + a k x +k z = f ()z. The a 0 x z + a 1 z x +1 z a k x z k +k z +k = f ()z.

14 5.. Geeratig fuctios ad recurrece equatios 19 Let The equatio ca be writte as a 0 X(z) + a 1 z X(z) = x z ad F(z) = f ()z. ( X(z) x0 ) + + a k z k ( X(z) x0 x 1 z x k 1 z k 1) = F(z). This ca be solved for X(z). If X(z) is a ratioal fractio, the it ca be decomposed ito partial (elemetary) fractios which, after expadig them ito series, will give us the geeral solutio x of the origial recurrece equatio. We ca also try to use the expasio ito series i the case whe the fuctio is ot a ratioal fractio. Example 5.11 Solve the followig equatio usig the above method x +1 x = +1, ha 0 és x 0 = 0. After multiplyig ad summig we have 1 x +1 z +1 x z = z z, z ad 1 ( ) X(z) x0 X(z) = z 1 z 1 z. Sice x 0 = 0, after decomposig the right-had side ito partial fractios 1 ), the solutio of the equatio is z X(z) = (1 z) + 1 z 1 z. After differetiatig the geeratig fuctio 1 1 z = z term by term we get Thus therefore (1 z) = z 1. ( X(z) = z + z z = ( ) + ) z, 1 x = ( ) +. The umber of biary trees Let us deote by b the umber of biary trees with vertices. The b 1 = 1, b =, b 3 = 5 (see gure 5.). Let b 0 = 1. (We will see later that this is a good choice.) I a biary tree with vertices, with the exceptio of the root, there are altogether 1 vertices i the left ad right subtrees. If the left subtree has k vertices ad the right subtree has 1 k vertices, the there exists b k b 1 k such biary trees. Summig over 1 For decomposig the fractio ito partial fractios we ca use the Udetermied Coefficiets Method.

15 0 5. Recurreces = = 3 Figure 5.. Biary trees with two ad three vertices. k = 0, 1,..., 1, we obtai exactly the umber of biary trees, b. Thus for ay atural umber 1 the recurrece equatio i b is b = b 0 b 1 + b 1 b + + b 1 b 0. (5.15) This ca also be writte as 1 b = b k b 1 k. k=0 Multiplyig both sides by z, the summig over all, we obtai Let B(z) = 1 b z = b k b 1 k z. (5.16) 1 1 k=0 b z be the geeratig fuctio of the umbers b. The left-had side of (5.16) is exactly B(z) 1 (because b 0 = 1). The right-had side looks like a product of two geeratig fuctios. To see which fuctios are i cosideratio, let us use the otatio A(z) = zb(z) = b z +1 = b 1 z. The the right-had side of (5.16) is exactly A(z)B(z), which is zb (z). Therefore 1 B(z) 1 = zb (z), B(0) = 1. Solvig this equatio for B(z) gives B(z) = 1 ± 1 4z z.

16 5.. Geeratig fuctios ad recurrece equatios 1 We have to choose the egative sig because B(0) = 1. Thus B(z) = = = = = 1 ( ) 1 ( ) 1 1 4z = 1 (1 4z) 1/ z z 1 ( ) 1/ z 1 ( 4z) = 1 ( ) 1/ z 1 ( 1) z ( ) 1 1/ 0 z z 0 ( ) 1/ ( 0 z + z 1/ )( 1) 1 z z + z ( ) ( ) ( ) 1/ 1/ 1/ 3 z + ( 1) 1 z ( ) 1/ ( ) ( 1) +1 z 1 = z Therefore b = 1 ( ). The umbers b are also called the Catala umbers. + 1 Remark. I the previous computatio we used the followig formula that ca be proved easily ( ) 1/ ( 1) ( ) = ( + 1) The umber of leaves of all biary trees of vertices Let us cout the umber of leaves (vertices with degree 1) i the set of all biary trees of vertices. Deote this umber by f. We remark that the root is ot cosidered leaf eve if it is of degree 1. It is easy to see that f =, f 3 = 6. Let f 0 = 0 ad f 1 = 1, covetioally. Later we will see that these values are good. As i the case of umberig the biary trees, cosider the biary trees of vertices havig k vertices i the left subtree ad k 1 vertices i the right subtree. There are b k such left subtrees ad b 1 k right subtrees. If we cosider such a left subtree ad all such right subtrees, the together there are f 1 k leaves i the right subtrees. So for a give k there are b 1 k f k + b k f 1 k leaves. After summig we have By a easy computatio we get 1 f = ( f k b 1 k + b k f 1 k ). k=0 f = ( f 0 b 1 + f 1 b + + f 1 b 0 ),. (5.17) This is a recurrece equatio, the solutio of which is f. Let F(z) = f z ad B(z) = b z. Multiplyig both sides of (5.17) by z ad summig gives 1 f z = f k b 1 k z. k=0

17 5. Recurreces Sice f 0 = 0 ad f 1 = 1, Thus ad sice F(z) z = zf(z)b(z). F(z) = B(z) = 1 z z 1 zb(z), ( 1 1 4z ), we have F(z) = After the computatios z = z(1 4z) 1/ = z 1 4z ( ) F(z) = z +1 = 1 ( 1/ ( ) z, 1 ) ( 4z). ad f = ( ) 1 or f +1 = ( ) = ( + 1)b. The umber of biary trees with vertices ad k leaves A bit harder problem: how may biary trees are there with vertices ad k leaves? Let us deote this umber by b (k). It is easy to see that b (k) = 0, if k > ( + 1)/. By a simple reasoig the case k = 1 ca be solved. The result is b (1) = 1 for ay atural umber 1. Let b (0) 0 = 1, covetioally. We will see later that this is a good choice. Let us cosider, as i the case of previous problems, the left ad right subtrees. If the left subtree has i vertices ad j leaves, the the right subtree has i 1 vertices ad k j leaves. The umber of these trees is b ( j) (k j). Summig over k ad j gives i b i 1 b (k) = b (k) 1 + k 1 b ( j) (k j) i b i=1 j=1 For solvig this recurrece equatio the geeratig fuctio B (k) (z) = b (k) z, where k 1 i 1. (5.18) will be used. Multiplyig both sides of equatio (5.18) by z ad summig over = 0, 1,,..., we get k 1 b (k) z = b (k) 1 z + b ( j) (k j) i b i 1 z. 1 1 Chagig the order of summatio gives 1 b (k) z = 1 i=1 k 1 b (k) 1 z + 1 j=1 1 j=1 b ( j) (k j) i b i 1 i=1 z.

18 5.. Geeratig fuctios ad recurrece equatios 3 Thus or k 1 B (k) (z) = zb (k) (z) + z B ( j) (z)b (k j) (z) B (k) (z) = z 1 z Step by step, we ca write the followig: B () (z) = B (3) (z) = B (4) (z) = Let us try to d the solutio i the form B (k) (z) = j=1 k 1 B ( j) (z)b (k j) (z). (5.19) j=1 z ( B (1) (z) ), 1 z z (1 z) ( B (1) (z) ) 3, 5z 3 (1 z) 3 ( B (1) (z) ) 4. c k z k 1 (1 z) k 1 ( B (1) (z) ) k, where c = 1, c 3 =, c 4 = 5. Substitutig i (5.19) gives a recursio for the umbers c k k 1 c k = c i c k i. i=1 We solve this equatio usig the geeratig fuctio method. If k =, the c = c 1 c 1, ad so c 1 = 1. Let c 0 = 1. If C(z) = c z is the geeratig fuctio of the umbers c, the, usig the formula of multiplicatio of the geeratig fuctios we obtai thus C(z) 1 z = (C(z) 1) or C (z) 3C(z) + z + = 0, C(z) = 3 1 4z. Sice C(0) = 1, oly the egative sig ca be chose. After expadig the geeratig fuctio we get ( ) z C(z) = 3 1 (1 4z)1/ = = 3 ( ) + 1 z = 1 + ( 1) 1 1 ( 1) ( ) z. From this c = ( ) 1, 1. ( 1)

19 4 5. Recurreces Sice b (1) = 1 for 1, it ca be proved easily that B (1) = z/(1 z). Thus ( ) B (k) 1 k z k 1 (z) = (k 1) k (1 z). k 1 Usig the formula therefore Thus or B (k) (z) = = 1 (1 z) m = ( + m 1 ) z, ( ) 1 k ( ) k + z k+ 1 (k 1) k ( ) 1 k ( ) 1 k+1 z. (k 1) k k + 1 b (k) = b (k) The Z-trasform method k 1 ( )( ) 1 k 1 k 1 k k = 1 ( )( ) k k. k k 1 k Whe solvig liear ohomogeeous equatios usig geeratig fuctios, the solutio is usually doe by the expasio of a ratioal fractio. The Z-trasform method ca help us i expadig such a fuctio. Let P(z)/Q(z) be a ratioal fractio, where the degree of P(z) is less tha the degree of Q(z). If the roots of the deomiator are kow, the ratioal fractio ca be expaded ito partial fractios usig the Udetermied Coefficiet Method. Let us rst cosider the case whe the deomiator has distict roots α 1, α,..., α k. The It is easy to see that But P(z) Q(z) = A 1 z α A i z α i + + A k z α k. A i = lim(z α i ) P(z), i = 1,,..., k. z αi Q(z) A i z α i = A i α i (1 1 ) = A iβ i 1 β i z, z α i where β i = 1/α i. Now, by expadig this partial fractio, we get A i β i 1 β i z = A iβ i ( 1 + βi z + + β i z + ). Deote the coefficiet of z by C i (), the C i () = A i β +1 i, so C i () = A i β +1 i = β +1 i lim(z α i ) P(z) z αi Q(z),

20 5.. Geeratig fuctios ad recurrece equatios 5 or C i () = β +1 i (z α i )P(z) lim. z αi Q(z) After the trasformatio z 1/z ad usig β i = 1/α i we obtai ( ) 1 p(z) C i () = lim (z β i )z, z βi q(z) where p(z) q(z) = P(1/z) Q(1/z). Thus i the expasio of X(z) = P(z) Q(z) the coefficiet of z is C 1 () + C () + + C k (). If α is a root of the polyomial Q(z), the β = 1/α is a root of q(z). E.g. if P(z) Q(z) = z z (1 z)(1 z), the p(z) q(z) = (z 1)(z ). If the root is multiple, e.g. if β i has multiplicity p, the its correspodig i the solutio is 1 C i () = (p 1)! lim d p 1 ( ) (z β z β i dz p 1 i ) p 1 p(z) z. q(z) Here dp f (z) is the derivative of order p of the fuctio f (z). dzp All these ca be summarised i the followig algorithm. Let us cosider that the coefficiets of the equatio are i array A, ad the costats of the solutio are i array C. LINEAR-NONHOMOGENEOUS(A, k, f ) 1 let a 0 x + a 1 x a k x +k = f () be the equatio, where f () is a ratioal fractio; multiply both sides by z, ad sum over all trasform the equatio ito the form X(z) = P(z)/Q(z), where X(z) = x z, P(z) ad Q(z) are polyomials 3 use the trasformatio z 1/z, ad let the result be p(z)/q(z), where p(z) are q(z) are polyomials 4 deote the roots of q(z) by β 1, with multiplicity p 1, p 1 1, β, with multiplicity p, p 1,... β k, with multiplicity p k, p k 1; the the geeral solutio of the origial equatio is x = C 1 () + C () + + C k (), ( where d C i () = 1/((p i 1)!) lim p i 1 z βi dz p i (z βi ) p i z 1 (p(z)/q(z)) ), i = 1,,..., k. 1 5 retur C

21 6 5. Recurreces If we substitute z by 1/z i the geeratig fuctio, the result is the so-called Z- trasform, for which similar operatios ca be deed as for the geeratig fuctios. The residue theorem for the Z-trasform gives the same result. The ame of the method is derived from this observatio. Example 5.1 Solve the recurrece equatio or Thus x +1 x = +1, ha 0, x 0 = 0. Multiplyig both sides by z ad summig we obtai x +1 z x z = +1 z z, 1 z X(z) X(z) = 1 z 1 z, where X(z) = x z. After the trasformatio z 1/z we get X(z) = z (1 z)(1 z). p(z) q(z) = z (z 1)(z ), where the roots of the deomiator are 1 with multiplicity 1 ad with multiplicity. Thus C 1 = lim z 1 ( d z C = lim z dz z 1 Therefore the geeral solutio is z (z ) = ad ) = lim z z 1 (z 1) z (z 1) = ( ). x = ( ) +, 0. Example 5.13 Solve the recurrece equatio so that is The x + = x +1 x, if 0, x 0 = 0, x 1 = 1. Multiplyig by z ad summig gives 1 x z + z + = x +1 z +1 x z, z 1 ( ) F(z) z = F(z) F(z), z z ( 1 F(z) z ) z + = 1 z. F(1/z) = z z z +.

22 5.3. Numerical solutio 7 The roots of the deomiator are 1 + i ad 1 i. Let us compute C 1 () ad C (): Sice z +1 i(1 + i) C 1 () = lim = z 1+i z (1 i) C () = lim z 1 i 1 + i = ( cos π 4 + i si π 4 z +1 ad i(1 i) =. z (1 + i) ), 1 i = ( cos π 4 i si π 4 raisig to the th power gives (1 + i) = ( ) ( cos π 4 + i si π ), (1 i) = ( ) ( cos π 4 4 i si π ), 4 x = C 1 () + C () = ( ) si π 4. ), Exercises 5.-1 How may biary trees are there with vertices ad o empty left ad right subtrees? 5.- How may biary trees are there with vertices, i which each vertex which is ot a leaf, has exactly two descedats? 5.-3 Solve the followig recurret equatio usig geeratig fuctios. H = H 1 + 1, H 0 = 0. (H is the umber of moves i the problem of the Towers of Haoi.) 5.-4 Solve the followig recurret equatio usig the Z-trasform method. F + = F +1 + F + 1, ha 0, és F 0 = 0, F 1 = Solve the followig system of recurrece equatios: u = v 1 + u, v = u + u 1, where u 0 = 1, u 1 =, v 0 = Numerical solutio Usig the followig fuctio we ca solve the liear recurret equatios umerically. The equatio is give i the form a 0 x + a 1 x a k x +k = f (),

23 8 5. Recurreces where a 0, a k 0, k 1. The coefficiets a 0, a 1,..., a k are kept i array A, the iitial values x 0, x 1,..., x k 1 i array X. To d x we will compute step by step the values x k, x k+1,..., x, keepig i the previous k values of the sequece i the rst k positios of X (i.e. i the positios with idices 0, 1,..., k 1). RECURRENCE(A, X, k,, f ) 1 for j k to do v A[0] X[0] 3 for i 1 to k 1 4 do v v + A[i] X[i] 5 v ( f ( j k) v ) /A[k] 6 if j 7 the for i 0 to k 8 do X[i] X[i + 1] 9 X[k 1] v 10 retur v Lies 5 compute the values x j ( j = k, k + 1,..., ) (usig the previous k values), deoted by v i the algorithm. I lies 7 9, if is ot yet reached, we copy the last k values i the rst k positios of X. I lie 10 x is obtaied. It is easy to see that the computatio time is Θ(k), if we do ot cout the time to compute the value of the fuctio. Exercises How may additios, subtractios, multiplicatios ad divisios are required usig the algorithm RECURRENCE, while it computes x 1000 usig the data give i Example 5.4? Problems 5-1. Existece of a solutio of homogeeous equatio usig geeratig fuctio Prove that a liear homogeeous equatio caot be solved usig geeratig fuctios (because X(z) = 0 is obtaied) if ad oly if x = 0 for all. 5-. Complex roots i the case of Z-trasform What happes if the roots of the deomiator are complex whe applyig the Z-trasform method? The solutio of the recurrece equatio must be real. Does the method esure this? Chapter otes The recurrece equatios are discussed i detail by Elaydi [1], Flajolet ad Sedgewick [8], Greee ad Kuth [3], Mickes [7]. Kuth [4] ad Graham, Kuth ad Patashik [] deal with geeratig fuctios. I the book of Vileki [9] there are a lot of simple ad iterestig problems about recurreces ad geeratig fuctios.

24 5. Megjegyzések a fejezethez 9 I [6] Lovász also presets problems o geeratig fuctio. Coutig the biary trees is from Kuth [4], coutig the leaves i the set of all biary trees ad coutig the biary trees with vertices ad k leaves are from Z. Kása [5].

25 Bibliography [1] S. N. Elaydi. A Itroductio to Differece Equatios. Spriger-Verlag, 1999 (. editio). 8 [] R. L. Graham, D. E. Kuth, O. Patashik. Cocrete Mathematics. Addiso-Wesley, 1994 (. editio). 8 [3] D. H. Greee, D. E. Kuth. Mathematics for the Aalysis of Algorithms. Birkhäuser, 1990 (3. editio). 8 [4] D. E. Kuth. Fudametal Algorithms, The Art of Computer Programmig 1. kötete. Addiso-Wesley, 1968 (3. updated editio). 8, 9 [5] Z. Kása. Combiatorica cu aplicaţii (Combiatorics with Applicatios). Presa Uiversitara Clujeaa, [6] L. Lovász. Combiatorial Problems ad Exercises. Akadémiai Kiadó, [7] R. E. Mickes. Differece Equatios. Theory ad Applicatios. Va Nostrad Reihold, [8] R. Sedgewick, P. Flajolet. A Itroductio to the Aalysis of Algorithms. Addiso-Wesley, [9] N. J. Vileki. Combiatorial Mathematics for Recreatio. Mir,

26 Subject Idex B biary trees coutig, 19 biomial formula geeralisatio of the, 17 C characteristic equatio, 07 F Fiboacci umber, 14 fudametal solutio, 07 G geeral solutio, 06 geeratig fuctio, 14 geeratig fuctios coutig biary trees, 19 operatios, 15 H harmoic umbers, 16 I, Í iitial coditio, 06 L LINEAR-HOMOGENEOUS, 1 LINEAR-NONHOMOGENEOUS, 5 M method of variatio of costats, 13 P particular solutio, 06 R RECURRENCE, 8 recurrece equatio, 06 liear, 07, 1, 18, 4 liear homogeeous, 07, 1 liear ohomogeeous, 1, 4, 5 solvig with geeratig fuctios, 18 residue theorem, 6 T Towers of Haoi, 13gy Z Z-trasform method, 4

27 Name idex E, É Elaydi, Saber N., 8, 30 L Lovász, László, 9, 30 F Fiboacci, Leoardo Pisao ( ), 06, 08, 14, 17 Flajolet, Philippe, 8, 30 G Graham, Roald Lewis, 8, 30 Greee, Daiel H., 8, 30 K Kása, Zoltá, 9, 30 Kuth, Doald Ervi, 8, 30 M Mickes, Roald Elbert, 8, 30 P Patashik, Ore, 8, 30 S Sedgewick, Robert, 8, 30 V Vileki, Naum Yakovlevich ( ), 8, 30

28 Cotets 5. Recurreces (Zoltá Kása) Liear recurrece equatios Liear homogeeous equatios with costat coefficiets Liear ohomogeeous recurrece equatios with costat coefficiets Geeratig fuctios ad recurrece equatios Deitio ad operatios Solvig recurrece equatios with geeratig fuctios Liear ohomogeeous recurrece equatios with costat coefficiets The umber of biary trees The umber of leaves of all biary trees of vertices The umber of biary trees with vertices ad k leaves The Z-trasform method Numerical solutio Bibliography Subject Idex Name idex