LINEAR RECURRENCE RELATIONS - PART US JAMES A. JESKE San Jose State College, San Jose, California. INTRODUCTION We continue our study to acquaint the beginner with l i n e a r r e - c u r r e n c e relations and the method of generating functions for solving them (see a r t i c l e s in [ l ] and [ z ] ). In this concluding article we shall consider r e c u r r e n c e relations in which there is m o r e than one independent variable.. DEFINITION A partial linear r e c u r r e n c e relation in two independent variables m, n is as equation of the form k p (.) S X a (m, n) y(m+i, n+j) = b(m, n) where a., and b a r e given functions of the discrete variables m and n over the set of non-negative integers. P a r t i a l r e c u r r e n c e r e - lations in three or m o r e independent variables may be defined in a s i m i l a r way. If b(m, n) = 0, relation (. ) is called homogeneous. The equation contains (k+l)(p+l) possible t e r m s and. is said to be of order k with r e s p e c t to m and of order p with respect to n. To solve c e r - tain r e c u r r e n c e relations we find it convenient to apply a generating function t r a n s f o r m.. A SERIES TRANSFORM The exponential generating function for the sequence { y(m, n)}, (m, n = 0,,,... ) is defined by the double infinite s e r i e s m n (.) Y Y(s,t) ( s, t ) = ~- I y(m,n) t ^ Ljm = 0 n= 0 97
9 L I N E A R R E C U R R E N C E R E L A T I O N S P A R T III O c t o b e r If the s e r i e s (. ) c o n v e r g e s w h e n s < a a n d t </ s i m u l t a n e o u s l y, t h e n a l l of t h e d e r i v e d s e r i e s of (. ) w i l l a l s o c o n v e r g e in the s a m e r e g i o n. T h u s, we h a v e (.) o o c o m _ l n BY ^ ^, x s t -^~- a s ~ * m y 9 "( m, n ' m! n! m = l n=0 o«cvt> m n y(m+l,n) m : n: a n d it i s s e e n t h a t ( Y / d s ) i s the e x p o n e n t i a l g e n e r a t i n g function of t h e s e q u e n c e y ( m + l, n ). S i m i l a r l y, one e a s i l y o b t a i n s the e q u a t i o n s oo oo m n (.) 4 V = y(m, n+) ^ V d t ^ J ^ m i n i a n d - oo co m n < - 4 > & = S. I y(m + l.n+l) ^ ^. w h i c h f u r n i s h e x p o n e n t i a l g e n e r a t i n g f u n c t i o n s for the s e q u e n c e s { y ( m, n+)} a n d { y ( m + l, n+)} r e s p e c t i v e l y. In g e n e r a l, the r e l a t i o n ' i+j Y oo oo m n (.) YTTZr- y(»+i.n + j) ^ - r ds d r n n i s the e x p o n e n t i a l g e n e r a t i n g function of the s e q u e n c e { y ( m + i, n + j ) }. T h i s e q u a t i o n p e r m i t s u s to t r a n s f o r m l i n e a r p a r t i a l r e c u r r e n c e r e - l a t i o n s (. ) w h e r e the c o e f f i c i e n t s a..(m, n) a r e a l l a s s u m e d to be c o n s t a n t s ( i. e., not a function of m a n d n ). If we t h e n m u l t i p l y both m n s t s i d e s of (. ) by J r r a n d s u m on m a n d n f r o m z e r o to infinity, J m l n! w e g e t the t r a n s f o r m e d e q u a t i o n
964 LINEAR RECURRENCE RELATIONS PART III 99 (.6) k P a. a i+j Y ~ ij ds i at j = B(s,t), where (.7) &J &o B(s,t) = b(m,n) m s n t After the transformed equation is solved for sequence { y(m, n) } either from the relation Y, we then obtain the (.) y(m, n) a m + n Y a s m a t n s = 0' t = o or by expanding the function Y(s t) in the form (. ). We illustrate the procedure with two simple examples in which b(m, n) = 0. 4. EXAMPLES Consider, for instance, the partial r e c u r r e n c e relation (4.) y ( m + l, n + l ) - y(m + l, n ) - y(m, n) = 0 with the given conditions (4.) y(m, 0) = 0 y(0,n) = y(m, n) = 0 if m i 0 if m < n The" transformed equation for (4. ) is then (4.) d Y as at ay as - Y = 0
00 LINEAR RECURRENCE RELATIONS PART III October with the conditions 4.4). Y(s, 0) =, Y(0,t) = e t. Now, a particular solution of (4. ) is (4.) Y(s,t) = e y I Q ( s/~if), where by I n (z) is the modified Bessel function of the first kind defined (4.6) I 0 (z) = i i ^ ). n ( m - ) m=0 x ' We can obtain the sequence {y(m, n)} by expanding (4.). Thus, DO m n Y(s,t) = e* - S (4.7) tj v v t z V * 7T j=0 m=0 Letting n = m+j, we then have (4.) oo m m oo n - m, *<" '> = ^ ~ > ^ n T T nt n ( m ) m = 0 Hence, from (. ) it is clear that n = m oo m o& ^n * mi ^ \ m / n! m=0 n=m (4.9) y(m,n) = ( ^ ), (m = 0,,..., n) = 0, n < m,
964 LINEAR RECURRENCE RELATIONS PART III 0 which simply r e p r e s e n t s the elements of P a s c a l ' s triangle (that is, the binomial coefficients). As a second example, we take the partial r e c u r r e n c e relation (4.0) y ( m + l, n + l ) - y(m, n+) - y(m, n) = 0 with the conditions (4.) y(0,n) = F n ; y(m,0) = F m, where F denotes the nth Fibonacci number. Transformation of n equation (4. 0) yields (4.) with the conditions a Z Y dy ds at " at Y = 0 (4.) where Y(s,0) = - x/ Y(0, t) = V sa, s a? e - e ta, t a? e - e tl - U l + y F ) > a z = H - ^ ) The solution of equation (4. ) is (4.) Y(s,t) = 7 a (t+s) e a (t+s) - e Now, employing the inverse transform (. ) yields (4.6) y(m,n) a m + n Y " a m _,n as at s=0 t=0 m+n m + n, >/ " \ a l - a
0 LINEAR RECURRENCE RELATIONS PART III October which is the solution of (4. 0) and r e p r e s e n t s a Fibonacci a r r a y shown in the following table V \ 0 0 0 4 4 4 4 Fibonacci a r r a y s of higher dimension can also be obtained. These involve the solutions of partial r e c u r r e n c e relations in three or m o r e independent variables.. CONCLUDING REMARKS The above examples involved the solution of two partial r e c u r - rence relations having only constant coefficients. R e c u r r e n c e r e l a - tions with polynomial coefficients may also be transformed by the method of generating functions. For instance, it is easy to show that the r e c u r r e n c e relation k P (.) ( a + m/ + n y ) y(m+i, n+j) = b(m, n), having linear coefficients, can be transformed to the equation (.) (a +0 4>+Y +) f * Y = B(s,t), z
964 LINEAR RECURRENCE RELATIONS PART III 0 where B(s, t) is given by (.7), and 0 and ^ are the differential operators (.) += s, *= t f t. I wish to thank Prof. Paul F. Byrd for his many helpful suggestions during the preparation of this article and the two previous ones. REFERENCES. J. A. Jeske, "Linear Recurrence Relations Part I, " The Fibonacci Quarterly, Vol., No., pp. 69-74.., "Linear Recurrence Relations Part II, " The Fibonacci Quarterly, Vol., No. 4, pp. -9. XXXXXXXXXXXXXXX ASSOCIATIVITY AND THE GOLDEN SECTION H. W. GOULD West Virginia University, Morgantown, West Virginia E. T. Bell, A functional equation in arithmetic, Trans. Amer. Math. Soc., 9(96), 4-44, gave a discussion of some matters suggested by the functional equation of associativity <p(x, <p(y, z)) = <p(<p(x, y), z). As a prelude, Bell noted the following theorem. THEOREM. The only polynomial solutions of <p(x, <p{y, z))=<p(<p(x, y), z) in the domain of complex numbers are the unsymmetric solutions <p(x, y) = x, <p(x, y) = y, and the symmetric solution <p(x, y) = a + b(x + y) + cxy, Z in which a, b, c, are any constants such that b - b - ac = 0. It is amusing to note a special case. The operation defined by x * y = + b(x + y) + xy is associative only if b = j ( ± \T$) XXXXXXXXXXXXXXX