ON POLYNOMIALS GENERATED BY TRIANGULAR ARRAYS

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1 ON POLYNOMIALS GENERATED BY TRIANGULAR ARRAYS KR1SHWASWAIVJ! ALLADI* Vivekananda College, Madras , India In this paper we study a class of functions which we call Pascal functions, generated by the diagonals of triangular arrays, and discuss some of their properties. The Fibonacci polynomials become particular cases of Pascal functions, and so our results are of a fairly general nature. 1. DEFINITIONS AMD GENERAL PROPERTIES Consider a polynomial function in two variables, p(x,y). It is defined to be a Pascal function of (k - 1) st order if [n/kj (1) p(x,y) = amx n ~ km y m, m=0 where the a m are non-zero constants, and [x] represents, for real x, the largest integer not exceedingx. Let us denote the set of all Pascal functions (polynomials) of k th order by I I ^. (Note: k is a positive integer.) One generalization of the famous Fibonacci polynomials is F 0 (x,y) = 0, F^(x,y) = 1, F n+2 (x,y) = xf n+1 (x,y) + yf n (x,y), n = 0, 1,2, We find that F n (x,y) e n 7 / n = 0, 1,2,3, -. See Hoggatt and Long [1]. It is interesting to note that the following properties hold: Lemma 1. \ip(x,y) mdp*(x,y) are in U^, then q(x,y) is in 11^, where q(x,y) = p(x,y)p*(x,y). This is the same as saying that H^ is closed under multiplication. If p(x,y)^ Uk-i, and has an expansion as given in (1), then tetd(p) = n. We then have Lemma 2. \\p(x,y) m&p*(x,y) are in H^, then is in II ^ if and only if D(p) = D(p*l Lemma 3. \ip(x,y) is in Ily^, then q(x,y) = p(x,y)+p*(x,y) dp(x,y) a n d dp(x,y) dy are in 11^. The three lemmas given above can be proved easily. We define a sequence of functions to be proper if (2) D(p n+1 ) = D(p n )+1 with D(p 0 ) = D( P1 ) = 0. By a Pascal array we mean a triangular array of numbers represented in Fig. 1 below: *Graduate Student, UCLA, Los Angeles, Calif. 481

2 462 ON POLYNOMIALS GENERATED BY TRIANGULAR ARRAYS [DEC. co,o C 1,0 C1.1 C2,0 C2,1 C2,2 C3,0 C3,1 C3,2 c 3,3 Figure 1 If now we replace every c/j by c,jx'y J, and take the rising diagonal sums, where the rising diagonals have aslope k, we get a proper sequence in 1 1 ^. Conversely, to every proper sequence in I I ^, we can associate a triangular array as in Fig. 1. Note that we can get infinitely many proper sequences from Fig. 1 as k varies, and all of these sequences for different values of k, we call "associated sequences." The triangular array which generates these sequences, is called their "associated array." We now discuss some special properties afp(x,y) e 11^. Theorem 2. SOME SPECIAL PROPERTIES OF PASCAL FUNCTIONS 1. Consider the proper sequence of Pascal functions \p n (x,y)\li=o e Rk satisfying (3) Pn+l(*,y) = axp n (x,y) + ayp n - k (x,y), n > k, with P 0 (x,y) = 0, Pi(x,y) = a, p 2 (x,y) = a 2 x, -, Pk(x,y) = a k x k ~ 1. Then,M\ zp n (x,y) dp n+k (x f y) * - *,, /, (4) = ~ f = 2-f Pk(x,Y)Pn-k(x,y)- v k=o One can establish the first part of (4) by induction. It is clear from (3) that (5) aw^w = gx a ^ M +apn( x,y, +ay IBiddHl and IR\ ZPn+k+l(x,y) _ Wn+k(x,y). /., f l l / *Pn(x,Y) (6) = ax + ap n [x,y) +ay. dy dy dy The form of (5) and (6) together with the fact that the first part of (4) holds for/7 = 1, 2, 3,, k, proves it by induction. We now want to show (7) ^ ^ - p k (x,y)p n. k (x,y). Consider the generating function We have GM = E Pn(x,y)t n = : - ^ n=0 1-axt-ayt k+1 ^ d Pn (X,y) n = ggft) = g2 _ = fr/, / 2 dx h ~ dx ' ~ d-axt-ayt k+1 f~ l ' This proves (7) and so we have established Theorem 1. Corollary. For the Fibonacci polynomials defined before, *>F n (x,y) df n+1 (x,y) \^» r / j r / J -Tx = = Tv h F k (x,y)f n k (x,y). dy

3 1976] ON POLYNOMIALS GENERATED BY TRIANGULAR ARRAYS 463 The corollary follows by taking k = 1 in Theorem 1. Theorem 2. If?>Pn (x,y) f. - = Pn,i(x,y), then define n (8) Pnjx,y) = J2 Pk,r-l(x,y)Pn-k(*,y)- Now Theorem Pnjx,y) 7 d r p n (x,y) r! r Differentiate the generating function G(t) in the proof of Theorem 1, r times. Theorem 2 follows. 3. If a proper sequence of Pascal functions {Pn(x,y)\n=0 satisfy (4), then they satisfy (3). (Converse of Theorem 1.) Consider the first (k + 1) members of the sequence Because of (4) we have and a x f 0, which gives a 0 = 0. Further, Similarly, one may show G Rk a 0,a 1/ a2x,a 3 x 2, -,a k x k 1. ^ (a 0 ) = 2a 0 a l, (a 2 x) = a 2 = a\. a r = a^ = a r, r = 7, 2, -, k. Now assume that (3) holds for n = 0, /, 2, 3,-,/7?. Let now m k=1 Clearly, by Lemmas 1 and 2, we have/7^7 (x,y) e I I ^. Now, denote P%+j(x,y) = axp m {x,y) + We have because of Theorem 1 But we know, because p 0 (x,y) = 0, and this gives dp +1 fx,y) p* m+1 (x f y) = Y^ Pk(*,y)Pm-k+i(x,y)- Pm+ifay)- *Pm+i(x,y) _» f j Vx Pm+ltx>V) Pm+ifcy) = Pm*+i(x,y) ayp m - k (x,y). by (1) and by Lemma 3. This proves that (3) holds, by mathematical induction. Hence we get Theorem PASCAL FUNCTIONS WHICH CAN BE PASCALISED We now shift our attention to Pascal functions which can be "pascalised." Given a proper sequence of Pascal polynomials

4 464 ON POLYNOMIALS GENERATED BY TRIANGULAR ARRAYS [DEC. \Pn(x,y)}n=0 form the associated array J a/j r= A. Now take Qn(x,y) e n * > Wn(x,v) to get a new proper sequence in Ft/.. Let j b/j \ = B be the associated Pascal array to this sequence. If we have the relation (9) bil^'h^v) we say p n (x,y) i can be "pascalised" to the first order. If,. d r p n (x,y) Qnfay) = ~ T - 7 r and do) b ij = a ii [ i ; r )n we say that the sequence ip n (x,y)l can be pascalised to the r th order. Theorem 4. A necessary and sufficient condition that a proper sequence of (k - 1) st order Pascal functions ip n (x,y) \^=o can be pascalised to the first order is that fn/kj (11) Pn(x,y)='E aj[ n - (k J 1jj - 1 )x n ' kh y j for some sequence of constants aj. 1=0 We will first prove the theorem for the case k = 2. Consider the sequence i p n (x,y)\ ^=o f and assume that the identity holds for/7 = ft 7, 2, -,m. We have then (12) Now let which gives (13) [m/2] Pm(x.y)* J=o j=o [(m+1)/2] ajfn-j-^x^t-'yj'. Pm + l(x fv )= J2 *fsn( m r J )x m - 2J y J *Pm+i(x,y) i-0 [m/2] = E af,m( m r / )x m - 2 <'- 1 yi(m-2j) 1=0 Now comparing coefficients in (12) and (13) and using (9) we get which gives a j,m ~ a j establishing part of the theorem for k = 2. The converse can be proved by retracing the steps. Now, once the theorem is proved for the first order (k= 2), it holds for any k > 1, for given a proper sequence of Pascal functions of (k - 1) st order, we can find its associated sequence of first order. The Pascal arrays for the derivatives of these two sequences is the same since the operator d/ will operate independently in the expansion of p n (x,y) with respect to coefficients in the associated Pascal array. This completes the proof of the theorem.

5 1976] ON POLYNOMIALS GENERATED BY TRIANGULAR ARRAYS 465 Theorem 5. If a proper sequence of k th order Pascal functions can be pascalised to the first order, then all their associated sequences can be pascalised to first order. Given in the last paragraph of the proof of Theorem 4. Theorem 6. If a proper sequence of k th order Pascal functions can be pascalised to first order, they can be pascalised to any order. By arguments similar to the above, it is enough if we prove it for k = 1. Furthermore, it is enough to prove the theorem for the special case ay = 1 for differential operators are unaffected by constant multiples. We know from Theorem 4 that the first-order proper sequence of Pascal functions which can be pascalised to first order can be put in the form Now, as mentioned, aj~ 1, so thatp n (x,y) [n/2] Pn(*>Y) = JL l n -j- 1 )x n - 2 f-'y<' aj. 1=0 X = F n (x,y), the Fibonacci polynomials. We then have,. [(n+r)/2] [ n + r - j \ yn+r-2jj J_ Wn+r+l(x,V) = _/_ y» _9_ \ j I X V r! *v r r! *-*, v r r! j=0 ax n+r-2j>0 which resembles (9) proving our theorem for Fibonacci polynomials, and so for Pascal functions. We demonstrate our result with the following: Pascal Array for F n (x,y) Pascal Array for - - I 1 I I K ' N o t e l : (2,2) = 2(1,1); (3,6,3) = 3(1,2,1); (4,12,12,4) = 4(1,3,3,1); -. Each row has a common factor. Note 2: Theorem 4 also says that each column has a common factor ay. In the above all the ay = 1. Note 3: The Pascal array for [df n (x,y)]/ax is also the Pascal array for for both are equal by Theorem 1. n YJ F k (x,y)f n.. k (x,y) ACKNOWLEDGEMENTS I would like to thank Professor V. E. Hoggatt, Jr., for his encouragement. His suggestion to make use of the generating function in Theorem 1 did much to shorten the proof. I also enjoyed the award of the Fibonacci Association Scholarship during the tenure of which this work was done. REFERENCE 1. V. E. Hoggatt, Jr., and Calvin T. Long, "Divisibility Properties of Generalized Fibonacci Polynomials," The Fibonacci Quarterly, Vol. 12, No. 3 (April 1974), pp

UNIT DETERMINANTS IN GENERALIZED PASCAL TRIANGLES

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