Linerly Similr Polynomils rthur Holshouser 3600 Bullrd St. Chrlotte, NC, US Hrold Reiter Deprtment of Mthemticl Sciences University of North Crolin Chrlotte, Chrlotte, NC 28223, US hbreiter@uncc.edu stndrd technique for solving the recursion x n+1 g (x n ), where g : C C is complex function, is to first find firly simple function g : C C nd bijection (i.e., 1-1 onto function) f : C C such tht g f 1 g f where is the composition of functions nd f 1 is the inverse function of f. Then x n g n (x 0 ) (f 1 g n f)(x 0 ) where g n nd g n re the n-fold compositions nd g n is firly esy to compute. With this motivtion we re in generl interested in studying ll pirs of rtionl functions g, g such tht for some, b, c, d C g ( ) 1 x + b g ( ) x + b where, b, c, d stisfy b c d 0. In this pper we study n intermedite problem by finding ll pirs of qudrtic polynomils g, g nd lso ll pirs of cubic polynomils g, g such tht for some, b C, 0, g (x + b) 1 g (x + b). We denote such pirs by g g nd we show tht g g if nd only if g nd g hve the sme signture which we define s those invrints tht qudrtic nd cubic polynomils hve under the liner trnsformtion g (x + b) 1 g (x + b). 1 Introductory Concepts Let C be the set of complex numbers. If g : C C nd g : C C re rbitrry functions from C into C, we sy tht g nd g re similr (denoted by g g) if there exists bijection (i.e., 1-1 onto function) f : C C such tht g f 1 g f where is the composition of functions. From elementry set theory, we know tht the word similr mens exctly wht it sys. Thus, for exmple, suppose tht g nd g re similr bijections on C. If we brek ech of g nd g down into its cycles..., g 2 (x), g 1 (x), g 0 (x) x, g (x), g 2 (x),..., then the types of these cycles will be the sme in both g nd g. For exmple, if ll of the cycles in g re the 3-cycles x, g (x), g 2 (x), g 3 (x) x, then ll of the cycles in g will lso be the 3-cycles x, g (x), g 2 (x), g 3 (x) x. If g nd g re rtionl functions (i.e., the quotient of polynomils) nd g ( ) 1 x + b g ( ) x + b, 1
then we encounter stronger type of similrity which we will cll lgebric similrity. Intuitively lgebric similrity mens tht g nd g hve similr lgebric properties. However, we hve not been ble to define exctly wht this mens. None-the-less we hve still been ble to use this intuitive concept to heuristiclly derive technique for computing ll of the invrints tht g nd g must hve when g nd g re either polynomils or rtionl functions of ny rbitrry degree nd g ( ) x+b 1 ( cx+d g x+b ) b cx+d, c d 0. We will show the reder how to do this when g (x + b) 1 g (x + b) nd g, g re polynomils of degree 2 or degree 3. These sme ides lso generlize for higher degree polynomils. However, the invrints quickly become so complex tht they must ll be proved with computer. Indeed, one of the three invrints used in this pper ws (for convenience only) computer proved. 2 Very Linerly Similr Qudrtic Polynomils Definition 1 Suppose g nd g re nth degree complex polynomils. We sy tht g nd g re very linerly similr (denoted by g g) if there exists, b C, 0, such tht g (x + b) 1 g (x + b). Theorem 1 The reltion is n equivlence reltion on the collection of ll complex polynomils. Proof. We let the reder verify the following three conditions which define n equivlence reltion.. g g.(reflexive condition). b. g g implies g g. (symmetric condition). c. g g nd g g implies g g. (trnsitive condition). Problem 1 Suppose g (x) x 2 + Bx + C, 0, is qudrtic polynomil. Define b ) (x 2 + Bx + C) (x + b) g (x) x 2 + Bx + C where 0, b re rbitrry complex numbers nd (x + b) 1 x b. We wish to compute n invrint for g (x) nd g (x). This mens tht we wish to find n expression involving, B, C tht remins unchnged when we substitute, B, C for, B, C. Solution 1 Suppose tht x + Bx + C nd x 2 + Bx + C, 0, 0 re ny rbitrry complex qudrtics. We now show tht we cn linerly trnsform x 2 + Bx + C into x 2 + Bx + C by ( ) x 2 + Bx + C b ) 2 + Bx + C ) (x + b) 2
(where 0) if nd only if certin condition is met, nd this condition will define the invrince reltion tht (, B, C) nd (, B, C ) hve under the bove liner trnsformtion ( ). Now b ) 2 + Bx + C ) (x + b) b ) [ (x + b) 2 + B (x + b) + C ] b ) ( 2 x 2 + (2b + B) x + b 2 + Bb + C ) x 2 + (2b + B) x + b2 + (B 1) b + C x 2 + Bx + C. Now, 2b + B B implies, b B B 2. Therefore, Therefore, C 1 [ b 2 + (B 1) b + C ] [ ( ) 2 ( ) ] B B B B + (B 1) + C. 2 2 4C ( B B ) 2 + 2 (B 1) ( B B ) + 4C B 2 2BB + B 2 + 2BB 2B 2B 2 + 2B + 4C. ( ) This implies (B 2 2B) 4C B 2 2B 4C. Therefore, (B 1) 2 4C ( ) 2 B 1 4C, nd this expression must be the invrint reltion when x 2 + Bx + C x 2 + Bx + C. This invrint cn lso esily be checked (by hnd) by substituting, B 2b + B, C b2 +(B 1)b+C for, B, C in (B 1) 2 4C nd showing tht (B 1) 2 4C remins unchnged. Definition 2 We cll θ (B 1) 2 4C the signture of x 2 + Bx + C. In solving Problem 1, we hve lso proved Theorem 2. Theorem 2 Suppose x 2 + Bx + C, x 2 + Bx + C, 0, 0, re rbitrry complex qudrtics. Then x 2 +Bx+C x 2 +Bx+C if nd only if x 2 +Bx+C nd x 2 +Bx+C hve the sme signture θ. 3
2.1 note on the Discriminnt of Polynomil If P (x) 0 x n + 1 x n 1 + + 0 is ny single vrible polynomil, then from lgebr we know tht there is stndrd expression (denoted by ρ (P (x), P (x)) which is clled the discriminnt of P (x). See p. 99, [1]. When the discriminnt of P (x) is zero, we know from lgebr tht P (x) hs repeted root, nd when the discriminnt of P (x) is non-zero, we know tht the n roots of P (x) re ll distinct. Of course, the discriminnt of the qudrtic P (x) 0 x n + 1 x + 2 is 2 1 4 0 2. We observe tht the invrint of the qudrtic x 2 + Bx + C, 0, tht ws derived in the lst section is lso the discriminnt of b 2 + (B 1) b + C, when we substitute b x for b nd where x 2 + Bx + C x 2 + (2b + B)x + b2 +(B 1)b+C. This simple observtion (with one slight modifiction) ppers to generlize for rbitrry degree polynomils, nd it will soon be used to compute the second invrint of cubic polynomils. 2.2 Computing the Liner Trnsformtion of Cubic Polynomils By strightforwrd clcultions we see tht b ) 3 + Bx 2 + Cx + D ) (x + b) b ) ( (x + b) 3 + B (x + b) 2 + C (x + b) + D ) b ) [ 3 x 3 + (3 2 b + B 2 ) x 2 + (3b 2 + 2Bb + C) x + b 3 + Bb 2 + Cb + D 2 x 3 + (3b + B) x 2 + ( 3b 2 + 2Bb + C ) x + (b 3 + Bb 2 + (C 1) b + D)/ x 3 + Bx 2 + Cx + D. 2.3 Two Invrints of Cubic Polynomils under Liner Trnsformtion Problem 2 Suppose g (x) x 3 + Bx 2 + Cx + D, 0, is cubic polynomil. Define b ) (x 3 + Bx 2 + Cx + D) (x + b) g (x) x 3 + Bx 2 + Cx + D where 0, b re rbitrry complex numbers. We wish to compute two invrints for g (x) nd g (x). This mens tht we wish to find two expressions involving, B, C, D tht remin unchnged when we substitute, B, C, D for, B, C, D. Solution 2 The first invrint cn be computed exctly s we did in solving Problem 1. From the previous section; we must hve 2, 3b + B B, 3b 2 + 2Bb + C C nd b3 +Bb 2 +(C 1)b+D D. Now 2 /, b B B 3, from which it follows tht C 3 [ C (B B)2 3 2 + 2B(B B) 3 ] ] 2 [ B B 3 + 2B ] B B + 3 + C. Hence, C B2 2BB+B 2 2 +2BB 2B 2 2 +3 2 C 3 2. Thus 3 2 C 4
( B 2 B 2 2 + 3 2 C. Therefore, 3 B 2 B2 + 3C. Finlly it follows tht 3C We will cll θ C B2 ) C B 2 B 2 ( B2 3C. ) ( + 3 ) C. Thus 3C B 2 the first invrint of g (x) 3 x3 + Bx 2 + Cx + D under the liner trnsformtion g (x) (x + b) 1 g (x) (x + b). We observe tht θ is lso equivlent to ρ (3x 2 + 2Bx + C) / where we substitute b x in C 3b 2 +2Bb+C. This invrint θ cn lso esily be verified by showing tht θ remins unchnged when we substitute 2, B 3b + B nd C 3b 2 + 2Bb + C in θ for, B, C respectively. This cn esily be done by hnd. When we try to compute the second invrint for x 3 + Bx 2 + Cx + D x 3 + Bx 2 + Cx + D in this exct sme wy, we run into insurmountble difficulty. Therefore, we will simply conjecture tht the second invrint clled φ ρ (x 3 + Bx 2 + (C 1) x + D) / k where k is decided by using specific exmple. This division by k is the modifiction tht we referred to erlier. Now the stndrd discriminnt for the cubic polynomil P (x) 0 x 3 + 1 x 2 + 2 x + 3 is ρ (P (x), P (x)) 27 2 0 2 3 + 18 0 1 2 3 4 3 1 3 4 0 3 2 + 2 1 2 2. Therefore, φ ( 27 2 D 2 + 18B (C 1) D 4B 3 D 4 (C 1) 3 + B 2 (C 1) 2 )/ since specific exmple shows tht k must be true. Professor Ben Klein of Dvidson College hs verified tht φ is n invrint by using the Mthemtic softwre. Definition 3 We define the signture of the cubic polynomil x 3 + Bx 2 + Cx + D, 0, to be the ordered pir (θ, φ) where θ nd φ re the invrints specified bove. The rest of this pper is devoted minly to proving tht two cubic polynomils g (x) nd g (x) re very linerly similr if nd only if g (x) nd g (x) hve the sme signture (θ, φ). 3 Proving our Min Results for Cubic Polynomils Theorem 3 The signture of the cubic polynomil x 3 +Cx+D is (θ, φ) ( C, 27D 2 4 (C 1) 3). Proof. Using 1, B 0, C C, D D in the formuls for θ, φ gives the signture (θ, φ). Corollry 1 Suppose x 3 + Cx + D hs signture (θ, φ). { } φ 4(θ 1) Then C θ nd D 3 φ 4(θ 1), 3. 27 27 Proof. Follow from Theorem 3. Theorem 4 Suppose x 3 + Bx 2 + Cx + D, 0, is ny rbitrry cubic polynomil. Then there exists cubic polynomil x 3 + Cx + D such tht x 3 + Bx 2 + Cx + D x 3 + Cx + D. 5
Proof. Now b ) 3 + Bx 2 + Cx + D ) (x + b) 2 x 3 + (3b + B) x 2 + ( 3b 2 + 2Bb + C ) x + b3 + Bb 2 + (C 1) b + D x 3 + 0 x 2 + Cx + D. Let, b be defined so tht 2 1, 3b + B 0. Therefore, ± completes the proof. 1, b B 3 which Theorem 5 Suppose x 3 + Cx + D nd x 3 + Cx + D hve the sme signture (θ, φ). Then x 3 + Cx + D x 3 + Cx + D. { Proof. From Corollry 1} we know tht C C θ. lso, from Corollry 1, D, D. φ 4(θ 1) 3 27 φ 4(θ 1) 3 27 Therefore, D D or else D D. Therefore, we cn complete the proof by ssuming tht D D. Using 1, B 0, C C, D D, 1, B 0, C C, D D in the proof of Theorem 4, we hve ±1, b 0. Let us use 1. Therefore, x+b x nd we hve ( x) (x 3 + Cx + D) ( x) ( x) ( x 3 Cx + D) x 3 + Cx D. Therefore, x 3 + Cx + D x 3 + Cx D x 3 + Cx + D. Min Theorem 6. Two cubic polynomils g (x) nd g (x) re very linerly similr if nd only if g (x) nd g (x) hve the sme signture (θ, φ). Proof. Of course, if g (x) nd g (x) re very linerly similr, then they must hve the sme signture (θ, φ) since θ nd φ re invrints under. Conversely, if g (x) nd g (x) hve the sme signture (θ, φ), then we know from Theorems 4, 5 nd from the equivlence reltion properties of tht were proved in Theorem 1 tht g (x) g (x) must be true. 4 Some Concluding Remrks We note tht the signture of g (x) x 3 1 x 3 + 0 x 2 + 0 x + 0 is (θ, φ) (0, 4). lso, we note tht the recursion x n+1 x 3 n cn esily be solved in closed form. Suppose g (x) x 3 + Bx 2 + Cx + D, 0, is ny cubic polynomil tht hs signture (θ, φ) (0, 4). From this we know tht g (x) x 3, nd we cn now esily solve the recursion x n+1 g (x n ) in closed form. lso this pper cn be generlized s follows. Suppose g (x) 0 x n + 1 x n 1 + + n nd g (x) 0 x n + 1 x n 1 + + n re very linerly similr n th degree complex polynomils. Tht is, g (x) (x + b) 1 g (x) (x + b) for some, b C, 0. Then we cn lmost certinly derive the n 1 invrints tht g (x) nd g (x) must hve. We do this exctly s we did in this pper, nd we then prove tht the invrints re correct by using computer progrm such s Mthemtic. 6
References [1] Brbeu, E. J. Polynomils, (Problem Books in Mthemtics) (Pperbck), Springer Verlg, New York, 1989. [2] Weisner, Louis, Introduction to the Theory of Eqution, The McMilln compny, New York, 1949. 7