Generalized Relation between the Roots of Polynomial and Term of Recurrence Relation Sequence

Many researchers have been working on recurrence relation which is an important topic not only in mathematics but also physics, economics and various applications computer science. There are many useful results sequence there main problem to find any term of we need all previous terms sequence. were theorems obtained relations. In this paper given special identity for generalized kth order relation. These identities very finding Authors define a formula by can direct so result important. property between coefficients roots polynomial second paper, gave same higher So finally, say that theorem valid condition distinct. So, generalization polynomial. Theorem: - Let C1 C2 arbitrary real numbers suppose the equation (1) Has X1 X2 distinct roots. Then src=image/13422456_25.gif> solution src=image/13422456_02.gif> (2) src=image/13422456_03.gif>. For n= 0, 1, 2 …where β1 β2 constants. Proof: First type src=image/13422456_04.gif> shall prove (2). Since X1, X3 satisfied have, src=image/13422456_06.gif>. Consider src=image/13422456_07.gif>. This implies src=image/13422456_09.gif>. Now will part theorem. src=image/13422456_10.gif> with three src=image/13422456_11.gif>. src=image/13422456_12.gif>. src=image/13422456_13.gif> (3). src=image/13422456_14.gif> (4). Multiply (3) subtracts from We src=image/13422456_15.gif> similarly src=image/13422456_16.gif>. values defined as non- trivial ofβ1 valid. Example: be such src=image/13422456_17.gif> n≥3 a0=0, a1=1, a2=2. a10 above Solution: The src=image/13422456_18.gif>. Solving 2, 3 using src=image/13422456_19.gif> (7). Using a2=2 (7) β1+β2+β3=0 (8). β1+2β2+3β2=1 (9).β1+4β2+9β3=2 (10) (8), (9) src=image/13422456_20.gif>, src=image/13422456_21.gif>, src=image/13422456_22.gif>. src=image/13422456_23.gif>. put n=10 a10=-27478. Recurrence mathematics, problems life may solved relations, major difficulty If want 100th sequence, then 99 get if satisfies theorem, apply without terms.