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Proof. If y(n, no, c) is the solution with initial conditions c\, 0 2 , . . 1 and also satisfies the same initial conditions zno = ci,z n o + i — C 2 , . . ,2 no +fc-i = Cfc. 3 The functions fi(n},i = 1, 2 , . . 15) implies c*7; — 0, i — 1, 2 , . . , k. Copyright © 2002 Marcel Dekker, Inc. 40 CHAPTER 2. 2 A sufficient condition for the functions f i ( n ) , i — 1, 2, . . , k, to be linearly independent is that there exists n> UQ such that detK(n) 7^ 0. Proof. 16) This linear homogeneous system of k equations in k unknowns has the coefficient matrix K(n).

F c - 1, i=0 whose determinant ^4fc(n) is not zero by hypothesis. It follows that the solution is Ap z (n) = 0, that is the pi are constant with respect to n, and then the conclusion follows. D Example 13 Consider the equation yn+l -ayn = Q, a 6 C. 44) The characteristic polynomial is p ( z ) = z — a, and its unique root is z = a. 44) is then yn — can. Example 14 Consider the equation yn+2 - yn+\ - yn = 0. 45) is + V/5V + C2 which is known as the Fibonacci sequence. Example 15 Consider the equation 7 Ti -\-1 J Ti ~~~~ jTl' From Example 13, it follows that the general solution of the homogeneous equation is can.

It follows then that yn = cipvTn(q} -f c'2pnUn-i(q} is the general solution. 5 Use of Operators A and E The method of solving difference equations with constant coefficients becomes simple and elegant when we use the operators A and E. Using the operator E. 51) '. 53) z=l where z\, z^, . . Zk are the zeros of p ( z ) . If there arc s distinct roots with multiplicity m,j,j = l , 2 , . . •= 1 from which it is seen that the homogeneous equation can be split into s difference equations of order nij.

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