Theorem 1.6. For A# 1, Pk- (1.249) - 1 A - 1 (A – 1)2 Proof. Let F, be a function of k. Therefore AX*Fx = \k+1 Fi+1 – X* Fr xk+1 EF% – X* F. = X*(\E – 1)Fr. (1.250) - Now, set (AE – 1)F; = Pk;...

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Extracted text: Theorem 1.6. For A# 1, Pk- (1.249) - 1 A - 1 (A – 1)2 Proof. Let F, be a function of k. Therefore AX*Fx = \k+1 Fi+1 – X* Fr xk+1 EF% – X* F. = X*(\E – 1)Fr. (1.250) - Now, set (AE – 1)F; = Pk; consequently F = (AE – 1)-1Pr.- (1.251) Therefore, from equation (1.250) AX* F = X* Pr, (1.252) and A-1* Px = \* Fx = X*(\E – 1)-'Px k 1 = \*. A(1+A) -P: - 1 1 (1.253) A- 11+ XA/(A – 1) P. A- 1 A- 1 (A – 1)2