A-l1* Px = \* F« = X*(\E – 1)-' Px 1 = A*. X(1+A) – 1° -Pk 1 (1 A - 11+ XA/(A – 1) 1 1 X- 1 (A – 1)² Pk.

College Algebra
1st Edition
ISBN:9781938168383
Author:Jay Abramson
Publisher:Jay Abramson
Chapter3: Functions
Section3.3: Rates Of Change And Behavior Of Graphs
Problem 2SE: If a functionfis increasing on (a,b) and decreasing on (b,c) , then what can be said about the local...
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Explain the determaine
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
Transcribed Image 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
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