4.2.4 Example D The Fibonacci difference equation is Yk+2 = Yk+1+ Yk- (4.40) The Fibonacci sequence is the solution that has yo = 0, y1 = 1. The charac- teristic equation is p2 – r – 1 = 0, (4.41) and has the two roots 1+ V5 1- V5 r2 = (4.42) ri = 2 2 Therefore, the general solution to the Fibonacci difference equation is k k 1+ V5 - V5 1- Yk = C1 + C2 (4.43) where ci and c2 are arbitrary constants. The general member of the Fibonacci sequence can be determined by solv- ing for c1 and c2 from the equations Yo = C1 + c2 = 0, 1+ V5 V5 (4.44) 1 - + C2 2 = 1. Y1 = C1 2 The solutions are c1 = -C2 = 1/V5. Therefore, the general member of the Fibonacci sequence is given by 1+ V5 V5 1 Yk = V5 1- (4.45) 2 2 We have {yk} = 0, 1, 1, 2, 3, 5, 8, 13, 21, 34, ... %3D

Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter10: Sequences, Series, And Probability
Section10.2: Arithmetic Sequences
Problem 68E
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4.2.4
Example D
The Fibonacci difference equation is
Yk+2 = Yk+1 + Yk.
(4.40)
The Fibonacci sequence is the solution that has yo = 0, Y1 = 1. The charac-
teristic equation is
p2 – r – 1 = 0,
(4.41)
and has the two roots
1+ V5
1- V5
r2
(4.42)
r1
2
Therefore, the general solution to the Fibonacci difference equation is
k
k
1+ V5
1
+ C2
V5
(4.43)
Yk = C1
2
2
where
C1
and
C2 are arbitrary constants.
The general member of the Fibonacci sequence can be determined by solv-
ing for c1 and c2 from the equations
Yo = C1 + C2 = 0,
(4.44)
1+ V5
1– V5
+ c2
Y1 = C1
= 1.
2
The solutions are (c1 = -c2 = 1//5. Therefore, the general member of the
Fibonacci sequence
given by
k
1+ V5
(4.45)
Yk
V5
2
2
We have {yk} = 0, 1, 1, 2, 3, 5, 8, 13, 21, 34, ....
Transcribed Image Text:4.2.4 Example D The Fibonacci difference equation is Yk+2 = Yk+1 + Yk. (4.40) The Fibonacci sequence is the solution that has yo = 0, Y1 = 1. The charac- teristic equation is p2 – r – 1 = 0, (4.41) and has the two roots 1+ V5 1- V5 r2 (4.42) r1 2 Therefore, the general solution to the Fibonacci difference equation is k k 1+ V5 1 + C2 V5 (4.43) Yk = C1 2 2 where C1 and C2 are arbitrary constants. The general member of the Fibonacci sequence can be determined by solv- ing for c1 and c2 from the equations Yo = C1 + C2 = 0, (4.44) 1+ V5 1– V5 + c2 Y1 = C1 = 1. 2 The solutions are (c1 = -c2 = 1//5. Therefore, the general member of the Fibonacci sequence given by k 1+ V5 (4.45) Yk V5 2 2 We have {yk} = 0, 1, 1, 2, 3, 5, 8, 13, 21, 34, ....
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