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 r1 1- V5 r2 = (4.42) Therefore, the general solution to the Fibonacci difference equation is k k 1- V5 + C2 1+ V5 Yk = C1 (4.43) 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 V5 = 1. 1- Y1 = C1 + C2 2 The solutions are c1 = -c2 = 1//5. Therefore, the general member of the Fibonacci sequence is given by k 1 V5 (4.45) Yk V5 2 We have {yk} = 0, 1, 1, 2, 3, 5, 8, 13, 21, 34, ..

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 :
teristic equation is
1. The charac-
gn? – r – 1 = 0,
(4.41)
and has the two roots
1+ V5
1 – V5
r1
r2 =
(4.42)
2
2
Therefore, the general solution to the Fibonacci difference equation is
k
k
1+ V5
V5
Yk = C1
+ c2
(4.43)
2
2
where
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
C1
Yo = C1 + c2 = 0,
1+ V5
- V5
(4.44)
1 -
Y1 = C1
+ c2
1.
The solutions are c1 =
1/V5. Therefore, the general member of the
-C2
Fibonacci sequence is given by
1+ V5
1
Yk =
V5
(4.45)
-
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 : teristic equation is 1. The charac- gn? – r – 1 = 0, (4.41) and has the two roots 1+ V5 1 – V5 r1 r2 = (4.42) 2 2 Therefore, the general solution to the Fibonacci difference equation is k k 1+ V5 V5 Yk = C1 + c2 (4.43) 2 2 where 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 C1 Yo = C1 + c2 = 0, 1+ V5 - V5 (4.44) 1 - Y1 = C1 + c2 1. The solutions are c1 = 1/V5. Therefore, the general member of the -C2 Fibonacci sequence is given by 1+ V5 1 Yk = V5 (4.45) - 2 2 We have {yk} = 0, 1, 1, 2, 3, 5, 8, 13, 21, 34, ...
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