Explain the determine purple 7.2.2Fibonacci NumbersDenote the kth Fibonacci number as F. These numbers are defined by therule that the kth number is equal to the sum of the two preceding ones withthe initial numbers being Fo = 0 and F1 = 1. They satisfy the followingdifference equation:FR = Fk-1+ Fk-2.(7.36)This is a second-order, linear difference equation with constant coefficientsand can be easily solved.Setting F = pk gives the characteristic equationp2 – r – 1 = 0,(7.37)whose solutions are1+ V51- V5r2 =(7.38)ri =2Consequently, the general solution of equation (7.36) is1+ V5Fr = C1+ C2(7.39)22The arbitrary constants C1 and C2 may be found by imposing the initialconditions Fo = 0 and F1 =1. Doing this givesC1 + C2 = 0,(1+ V5)C1 + (1 – V5)C2 = 2,(7.40)which, when solved for C1 and C2, gives1C1 = -C2 =V5(7.41)APPLICATIONS219Putting these values in equation (7.39) gives, for the kth Fibonacci number,the result+ V51F. =V51- V5(7.42)2In the remainder of this section, we will examine some of the properties ofthese numbers.First, consider the limitFR+1lim(7.43)From equation (7.42), we find(*-()1 – ((1 – v5/(1+ v5)]k+1 \-- [(1 – V5)/(1+ v5)]*Fr+11+-(7.44)FrSince1- V5< 1,1+ V5(7.45)thenFr+1lim1+ V5= 1.618033988....(7.46)This number, called the golden mean, represents the ratio of the sides of arectangle that is supposed to be "most pleasing" to the eye.Consider now the values of the Fibonacci numbers for the index k being anegative integer. The most direct way of obtaining these values is to substitute-k for k in equation (7.42). Doing this gives-k1+ V51F-k =V522(7.47)1V5However,--(누)21- V5(7.48)1+ V52and(7.49)1- V5

The purple expressions are explained in detailed in below strps.

7.2.2 Fibonacci Numbers Denote the kth Fibonacci number as Fk. These numbers are defined by the rule that the kth number is equal to the sum of the two preceding ones with the initial numbers being Fo = 0 and Fi = 1. They satisfy the following difference equation: FR = Fk-1+ Fk-2- (7.36) This is a second-order, linear difference equation with constant coefficients and can be easily solved. Setting F = rk gives the characteristic equation p2 - r -1= 0, (7.37) whose solutions are 1+ V5 1- V5 r2 = (7.38) 2 2 Consequently, the general solution of equation (7.36) is F = C1 1+ V5 1- V5 + C2 (7.39) 2 The arbitrary constants C1 and C2 may be found by imposing the initial conditions Fo = 0 and F1 = 1. Doing this gives C1 + C2 = 0, (1+ V5)C1 + (1 - V5)C2 = 2, (7.40) which, when solved for C1 and C2, gives 1 C = -C2 = V5 (7.41) APPLICATIONS 219 Putting these values in equation (7.39) gives, for the kth Fibonacci number,

7.2.2 Fibonacci Numbers Denote the kth Fibonacci number as Fk. These numbers are defined by the rule that the kth number is equal to the sum of the two preceding ones with the initial numbers being Fo = 0 and Fi = 1. They satisfy the following difference equation: FR = Fk-1+ Fk-2- (7.36) This is a second-order, linear difference equation with constant coefficients and can be easily solved. Setting F = rk gives the characteristic equation p2 - r -1= 0, (7.37) whose solutions are 1+ V5 1- V5 r2 = (7.38) 2 2 Consequently, the general solution of equation (7.36) is F = C1 1+ V5 1- V5 + C2 (7.39) 2 The arbitrary constants C1 and C2 may be found by imposing the initial conditions Fo = 0 and F1 = 1. Doing this gives C1 + C2 = 0, (1+ V5)C1 + (1 - V5)C2 = 2, (7.40) which, when solved for C1 and C2, gives 1 C = -C2 = V5 (7.41) APPLICATIONS 219 Putting these values in equation (7.39) gives, for the kth Fibonacci number,

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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Question

Explain the determine purple

7.2.2
Fibonacci Numbers
Denote the kth Fibonacci number as F. These numbers are defined by the
rule that the kth number is equal to the sum of the two preceding ones with
the initial numbers being Fo = 0 and F1 = 1. They satisfy the following
difference equation:
FR = Fk-1+ Fk-2.
(7.36)
This is a second-order, linear difference equation with constant coefficients
and can be easily solved.
Setting F = pk gives the characteristic equation
p2 – r – 1 = 0,
(7.37)
whose solutions are
1+ V5
1- V5
r2 =
(7.38)
ri =
2
Consequently, the general solution of equation (7.36) is
1+ V5
Fr = C1
+ C2
(7.39)
2
2
The arbitrary constants C1 and C2 may be found by imposing the initial
conditions Fo = 0 and F1 =1. Doing this gives
C1 + C2 = 0,
(1+ V5)C1 + (1 – V5)C2 = 2,
(7.40)
which, when solved for C1 and C2, gives
1
C1 = -C2 =
V5
(7.41)
APPLICATIONS
219
Putting these values in equation (7.39) gives, for the kth Fibonacci number,
the result
+ V5
1
F. =
V5
1- V5
(7.42)
2
In the remainder of this section, we will examine some of the properties of
these numbers.
First, consider the limit
FR+1
lim
(7.43)
From equation (7.42), we find
(*-()
1 – ((1 – v5/(1+ v5)]k+1 \
-- [(1 – V5)/(1+ v5)]*
Fr+1
1+
-
(7.44)
Fr
Since
1- V5
< 1,
1+ V5
(7.45)
then
Fr+1
lim
1+ V5
= 1.618033988....
(7.46)
This number, called the golden mean, represents the ratio of the sides of a
rectangle that is supposed to be "most pleasing" to the eye.
Consider now the values of the Fibonacci numbers for the index k being a
negative integer. The most direct way of obtaining these values is to substitute
-k for k in equation (7.42). Doing this gives
-k
1+ V5
1
F-k =
V5
2
2
(7.47)
1
V5
However,
--(누)
2
1- V5
(7.48)
1+ V5
2
and
(7.49)
1- V5

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