4.1 Case (1)
In this section we study the following special case of Equation 1
(7)
where the initial conditions x
-1, x
0 are arbitrary positive real numbers.
Theorem 5
Let be a solution of Equation 7. Then for n = 0, 1,...
where x
-1 = k, x
0 = h and F
n-1
, F
n-2are the Fibonacci terms.
Proof: For n = 0 the result holds. Now suppose that n > 0 and that our assumption holds for n-1, n-2. That is;
Now, it follows from Equation 7 that
Hence, the proof is completed.
For confirming the results of this section, we consider numerical example for x
-1 = 11, x
0 = 4 (see Figure 1), and for x
-1 = 6, x
0 = 15 (see Figure 2), since the solutions take the forms {6, -12, 4, -3, 1.714286, -1.090909, .6666667, -.4137931, .2553191,....}, {-60, 10, -8.571428, 4.615385, -3, 1.818182, -1.132075, .6976744,...}.
4.2 Case (2)
In this section we study the following special case of Equation 1
(8)
where the initial conditions x
-2, x
-1, x
0 are arbitrary positive real numbers.
Theorem 6
Let be a solution of Equation 8. Then , for n = 1, 2,...
where x
-2 = r, x
-1 = k, x
0
= h, i.e., g
m
= g
m-2+ g
m-3, m ≥ 0, g
-3 = 0, g
-2 = -1, g
-1 = 1.
Proof: For n = 1, 2 the result holds. Now suppose that n > 1 and that our assumption holds for n - 1, n - 2. That is;
Now, it follows from Equation 8 that
Hence, the proof is completed.
Assume that x
-2 = 8, x
-1 = 15, x
0 = 7, then the solution will be {17.14286, -13.125, 11.83099, 7.433628, -6.222222, -20, 3.387097, -9.032259,...}(see Figure 3).
The proof of following cases can be treated similarly.
4.3 Case (3)
Let x
-2 = r, x
-1 = k, x
0 = h, and F
2i-1, F
2i
, F
2i+1(where i = 0 to n) are the Fibonacci terms. Then the solution of the difference equation
(9)
is given by
Figure 4 shows the solution when x
-2 = 9, x
-1 = 12, x
0 = 17.
4.4 Case (4)
Let x
-2 = r, x
-1 = k, x
0 = h. Then the solution of the following difference equation
(10)
is given by
Figure 5 shows the solution when x
- 2= 21, x
- 1= 6, x
0 = 3.
4.5 Case (5)
Let x
- 2= r, x
- 1= k, x
0 = h. Then the solution of the following difference equation
(11)
is given by
Figure 6 shows the solution when x
- 2= 9, x
- 1= 5, x
0 = 4.
Figure 7 shows the solution when x
- 2= .9, x
- 1= 5, x
0 = .4.
4.6 Case (6)
Let x
- 2= r, x
- 1= k, x
0 = h, Then the solution of the following difference equation
(12)
is given by
Where i. e. u
m
= u
m-1- u
m-3, m ≥ 0, u
-3 = 0, u
-2 = 0, u
-1 = 1.
Figure 8 shows the solution when x
-2 = 11, x
-1 = 6, x
0 = 17.
4.7 Case (7)
Let x
-2 = r, x
-1 = k, x
0 = h and F
n-1F, n-2, F
n
are the Fibonacci terms.
Then the solution of the following difference equation
(13)
is given by
Figure 9 shows the solution when x
- 2= 8, x
- 1= 5, x
0 = 0.9.