Description
Fibonacci numbers πΉπΉ0, πΉπΉ1, πΉπΉ2, β¦ satisfy the following recurrence relation:
πΉπΉππ+1 = πΉπΉππ + πΉπΉππβ1 (ππ β₯ 1, πΉπΉ0 = 0, πΉπΉ1 = 1)
The sequence begins with 0,1,1,2,3,5 β¦. Now, consider the polynomial defined by
ππ(π‘π‘) β πΉπΉ0
0!
+
πΉπΉ1
1!
π‘π‘ +
πΉπΉ2
2!
π‘π‘2 +
πΉπΉ3
3!
π‘π‘3 β¦ = οΏ½πΉπΉππ
ππ!
β
ππ=0
π‘π‘ππ
This is also known as exponential generating function (EGF) of Fibonacci sequence.
1. (10 pts) Show the ππth Fibonacci number can be obtained as
πΉπΉππ = ππππ
πππ‘π‘ππ ππ(π‘π‘)|π‘π‘=0
2. (10 pts) By using the recurrence relation, prove that the following relationships holds:
ππ2ππ(π‘π‘)
πππ‘π‘2 = ππππ(π‘π‘)
ππππ
+ ππ(π‘π‘)
3. (20 pts) Calculate the Laplace transform of ππ(π‘π‘) from the differential equation above.
4. (20 pts) Solve ππ(π‘π‘) by using partial fractions expansion of πΉπΉ(π π ) and identify the zeros and poles of πΉπΉ(π π ).
5. Plot ππ(π‘π‘) in MATLAB between for π‘π‘ β [β2,2]:
a. (2 pts) By using EGF expression above by using terms πΉπΉ0, β¦ , πΉπΉ5
b. (8 pts) By using the exact expression obtained by using the inverse Laplace transform.
6. (20 pts) Obtain a closed-form expression for πΉπΉππ by using the closed-form expression obtained from the inverse
Laplace transform and πΉπΉππ = ππππ
πππ‘π‘ππ ππ(π‘π‘)|π‘π‘=0 (note that π’π’(π‘π‘) = 1,π‘π‘ β₯ 0, π’π’(π‘π‘) = 0,π‘π‘ < 0).
7. (10 pts) Express πΉπΉ19 by using the closed-form expression and calculate its value based on this expression in
MATLAB.

