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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.