This question is 4 parts and I think I did Parts 1-3 right, I am just unsure of what or how to do part 4. If I messed up prior parts I wouldn't mind some input on them as well
- Choose four integers a, b, c, d whose absolute values are all distinct and all greater than 1. (For instance, -2, 4, 5,and -8 are a valid set of numbers.)
I chose the integers: a = -2 , b = 3 , c = 4 , d = -5
- With these, create a sequence with closed form an = a(b)^n + c(d)^n. Evaluate a0, a1, a2, and a3.
a0 = −2⋅(3 ^0)+4⋅(−5^0 ) = -2 + 4 = 2
a1 = −2⋅(3 ^1)+4⋅(−5^1 ) = -6 + -20 = -26
a2 = −2⋅(3 ^2)+4⋅(−5^2 ) = -18 + 100 = 82
a3 = −2⋅(3 ^3)+4⋅(−5^3 ) = -54 + -500 = -554
- Now, find a recurrence relation for which together with initial conditions a0 and a1, is solved by the closed form you found above.
To find a recurrence relation for the sequence: an = −2⋅(3^n) + 4⋅(−5^n) with initial conditions a0 = 2 and a1 = −26 I used a(n-1) and a(n-2) for a possible recurrence relation of:
an = 3a(n-1) - 5a(n-2)
- Now form a new sequence by swapping the values of a0 and a1. (For instance, if your initial choice of numbers produced a0 = 6 and a1 = 10, make a new sequence bn with b0 = 10 and b1 = 6.) With these values and the recurrence relation, create a new closed form for this sequence.
I am unsure of what to do for this part...
So far, you've created the sequence $a_n$ with the recurrence relation
$$\begin{eqnarray} a_0 & = & 2 \\ a_1 & = & -26 \\ a_n & = & 3 a_{n-1} - 5 a_{n-2}\end{eqnarray}$$
Now the question is asking you to consider the sequence
$$\begin{eqnarray} b_0 & = & -26 \\ b_1 & = & 2 \\ b_n & = & 3 b_{n-1} - 5 b_{n-2}\end{eqnarray}$$
and find a closed form for it.