recurrence relation

Question

Find the solution of recurrence relation  a_n+1-a_n-1 = 2a_n for n ≥ 0 and satisfying initial conditions of a₀=3 and a₁=3.

Answer 1

I believe the solution is a summation for n ≥ 0 rather than the values of each term.

 

a_n+1 - a_n-1 = 2a_n    and be rearranged to

 

a_n+1 = 2a_n + a_n-1      The subscripts can be changed to n, n-1, and n-2 so we have an expression for a_n in terms of two previous values in the sequence.

 

 

Starting with a_n = 2a_n-1 + a_n-2 you can make successive substitutions for a_n-1, then a_n-2, and so on until you get to a_0.

 

a₄ has the value 12a₁ + 5a₀

 

a₆ has the value 70a₁ + 29a₀

 

What I am unable to get is a similar expression for the general term a_n

 

Maybe we are supposed to stop at a₂ given a₀ and a₁ are defined initial values.

 

I am hoping someone can clarify this.

 

Answer 2

     The way a recurrence relation works is that each term in a sequence is besed on the terms before it. It is usually solved fot the highest numbered term, but in your case, it is not. I would therefore solve it for  a(n+1). I'll do the first two using your definition.

     So you pick the first value of n where all the terms exist. In your case, when you pick n=0 you'd get a(1) - a(-1) = 2a(0). a(-1) doesn't exist so try the next value of n. Notice that you cant skip around in choosing n. Let's say n=10. I'd have to then figure out 

a(11) - a(9) = 2a(10). This is perfect but wont help us with anything since w dont know any of these terms. So, you must go in order.

 

n=1       a(2) - a(0) = 2a(1)  but we were given a(0) and a(1).     a(2) - 3 = 2*3    a(2) - 3 = 6      a(2) =9

n=2       a(3) - a(1) = 2a(2)  replacing what we know we get        a(3) - 3 = 2*9    a(3) - 3 = 18    a(3) =21

n=3       a(4) - a(2) = 2a(3)  we know a(2) and a(3)                     a(4) - 9 = 2*21  a(4) - 9 = 42    a(4) =51

 

Instead of solving an equation each time you could have rewritten this relation as a(n+1) = 2a(n) + a(n-1). so a(n+1) is clearly defined by combining the terms before it. Using this we get

n=4       a(5) = 2a(4) +a(3)      a(5) = 2*51+21 = 123

n=5       a(6) = 2a(5) + a(4)     a(6) = 2*123+51 = 297

 

Hope this helps.