While working with a non-zero sequence recently, I noticed that when I found the second difference of the sequence, the result was identical to the original sequence. Specifically, the first term of the 2nd difference sequence was the same as the first term of the original sequence. And so on.
What is the limit (as n goes to infinity), of the ratio of the n-th term to the previous term?

Part 2
Another sequence has the property that each term of the 2nd difference sequence is equal to the corresponding term of the original sequence multiplied by "k", where k is a positive real number, not necessarily an integer.
For the original sequence, what is the ratio (in the limit) of the n-th term to the previous term?

Definition of 1st difference sequence:
For sequence: a(1), a(2), ..., a(n),...
1st difference is: a(2)-a(1), a(3)-a(2), ... a(n+1)-a(n),...

The most obvious thing to try first is to try a little algebra.

Seq.   1st dif.     2nd dif
f(n)   f(n+1)-f(n)  f(n+2)-2f(n+1)+f(n)

So for part 1:
f(n) = f(n+2) - 2f(n+1) + f(n)
f(n+2) = 2f(n+1)

Therefore, each term is double the previous one, at least for N>2.

For part 2:
f(n) = kf(n+2) - 2kf(n+1) + kf(n)
(k-1)f(n) - 2kf(n+1) + kf(n+2) = 0
k-1 - 2kf(n+1)/f(n) + kf(n+2)/f(n) = 0  OR  f(n) = 0

Assuming that the ratio has a limit, f(n+1)/f(n) approaches f(n+2)/f(n+1) as n approaches infinity.

k-1 - 2k ( f(n+1)/f(n) ) + k ( f(n+1)/f(n) )^2 = 0
f(n+1)/f(n) = [ 2k +/- sqrt( 4k^2 - 4k^2 + 4k ) ] / 2k
f(n+1)/f(n) = 1 +/- sqrt(k)/k

So there are 3 final answers for the limit of the ratio:
0
1 + sqrt(k)/k
1 - sqrt(k)/k

I'm not yet sure of the origin of two roots, but I don't yet have a reason to think either of them is an extraneous root.

I may have made mistakes.  Please correct me if you catch any.

Edit: I believe that I've made an error somewhere, since these ratios give 1/k rather than k.  I don't have the time or will to make corrections now though.

Edited on October 26, 2005, 2:05 am

Edited on October 26, 2005, 2:15 am

Posted by Tristan on 2005-10-25 19:14:10