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 2nd difference sequence (Posted on 2005-10-22)
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),...

 See The Solution Submitted by Larry Rating: 3.5000 (2 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
 re: Solution? ... and a suggestion | Comment 4 of 11 |
(In reply to Solution? by goFish)

I tried your formula:
"In general a(n)=2 a(n-1)+a(n-2) (k-1) works for arbitrary values a(1),a(2), multiplier k and n>2;"

I arbitrarily chose the first value in the sequence to be 1, and I picked k to be 4, just to test it out.  If I interpret your formula correctly, I got for
the sequence:    1, 2, 7, 20, 61, 182
the 1st difference:  1, 5, 13, 41, 121
the 2nd difference:     4,  8, 28, 80

So this doesn't satisfy the condition ( the 2nd difference should have computed out to 1, 2, 7, 20, etc).  Besides, the question is to find the RATIO of a(n)/a(n-1).  (For the sequence in this comment, that ratio appears to be about 3)

Suggestion:  rather than trying to find a sequence that works, focus on the formulae for the 1st and 2nd differences.

Edited on October 23, 2005, 4:11 pm

Edited on October 23, 2005, 4:12 pm
 Posted by Larry on 2005-10-23 16:05:33

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