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I got this interesting sequence from a friend and I wish to know more about its behavior.

I have a sequence of $10$, $-10$, and $198$ zeroes. Suppose we, for every number in the sequence, replace it with the sum of that number and the number after it in the sequence, last number being replaced with the sum of first and last numbers. We repeat this process indefinitely.
What is the behavior of such a series?

I think that the series will eventually increase indefinitely, looking at smaller case with smaller numbers of zeroes.

Calculation example:

original: $10 \ -10 \ \ \ 0 \ \dots \ 0$,
1st time: $0 \ -10 \ \ \ 0 \ \dots \ 10$,
2nd time: $-10 \ -10 \ \ \ 0 \ \dots \ 10 \ \ \ 10$,
3rd time: $-20 \ -10 \ \ \ 0 \ \dots \ 10 \ \ 20 \ \ \ 0$

and so on.

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  • $\begingroup$ can you give the calculation for few terms? $\endgroup$ – Arjang Jan 3 '16 at 11:37
  • $\begingroup$ I added a couple of sequences. $\endgroup$ – user298322 Jan 3 '16 at 11:41
  • $\begingroup$ Welcome to MSE! For some basic information about writing math at this site see e.g. here, here, here and here. $\endgroup$ – KittyL Jan 3 '16 at 11:47
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One iteresting way to look at it is to consider your intitial sequence to be a 200-vector and after that to notice that every step can be described as an application of a matrix $P$: $$\begin{bmatrix} 1 & 1 & 0 & \dots & 0\\ 0 & 1 & 1 & \dots & 0\\ \vdots & \vdots & \vdots & \ddots & \vdots\\ 1 & 0 & 0 & \dots & 1 \end{bmatrix}$$ Then behaviour of your sequence can be researched through properties of $P$, particularly it's eigenvalues, since via transition to Jordan basis you can basicaly look at behaviour of $A^n x'$, where $A$ is a normal Jordan form of $P$. For example, without much calculation, we can see that one of the eigenvalues is 2 with eigenvector, consisting of 200 ones. Since projection of your initial sequence on this vector is not null we can conclude that at least some component of your sequence will increase indefinitely as a consequence of $P$ acting on it's eigenvector with eigenvalue greater than 1.

Some less general results can be obtained through calculating $P^n$ up to n = 199 since it seems that every time rows of $P^n$ would contain $(n + 1)'s$ row of Pascal triangle moved to the right by one position each time we descend. This observation gives, for example, value of a first entry in a sequence up to $n = 199$: $a_n = 10 - 10*n$.

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  • $\begingroup$ I believe the original sequence is: 10, -10, 0, ..., 0 and after 1st time: 0 -10 0 ... 10. I'm assuming 1 is used for simplicity's sake. $\endgroup$ – user298322 Jan 3 '16 at 11:52
  • $\begingroup$ May be so, but that doesn't change the matrix that should be applied to compute next step. $\endgroup$ – Sergey Rusakov Jan 3 '16 at 11:56

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