Creating an alternating sequence of positive and negative numbers

Question

TL; DR -> How does one create a series where at an arbitrary $nth$ term, the number will become negative.

I'm learning a lot of mathematics again, primarily because there are such wonderful resources available on the internet to learn. On this journey, I've stumbled across some very interesting sequences, for example:

$$ a_n = \{ 1, -1, 1, -1, 1, -1, ... \tag{1} \}$$

And this is one example of an interesting diverging sequence, and this can be created using this function:

$$f(x) = x^{n+1} \tag{2}$$

Now this is a sequence that you can easily create, how would one create a series where you can have the $-1$ appear at an arbitrary $nth$ term?

For example:

$$ a_n = \{ 1, 1, -1, 1, 1, -1, ... \tag{3}\}$$

How would one attempt to define the series on $(3)$?

@Casteels Thank you so much for that, it was truly something interesting to read. — Nafiul Islam, Oct 04 '13 at 10:29

score 3 · Answer 1 · answered Oct 02 '13 at 12:08

3

$a_n=\frac13-\frac23\cos\frac{2n\pi}3-\frac23\cos\frac{4n\pi}3$

answered Oct 02 '13 at 12:08

bof

6,376

Could you make the font a little bigger? But this was indeed very interesting. – Nafiul Islam Oct 04 '13 at 10:26

score 2 · Answer 2 · answered Oct 02 '13 at 10:45

There are basically two approaches as I see it. Either you just say $$ a_n = \begin{cases}-1 & \text{if }3|n\\\\1 & \text{otherwise}\end{cases} $$ or, if you want a specific formula, the best way (I think) to go about it would be to find some sequence which is odd / even wherever you want $-1$ and even / odd wherever you want $1$, and raise $(-1)$ to that power. As an example, the Fibonacci sequence $F_n$ defined by $$ F_n = F_{n-1} + F_{n-2}, \quad F_1 = 1, \quad F_2 = 1 $$ is even at every third term, which means that the sequence $$ a_n = -(-1)^{F_n} $$ would be what you're looking for. This quickly gets pretty darn dirty, though, especially if you want to swap $F_n$ for the acual expression. I'd go with the top option, personally.

score 1 · Accepted Answer · answered Oct 02 '13 at 10:45

I know this might seem like a bit of a disappointing answer, but usually you would just define it exactly how you'd explain it (and hence piecewise).

$a_n= \begin{cases} -1 & 3\mid n \\ 1 & 3\nmid n \end{cases}$

or if you want $-1$ to appear every $k$'th term,

$a_n= \begin{cases} -1 & k\mid n \\ 1 & k\nmid n \end{cases}$

score 0 · Answer 4 · edited Apr 02 '19 at 12:48

0

what about $(-1)^{n(n-1)/2}$?

This seems to work if the sequence starts at $n=0$.

edited Apr 02 '19 at 12:48

Javi

6,541

answered Apr 02 '19 at 12:11

user660153

1

Welcome to Math.SE! Please format your posts using MathJax. – csch2 Apr 02 '19 at 12:20

Creating an alternating sequence of positive and negative numbers

4 Answers4

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