Sequences! problem help

[Zhalfirin88]

Homework Statement

Find the expression for the nth term for the sequence {1,1,-1,-1,1,1,-1,-1,...}

The Attempt at a Solution

No idea.

 

[Mark44]

Do you know what the next term (the ninth) is going to be? There's a very definite pattern in this sequence.

 

[Zhalfirin88]

Yes, but my instructor wants a formula for it. The pattern is very easy to see but I can't think of a way to express it mathematically.

 

[Mark44]

Let's take first things first, and not worry about the formula just yet. What are the next four terms in the sequence?

 

[Zhalfirin88]

1,1,-1,-1

 

[Mark44]

Yes, that's what I get. Here's what you have:
a₁ = 1
a₂ = 1
a₃ = -1
a₄ = -1
a₅ = 1
a₆ = 1
a₇ = -1
a₈ = -1
a₉ = 1
a₁₀ = 1
a₁₁ = -1
a₁₂ = -1

I can see that a_{4n + 1} = 1 for n = 0, 1, 2, 3, ... Can you continue from there? This would be one representation; there are probably others as well.

 

[Zhalfirin88]

I don't exactly understand your notation. For n = 1 you're saying that a₅ = 1? What does that mean?

 

[Zhalfirin88]

Bump. I'm not exactly sure that is what my instructor is looking for. I think he just wants one formula for the whole sequence which is why I'm having so much trouble with this. He specifically said that he won't accept anything obvious.

 

[Mathnerdmo]

There are a couple ways to do this, even if you don't want to do the "obvious".

Whenever things are periodic, a good starting point is looking at sine functions, so you might start there.

Also, I found another solution via generating functions. If you know that is, you might look at the generating function of this sequence. If you work on it, I'll show this solution later...

 

[Zhalfirin88]

No, I don't know what generating function are I don't think, at least I've never heard of them. I was thinking about the sine/cosine functions, but how do you make it repeat twice?

Just feel like this is a over my head a little bit.

 

[Mathnerdmo]

Just ignore my comment about the generating functions then, and let's investigate sine/cosine...

Well, if you look at the graph of y=sin(x), you'll see that it crosses the line $$y = \frac{\sqrt{2}}{2}$$ twice, then it crosses the line $$y = -\frac{\sqrt{2}}{2}$$ twice, then it repeats...

Try to use this fact to find your function.

 

[Mark44]

I don't exactly understand your notation. For n = 1 you're saying that a₅ = 1? What does that mean?

Yes, in part. I'm saying that a₁ = 1, a₅ = 1, a₉ = 1, and so on.

Thinking about this a little more, I would like to number the elements of the sequence starting at index 0: {a₀, a₁, a₂, ...}.

So you have
a₀ = 1
a₁ = 1
a₂ = -1
a₃ = -1
a₄ = 1
a₅ = 1
a₆ = -1
a₇ = -1
etc.

With this change it's possible to write a single formula for the members of this sequence, as
a_2m+n = <something>, for m = 0, 1, 2, ..., and n = 0, 1.

The <something> is a very simple expression that depends only on m, and gives the right sign alternation.

 

[Zhalfirin88]

@Mathnerdmo: Well that is equal to pi/4

Wasn't there something like sin((n-1) * pi/4) = 1 My teacher showed us something like that really quick and I didn't get to write it down. I know that's wrong, but do you at least know what I am trying to say?

@Mark44:
Can you explain that to me again? :) How can you say "for m = 0, 1, 2, ..., and n = 0, 1." Maybe we haven't covered that far, cause we've only used for n = 0, 1, 2, ... so far.

And give another hint ! ;)

 

[Mark44]

Mathnerdmo, IMO the advice to work with sine and cosine and the line y = sqrt(2)/2 adds a lot of needless complexity.

Zhalfirin88,
I can't give you much more of a hint without doing the problem. What I wrote was a compact way of writing the index on a term in the sequence. That's all that a_{2m + n} is. If m = 3 and n = 1, which term in the sequence are we talking about? If m = 4 and n = 0, which term are we talking about?

 

[Zhalfirin88]

Yes I know that, I just never saw it written like that before. Anyways I don't see the <something> is equal to. Thanks for your help though.

 

[Mathnerdmo]

Mark44: It really depends whether we're okay with a two-parameter solution or whether we want a single-parameter solution. How I read that the teacher "won't accept anything obvious" is that he wanted to see the latter.

Unfortunately, I don't know any way to defend my approach other than to show the solution and spoiler tags don't seem to work with tex, so:

$$\sqrt{2}\mbox{ sin}(\frac{\pi}4) = 1$$
$$\sqrt{2}\mbox{ sin}(\frac{3\pi}4) = 1$$
$$\sqrt{2}\mbox{ sin}(\frac{5\pi}4) = -1$$
$$\sqrt{2}\mbox{ sin}(\frac{7\pi}4) = -1$$
etc.

and
$$a_n = \sqrt{2}\mbox{ sin}(\frac{(2n-1)\pi}4)$$
doesn't seem very complicated to me (especially if they've been talking about sine/cosine in class recently)

 

[Mark44]

Mathnerdmo,
I think that both of our solutions would satisfy the "won't accept anything obvious" requirement. The instructor's instructions against submitting obvious solutions are vague to the point of meaninglessness, I doubt very much that the instructor has any concern about one-parameter vs. two-parameter solutions.

Your representation works, but is more complex than mine by at least one measure: my representation can be written more compactly.

Zhalfirin88, what number multiplied by itself repeatedly produces 1 and -1?

 

[Mathnerdmo]

Mathnerdmo,
I think that both of our solutions would satisfy the "won't accept anything obvious" requirement. The instructor's instructions against submitting obvious solutions are vague to the point of meaninglessness, I doubt very much that the instructor has any concern about one-parameter vs. two-parameter solutions.

Your representation works, but is more complex than mine by at least one measure: my representation can be written more compactly.

My opinion differs from yours on the meaning of "obvious" here, but I won't create any more of an argument from my speculation.

Also, I personally prefer one-parameter solutions, which is why I went that direction *shrugs*.

 

[Zhalfirin88]

Honestly, I think that my teacher was looking for the solution Mathnerdmo gave. But he won't take off if it's still correct.

@Mark44: -1 :)

Are you saying a_{2m + n} = (-1)^m ?

 

[Mark44]

Honestly, I think that my teacher was looking for the solution Mathnerdmo gave. But he won't take off if it's still correct.

@Mark44: -1 :)

Are you saying a_{2m + n} = (-1)^m ?

Yes, that's exactly it.