What are the partial limits of cos(pi*n/3)?

[peripatein]

Hi,
Trying to find all partial limits of cos(pi*n/3), I separated it into:
a_3k -> -1
a_6k -> 1

Is this a valid approach? Are there any other partial limits?

 

[HallsofIvy]

Essentially you are looking at n= 3 (mod 6) and n= 0 (mod 6), which is a very good idea, but those two are not the only possibilities. Suppose n is equal 1 (mod 6). That is, n= 1+ 6k. In that case, $\pi n/3= \pi/3+ 2\pi k$. Now, $cos(A+ B)= cos(A)cos(B)- sin(A)sin(B)$ so $cos(\pi/3+ 2\pi k)= cos(\pi/3)cos(2\pi k)+ sun(\pi/3)sin(2\pi k)= cos(\pi/3)$ for all k.

Similarly, look at n= 2+ 6k, n= 4+ 6k, and n= 5+ 6k.

 

[peripatein]

Thanks! Hence this sequence has only four partial limits. Is that correct?

 

[peripatein]

And what about the sequence ncos(pi*n/4). I believe it has only two partial limits, converging to +-infinity. Is that true?

 

[haruspex]

In all sequences cos(n*pi/m), constant integer m, each value that the sequence can take occurs infinitely often. So every value is the limit of a subsequence. So, yes, for m=3 there are 4 values.
For n*cos(n*pi/4), what are the possible values in the sequence?

 

[peripatein]

Would it be incorrect to say that n*cos(n*pi/4) has merely two partial limits (+-infinity) as cos(n*pi/4) is bounded whereas n isn't?

 

[haruspex]

Would it be incorrect to say that n*cos(n*pi/4) has merely two partial limits (+-infinity)

I could be wrong, but I wouldn't have thought infinity qualified as the limit of a sequence. It isn't a number, so you can't converge to it.
OTOH, there is one finite limit for this sequence. What are all the possible values of cos(n*pi/4) ?

 

[peripatein]

Infinity, plus and minus, could well serve as PL's in this case, as thus the question is formulated in the textbook. In any case, the possible values are +-1/sqrt(2), 0 and 1. But that multiplied by infinity would either be infinity or, in the case of 0, undetermined, would it not?

 

[haruspex]

Infinity, plus and minus, could well serve as PL's in this case, as thus the question is formulated in the textbook. In any case, the possible values are +-1/sqrt(2), 0 and 1. But that multiplied by infinity would either be infinity or, in the case of 0, undetermined, would it not?

No, you're not multiplying cos(n*pi/4) by infinity, you're taking a limit. In particular, what is the limit of n*cos(n*pi/4) for the subsequence n = 4k+2?

 

[peripatein]

I still don't see why it wouldn't be infinity. Could you help me see it?

 

[haruspex]

I still don't see why it wouldn't be infinity. Could you help me see it?

(4k+2)cos((4k+2)pi/4) = (4k+2)cos(pi*k+pi/2) = 0 for all k. Therefore the sequence converges to 0.

 

[peripatein]

Thank you. Are there are any other partial limits?

 

[vela]

You tell us.