limit points of a_n = sin(pi n^2 cos(1/n)), n=1,infty

[micke]

Find all the limit points of {a[sub:1sm5mxh4]n[/sub:1sm5mxh4]} (n = 1 to infinity)
where a[sub:1sm5mxh4]n[/sub:1sm5mxh4] = sin(pi*n[sup:1sm5mxh4]2[/sup:1sm5mxh4]cos(1/n))

Any help is appreciated.

 

[tkhunny]

Re: limit points

Isn't that just oscillating?

 

[micke]

Re: limit points

Isn't that just oscillating?

Well, the answer is supposed to be -1 and 1. One subsequence converges to -1 and another to 1.
But I don't understand why exactly. Somehow pi*n[sup:2iojp7qq]2[/sup:2iojp7qq]cos(1/n) must equal pi*n+pi/2 or something like that.

 

[pka]

Re: limit points

Well, the answer is supposed to be -1 and 1.

I just don't see that.
I think that someone looked at a graph and concluded something that is just not true.
But I am open to a proof of the above, although I know that this is beyond undergraduate calculus.

 

[micke]

Re: limit points

:idea:
I checked up on Taylor series in the textbook, where it says that cos x = 1 - x[sup:36e0b57n]2[/sup:36e0b57n]/2! + x[sup:36e0b57n]4[/sup:36e0b57n]/4! and so on.
If x = 1/n we get 1 - 1/(2!n[sup:36e0b57n]2[/sup:36e0b57n]) + 1/(4!n[sup:36e0b57n]4[/sup:36e0b57n])
Multiplying with pi*n[sup:36e0b57n]2[/sup:36e0b57n] we get pi*n[sup:36e0b57n]2[/sup:36e0b57n] - pi/2 + pi/(4!n[sup:36e0b57n]2[/sup:36e0b57n])
and the tail starting from the third term will approach zero as n approaches infinity.
So we're left with sin(pi*n[sup:36e0b57n]2[/sup:36e0b57n]-pi/2) which of course is either 1 or -1!

I also managed to solve another excercise I was stuck at by using the same method.

 

[pka]

Re: limit points

I think that since you posted this in the “Calculus Forum” both tkhunny and I assumed that the question was about limits of a sequence. I now realize that it is more of a topological question. Yes there are two subsequences from that set that do have 1 and –1 as sequential limit points.

 

[tkhunny]

Re: limit points

I thought the term "limit points" suspicious, but not suspicious enough.

 

[micke]

Re: limit points

I think that since you posted this in the “Calculus Forum” both tkhunny and I assumed that the question was about limits of a sequence. I now realize that it is more of a topological question.

Ok, sorry for the confusion.
The course is called "Several Variable Calculus", but being at the very end of the course we've started using a booklet called "Topology and Convergence".

I thought the term "limit points" suspicious, but not suspicious enough.

Hm, I looked up three possible translations for the word we use in Swedish. Maybe I should have used cluster point or accumulation point instead.

 

[pka]

Re: limit points

Ok, sorry for the confusion.
The course is called "Several Variable Calculus", but being at the very end of the course we've started using a booklet called "Topology and Convergence".

No problem. But in the future please post such questions in that “Advanced Forum”.
We, at least I do, tend to read questions there with more care.