Math Is Fun Forum

luca-deltodesco

Member

Registered: 2006-05-05

Posts: 1,470

limit of a cosine progression

is there anyway to find an exact value for the following progression:

with a calculator, i get the limit to be 0.739085133215161

is there anyway of finding an exact value for it, i.e. in terms of a mathematical expression, like cos 0.25pi = 1/sqrt(2) etc.

Last edited by luca-deltodesco (2007-02-09 03:54:48)

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Ricky

Moderator

Registered: 2005-12-04

Posts: 3,791

Re: limit of a cosine progression

So L = cos(L).  Solving this equation will give you your solution.

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luca-deltodesco

Member

Registered: 2006-05-05

Posts: 1,470

Re: limit of a cosine progression

any idea how i can solve that

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mathsyperson

Moderator

Registered: 2005-06-22

Posts: 4,900

Re: limit of a cosine progression

You might be able to approximate by rewriting cos x as a Maclaurin expansion, but as you've already got an accurate approximation that probably wouldn't be very helpful.

As for finding the exact answer, I haven't any idea.

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Ricky

Moderator

Registered: 2005-12-04

Posts: 3,791

Re: limit of a cosine progression

I'm fairly certain that there isn't one, at least, in the tradition sense.  It's like coming up with an indefinite integral for e^(-x^2).

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"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

George,Y

Member

Registered: 2006-03-12

Posts: 1,379

Re: limit of a cosine progression

Well, I wonder whether the series is convergent. Of course, convergence is given by the question setting because the limit notion is used.

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Zhylliolom

Real Member

Registered: 2005-09-05

Posts: 412

Re: limit of a cosine progression

That is a good point to bring up George; it seems that we have merely assumed the limit exists and have not addressed the fact that the sequence may "oscillate" too much to converge. So let us determine if {a[sub]n[/sub]} is convergent. Clearly, {a[sub]n[/sub]} is bounded, since for any x ∈ R, -1 ≤ cos x ≤ 1, and thus |a[sub]n[/sub]| ≤ sup{1, |a[sub]0[/sub]|}. Then by the Bolzano-Weierstrass Theorem, {a[sub]n[/sub]} has a convergent subsequence. Now I leave the rest of this to someone else, and perhaps I will post a solution soon if no one seems to want to step up to the plate: {a[sub]n[/sub]} is convergent if every convergent subsequence of {a[sub]n[/sub]} has the same limit. Can you prove that this sequence has that property? I am very tired so I may be giving a difficult approach to the problem, but I am confident that it does in fact converge.

If we assume that the sequence does converge, then Ricky's solution is valid, since for any convergent sequence {a[sub]n[/sub]}, any tail of the sequence is also convergent to the same limit; in particular, lim(a[sub]n[/sub]) = lim(a[sub]n+1[/sub]).

luca-deltodesco

Member

Registered: 2006-05-05

Posts: 1,470

Re: limit of a cosine progression

well im not sure if i can prove its convergent, however, i did try it with over 1000 values of x from 0 to 2pi, iterating to a[sub]10000[/sub] and all of them did converge to the same value

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George,Y

Member

Registered: 2006-03-12

Posts: 1,379

Re: limit of a cosine progression

convergence just because |an| ≤ sup{1, |a0|}?
|a[sub]n[/sub]| ≤ Max{1, |a[sub]n-1[/sub]|} seems a better reason

if every convergent subsequence of {an} has the same limit.
-first off, we shall list all the convergent subsequences

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X'(y-Xβ)=0