College Math Teaching

October 25, 2013

Why the sequence cos(n) diverges

We are in the sequences section of our Freshman calculus class. One of the homework problems was to find whether the sequence a_n = cos(\frac{n}{2}) converged or diverged. This sequence diverges, but it isn’t easy for a freshman to see.

I’ll discuss this problem and how one might go about explaining it to a motivated student. To make things a bit simpler, I’ll discuss the sequence a_n = cos(n) instead.

Of course cos(x) is periodic with a fundamental region [0, 2\pi] so we will work with that region. Now we notice the following:

n (mod 2 \pi) is a group with the usual operation of addition.

By n (mod 2 \pi) , I mean the set n + k*2\pi where k \in \{..-2, -1, 0, 1, 2, 3,...\} ; one can think of the analogue of modular arithmetic, or one might see the elements of the group \{ r| r \in [0, 2 \pi), r = n - k 2\pi \} .

Of course, to get additive inverses, we need to include the negative integers, but ultimately that won’t matter. Example: 1, 2, 3, 4, 5, 6 are just equal to themselves mod 2 \pi. 7 = 7 - 2\pi (mod 2\pi), 13 = 13 - 4 \pi (mod 2\pi) , etc. So, I’ll denote the representative of n (mod 2\pi) by [n] .

Now if n \ne m then [n] \ne [m] ; for if [n]=[m] then there would be integers j, k so that n + j2\pi = m +k2\pi which would imply that |m-n| is a multiple of \pi . Therefore there are an infinite number of [n] in [0, 2\pi] which means that the set \{[n]\} has a limit point in the compact set [0, 2\pi] which means that given any positive integer m there is some interval of width \frac{2\pi}{m} that contains two distinct [i], [j] (say, j greater than i .)

This means that [j-i] \in (0, \frac{2\pi}{m}) so there is some integers k_2, k_3, so that k_2[j-i] \in (\frac{2\pi}{m}, \frac{2*2\pi}{m}), k_3[j-i] \in (\frac{2*2\pi}{m}, \frac{3*2\pi}{m})  , etc. Therefore there is some multiple of [j-i] in every interval of width \frac{2\pi}{m} . But m was an arbitrary positive integer; this means that the [n] are dense in [0,2\pi] . It follows that cos([n]) = cos(n) is dense in [-1,1] and hence a_n = cos(n) cannot converge as a sequence.

Frankly, I think that this is a bit tough for most Freshman calculus classes (outside of, say those at MIT, Harvard, Cal Tech, etc.).


May 29, 2013

Thoughts about Formal Laurent series and non-standard equivalence classes

I admit that I haven’t looked this up in the literature; I don’t know how much of this has been studied.

The objects of my concern: Laurent Series, which can be written like this: \sum^{\infty}_{j = -\infty} a_j t^j ; examples might be:
...-2t^{-2} + -1t^{-1} + 0 + t + 2t^2 ... = \sum^{\infty}_{j = -\infty} j t^j . I’ll denote these series by p(t) .

Note: in this note, I am not at all concerned about convergence; I am thinking formally.

The following terminology is non-standard: we’ll call a Laurent series p(t) of “bounded power” if there exists some integer M such that a_m = 0 for all m \ge M ; that is, p(t) = \sum^{k}_{j = -\infty} j t^j for some k \le M .

Equivalence classes: two Laurent series p(t), q(t) will be called equivalent if there exists an integer (possibly negative or zero) k such that t^k p(t) = q(t) . The multiplication here is understood to be formal “term by term” multiplication.

Addition and subtraction of the Laurent series is the usual term by term operation.

Let p_1(t), p_2(t), p_3(t)....p_k(t).... be a sequence of equivalent Laurent series. We say that the sequence p_n(t) converges to a Laurent series p(t) if for every positive integer M we can find an integer n such that for all k \ge n, p(t) - p_k = t^M \sum^{\infty}_{j=1} a_j t^j ; that is, the difference is a non-Laurent series whose smallest power becomes arbitrarily large as the sequence of Laurent series gets large.

Example: p_k(t) = \sum^{k}_{j = -\infty} t^j converges to p(t) = \sum^{\infty}_{j = -\infty} t^j .

The question: given a Laurent series to be used as a limit, is there a sequence of equivalent “bounded power” Laurent series that converges to it?
If I can answer this question “yes”, I can prove a theorem in topology. 🙂

But I don’t know if this is even plausible or not.

May 12, 2012

A simple demonstration of Cantor’s Diagonal Arugment

Filed under: advanced mathematics, infinity, logic, pedagogy, sequences — collegemathteaching @ 7:27 pm
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