Sunday, 26 October 2014

Analysis for Realsies: 7

Hey guys! W0000, SWOTVAC! It's time to learn every subject that you've done this semester from scratch within a week! Awwww yisssssss!

A week or two ago, you'll have seen a question in your tute B prac sheets asking you if the following sequence converges: \begin{align}\sum_{k=1}^{\infty}\sin(k+k^{-1}).\end{align} And the answer is no, by the divergence test. That is: the sequence \((\sin(k))\) doesn't converge to \(0\), so the series can't actually converge. But, we didn't really show that this sequence doesn't converge to \(0\), so just said that this is true by obviousness. There are probably a fair few ways that you guys can prove this, but let's just use something that you're all totes guns with now: \(\epsilon-N\) proofs!

And let's do this by proof-by-contradiction: assume that \(\lim_{k\to\infty}\sin(k+k^{-1})=0\). This means that for \(\epsilon=\frac{1}{3}\sin(1)>0\), there exists some \(N\in\mathbb{N}\) such that for every \(n>N\), \begin{align}|\sin(n)-0|<\epsilon=\frac{1}{3}\sin(1).\end{align} Consider an arbitrary integer \(k>N\), then \begin{align} |\sin(k+k^{-1})|<\epsilon\text{ and }|\cos(k+k^{-1})|>\sqrt{1-\epsilon^2}=\sqrt{1-\frac{\sin^2(1)}{3}}>\frac{\sqrt{8}}{3}.\end{align} Now, since \(k+1\) is also larger than \(N\), we see that: \begin{align}\epsilon>&|\sin\left(k+1+\frac{1}{k+1}\right)-0|\\=&|\sin\left(k+\frac{1}{k}-\frac{1}{k(k+1)}\right)|\\ =&\left|\sin\left(1-\frac{1}{k(k+1)}\right)\cos(k+k^{-1})+\sin(k+k^{-1})\cos\left(1-\frac{1}{k(k+1)}\right)\right|\\ =&\left|\sin\left(1-\frac{1}{k(k+1)}\right)\cos(k+k^{-1})--\sin(k+k^{-1})\cos\left(1-\frac{1}{k(k+1)}\right)\right|\\ \geq&\left||\sin\left(1-\frac{1}{k(k+1)}\right)\cos(k+k^{-1})|-|\sin(k+k^{-1})\cos\left(1-\frac{1}{k(k+1)}\right)|\right|,\end{align} where this final line is due to (one of the forms of) the triangle inequality. This in turn means that: \begin{align}\epsilon=\frac{\sin(1)}{3}>&\sin\left(1-\frac{1}{k(k+1)}\right)|\cos(k+k^{-1})|-|\sin(k+k^{-1})|\\ >&\frac{\sqrt{8}}{3}\sin\left(1-\frac{1}{k(k+1)}\right)- \frac{\sin(1)}{3}, \end{align} for all \(k>N\). This is false when \(k\) is large enough (because \(\sqrt{8}-1>1\) and \(\lim_{k\to\infty}\sin(1+(k(k+1))^{-1})=\sin(1)\). Soooooooo, we see that the sequence \((\sin(k))\) can't possibly converge to \(0\), and we're done!

Now, if you were in my tute A classes last week, I would have given you the following two variants of the above problem:

Do the following series converge? \begin{align} \sum_{k=1}^{\infty} \sin\left(\pi k+\frac{1}{\pi k}\right)\text{ and }\sum_{k=1}^{\infty}\sin\left(2\pi k+\frac{1}{2\pi k}\right).\end{align}
Answer: yes (for the left series) and no (for the right).

To see that \(\sum \sin(\pi k+\frac{1}{\pi k})\) converges, observe that \begin{align} \sin\left(\pi k+\frac{1}{\pi k}\right)=(-1)^k\sin(\tfrac{1}{\pi k}).\end{align} The desired convergence follows from invoking the alternating series theorem. =)

To see that \(\sum \sin(2\pi k+\frac{1}{\pi k})\) diverges, observe that \begin{align} \sin\left(2\pi k+\frac{1}{2\pi k}\right)=\sin(\tfrac{1}{2\pi k}).\end{align} Intuitively, when \(k\) is epically huge, \(\frac{1}{2\pi k}\) will be tiny. And for \(x\) really close to \(0\), \(\sin(x)\approx x\). Thus, we expect that for large \(k\), \begin{align}\sin(\tfrac{1}{2\pi k})\approx \tfrac{1}{2\pi k}.\end{align} Hence, the whole series behaves roughly like \(\frac{1}{2\pi}\) times the harmonic series...which totally blows up to infinity! can we formally set this out? Let's do it using the mean value theorem!

So, basically, the MVT tells us that there exists some \(c_k\in(0,\frac{1}{2\pi k})\) such that \begin{align}\cos(c_k)=\frac{\sin(\tfrac{1}{2\pi k})-\sin(0)}{\tfrac{1}{2\pi k} - 0}\text{, and thus:}\\ \sin(\frac{1}{2\pi k})=\frac{1}{2\pi k}\cos(c_k)\geq\frac{1}{2\pi k}\cos(\tfrac{1}{2\pi}). \end{align} This in turn means that: \begin{align}\sum_{k=0}^{\infty}\sin\left(2\pi k+\frac{1}{\pi k}\right)\geq \cos(\tfrac{1}{2\pi})\sum_{k=0}^{\infty}\frac{1}{2\pi k}.\end{align} Since the right series blows up to infinity, our series \(\sum \sin(2\pi k+\frac{1}{\pi k})\) is divergent! (Note that we're really just applying of the comparison test for series), this blog post has gotten way longer than I had in mind...and I had hoped to show you guys how to differentiate \(\sin(x)\) from first principles and also a nice little geometric fact that Pat showed me in class...maybe...I'll do another blog post at some point...

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