This blog may be moving back!

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In 2008 this blog moved to http://explainingmaths.wordpress.com
At the time WordPress offered better LaTeX support. But if MathJax works here, that may be an improvement on the standard LaTeX support on WordPress.com
Let's see how it goes!

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A rule of thumb for when to use the Ratio Test

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The Ratio Test is a powerful test for both sequences and series of non-zero real numbers. It comes in various forms, but here is one commonly used version. Theorem (Ratio Test) Let $(x_n)_{n\in\N}$ be a sequence of non-zero real numbers. Suppose that \[\frac{|x_{n+1}|}{|x_n|}\to L \text{ as } n\to\infty\,,\] where $L$ is either a non-negative real number or $+\infty$. (a) If $L>1$, then $|x_n| \to +\infty$ as $n \to + \infty$, the sequence $(x_n)$ is divergent, and the series $\displaystyle \sum_{n=1}^\infty x_n$ diverges. (b) If $L \in [0,1)$, then the series $\displaystyle \sum_{n=1}^\infty x_n$ is absolutely convergent (and hence convergent), and the sequence $(x_n)$ converges to $0$. (c) If $L=1$, then the Ratio Test is inconclusive , and you need to use a different test . (The Ratio Test tells you nothing in this case about the convergence or otherwise of either the sequence or the series.) Notes Here (b) does not tell you anything about the value of the sum of the series ...
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The Extreme Value Theorem: a bootstrap approach

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At the very end of last week I was teaching the first-year mathematics students about the Extreme Value Theorem. Here is a reminder of what the theorem says. Theorem (Extreme Value Theorem) Let $a,b$ be real numbers with $a<b$, and let $f:[a,b]\to\R$ be continuous. Then there exist points $c$ and $d$ in $[a,b]$ such that, for all $x \in [a,b]$, we have $f(c)\leq f(x) \leq f(d)$. Here we have \[f(c)=\min\{f(x):x\in[a,b]\}=\inf\{f(x):x\in[a,b]\}\] and  \[f(d)=\max\{f(x):x\in[a,b]\}=\sup\{f(x):x\in[a,b]\}\,.\] Unfortunately I only had a few minutes left for this, ran out of time, and didn't finish the proof. As I am on a very tight schedule this year, I had to refer the students to the the typed notes for the rest of the proof. On the other hand, I did finish the first part of the proof, showing that a continuous real-valued function $f$ on a closed and bounded interval $[a,b]$ is always bounded above, i.e., the set $f([a,b])$ is bounded above. In this post I want to see how you ca...
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Some musings on Cauchy-Schwarz, Part Vb

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 Recall our claim from Part I Cauchy-Schwarz in complex inner product spaces can be deduced from Cauchy-Schwarz in real inner product spaces. This is essentially because every complex inner product space can also be regarded as a real inner product space with the same norm, but where the real inner product is the real part of the complex inner product. The last bit of the deduction uses a fairly standard "rotation trick": for any complex number \(z\), we have  \[|z|=\max\{\Re(\alpha z):\alpha \in \C, |\alpha|=1\,\}\,.\qquad(*)\] Let's check $(*)$ first. Let $z\in\C$, and set \[A=\{\Re(\alpha z):\alpha \in \C, |\alpha|=1\,\} \subseteq \R\,.\]  Clearly $A$ is non-empty.  Let $\alpha \in \C$ with $|\alpha|=1$. Then certainly we have \[\Re(\alpha z) \leq |\alpha z| = |z|\,.\] Thus we have \[\sup A \leq |z|\,.\] We now need to prove that equality holds, and that the supremum is actually a maximum. For this we just need to show that $|z| \in A$. This is trivial if $z=0$, b...
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