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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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 ...
The Extreme Value Theorem: a bootstrap approach
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...
Discussion of the proof that the uniform norm really is a norm
In response to a request on Piazza, I gave a detailed proof that the uniform really is a norm, including some comments and warnings on possible pitfalls along the way. Here (essentially) is what I said. Let’s have a look at \(\|\cdot\|_\infty\) on the bounded, real-valued functions on \([0,1]\) (but you can also work with bounded functions on any non-empty set you like, and the same proof will work). We define \(\ell_\infty([0,1]) = \{f: f \textrm{ is a bounded, real-valued function on }[0,1]\,\}\,,\) and this is the set/space we will look at. So we set \(Y=\ell_\infty([0,1])\). Then you can check for yourselves that \(Y\) really is a vector space over \(\mathbb{R}\) (using the usual "pointwise" operations for functions). Note that the zero element in this vector space is the constant function \(0\) (from \([0,1]\) to \(\mathbb{R}\)), which I'll denote by \(\mathbf{0}\) here for clarity. (We also used \(Z\) for this function in one of the exercises.) I'll first give...
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