So I now have two blogs: this one, and my WordPress blog at https://explainingmaths.wordpress.com/
My WordPress blog currently has far more traffic than this one, and I am only duplicating selected posts here. However, I do think that the Maths looks better here (using the MathJax version of LaTeX for web pages) than it looks on WordPress (using a less modern version of LaTeX for web pages). So I will continue to duplicate my new maths-filled posts and some of my older maths-filled posts here. But posts which are primarily text and images will just appear on the WordPress blog.
Popular posts from this blog
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 Christmas Equation
I posted the following in December 2021 on my first-year pure module's Piazza forum. Hi everyone I saw this a few years ago on QI, and found another version on the web at http://mathandmultimedia.com/2014/12/02/merry-christmas-equation/ If we are told that \(\qquad\displaystyle y=\frac{\ln(\frac{x}m-sa)}{r^2}\) multiplying by \(r^2\) gives \(\qquad yr^2=\ln(\frac{x}m-sa)\,.\) Then taking \(e\) to the power of both sides give \(\qquad\displaystyle e^{yr^2}=\frac{x}m-sa\,,\) and multiplying both sides by \(m\) gives \(\qquad me^{yr^2}=x-msa\,,\) which can be rewritten as \(\qquad me^{rry}=x-mas\,.\) (And a Happy New Year!) Best wishes, Dr Feinstein
Measures, outer measures and measurable sets I
Although I have taught Measure and Integration many times over the years, typically I don't have time to prove all of the details of the Carathéodory theory of outer measures and their associated $\sigma$-algebras and measures. I've been thinking about how we prove some of the fiddly bits at the end, and trying to decide how to make some of the final bootstrapping from finite unions to infinite unions a bit more natural. One possibility I have thought of is to change the order in which we prove some of the results, essentially in order to avoid proving some facts twice. However, this might be at the cost of having to prove some results again for general measures that we first prove for measures coming from outer measures. On the other hand, every measure on a $\sigma$-algebra is actually the restriction to that $\sigma$-algebra of a measure that comes from some outer measure! The same is also true for the more general "measures" on rings of sets and on semi-rings of...
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