Testing MathJax on Blogger

Get link
Facebook
X
Pinterest
Email
Other Apps

- June 11, 2021

Following suggestions on

http://holdenweb.blogspot.com/2011/11/blogging-mathematics.html

$\int_0^\infty \exp(-x) \mathrm{d} x = \left[-\mathrm{e}^{-x}\right]_{0}^\infty = 1\,.$

Looks OK!

Get link
Facebook
X
Pinterest
Email
Other Apps

Comments

Joel Feinstein23 July 2021 at 18:42
It would be good if I could edit my comments though!
ReplyDelete
Replies
Joel Feinstein23 July 2021 at 18:45
$\int_{-\infty}^\infty \exp(-x^2)\,\mathrm{d}x$
ReplyDelete
Replies

Add comment

An introduction to the Hahn-Banach extension theorem: Part VI

- April 07, 2024

In this post we sketch a second proof of the Hahn-Banach Extension Theorem for continuous linear functionals on real normed spaces, this time using transfinite induction. We omit some of the details, which the reader can fill in. In particular, some of the details are similar to those involved in proving the claim made in Part V (when we obtained an upper bound for a chain). For a non-zero ordinal

$\gamma$ , recall that, with ordinal interval notation, we have

$\gamma=[0,\gamma)$ . That is,

$\gamma$ is equal to the set of all ordinals which are strictly less than

$\gamma$ . However, we use ordinal interval notation throughout. If you prefer to use

$\gamma$ instead of

$[0,\gamma)$ that is fine, of course! Looking at the Wikipedia page on the Hahn-Banach Theorem , apparently both Hahn and Banach (independently) used transfinite induction in their proofs, rather than Zorn's Lemma. Theorem (Hahn-Banach Extension Theorem for linear functionals on normed spaces over

$\R$ ) Let

$(E,\jnorm)$ ...

A rule of thumb for when to use the Ratio Test

- December 11, 2024

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 ...

Discussion of the proof that the uniform norm really is a norm

- October 24, 2023

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...

Search This Blog

Explaining maths