## Archive for May 2nd, 2012

### Elsevier’s recent update to its letter to the mathematical community

May 2, 2012

Elsevier has recently put out a new statement giving details of some changes it has made. In their own words,

In February, we informed you of a series of important changes that we are making to how the Elsevier mathematics program will be run. In this letter, we would like to update you on where we currently stand, and inform you of some new initiatives we have undertaken based upon the feedback we have received from the community.

I have known for some time that they were going to make an announcement of this kind, and that it would involve something called “more flexible subject collections”. During that time I have become clearer in my mind what it is that I don’t like about bundling. So before Elsevier’s announcement, I had in mind some tests that I would apply, to see whether having these new collections would mitigate the problems with bundling. (Spoiler: they don’t.)
(more…)

### A look at a few Tripos questions IV

May 2, 2012

This post belongs to a series that began here. Next up is a question about integration.

11B. Let $f:[a,b]\to\mathbb{R}$ be continuous. Define the integral $\int_a^bf(x)dx$. (You are not asked to prove existence.)

Suppose that $m, M$ are real numbers such that $m\leq f(x)\leq M$ for all $x\in [a,b]$. Stating clearly any properties of the integral that you require, show that

$\displaystyle m(b-a)\leq\int_a^bf(x)dx\leq M(b-a)$.

The function $g:[a,b]\to\mathbb{R}$ is continuous and non-negative. Show that

$\displaystyle m\int_a^bg(x)dx\leq\int_a^bf(x)g(x)dx\leq M\int_a^bg(x)dx$.

Now let $f$ be continuous on $[0,1]$. By suitable choice of $g$ show that

$\displaystyle \lim_{n\to\infty}\int_0^{1/\sqrt{n}}nf(x)e^{-nx}dx=f(0)$,

and by making an appropriate change of variable, or otherwise, show that

$\displaystyle \lim_{n\to\infty}\int_0^1nf(x)e^{-nx}dx=f(0)$.
(more…)