What's all this, then?

Here is an interactive version of this diagram: (click on the image to show a larger version.)

Here we show how a Tschirnhausen transformation can be used to solve a quartic equation. The steps are: Ensure the quartic is missing the cubic term, and its initial coefficient is 1. We can do this by first dividing by the initial coefficient to obtain an equation \[ x^4+b_3x^3+b_2x^2+b_1x+b_0=0 \] and then replace the variable \(x\) with \(y=x-b_3/ 4\). This will produce a monic quartic equation missing the cubic term.

Test: \[ A^{3⁄2} \] A general cubic polynomial has the form \[ ax^3+bx^2+cx+d \] but a general cubic equation can have the form \[ x^3+ax^2+bx+c=0. \] We can always divide through by the coefficient of \(x^3\) (assuming it to be non-zero) to obtain a monic equation; that is, with leading coefficient of 1. We can now remove the \(x^2\) term by replacing \(x\) with \(y-a/3\): \[ \left(y-\frac{a}{3}\right)^{\negmedspace 3}+a\left(y-\frac{a}{3}\right)^{\negmedspace 2} +b\left(y-\frac{a}{3}\right)+c=0.

The date January 26 is one of immense current debate in Australia. Officially it’s the date of Australia Day, which supposedly celebrates the founding of Australia. To Aboriginal peoples it is a day of deep mourning and sadness, as the date commemorates over two centuries of oppression, bloodshed, and dispossession. To them and their many supporters, January 26 is Invasion Day. The date commemorates the landing in 1788 of Arthur Phillip, in charge of the First Fleet and the first Governor of the colony of New South Wales.

Recently we have seen senators behaving in ways that seem stupid, or contrary to accepted public opinion. And then people will start jumping up and down and complaining that such a senator only got a tiny number of first preference votes. One commentator said that one senator, with 19 first preference votes, “couldn’t muster more than 19 members of his extended family to vote for him”. This displays an ignorance of how senate counting works.

This evening I saw the Australia Brandenburg Orchestra with guest soloist Lixsania Fernandez, a virtuoso player of the viola da gamba, from Cuba. (Although she studied, and now lives, in Spain.) Lixsania is quite amazing: tall, statuesque, quite absurdly beautiful, and plays with a technique that encompasses the wildest of baroque extravagances as well as the most delicate and refined tenderness. The trouble with the viol, being a fairly soft instrument, is that it’s not well suited to a large concert hall.

Here’s an example of a transportation problem, with information given as a table: .transport table { width: 80%; table-layout: fixed;} th, td { text-align: center;} th { background: #DCDCDC;} Demands 300 360 280 340 220 750 100 150 200 140 35 Supplies  400 50 70 80 65 80 350 40 90 100 150 130 This is an example of a balanced, non-degenerate transportation problem.

For my elementary linear programming subject, the students (who are all pre-service teachers) use Excel and its Solver as the computational tool of choice. We do this for several reasons: Excel is software with which they’re likely to have had some experience, also it’s used in schools; it also means we don’t have to spend time and mental energy getting to grips with new and unfamiliar software. And indeed the mandated curriculum includes computer exploration, using either Excel Solver, or the Wolfram Alpha Linear Programming widget.

Here’s an example of a coloured tetrahedron: hello

There’s a celebrated elementary result which claims that: There are irrational numbers \(x\) and \(y\) for which \(x^y\) is rational. The standard proof goes like this. Now, we know that \(\sqrt{2}\) is irrational, so let’s consider \(r=\sqrt{2}^\sqrt{2}\). Either \(r\) is rational, or it is not. If it is rational, then we set \(x=\sqrt{2}\), \(y=\sqrt{2}\) and we are done. If \(r\) is irrational, then set \(x=r\) and \(y=\sqrt{2}\). This means that \[ x^y=\left(\sqrt{2}^\sqrt{2}\right)^{\sqrt{2}}=\sqrt{2}^2=2 \] which is rational.