Goldbach and Lewis

Tuesday 31 January 2006 at 9:41 pm | In Articles | 1 Comment

It’s not often that mathematics makes an appearance on British TV so I was suprised to see it in Lewis, ITV’s follow up to the Inspector Morse series (see for example Lewis, and the Ghost of Inspector Morse and Newsnight Review).

Oxford University often appeared in the Morse programmes and this time in Lewis it featured a very bright student who had shown that a Fields’ medallist work on the Goldbach Conjecture was flawed, leading to all sorts of murders. Along the way we were shown what a perfect number was, as perfect numbers were used for passwords.

It’s a pity that the mathematics shown on one of the whiteboards was elementary coordinate geometry :). The programme showed how much passion mathematics can arouse, though I wouldn’t have thought it has ever lead to murder!

Mathematics Today Article – Maths Blogs

Monday 30 January 2006 at 12:52 pm | In Articles | Post Comment

Craig Laughton, aka Gooseania, has written an article about mathematics blogs in Mathematics Today. Do go to his blog Exploring the blogosphere and read the article and provide the feedback and discussion he asks for.


Thursday 26 January 2006 at 9:26 pm | In Articles | 1 Comment

A problem that I give some of the students is to show that for a cubic f that if x=\alpha and x=\beta are the x-coordinates of its stationary points then f^{\prime\prime}( \alpha ) =-f^{\prime \prime}( \beta ). Hopefully, they will have already noticed this result in the exercises they have done.

I then ask them to show that the point of inflexion of f is at the midpoint of the turning points. This involves using the equations for the roots of a quadratic: \alpha + \beta=-\frac{b}{a} and \alpha \beta=\frac{c}{a} but the algebra isn’t particularly nice (details are here).

It dawned on me that there’s a better way which is not only simpler to prove but gives a more powerful result: a cubic is (rotationally) symmetric about its point of inflexion.

To see this, take the cubic and translate it so that the point of inflexion is at the origin. This clearly has no effect on the symmetry so it is sufficient to prove the result in this case.

If f(x)=ax^3+bx^2+cx+d has its point of inflexion at the origin then clearly d=0 and, as f^{\prime\prime}(0)=2b, it follows that b=0 so f(x)=ax^3+cx. Rotate this by \pi about the origin, for example by applying the usual matrix \begin{pmatrix} -1 & 0 \\ 0 & -1 \end{pmatrix}, and you get -f(x)=-ax^3-cx which is clearly the same cubic. Hence it has rotational symmetry about the origin, and in particular, the point of inflexion at the origin is the midpoint of the stationary points.

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