A dinner was held in Hall to mark the retirement of Michael Vaughan-Lee on 18th September 2010. Michael has been a dedicated tutor for many, many generations of Mathematics undergraduates at Christ Church.
Michael’s retirement was a moment to celebrate the long history of Mathematics at Christ Church. Charles Dodgson was of course a Lecturer then Student here, and the House has always had a strong presence in the subject. The Christ Church Mathematics community now numbers some 25 undergraduates, 30 graduates, two Junior Research Fellows (Becky Shipley and David Craven), two Lecturers (Chris Breward and John Wright), a Career Development Fellow (to be appointed), two Students (Sam Howison and Michael Vaughan-Lee, with Michael's replacement to be appointed) and, recently arrived, the Man Professor of Quantitative Finance (Thaleia Zariphopoulou, the first ever woman to hold a Statutory Professorship in Mathematics in Oxford). The former Christ Church Mathematics undergraduates or JRFs, Nick Woodhouse, Peter Howell and Dominic Joyce are on the Mathematics faculty at other colleges, while others hold positions in universities around the world.
The 18th September was a very special night. The list of Mathematics alumni is a long one and a good many of you came to renew old friendships. We contacted as many potentially interested attendees as possible, but we are missing contact addresses for those Old Members on the list attached below. We would be most grateful if anybody out there can help us get in touch with them; please send details to firstname.lastname@example.org so that we can inform them about future events.
Here are answers to the questions in the little pamphlet we distributed at the dinner.
1. 100. You replace two numbers 1/i and 1/j with 1/i + 1/j + 1/ij ; when you combine this with a third number 1/k you get
1/i + 1/j + 1/k + 1/ij + 1/jk + 1/ki + 1/ijk.
The patterm becomes clear: when you make the last replacement, you will have
1/i + 1/j + 1/k +... + 1/ij + 1/jk + 1/ki +... + 1/ijk + ... + 1/(ijkl....)...
which is the same thing as
(1 + 1/i)(1 + 1/j)(1 + 1/k) ....- 1
and the product is 101!/100!. (The representation as a product holds for any set of starting numbers.)
Well, that was Sam Howison's (applied mathematician's) answer. Michael V-L looked at it and sent the following: typically incisive comment:
I also enjoyed the 4 problems, and I was intrigued to see the answer to Problem 1 on the web, on the Ch Ch e-matters pages. The reason I was intrigued was that the problem [...] is really a problem in Pure Mathematics!! The binary operation * on the real numbers defined by
a*b = a+b+ab
is commutative and associative. So it doesn't matter what order the numbers are in, or the bracketing, and you can just * them up in the order given in the problem. After the first two or three you quickly see the answer is going to be 100."
2. Write n = n + 1 - 1 in the numerator and the sum collapses to 1 - 1/(n + 1)!.
3. The water flows in for the top 1/3 of the tidal cycle and it flows in fast because the tide rises a relatively long way above the level of the lake. It flows out more slowly because the lake only rises by a small amount.
4. Didn't fall for that one I hope. One of D and E is inside the triangle and one is outside.
Chris Breward (Maths Lecturer) and Sam Howison (Maths Tutor)