T-shirt Slogan

Saturday 19 November 2005 at 6:57 pm | In Articles | 2 Comments

I can’t resist repeating a quote from Trevor Hawkes’ Blogging Mathematics

    We are number -e^{i\pi}

sometimes seen on mathematics students’ T-shirts.

Warwick blogs (who host Blogging Mathematics) have recently added \LaTeX using MimeTeX to their blogs so that students and staff can blog about mathematics.

Warwick was one of the first universities to offer blogs to their staff and students and, since they have such a strong mathematics tradition (and it’s my alma mater – hi Trevor :)), it’s nice to see a maths facility.

If you want to add mimeTeX support to your blog then LatexRender (used by this blog) has an option to do so (WordPress can use the plugin at Using LaTeX in WordPress). It is better to have the full \LaTeX support that LatexRender also offers but that may not be possible for many bloggers, as it requires \LaTeX, ImageMagick and Ghostscript on the server.

Another way is to install these programs on your own computer, generate the images, then upload to the server. LatexRender for Windows is a new program to help Windows users to do this.

Signature

Wednesday 9 November 2005 at 10:51 pm | In Articles | 4 Comments

Posters to forums often have signatures that are amusing and/or witty (though some are just plain daft). Few though are mathematical, but this one posted on Art Of Problem Solving’s Forum made me smile:

    The number you have dialled is imaginary. Please rotate your phone by 90 degrees and try again.

Cauchy-Riemann

Friday 28 October 2005 at 2:49 pm | In Articles | 1 Comment

The Cauchy-Riemann equations are one of the first results one comes across in Complex Analysis. A poster on S.O.S. Mathematics Cyberboard has pointed that that proofs like that at Cauchy-Riemann equations tend to take it for granted that if f(x+iy)=u+iv is analytic then the partial derivatives of u and v exist. Thus the proof at Cauchy-Riemann equations says

\displaystyle f^{\prime}(z)=\lim_{h\rightarrow 0}{\left[\frac{u(x+h,y)-u(x,y)}{h}+i\frac{v(x+h,y)-v(x,y)}{h}\right]} and then deduces that \displaystyle f^{\prime}(z)=\frac{\partial u}{\partial x} + i\frac{\partial v}{\partial x}

Looking at various textbooks this omission seems to the norm. Even Ahlfors Complex Analysis says: We remark that the existence of the … partial derivatives … is implied by the existence of f^{\prime}(z)

One excellent book A First Course in Complex Functions by G.J.O. Jameson does give a proper proof of this result. It defines differentiability for u:A \to \mathbb{R}^2 (where A is a subset of \mathbb{R}^2) at a point (a,b) in the interior of A if there exists real numbers \lambda,\mu such that, given \epsilon>0, there exists \delta>0 such that, for all real h,k with \sqrt{h^2+k^2}< \delta, |u(a+h,b+k)-u(a,b)-(\lambda h + \mu k)|\le\epsilon \sqrt{h^2+k^2}

Putting k=0 shows that \lambda = \dfrac{\partial u}{\partial x}; similarly \mu = \dfrac{\partial u}{\partial y}

If f^{\prime}(a+ib)=\lambda +i\mu then, given \epsilon>0, there exists \delta>0 such that for all real h,k with |h+ik|<\delta

|f((a+h)-i(b+k))-f(a+ib)-(\lambda + i\mu)(h+ik)|\le \epsilon|h+ik|

and taking real parts

|u(a+h,b+k)-u(a,b)-(\lambda h - \mu k)|\le\epsilon \sqrt{h^2+k^2} from which it follows that \dfrac{\partial u}{\partial x} and \dfrac{\partial u}{\partial y} exist. Taking imaginary parts gives the other 2 partial derivatives.

Times

Saturday 22 October 2005 at 9:17 pm | In Articles | 6 Comments

Along with other British mathematicians I am very lax about how I write the multiplication symbol. I will write

    \1 \times 2 \
1.2 \
1 \cdot 2

and then maintain that the context will tell the reader what is meant.

The problem comes from the fact that:
1. British students aren’t used to using the dot as a multiplication symbol – it doesn’t seem to be used in schools. They only want to use \times

2. The international students have never used \times (though it appears on their calculators which seem to be designed for the American market) and they are much more careful with the dot which must be on the centre of the line. They also are unhappy about the British habit of not worrying where the decimal point goes: 1.2 or 1 \cdot 2

The students love to tell me off for using the ‘wrong’ symbol on the grounds that whatever I do is wrong 😎 It makes for very interesting discussions about the international differences in mathematics and led to a wonderful note from some of the students:

    If I insist on using . for multiplication then asking them to do Q.1 on p.77 should be read as Q \times 1 \text{ on } p \times 77 😕

It’s nice to leave a class with a smile on my face.

PS For many years British students have used the word ‘times’ as in times by 3 or even worse ‘timesing’. It sounds horrible to me but I seem to have lost the battle to say ‘multiply’

Function Notation

Sunday 25 September 2005 at 6:17 pm | In Articles | 9 Comments

The notation f(x) is very common and is taught at A level (and sometimes earlier) in the UK. It’s well understood and used but has one well-known flaw – composition of functions. f\circ g or just fg is defined by fg(x)=f(g(x)). The problem here is that fg means do g first then f, rather than the other way round and is therefore counter-intuitive to the beginner.

One solution is to use a different notation for functions and use xf instead of f(x). It’s certainly more economical to write but, more importantly, the composite function fg is defined by x(fg)=(xf)g, which means do f then g. This seems to be much nicer if it weren’t for one thing. As far as I know, this notation is only taught in advanced courses (3rd year degree/postgraduate) and in books like Universal Algebra by P M Cohn, which is certainly not meant to be read by a novice. By this time, the f(x) is so engrained in a student’s mind it is quite difficult to change to xf – it certainly was for me 😕

Does anyone know if the xf notation is taught in more elementary mathematics courses; if so, how was it received?

A Further Five Numbers

Thursday 25 August 2005 at 12:34 pm | In Articles | Post Comment

BBC Radio 4 is broadcasting another series of programmes on numbers. This week’s programme features Benford’s Law which says that in naturally occurring data (so not random number tables) the number 1 appears as the first digit around 30% of the time, which is not what one would expect. It is used to detect forgeries.

You can listen to the programme at A Further Five Numbers and the rest of the series will look at 2,\ 6,\ 6.67 \times 10^{-11} \text{ and } 1729

The previous two series are also well-worth listening to; they feature 0,\ \pi,\ \phi,\ i,\ \infty,\ 4,\ 7,\ 2^{13,466,917} -1,\ 74\% and Game Theory. They can be found at Five Numbers and Another Five Numbers

(Thanks to Gooseania)

MathWorld

Wednesday 10 August 2005 at 11:47 pm | In Articles | 4 Comments

MathWorld is a wonderful mathematics resource that is well worth browsing. It’s nice to see it has a sense of humour too – see 2001 A Space Odyssey and 42.

Does anybody know of other similar humorous entries at MathWorld or at other such resources such as PlanetMath or the maths pages of Wikipedia?

Odd one out

Wednesday 3 August 2005 at 6:38 pm | In Articles | 5 Comments

You may well have seen those questions which ask “Which is the odd one out?”, and you often get them as part of a so-called intelligence test 😮
What the question really says is “Which one does the questioner think is the odd one out?”. So it becomes a rather more difficult test of trying to read someone else’s mind and nothing to do with your ability to solve a problem.

Here’s an example of a mathematical “odd one out” to show what I mean. Here are 4 ‘true ‘ statements:

    a.  1 + 1 = 0          true in \mathbb{Z}_2, arithmetic modulo 2
    b.  1 + 1 = 2          true in \mathbb{Z}, ring of integers
    c.  1 + 1 = 10         true in binary
    d.  1 + 1 = 11         true for string concatenation in Basic

The odd one out could be

    b.  because it is the only one that is true in \mathbb{Z}
    d.  because it is the only one that doesn’t involve addition
    a.  because it is the only one where the right-hand side is less than 1 (using ordering in \mathbb{Z})
    c.  because it is the only one where I can’t think of a reason (self-contradictory of course!)

Next time you come across one of those problems, see if you can give a reason why every one of the possibilities could be the odd one out.

Summer Reading

Monday 11 July 2005 at 5:25 pm | In Articles | 7 Comments

I finally succumbed and bought Martin Gardner’s Mathematical Games: The Entire Collection of His Scientific American Columns. Of course, as it had to be sent from America, the postage wasn’t cheap and it locked up my computer when I first tried it. Turning off CD autorun solved that problem.

Now I can spend the summer reading all those articles I’ve never read or even only half-remembered. The ability to search all the articles is a real boon. Did you know, for example, pi is mentioned 5645 times in the books?

In Fractal Music, Hypercards and More… Martin Gardner refers to the e^\pi>\pi^e problem (which I’ve mentioned before) and says:

    Dozens of proofs have already been published. One of the shortest is based on the fact from elementary calculus that x^{1/x} has a maximum value when x equals e. Hence e^{1/e} is greater than \pi^{1/\pi}}. Multiplying each exponent by \pi^e and canceling yields the inequality e^\pi>\pi^e.

The perils of publishing mathematics 😕 Can you spot the error and correct it?

Getting help

Monday 4 July 2005 at 11:08 pm | In Articles | Post Comment

At the S.O.S. Mathematics Cyberboard there’s a nice example of how students can spend all day struggling with a problem, but with a little help they can see their way to solving it in seconds. Such forums are great for getting those little nudges in the right direction that are so crucial in mathematics.

The student was trying to solve:

    For a,b,x,y \in \mathbb{R} show that if |x-a|< \varepsilon and |y-b|<\varepsilon then |xy-ab|<\left(|a|+|b|\right)\varepsilon+\varepsilon^2

If you’ve had a go and would like the ‘nudge’ the forum provided then click here Continue reading Getting help…

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