Mathematics Weblog

Who Wants to be a Millionaire?

Saturday 3 April 2004 at 8:33 pm | In Articles | 1 Comment

On tonight’s programme the following question was asked:
Which is not a prime number?
The choices are

The sad thing is that the contestant said he gave up as he would only be guessing (a wrong answer would lose him a lot of money). 😥

Coin tossing is not fair

Saturday 20 March 2004 at 11:02 pm | In Articles | Post Comment

If you toss a coin and want it to come up heads then start with it heads up. 51% of the time it will come up heads. This remarkable result was discovered by a mathematician Persi Diaconis and you can hear an interview with him here (then click on All Things Considered audio link) where he talks about the lack of randomness of coin tossing.

It turns out that any randomness is just due to sloppy humans who toss coins differently every time. So all those text books on tossing coins are wrong and until now nobody had noticed!

Division by zero shock!

Tuesday 16 March 2004 at 4:35 pm | In Articles | Post Comment

It is well known that division by zero is the ultimate mistake to make in mathematics (see Dividing by 0 and Wikipedia). Different rules may apply in other advanced number systems but for use in the set of real numbers it has to be avoided. Doing so is often not easy, but one has to live with the consequences.

It is thus a shock to see division by zero hidden in a question in the AQA A Level Mathematics Specimen P4 Paper. Question 7 starts with

$\begin{displaymath}\text{f} : x\mapsto\tan3x-4x \end {displaymath}$

$0\le x \le \frac{\pi}{6}$

Where’s the division by zero? It is implicit in the definition of $\tan\theta$ which is defined as $\dfrac{\sin\theta}{\cos\theta}$ , provided of course that $\cos\theta\not=0$ ; so what happens in the given domain? Thus doesn’t exist, which makes the rest of the question unanswerable, since it requires you to differentiate and integrate 😀

The examiners should have known better. Perhaps it is just as well that the syllabus (and hence the specimen paper) will be replaced, starting in 2004.

Thanks to one of my eagle-eyed students who spotted this horror!

Pi

Sunday 14 March 2004 at 6:35 pm | In Articles | Post Comment

The digits in the decimal of $\pi$ have always intrigued mathematicans. In the computer age $\pi$ has been calculated to billion of places. You can see the first 30 million digits at The Inverse Symbolic Calculator.

$\pi$ is known to be irrational (so the decimal expansion is infinite and non-recurring) and transcendental (so it is not a root of a polynomial equation with rational coefficients). You can see a proof on my page here. Yet it isn’t known if the digits are random in the statistical sense.

One of the things that doesn’t help in examining the decimal digits of $\pi$ is that if you want to know, say, the 1000^th decimal digit then you will have to calculate the previous 999.

Note the word decimal in the last paragraph. If you go over to hex digits (base 16) then you can calculate any digit without knowing the previous ones. This truly remarkable result was discovered in 1996 by David Bailey, Peter Borwein and Simon Plouffe (their paper can be found here). What is astonishing is that the formula for $\pi$ used to calcuate the n^th is simple enough to be shown here:

$\displaystyle \pi = \sum _{k=0} ^{\infty} \dfrac{1}{16^k} \left[\dfrac{4}{8k+1}-\dfrac{2}{8k+4}-\dfrac{1}{8k+5}-\dfrac{1}{8k+6}\right]$

A program listing implementing this formula can be found here.
See Pi on the Web for lots more about $\pi$ .

Whoops!

Monday 8 March 2004 at 1:34 pm | In Articles | Post Comment

In Advancing Maths For AQA: Pure Maths 2 (P2) there is the following question on page 121:

2. Find the gradient of the following curve at the point where x = 0

$y = 4\sqrt{x}+2e^{-x}$

and gives an answer of 2 in the Answers section.

Spot the error.

Just for fun

Sunday 7 March 2004 at 10:41 pm | In Articles | Post Comment

$\displaystyle\int\frac{\mathrm{d}\left(\mathrm{cabin}\right)}{\mathrm{cabin}}=\log\left(\mathrm{cabin}\right)+\mathrm{sea}$

Zero

Sunday 29 February 2004 at 12:48 pm | In Articles | Post Comment

Why is ?
By looking at $\displaystyle\lim_{x\to 0}0^x$ and at $\displaystyle\lim_{x\to 0}x^0$ what can you say about ?
Can $\sqrt[0]{x}$ (the zero^th root of x) be defined? If so, how; if not, why?
Criticise the following ‘proof’: If $n\in\mathbb{Z}$ then $\displaystyle\lim_{n\to 0}\frac{n}{n}=\frac{\displaystyle\lim_{n\to 0}n}{\displaystyle\lim_{n\to 0}n}=\frac{0}{0}=1$

Powers and Logs

Monday 23 February 2004 at 8:10 pm | In Articles | Post Comment

Prove that $2^{\ln3}=3^{\ln2}$
The proof should show you how this generalises. If you have studied group theory you can extend this even further by showing that $\left(\mathbb{R}^{+}-\left\{1\right\},\circ\right)$ , where $\circ$ is defined by $\begin{displaymath} x\circ y=x^{\ln y}\end{displaymath}$ , is an abelian group.
Why can’t 1 be an element of this group?

Factorials

Saturday 21 February 2004 at 11:06 pm | In Articles | Post Comment

Inspired by a test posting on S.O.S. Mathematics CyberBoard

Factorials are fascinating. They are obtained by multiplying the numbers 1, 2, 3, 4, 5 … together and are written using the ! sign. Thus

$\begin{array}{llllr} 1! &=&1&=&1 \\ 2!&=&1\times2&=&2 \\ 3!&=&1\times2\times3&=&6 \\ 4!&=&1\times2\times3\times4&=&24 \\ 5!&=&1\times2\times3\times4\times5&=&120 \\ &&\cdots \\ 10!&=&1\times2\times3\times\ldots\times10&=&3628800 \\ &&\cdots \end{array}$

Factorials get large very quickly so 25! = 15 511 210 043 330 985 984 000 000 and as it gets larger there are more and more zeros.

Question: How can you find the number of zeros in 102! without calculating it?
Answer: All you do is count the number of times powers of 5 divide into 102 and add them up. So
$\begin{array}{llrl} 102\div5&=&20& \textrm{plus a remainder} \\ 102\div5^2&=&4& \textrm{plus a remainder} \\ 102\div5^3&=&0& \textrm{plus a remainder - get 0 so no higher powers needed} \end{array}$

Throw away the remainders and we get that there are 20+4=24 zeros in 102!
Now find how many zeros there are in 2004!

Can you show why this method works? Can you generalise it? Have a go, then look for answers on the internet.

Mathematicians will recognise this as a special case of
$\par\boxed{\begin{displaymath} \textit{The highest power of a prime p dividing n! is } \sum_{k=1}^\infty \left[\frac{n}{p^k}\right] \hspace{0.5em} \end{displaymath}}$
which is not difficult to prove by induction.