Mathematics Weblog

Amazing Formula

Monday 11 October 2004 at 9:47 pm | In Articles | 3 Comments

There are many interesting formulae in mathematics;

$\displaystyle \sum _{i=1}^\infty \frac{1}{n^2} = \frac {\pi^2}{6}$

must be one of the most amazing of all.

The first reaction is where did that $\pi$ come from? You can find 14 different proofs of this in a paper on Robin Chapman’s Home Page [look for Evaluating zeta(2)]

Given this result can you prove another amazing result?
If you pick two positive integers at random, the probability of them having no common divisor is $\dfrac{6}{\pi^2}$

$\pi$ gets everywhere! See Wikipedia for more such as

$\displaystyle \left(\frac{1}{2}\right)!=\frac{\sqrt{\pi}}{2}$

$\displaystyle \frac{2}{\pi}=\frac{\sqrt{2}}{2}\frac{\sqrt{2+\sqrt{2}}}{2}\frac{\sqrt{2+\sqrt{2+\sqrt{2}}}}{2} \ldots$

$\displaystyle \int_{-\infty}^{\infty}e^{-x^2}\;dx=\sqrt{\pi}$

Primes

Sunday 10 October 2004 at 6:42 pm | In Articles | Post Comment

Euclid’s proof that there are an infinite number of primes is a classic and as such appears as the first proof in Proofs from The Book.
Equally well-known is the formula (known as The Prime Number Theorem) which tells you that the number of primes $\pi(x)$ less than is given by $\pi(x)\sim \dfrac{x}{\log x}$ which means that the larger the value of the closer (in a well-defined mathematical sense) $\dfrac{x}{\log x}$ is to $\pi(x)$ . This is quite hard to prove.

An easier, but non-trivial result, is Bertrand’s postulate which says that there is always a prime between and .

The fact that there are arbitrarily large gaps between successive primes is not difficult to prove. Suppose we want to find a gap between successive primes which is at least of size . Then we look at the numbers

$N!+2,N!+3,N!+4,\ldots,N!+(N-1),N!+N$

Then each of these numbers is not prime. Why? Look at N!+a where $2 \leq a \leq N$ . Then divides both and and so divides N!+a . Clearly a<N!+a so $N!+a=a \times \frac{N!+a}{a}$ shows N!+a is not prime.
So we have a series of N-1 numbers all of which are not prime; thus the gap between a prime less than N!+2 and a prime more than N!+N is at least .

Error

Saturday 2 October 2004 at 10:04 pm | In Articles | 2 Comments

Six months ago in an article on the LambertW function I wrote:

$\ln$

$\displaystyle \ln x=\int ^x _0 \frac{dt}{t}\text{ for } x>0$

$\ln 1=0$

There’s a serious error in there which also completely invalidates it is then clear that …. Going back to the article the mistake leapt out at me – is it obvious to you? It’s strange how you read what you want to read rather than what is actually there 😕

Two maths problems solved??

Tuesday 7 September 2004 at 5:21 pm | In Articles | 1 Comment

The Guardian reports today, in an article with an over-the-top title of Maths holy grail could bring disaster for internet, that 2 of the seven millenium problems, the Riemann Hypothesis and the Pincar&#233 Conjecture may have been solved.

The article links a proof of the Riemann Hypothesis, to knowing more about primes, to cracking of codes based on prime numbers. This is tenuous at best. Yes, the proof may lead to further understanding of primes, but why that means that the codes become useless is beyond me. There are many results which depend on the Riemann Hypothesis (they say things like assuming the Riemann Hypothesis is true then …) so knowing the Riemann Hypothesis is true just makes these results true. If one of these results had led to cracking of codes it would be known about now.

On the other hand if someone shows that the Riemann Hypothesis is false, then that would cause a drastic rethink in many areas of mathematics.

Fascinating fact

Sunday 18 July 2004 at 11:35 am | In Articles | Post Comment

$2^{2^{2^{2^{2^{-\infty}}}}}=16$

seen in announcement for EuroTeX 2005

Mathematics Blogs

Thursday 15 July 2004 at 5:56 pm | In Articles | 2 Comments

It’s a shame there aren’t many maths blogs around, but Isabel’s math blog is an exception and is worth exploring.

I have put a newsfeed to this blog at Maths & Science Newsfeeds along with other resources.

Do you know any other good maths blogs/newsfeeds?

Powers

Friday 25 June 2004 at 3:37 pm | In Articles | 3 Comments

I was asked recently why 2^0=1 . Remember that if n is a positive whole number then $2^n= \underset{n \text{ times}}{\underbrace{2 \times 2 \times 2\times \dots \times 2}}$ . Clearly you can’t multiply 2 by itself 0 times 😕

The key, when extending properties of the number system, is to use definitions that work for every number. So, for example

$\dfrac{2^5}{2^2}=\dfrac{2 \times 2 \times 2 \; \times \! \not{2} \; \times \! \not{2}}{\not{2}\; \times \! \not{2}}=2^3=2^{5-2}$

which gives you the rule that

indices

This leads to

$1=\dfrac{2^3}{2^3}=2^{3-3}=2^0$

Similarly, a^0=1 for any positive real number.

What about zero powers of non-negative powers? 0^0 is a controversial case I have mentioned on 29 February (Q2.) and see Dr Math FAQ for more on this.

And if the number is negative? Great care is needed in this case. For example, using only real numbers, $(-1)^{1/3}=-1$ but $(-1)^{1/2}$ is not a real number. The problem arises because the general definition of a power is given by $a^x=e^{x\ln a}$ and $\ln a$ is undefined if a is negative or 0. Using complex numbers (which helps with $(-1)^{1/2}$ ) just makes things more complex 😕 – see Log of Complex Number

Misunderstanding

Wednesday 16 June 2004 at 4:23 pm | In Articles | 1 Comment

When teaching maths that you are familiar with, it is not easy to see why students struggle with it – indeed once the student has understood the problem, they can’t see why they had difficulty before! This means that you have to be very careful what you say in case an attentive student takes it literally.

How many times has a lecturer said “the integral of e is itself ” ? So this happens:

$\displaystyle \int_e^{e^2}\left(\frac{5}{x}+e\right)\ dx$

A. $\displaystyle \int_e^{e^2}\left(\frac{5}{x}+e\right)\ dx=\left[5\ln |x| + e\right]_e^{e^2}=\dots=5$

When the error was pointed out to the student, they responded with

Why does integrate to $ex\;$ ? integrates to and as surely the integral of is also $e\;$

How would you respond to this?

Generalisation of derivative

Sunday 6 June 2004 at 2:25 pm | In Articles | Post Comment

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

Most students will be familiar with the definition of the derivative of a real-valued function of a real variable defined on some interval (a,b):

$x \in (a,b)$

$\displaystyle \lim_{h \to 0}\frac{f(x+h)-f(x)}{h}$

$f^{\prime}(x)$

It is also clear that for this to make sense must be defined at (and of course it is a well-known consequence of the definition that is also continuous at ). But what if is defined on (a,b) but not at , can we do anything then? Yes, we can define a pseudo-derivative $\Lambda f$ of provided is defined on a neighbourhood of :

$\displaystyle \Lambda f(x)=\lim_{h \to 0}\frac{f(x+h)-f(x-h)}{2h}$

This pseudo-derivative has similar properties to the derivative and indeed it has the same values where is differentiable but there are significant differences as the following exercises show:

If is differentiable at show that $\Lambda f(x)=f^{\prime}x)$

If show that $\Lambda f(0)$ exists although $f^{\prime}(0)$ does not

If $f(x)=\left\{\begin{array}{ll}x & \mbox{ if } x<0\\-x^2 & \mbox{ if } x \geq 0\end{array}\right.$ show that has a local maximum at 0 but $\Lambda f(0) \neq 0$

Suppose is differentiable on , except at a point in , with $f^{\prime}(x)\geq 0$ for $x \neq x_0$ .
If $\Lambda (f)(x_0)$ exists show that $\Lambda (f)(x)\geq0$

New Mersenne Prime

Monday 31 May 2004 at 2:32 pm | In Articles | Post Comment

It’s not often that one can report mathematics news but two weeks ago a new Mersenne prime was discovered. A Mersenne prime is a prime number of the form 2^p-1 ie one less than a power of two. It is easy to show that itself must also be prime. The new prime number is $2^{24,036,583}-1$ has 7,235,733 decimal digits and was found on a 2.4 GHz Pentium 4 computer running Windows XP.

You can help search for larger primes and possibly win $100,000 for discovering the first 10-million-digit prime. See the Great Internet Mersenne Prime Search (GIMPS)

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