### Euler’s formula proof

Reading An Introduction to Laplace Transforms and Fourier Series I reach the point where it is stated, rather axiomatically, that: $e^{ix} = \cos x + i \sin x$.

This is a beautiful formula and has always suggested to me some sort of mystical inner mathematical harmony (yes, I am a materialist, but I cannot help it).

But these days I also want to see the proof, so here is one:

We know that complex numbers can be described in polar co-ordinates:

$z = |z| (\cos \theta + i\sin \theta)$

So too $e^{ix} = r(\cos \psi + i \sin \psi)$ where $r$ and $\psi$ depend on $x$.

Now (and applying the product rule) $\frac{d}{dx}e^{ix} = ie^{ix} = \frac{dr}{dx}(\cos \psi + i\sin \psi) + \frac{d \psi}{dx}r(-\sin \psi + i \cos \psi)$

So we equate the real and imaginary sides of both sides of this equality we have:

$ie^{ix} = (\cos \psi \frac{dr}{dx} - r \sin \psi \frac{d \psi}{dx}) + i(\sin \psi \frac{dr}{dx} + r \cos \psi \frac{d\psi}{dx})$

Then, recalling $e^{ix} = r(\cos \psi + i \sin \psi)$, we have $ir (\cos \psi + i \sin \psi) = ir \cos \psi - r sin \psi = ( \cos \psi \frac{dr}{dx} - r \sin \psi \frac{d\psi}{dx}) + i( \sin \psi \frac{dr}{dx} + r \cos \psi \frac{d\psi}{dx})$

By inspection we can see that $\frac{dr}{dx} = 0$ and $\frac{d\psi}{dx} = 1$, giving us:

$ie^{ix} = - r \sin \psi + ir \cos \psi$ and multiplying both sides by $i$ we have: $-e^{ix} = -r \cos \psi - i r \sin \psi$

Reversing the signs: $e^{ix} = r \cos \psi + i r \sin \psi$

But what of $r$ and $\psi$? Well, we have $\frac{dr}{dx} = 0$ and $\frac{d \psi}{dx} = 1$.

So $r$ is constant with respect to $x$ while $\psi$ varies as $x$.

If we set $x = 0$ then $e^{i \times 0} = 1$ – a wholly real number, so $\sin \psi = 0$ and $\psi = 0$. Thus $r ( \cos 0 ) = e^0 = r = 1$ and we can replace $\psi$ with $x$ throughout.

Hence: $e^{ix} = \cos x + i \sin x$.

### Centenary of Paul Erdős

Paul Erdős was born 100 years ago today.

picture of paul erdös at student seminar in Budapest (fall 1992) (Photo credit: Wikipedia)

As the most prolific mathematician of all time I am rather disappointed he hasn’t been deemed worth a “Google doodle“.

Erdős was victimised in the McCarthy era in the US and certainly seems to have had a warm relationship with the Hungarian Communist authorities (though there is nothing to suggest he was a spy), though eventually boycotted the country over its treatment of Israeli citizens.

To see some of his work have a look at the Erdős-Straus conjecture.

### Derivative of any number raised to the power of x

I know this is a piece of elementary calculus but I just worked it out from (more or less) first principles (as I knew what the answer was but did not know why).

Let $y = r^x$ what is $\frac{dy}{dx}$?

$r = e. k$ where $k$ is some constant. Hence $y=(ke)^x$ or $k^xe^x$

$\frac{r}{e} = k$ and so $\frac{e^{ln(r)}}{e} = k$ and so $e^{ln(r) - 1} = k$.

And $e^{x(ln(r) - 1)}e^x = r^x$, so $e^{xln(r)} = r^x$

Applying the chain rule $u = xln(r)$ and $y = e^u$ so $\frac{dy}{du}\frac{du}{dx} = e^u \times ln(r)$

Which is $e^{xln(r)}ln(r) = r^xln(r)$

Now I have written it all out I know there are redundant steps in there, but this is how I (re)discovered it…

### Finding “new” primes

To accompany How to Solve it, I also bought How to Prove It: A Structured Approach which deals with the construction of proofs.

I am puzzled, though, by its treatment of Euclid’s famous proof of the infinite order of the set of primes.

Not because it gets the proof wrong – but because I do not understand the answer it gives to one part of one exercise.

Now the proof (this is my summary not the book’s) runs like this:

Suppose you decide that the number of primes is finite, the set $p_1 ... p_n$ (where $p_1$ is 2 and so on). You then form a number $m$ from the product of these primes plus 1 – $m = p_1 \times p_2 \times ... \times p_n + 1$. This number cannot be a prime and yet is not divisible by any of the primes. Hence we have a contradiction and the set of primes must be infinite.

Now the book asks the following:

The proof of [the theorem] gives a method for finding a prime number from any in a given list of prime numbers:
(a) Use this method to find a prime different from 2, 3, 5 and 7
(b) Use this method to find a prime different from 2, 5 and 11

Now, the first seems easy enough:

$p_{new} = 2 \times 3 \times 5 \times 7 + 1 = 211$

and the book agrees with my calculation.

But what of the second?

Well, $2 \times 5 \times 11 + 1 = 111$, but that’s not a prime at all. The book gives the answer as “3 and 37″ (which obviously are primes) but I cannot see what formal method is used from the list of “2, 5 and 11″ to generate these numbers.

The obvious suggestion is that I have missed the “method” in the theorem as set out in the book, but I have reread it many times and I cannot see where I have missed anything (it is really just a longer version of my summary).

Can someone set out the formal method that would work with a partial list such as 2, 5 and 11?

### The great dumbing down debate

Standard matrix in mathematics (Photo credit: Wikipedia)

I was a bit surprised to find this blog linked from a discussion forum via an entry that said I had proved that Maths ‘A’ level had dumbed down. I don’t think I have done any such thing – as I am not equipped to do so.

Of course, I’d love to be able to suggest my B grade from 1984 was equivalent to an A* today but on what basis? (And, of course, another bit wants me to say today’s kids, like my own daughters, are way ahead of where we were, but that’s not backed by evidence either.)

The only comments I had – on the lower marked questions – were that they were very similar indeed to questions that might appear on today’s papers. Nobody has yet passed comment on the higher marked question.

As to my own “dumbing down” – actually once I studied the questions a bit and despite the mess I made of the higher marked question – I came to realise that while I was very rusty none of this was really over my head (as an actual maths graduate said she found one of the easier questions beyond her without the aid of an external reference I don’t feel too disheartened by that.)

If I hadn’t been accepted on to the PhD I probably would have applied to do Maths at undergrad level at Birkbeck – my interest was sufficiently piqued by the MSc in Computer Science – and part of me even now thinks that’s what I really should have done…

### Maths ‘A’ level: now for a tough one

From the same paper as the previous two questions, here is the hardest question (as measured by marks on offer – 12):

Sketch the curve given parametrically by $x = t^2$, $y = t^3$
Show that an equation of the normal to the curve at the point $A(4,8)$ is $x + 3y - 28 = 0$
This normal meets the x-axis at the point $N$. Find the area of the region enclosed by the arc $OA$ of the curve, the line segment $AN$ and the x-axis.

While I can see the ideas behind each part of this question I am not at all sure where to begin with it.

For background, by the marking scheme you should take about 16 minutes to answer this – (and about 5 minutes per question for the earlier two). No pressure then.

### Venn diagrams for 11 sets

Ask most people about set theory and you will get a blank look, but ask them about a Venn diagram and they are much more likely to understand: indeed Venn diagrams are so well grasped that Mitt Romney’s campaign for the US Presidency recently attempted to make use of them (though I am not sure it was much of a success, but that’s another story…)

So called 2-Venn (two circles) and 3-Venn diagrams are very familiar. But higher dimension Venn diagrams that are (relatively) easy to grasp (I’ll explain what I mean by that below) are actually difficult to produce – and until last month nobody had managed to get beyond 7.

So, let’s state a few basic properties of any Venn diagram (here is a good general survey of Venn diagrams)- firstly – each region (face) is unique – there is only one region where the bottom curve intersects with the right curve alone, and only one where it intersects with the left curve alone and only one where all three curves intersect (the grey region) and so on.

This image (taken from that survey, apologies) – shows a series of set intersections which are not a Venn diagram:

For instance, we can see the two shaded areas both represent intersections of the ‘blue’ and ‘red’ sets.

A second point is that there is a finite number of intersections. In other words segments of curves cannot lie on top of one another (in fact this rule means the intersections must be in the form of Eulerian points of zero length – as, following on from the last post about Aristotle’s Wheel Paradox, any segment of a curve is continuous and has an uncountable infinite number of points).
The 3-Venn example above illustrates some of the key points about easier to understand Venn diagrams – firstly it is simple: no intersection is of more than two curves and secondly it is symmetric. In fact, if we are willing to ignore these points we can draw Venn diagrams of any number of sets, each with less intelligibility than the last.

Drawing higher number simple and symmetric Venn diagrams is exceptionally difficult and it has been proved that such $n-Venn$ diagrams only exist when $n$ is a prime.

So we have 2-Venns and 3-Venns, and mathematicians have managed 5-Venns:And 7-Venns:

But, until now, simple symmetric 11-Venns have been elusive. Certainly 11-Venn’s have been around – as the example below shows:

This example is symmetric but it is not simple.

Now, though, a breakthrough has been made. Named newroz – the Kurdish name for the new year – the first simple, symmetric 11-Venn has come from Khalegh Mamakani and Frank Ruskey, both of the Department of Computer Science at the University of Victoria, Canada.

And it is beautiful:

That said, I don’t think it will be featuring in any presidential campaigns just yet – by definition there are $2^{11} - 1= 2047$ intersecting regions, probably a bit more than even the keenest voter would care for.

Update: hello to students from F. W. Buchholz High School, hope you find the page useful.

### Volume of a ball

As ane fule kno the volume of a ball (ie., the interior of a sphere) is $\frac{4}{3}\pi r^3$.

But I have just had one of those “why’s that then” moments and sought to prove to myself, using integration, why this would be so.

But my logic is flawed and I get the wrong result – so risking looking stupid – I am asking for someone to correct the error in my logic.

Starting from the start and a familiar (I hope!) lemma:

$\pi$ is the ratio of a circle’s circumference ($c$) to its diameter ($d$), hence $c = \pi d$ or, more familiarly, $c = 2 \pi r$ where $r$ is the radius of the circle.

Now the area (a) of a circle can then be found by integration:

$a = \int_0^r 2\pi x dx$, giving the familiar $\pi r^2$

So, my reasoning runs, the volume of a ball would then be:

$2 \int_0^r \pi x^2 dx$ or $\frac{2}{3}\pi r^3$, which is precisely half the figure it should be – so where have I gone wrong?

Update: with thanks to Hugh in the comments – of course what I have described is a (double) cone with a height equal to the radius of the base. Cones and circles are not the same, obviously.

One thing has occupied my free time more than anything else these last few days – Francis Spufford‘s marvellous work of history and imagination, Red Plenty.

The book is a marvel in joining linear programming, economics, mathematics, cybernetics, computing, chemistry, textiles, politics, sociology, popular music, genetics and history all in one long fabric. The book is not quite a novel but nor is it history, the author himself calls it a “fairy tale”.

The ground on which it works is the Soviet Union between Stalin’s death in 1953 and what might be considered as the cementing in of what was later called the “era of stagnation” in 1970. The main characters are the scientists and engineers who saw, in that time, a new hope for the USSR in Khrushchev‘s claims that “this generation will know communism” – with a 1980 deadline – and who was willing to indulge their hopes of a rational, mathematical reshaping of the Soviet system.

Novelisations of actual events and the actions of real and fictional people are interwoven with passages of historical and scientific commentary and the effect is that we can sympathise with the hopes and dreams of the scientists but also know that they are destined for heart-breaking (for many at least) failure as the essential gangsterised and cynical nature of the state created by Stalin crushes their hopes which, in any case, were always naive at best.

But along the way we get to understand why the Soviet Union excelled at maths (and to a lesser extent computing science) – as it was both free of the pollutant of Marxism-Leninism but also valued by the Marxist-Leninists – scared the west with epic economic growth in the 1950s and so failed its citizens economically – nobody lost their job or reputation by failing the consumer, but if you failed to deliver a capital good you risked both.

We also get a portrait of a society that is much more granulated than the simple riffs of anti-communism would let us believe.

At the top we see Khrushchev was a fool who had done many evil things but he also hoped to make amends, Brezhnev and Kosygin the champions of a new wave of repression and stultification but also men frightened by how earlier reforms led to massacres and desperate not to see that return.

But most of all we see the scientists and their hopes get ground down. They all begin as believers and have only three choices in the end: to rebel and lose every physical thing, to compromise and lose hope or to opt out of the real world and chose only science.

The bitterness of their defeat, and that of all those who hoped for a better world after Stalin, is summed up in the words of a (real) song, sung in the book by Alexander Galich, a writer of popular songs turned underground critic and, after the shock of the public performance, recounted here, of his satirical works, exile.

We’ve called ourselves adults for ages

We don’t try to pretend we’re still young

We’ve given up digging for treasure

Far away in the storybook sun.

(As a companion work I’d also recommend Khrushchev: The Man and His Era)

### Damian Green: venal or just poorly educated?

Image via Wikipedia

The UK’s immigration minister, Damian Green, made this claim:

“Do immigrants displace British workers in the labour market? The MAC [Migration Advisory Committee - official advisors to the UK government on immigration] research showed that in certain circumstances there can be displacement of British-born workers by non-EEA [European Economic Area] migrants, up to a level of 23 displaced for every 100 additional working age non-EEA migrants… This analysis gives us the basis for a more intelligent debate.  It supports a more selective approach to non-EU migration. The old assumption was that as immigration adds to GDP—national output—it is economically a good thing, and that therefore logically the more immigration the better, whatever the social consequences.”

The problem for Green is that the MAC research did not show what he claimed at all. Instead it showed that (I am quoting this excellent blog here) that for every 100 working-age non-EU immigrants who arrived in a particular UK region in a particular year, 23 fewer UK natives were employed in that region in that year.

Green appears to have made one of those classic post hoc ergo propter hoc mistakes of those with a poor grasp of maths or statistics – the fact that one event follows another does not mean they are actually causally linked. As the MAC professionals state:

“Our findings should therefore be considered as estimating the association between migration and the native employment rate rather than the impact of migration on the native employment rate”

Perhaps the Daily Telegraph ought to highlight Green’s mistake in their ongoing and well-timed campaign on numeracy – Make Britain Count – as Green’s mistake is all too common.

Of course, the Telegraph would first have to assure themselves that the former business news editor of the Times and Channel 4 News was so innumerate that he had indeed made a mistake.