From “The Foundations of Mathematics”

This is a fragment of a question from the first chapter of “The Foundations of Mathematics” (Amazon link):

Record whether you think the following statements are true or false:

(a) All of the numbers 2, 5, 17, 53, 97 are prime.

(b) Each of the numbers 2, 5, 17, 53, 97 is prime.

(c) Some of the numbers 2, 5, 17, 53, 97 are prime.

(d) Some of the numbers 2, 5, 17, 53, 97 are even.


For what-it’s-worth I thought all of these statements are true.

Could someone explain this contradiction to me?

Reading on with Julian Havil’s Gamma: Exploring Euler’s Constant and inspired by his discussion of the harmonic series, I come across this:

\frac{1}{1-e^x} = 1 + e^x + e^{2x} + e^{3x} + ...

Havil calls this a “non-legitimate binomial expansion” and it seems to me it can be generalised:

(1 - r^x)^{-1}= 1 + r^x + r^{2x} + r^{3x} + ...

as 1 = (1 - r^x)(1 + r^x +r^{2x}+r^{3x}+... )= 1 + r^x +r^{2x}+r^{3x}+...-r^x-r^{2x}-r^{3x}-...

And, indeed if we take x=-1, r=2 we get:

\frac{1}{1-2^{-1}} = 2 = 1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} +... at the limit.

But if we have x \geq 0 it is divergent and the identity, which seems algebraically sound to me, breaks down. E.g., r=2, x=2:

\frac{1}{1-4} = -\frac{1}{3} = 1 + 4 + 8 + 16 + ...

So what is the flaw in my logic?

In Pursuit of the Traveling Salesman

In Pursuit of the Traveling Salesman: Mathematics at the Limits of Computation

English: A representation of the relation amon...
English: A representation of the relation among complexity classes, which are subsets of each other. (Photo credit: Wikipedia)

This is a strange little book – and while I am happy to recommend it, I am not really clear in my own mind what sort of a book it is meant to be.

A popular description of a famous problem in the “NP” space and a gentle introduction to the whole issue of computational complexity and complexity classes? Not really, it assumes a bit too much knowledge for that.

So, a canter round the basic maths of the TSP and complexity? Not really that either, as it is light on mathematical specifics.

Still, it’s a decent book.

Was new maths really such a disaster?

English: Freeman Dyson
English: Freeman Dyson (Photo credit: Wikipedia)

On holiday now, so I fill my time – as you do – by reading books on maths.

One of these is Julian Havil’s Gamma: Exploring Euler’s Constant.

The book begins with a foreword by Freeman Dyson, in which he discusses the general failure, as he sees it (and I’d be inclined to agree), of mathematical education. He first dismisses learning by rote and then condemns “New Mathematics” as – despite its efforts to break free of the failures of learning by rote – an even greater disaster.

Now, I was taught a “new maths” curriculum up to the age of 16 and I wonder if it really was such a disaster. The one thing I can say is that it didn’t capture the beauty of maths in the way that the ‘A’ level (16 – 18) curriculum came close to doing. At times I really wondered what was it all about – at the age of 15 matrix maths seems excessively abstract.

But many years later I can see what all that was about and do think that my new maths education has given me quite a strong grounding in a lot of the fundamentals of computing and the applied maths that involves.

To the extent that this blog does have an audience I know that it is read by those with an interest in maths and computing and I would really welcome views on the strengths and weaknesses of the new maths approach.

A better demonstration of the product rule

Inspired by The Theoretical Minimum: What You Need to Know to Start Doing Physics: here’s a better proof/justification for the product rule in differential calculus than the one I set out here last month.

We will start with what we will treat as an axiomatic definition of the differential of the function y=f(x):

\frac{dy}{dx} = \frac{df(x)}{dx} = \frac{f(x+\Delta x) - f(x)}{\Delta x} as \Delta x \rightarrow 0

In this case we have y=f(x)g(x), so \frac{dy}{dx} = \frac{f(x + \Delta x)g(x +\Delta x) - f(x)g(x)}{\Delta x}

From our definition we can substitute for f(x+\Delta x) and g(x + \Delta x) and simplifying our notation for presentational reasons so that \frac{df(x)}{dx} = f^{\prime} etc:

f(x+\Delta x) = f^{\prime}\Delta x + f(x)

g(x+\Delta x) = g^{\prime}\Delta x + g(x)

Giving (after dividing through by \Delta x ):

y^{\prime} =f^{\prime}g^{\prime}\Delta x + g(x)f^{\prime} + \frac{f(x)g(x)}{\Delta x} + g^{\prime}f(x) - \frac{f(x)g(x)}{\Delta x}

=f^{\prime}g^{\prime}\Delta x + g(x)f^{\prime} +g^{\prime}f(x)

As \Delta x \rightarrow 0 the first term falls to zero and so we are left with:

y^{\prime}=f^{\prime}g(x) + g^{\prime}f(x)

Which, of course, is the product rule.

Update: See this most excellent comment from Professor Rubin.

Enhanced by Zemanta

Learnt this week … 31 January 2014

1. Roots (solutions) to a polynomial in a single variable

Quite why I was not taught this for ‘A’ level maths is beyond me (or more likely I was and have simply forgotten) but, if we have a polynomial in a single variable:

a_n x^n + a_{n-1} x^{n-1} + ... + n_0 = 0

Then the general form of its factorisation is:

a_n(x - q_n)(x - q_{n- 1}) ... (x - q_1)

and it has the roots:

q_n, q_{n - 1}, ..., q_1

Here’s an (very) outline proof…

Take a polynomial f(x) = 0 with a known root q then f(x) will divide evenly (no remainders) by x - q and so we can say f(x) = g(x)(x - q) where g(x) is of one degree less than f(x) . We can continue this until we are left with a function of degree 0 – i.e. the constant a_n (possibly 1) and then we have the form f(x) = a(x-q_n)...(x-q_1) .

(And I recommend Elliptic Tales: Curves, Counting, and Number Theory for those who want to know more – I am on my second iteration of trying to read this, but it is good fun.)

2. I can construct a curve that has no tangent at any point

At least, algebraically, it has no tangent.

For instance, the line defined by y^2 - 2xy + x^2 = 0.

We define the tangents for such lines, f(x,y) = 0 as, at the point (a, b) as being of the form: \frac {\partial f}{\partial x} x + \frac{\partial f}{\partial y}y = f(a, b)

Now, f(a, b) = 0, but note that \frac {\partial f}{\partial x} and \frac {\partial f}{\partial y} are also equal to zero for all points, so we have 0 = 0 which describes no line.

But – geometrically – this line is the same as that described algebraically by y = x which has a tangent line of y = x .

Are they different lines? They are, algebraically.

(This comes from the same source as point 1).