# 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).