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).

2 thoughts on “Learnt this week … 31 January 2014

  1. I just think I must have forgotten this. I even a year of maths at university (though it was more “maths recipes” than theory) and so I cannot believe we weren’t taught this.

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