How to depress a general polynomial

This post showed how to do a change of variables to remove the quadratic term from a cubic equation. Here we will show that the technique works more generally to remove the xn−1 term from an nth degree polynomial.

We will use big-O notation O(xk) to mean terms involving x to powers no higher than k. This is slightly unusual, because typically big-O notation is used when some variable is tending to a limit, and we’re not taking limits here.

Let’s start with an nth degree polynomial

p(x) = ax^n + bx^{n-1} + cx^{n-2} + \cdots

Here a is not zero, or else we wouldn’t have an nth degree polynomial.
The following calculation shows that the change of variables

x = t - \frac{b}{na}

results in an nth degree polynomial in t with no term involving xn – 1.

 \begin{align*} p\left(t - \frac{b}{na}\right) &= a\left(t - \frac{b}{na}\right)^n + b\left(t - \frac{b}{na}\right)^{n-1} + {\cal O}(t^{n-2}) \\ &= a\left(t^n - \frac{b}{a}t^{n-1} + {\cal O}(t^{n-2})\right) + b\left( t^{n-1} + {\cal O}(t^{n-2})\right) + {\cal O}(t^{n-2}) \\ &= a t^n + {\cal O}(t^{n-2}) \end{align*}

Finite fields

This approach works over real or complex numbers. It even works over finite fields too, if you can divide by na.

I’ve mentioned a couple times that the Weierstrass form of an elliptic curve

y^2 = x^3 + a x + b

is the most general except when working over a field of characteristic 2 or 3. The technique above breaks down because 3a may not be invertible in a field of characteristic 2 or 3.

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One thought on “How to depress a general polynomial

  1. What properties does the depressed polynomial share with the original? Are they exactly identical? Surely they must differ in some ways, and are they ever interesting ways?

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