Here's a semi-visual, semi-algebraic approach.
A useful alternative perspective on (univariate) polynomials is to think of them as being represented by their sequence of coefficients – filling in zeroes for any missing monomials. With this perspective, multiplying by $z$ simply shifts the sequence over by one. Arranging the coefficients in a table makes the result obvious. Here it is for $n=3$ using the $z$ form.
| | 0 | 1 | 2 | 3 | Corresponding polynomial |
| - | - | - | - | - | - |
|(1) | $1$ | $1$ | $1$ | $0$ | $z^2 + z + 1$ |
|(2) = $z$ times (1) | $0$ | $1$ | $1$ | $1$ | $z(z^2 + z + 1)$ |
|(3) = $-1$ times (1) | $-1$| $-1$| $-1$| $0$ | $-(z^2 + z + 1)$ |
|(4) = (2) + (3) | $-1$| $0$ | $0$ | $1$ | $z^3 - 1$ |
This is, of course, the typical algebraic proof just written with a different notation for polynomials, but I think it makes it pretty obvious how to arrive at the solution and why it is correct and will generalize. I do find viewing polynomials as their sequences of coefficients does often make what is happening clearer. Theoretically, a benefit of defining operations in terms of this sequence perspective, is that you are always operating on a normal form representation of the polynomials, so you never end up with the same polynomial presented in different ways.
Derek Elkins
·
2023-08-22T21:31:45Z (about 1 year ago)