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> Proof: the polynomial of degree n going through n+1 points is unique. Suppose you have nine points lined up in an exact 3x3 grid. What, according to your pro

> Proof: the polynomial of degree n going through n+1 points is unique. Suppose you have nine points lined up in an exact 3x3 grid. What, according to your proposed definition, is the unique degree 8 polynomial going through them? There are two obvious cubic polynomials I can think of: One, cover the nine points by three horizontal lines and multiply the equations of the lines. Two, do the same with the vertical lines. I can't really think of any reasonable way to distinguish any one of these (or another) polynomial over other candidates. In contrast, with the definition that a cubic polynomial requires its coefficients to be of degree 3, you get things like Bezout's theorem: https://en.wikipedia.org/wiki/Bézout%27s_theorem https://en.wikipedia.org/wiki/Bézout%27s_theorem

thaumasiotes 11y ago
I was bothered as I was writing my comment by an issue related to this one. A curve covering a horizontal/vertical 3x3 grid of points cannot be a polynomial, as, if it is interpreted as a function, it has the range {true,false}, but some other domain, probably ℝ^2. The result I mention should be applicable to a set of points which can be represented as a function from ℝ to ℝ; no curve covering two different points which are vertically aligned can be represented that way (well, you could transpose the axes, but for a 3x3 grid that won't work either). I do have some questions, because I don't fully understand what you're saying: Suppose the 9 points are {-1,0,1} × {-1,0,1}. The horizontal line equations are y = -1, y = 0, and y = 1. If I multiply those together, I get y^3 = 0, which is a single horizontal line. I feel I must have done something wrong there. If I instead do (y + 1)(y - 0)(y - 1) = 0 I get y^3 - y = 0, which at least has y = -1, y = 0, and y = 1 as solutions. Since a cubic equation can't have four roots, this graph consists exactly of three horizontal lines. It's not a polynomial, since it has the " = 0" constraint embedded. (Using the definition on wikipedia, "[i]n mathematics, a polynomial is an expression consisting of variables (or indeterminates) and coefficients, that involves only the operations of addition, subtraction, multiplication, and non-negative integer exponents".) Is this what you meant? I know I've seen a diagram of exactly the example you're talking about somewhere, as something in the spirit of "math fun facts". But I can't find it; do you know of a writeup you could point me toward?
- impendia 11y ago
  I don't know of a good writeup, but y^3 - y = 0, or alternatively x^3 - x = 0, among other polynomials, is what you want. In "ordinary" algebra, typically we consider polynomials of one variable, which we write f(x). The graph of the polynomial represents the set of solutions to y = f(x), or alternatively y - f(x) = 0. In algebraic geometry, the solution set to y - f(x) = 0 is indeed a plane curve, but we think of y - f(x) as a special case of a polynomial in the two variables x and y. All such polynomials (except constant polynomials) also define plane curves. This is the concept that generalizes. In particular: y^3 - y = 0 and x^3 - x = 0 are both plane curves, they are both polynomials in x and y (i.e. we can think of y^3 - y = 0 as a polynomial in not only y but also x, even though x doesn't appear), and the intersection of these two curves is the set of nine points in question. Hope this helps.

thaumasiotes 11y ago

I was bothered as I was writing my comment by an issue related to this one. A curve covering a horizontal/vertical 3x3 grid of points cannot be a polynomial, as, if it is interpreted as a function, it has the range {true,false}, but some other domain, probably ℝ^2. The result I mention should be applicable to a set of points which can be represented as a function from ℝ to ℝ; no curve covering two different points which are vertically aligned can be represented that way (well, you could transpose the axes, but for a 3x3 grid that won't work either). I do have some questions, because I don't fully understand what you're saying: Suppose the 9 points are {-1,0,1} × {-1,0,1}. The horizontal line equations are y = -1, y = 0, and y = 1. If I multiply those together, I get y^3 = 0, which is a single horizontal line. I feel I must have done something wrong there. If I instead do (y + 1)(y - 0)(y - 1) = 0 I get y^3 - y = 0, which at least has y = -1, y = 0, and y = 1 as solutions. Since a cubic equation can't have four roots, this graph consists exactly of three horizontal lines. It's not a polynomial, since it has the " = 0" constraint embedded. (Using the definition on wikipedia, "[i]n mathematics, a polynomial is an expression consisting of variables (or indeterminates) and coefficients, that involves only the operations of addition, subtraction, multiplication, and non-negative integer exponents".) Is this what you meant? I know I've seen a diagram of exactly the example you're talking about somewhere, as something in the spirit of "math fun facts". But I can't find it; do you know of a writeup you could point me toward?

impendia 11y ago
I don't know of a good writeup, but y^3 - y = 0, or alternatively x^3 - x = 0, among other polynomials, is what you want. In "ordinary" algebra, typically we consider polynomials of one variable, which we write f(x). The graph of the polynomial represents the set of solutions to y = f(x), or alternatively y - f(x) = 0. In algebraic geometry, the solution set to y - f(x) = 0 is indeed a plane curve, but we think of y - f(x) as a special case of a polynomial in the two variables x and y. All such polynomials (except constant polynomials) also define plane curves. This is the concept that generalizes. In particular: y^3 - y = 0 and x^3 - x = 0 are both plane curves, they are both polynomials in x and y (i.e. we can think of y^3 - y = 0 as a polynomial in not only y but also x, even though x doesn't appear), and the intersection of these two curves is the set of nine points in question. Hope this helps.