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Take 27 lines that intersect in exactly one point (all coincide in the same). Would you tell that they define a smooth surface?

by enedil 11y ago

pklausler 11y ago
Good point.
duskwuff 11y ago
Certainly. Those lines could all lie in a plane, for instance -- that's very much a smooth surface.
- pklausler 11y ago
  I don't know why you're getting downvotes. Thanks for the example.
  - bcook 11y ago
    I would like to know the reason for the downvotes as well.
- impendia 11y ago
  It would not be cubic though (a plane is defined by a linear rather than a cubic equation). (p.s. I was not the downvoter)
  - thaumasiotes 11y ago
    It's a little odd to say a polynomial "isn't cubic" just because the degree-3 coefficient is zero. Ordinarily what matters is that every coefficient of degree higher than three is zero. Think of the result that you can fit a polynomial of degree n to go through n+1 points. It's easy to show that the polynomial indicated by that theorem which goes through a set of collinear points must be a line (degree-1 polynomial), no matter how many points are given[1]. But we don't say "as long as the points aren't collinear, this theorem holds"; we just accept that a line is a special case of cubic polynomial. [1] Proof: the polynomial of degree n going through n+1 points is unique. The line going through n+1 collinear points goes through all of those points, and therefore must be the unique polynomial of degree n to do so. I do see in the wikipedia article, though, that "cubic surface" appears to be defined to exclude polynomials which have any nonzero term of degree other than three. It's a weird world. :/
    - Sniffnoy 11y ago
      Well, it's defining it in projective space; in projective space, the polynomial has to be homogeneous to make sense. If you were working in affine space, you wouldn't have that requirement -- but then the theorem wouldn't be true without first passing to projective space, I imagine. Which would involve taking your polynomial and adding an extra variable so as to make it homogeneous.
    - impendia 11y ago
      > 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?

Take 27 lines that intersect in exactly one point (all coincide in the same). Would you tell that they define a smooth surface?

by enedil 11y ago

Take 27 lines that intersect in exactly one point (all coincide in the same). Would you tell that they define a smooth surface?

pklausler 11y ago

Good point.

duskwuff 11y ago

Certainly. Those lines could all lie in a plane, for instance -- that's very much a smooth surface.

pklausler 11y ago
I don't know why you're getting downvotes. Thanks for the example.
- bcook 11y ago
  I would like to know the reason for the downvotes as well.
impendia 11y ago
It would not be cubic though (a plane is defined by a linear rather than a cubic equation). (p.s. I was not the downvoter)
- thaumasiotes 11y ago
  It's a little odd to say a polynomial "isn't cubic" just because the degree-3 coefficient is zero. Ordinarily what matters is that every coefficient of degree higher than three is zero. Think of the result that you can fit a polynomial of degree n to go through n+1 points. It's easy to show that the polynomial indicated by that theorem which goes through a set of collinear points must be a line (degree-1 polynomial), no matter how many points are given[1]. But we don't say "as long as the points aren't collinear, this theorem holds"; we just accept that a line is a special case of cubic polynomial. [1] Proof: the polynomial of degree n going through n+1 points is unique. The line going through n+1 collinear points goes through all of those points, and therefore must be the unique polynomial of degree n to do so. I do see in the wikipedia article, though, that "cubic surface" appears to be defined to exclude polynomials which have any nonzero term of degree other than three. It's a weird world. :/
  - Sniffnoy 11y ago
    Well, it's defining it in projective space; in projective space, the polynomial has to be homogeneous to make sense. If you were working in affine space, you wouldn't have that requirement -- but then the theorem wouldn't be true without first passing to projective space, I imagine. Which would involve taking your polynomial and adding an extra variable so as to make it homogeneous.
  - impendia 11y ago
    > 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?