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A quadratic polynomial?

  1. May 8, 2012 #1
    Hello, I just wanna clear up a confusion.

    Is f(x,y) = a + bx + cy + dxy
    a quadratic polynomial?

    where x and y are variables and a,b,c,d are constants.
     
  2. jcsd
  3. May 8, 2012 #2

    Mark44

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    No. A quadratic polynomial in two variables looks like this:
    f(x, y) = ax2 + bxy + cy2 + dx + ey + f. See http://en.wikipedia.org/wiki/Quadratic_polynomial (two variables case).
     
  4. May 8, 2012 #3

    HallsofIvy

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    Mark44, I don't understand your response. You say "no" but then you give the general form for a quadratic in two variables which fits the one given by n0ya.
    [itex]a+ bx+ cy+ dxy[/itex]
    is the same as [itex]ax^2+ bxy+ cy^2+ dx+ ey+ f[/itex].

    Those are the same with your a= 0, b equal to nOya's d, c= 0, d equal to n0ya's b, e equal to n0ya's c, and your f equal to n0ya's a.
     
  5. May 8, 2012 #4

    mathman

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    The usual terminology would call this bilinear.
     
  6. May 8, 2012 #5

    Mark44

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    I was under the impression that a and c could not both be zero, but that b could be zero. Clearly, we can't have a = b = c = 0, or the function wouldn't be quadratic, so there need to be some restrictions on a, b, and c (but not on d, e, and f). Perhaps the restriction is that not all of a, b, c can be zero.
     
  7. May 8, 2012 #6
    To reduce this problem to its essence, take f(x,y) = xy. On the one hand there is no x2 or y2 term; but there is an xy term. So is this a quadratic or not?
     
  8. May 8, 2012 #7

    chiro

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    Hey n0ya and welcome to the forums.

    As a guiding rule for further questions like this, look at the highest term (in terms of the order) and use that as a basis to figure out what kind of equation/polynomial/blah something is on top of other things that need to be considered (for example polynomials only have integer powers so a x^(1/2) in the expression would make it a non-polynomial instantly).
     
  9. May 9, 2012 #8
    My problem is really whether f(x,y) = xy is a quadratic polynomial, as SteveL27 points out.

    I know that this polynomial is of degree 2, and according to Wikipedia, "a quadratic polynomial or quadratic is a polynomial of degree two", it is a quadratic polynomial. However, some argue that a x^2 or a y^2 term is needed in order to call it 'quadratic', so I just wanted to clear up this confusion.
     
  10. May 9, 2012 #9

    HallsofIvy

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    Yes, it is a quadratic polynomial. If you were to rotate the coordinate system through 45 degrees, calling the new coordinates x' and y', f(x,y)= xy would become [itex]x'^2- y'^2= 1[/itex]. Another way of looking at it is that the graph is a conic section, a hyperbola.
     
  11. May 9, 2012 #10

    HallsofIvy

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    No, it would not. "xy" is not linear.
     
  12. May 9, 2012 #11
    It's bilinear, meaning that

    x(y1 + y2) = xy1 + xy2

    and x(ay) = a(xy) for a constant a;

    and likewise in y. It's linear in each variable.
     
  13. May 9, 2012 #12

    mathman

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    In my experience there is a distinction between polynomials in one variable and polynomials in two variables. Highest order term x2y2 is called biquadratic and x3y3 is called bicubic.
     
  14. May 10, 2012 #13
    Thanks for the replies.

    Then, as far as I understand it, if you for example have: x = t; y = t, then f(x,y)=xy will be a parabola (t^2).

    Anyway, I think both quadratic and bilinear are both correct terms here. However, it might be more intuative to call it a bilinear mapping, instead of a quadratic polynomial, since we have no x^2 or y^2 terms.
     
  15. May 10, 2012 #14
    most of the time when you have an xy term the curve is a hyperbola
     
  16. May 10, 2012 #15

    Mark44

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    f(x, y) = xy represents a surface in three dimensions, not a curve in the plane, as are parabolas and hyperbolas. According to wolframalpha, the surface is a hyperbolic paraboloid.
     
  17. May 11, 2012 #16
    I think you found a bug, if you plot the old fashioned way n=xy, you get all of the real factors of n and it's a hyperbola. BTW, where's the third dimension variable?
     
  18. May 11, 2012 #17

    Mentallic

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    f(x,y) IS the 3rd dimension variable. It's basically equivalent to z=xy. For any fixed value z, then yes, you're dealing with a hyperbola, but if you consider the graph intersected by the plane y=mx, then you're dealing with a parabola z=mx2, hence why it's called a hyperbolic paraboloid.
     
  19. May 11, 2012 #18
    It must be too early, I was thinking f(x) not f(x,y)...so what is the answer to the original question? did we settle on bi-linear?
     
  20. May 11, 2012 #19

    Mentallic

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    I don't know. There seems to be quite a few different proposals and arguments for each term. I don't think we should settle on anything yet :tongue:
     
  21. May 11, 2012 #20
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