1. May 8, 2012

n0ya

Hello, I just wanna clear up a confusion.

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

where x and y are variables and a,b,c,d are constants.

2. May 8, 2012

Staff: Mentor

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).

3. May 8, 2012

HallsofIvy

Staff Emeritus
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.
$a+ bx+ cy+ dxy$
is the same as $ax^2+ bxy+ cy^2+ dx+ ey+ f$.

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.

4. May 8, 2012

mathman

The usual terminology would call this bilinear.

5. May 8, 2012

Staff: Mentor

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.

6. May 8, 2012

SteveL27

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?

7. May 8, 2012

chiro

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).

8. May 9, 2012

n0ya

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.

9. May 9, 2012

HallsofIvy

Staff Emeritus
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 $x'^2- y'^2= 1$. Another way of looking at it is that the graph is a conic section, a hyperbola.

10. May 9, 2012

HallsofIvy

Staff Emeritus
No, it would not. "xy" is not linear.

11. May 9, 2012

SteveL27

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.

12. May 9, 2012

mathman

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.

13. May 10, 2012

n0ya

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.

14. May 10, 2012

coolul007

most of the time when you have an xy term the curve is a hyperbola

15. May 10, 2012

Staff: Mentor

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.

16. May 11, 2012

coolul007

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?

17. May 11, 2012

Mentallic

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.

18. May 11, 2012

coolul007

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?

19. May 11, 2012

Mentallic

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:

20. May 11, 2012