# A quadratic polynomial?

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

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

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

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