A quadratic polynomial?

  • Thread starter n0ya
  • Start date
  • #1
7
0
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.
 

Answers and Replies

  • #3
HallsofIvy
Science Advisor
Homework Helper
41,833
961
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.
 
  • #4
mathman
Science Advisor
7,890
460
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.
The usual terminology would call this bilinear.
 
  • #5
34,549
6,259
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
795
7
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.
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
chiro
Science Advisor
4,790
132
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
7
0
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
HallsofIvy
Science Advisor
Homework Helper
41,833
961
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.
 
  • #10
HallsofIvy
Science Advisor
Homework Helper
41,833
961
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.
The usual terminology would call this bilinear.
No, it would not. "xy" is not linear.
 
  • #11
795
7
No, it would not. "xy" is not linear.
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
mathman
Science Advisor
7,890
460
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
7
0
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
coolul007
Gold Member
265
7
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.
most of the time when you have an xy term the curve is a hyperbola
 
  • #15
34,549
6,259
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).
most of the time when you have an xy term the curve is a hyperbola
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
coolul007
Gold Member
265
7
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.
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
Mentallic
Homework Helper
3,798
94
BTW, where's the third dimension variable?
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
coolul007
Gold Member
265
7
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.
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
Mentallic
Homework Helper
3,798
94
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:
 

Related Threads on A quadratic polynomial?

  • Last Post
Replies
5
Views
1K
Replies
25
Views
13K
Replies
4
Views
4K
  • Last Post
Replies
2
Views
1K
Replies
5
Views
1K
  • Last Post
Replies
6
Views
3K
  • Last Post
Replies
19
Views
3K
  • Last Post
Replies
4
Views
2K
  • Last Post
Replies
4
Views
5K
Top