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Proving Characteristics of Projections

  • #1
Can anyone think of a trivial proof for the fact that the projection of any polynomial onto the xy-plane itself gives a polynomial? My professor and I were speculating about this, but could not discover a trivial proof for the fact. Does such a thing exist?
 

Answers and Replies

  • #2
HallsofIvy
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First, it would help if you would explain what you are talking about. A "polynomial" is not a geometric figure and so has no "projection". Do you mean the graph of a polynomial in x, y, and z.
 
  • #3
Yes, that's exactly what I meant.
 
  • #4
Bacle2
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I am not sure I understand: you have , I guess, a polynomial P(x,y,z) . Then evaluate P(x,y,0). This is a polynomial in (x,y).
 
  • #5
But a projection onto the xy-plane isn't the same as letting the z-coordinate equal zero. It may even be that the polynomial doesn't intersect with the xy-plane.
 
  • #6
Bacle2
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I'm sorry, it may be just me, but I'm not sure I understand your layout. Maybe a formal definition and/or some examples may help, if you could provide them. I understand, e.g., the projection of the unit sphere x2+y2+z=1 into the xy-plane to be the circle x2+y2=1 , but I don't understand well what you're doing.
 
  • #7
HallsofIvy
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No, the projection of the unit sphere, [itex]x^2+ y^2+ z^2= 1[/itex] onto the xy-plane is NOT the circle, [itex]x^2+ y^2= 1[/itex]. It is the disk [itex]x^2+ y^2\le 1[/itex]. That is NOT a polynomial function, it is not even a function.

The only interpretation I can put on this, so that it is true, is that a polynomial path in three dimensions, given by parametric equations, x= p(t), y= q(t), z= r(t), where p, q, and r are polynomials, projected to the xy-plane, which would indeed be x= p(t), y= q(t), z= 0, is a polynomial path in the xy-plane since p and q are still polynomials.
 
  • #8
Bacle2
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Yes, my bad, H of Ivy, i was obviously wrong, I was thinking about something else, also trying to make sense of the OP.
 

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