Domain, why are they both not defined?

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Discussion Overview

The discussion revolves around the domains of two mathematical expressions involving square roots, specifically focusing on the conditions under which these expressions yield real values. The scope includes conceptual understanding of domains in the context of ordinary differential equations (ODE).

Discussion Character

  • Conceptual clarification, Technical explanation

Main Points Raised

  • One participant states that for the expression z=\sqrt{-(x^2 + y^2)}, the only solution occurs at (0,0), as any other values lead to a square root of a negative number.
  • Another participant agrees and elaborates that for the expression z=\sqrt{-(x^2 + y^2 + 1)}, the expression is always negative, thus undefined for real numbers.
  • Some participants express uncertainty about their interpretations and suggest there may be additional considerations or "catches" to the problems presented.
  • A later reply confirms the understanding that the first expression is defined only at (0,0) and that the second expression cannot yield a real number due to the nature of the terms involved.

Areas of Agreement / Disagreement

Participants generally agree on the interpretations of the domains for both expressions, but some express uncertainty about the completeness of their reasoning.

Contextual Notes

There is an assumption that x and y are real-valued, and the discussion does not explore potential complex solutions or other mathematical contexts that might apply.

Who May Find This Useful

Students studying ordinary differential equations or those interested in understanding the conditions for the domains of mathematical functions involving square roots.

rocomath
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I'm currently self-studying ODE, text by Morris Tenenbaum, and I am confused by these 2 problems.

[tex]z=\sqrt{-(x^2 + y^2)}[/tex]

D: (0,0)

[tex]z=\sqrt{-(x^2 + y^2 +1)}[/tex]

undefined
 
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the second one obviously, the first one no clue
 
I assume that x and y are real valued as is the value z. So, for the first equation, you can only have x=0 and y=0 as domain because you would end up with [tex]\sqrt{0}=0[/tex] which is a real value. In all other cases you would have a square root of a negative number giving no solution for the real. For the second one it is always negative due to the squares and the adding of 1.

I assume this is correct, however I have a feeling there is a a catch to it...
 
No, coomast, you are correct. These are obviously assumed to be real valued functions of real numbers.

In [itex]z= \sqrt{-(x^2+ y^2)}[/itex], if x= y= 0, then z= 0. If either x or y is non-zero, then [itex]x^2+ y^2[/itex] is positive and so [itex]-(x^2+ y^2)[/itex] is negative. We cannot take the square root of a negative number so z is defined only for x= y= 0.

In [itex]z= \sqrt{-x^2+ y^2+ 1)}[/itex], if x= y= 0 then [itex]z= \sqrt{-1}[/itex] which is undefined for real numbers. If either x or y is non-zero, then [itex]x^2+ y^2+ 1[/itex] is larger than 1 so [itex]-(x^2+ y^2+ 1)[/itex] is less than -1, still negative and z is not defined for any x or y.
 
o wow i misread the problem
 
AH! I get it.

The first one, the only time it is defined is if it's restricted to (0,0).

For the second, since x^2 and y^2 will always give positive values, thus -(x^2 + y^2 + 1) will never be 0.

I appreciate the help, but be prepared for more! Lol, I'm on winter-break so I don't have a teacher to go to :-]
 

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