How are even and odd functions defined in n-dimensions?

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Homework Help Overview

The discussion revolves around the definition of even and odd functions in the context of multiple dimensions, specifically focusing on a function f(x,y,z) and its integration over a symmetric region, such as a sphere centered at the origin.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to define odd functions in n-dimensions and questions whether their assumption about the function being odd is valid. Other participants explore the implications of parity transformations and the ambiguity in defining oddness for multivariable functions.

Discussion Status

Participants are actively discussing the definitions and implications of even and odd functions in multiple dimensions. Some guidance has been offered regarding the need to specify the type of transformation when discussing odd functions, and the relationship between symmetry in the region of integration and the result of the integral is being explored.

Contextual Notes

The original poster is working under the constraints of a homework assignment and is seeking clarification on their reasoning before receiving formal feedback on their submission.

PeteyCoco
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How are even and odd functions defined in n-dimensions?

In my homework, we had to integrate f(x,y,z)= (x^2)z + (y^2)z + z^3 over a sphere centered at the origin. My answer came out to be 0 and I made the guess that it might be because f(x,y,z) was an odd function. Now, I don't know if this is true. Does this assumption hold:

f(x) is odd if f(-x) = -f(x)
f(x,y,z) is odd if f(-x,-y,-z) = -f(x,y,z) ?

I'm thinking this might just mean it's odd across a plane.I just handed in my assignment and will have to wait 2 weeks before I know if what I did was right or wrong, so I'm asking here to speed things up!
 
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PeteyCoco said:
How are even and odd functions defined in n-dimensions?

In my homework, we had to integrate f(x,y,z)= (x^2)z + (y^2)z + z^3 over a sphere centered at the origin. My answer came out to be 0 and I made the guess that it might be because f(x,y,z) was an odd function. Now, I don't know if this is true. Does this assumption hold:

f(x) is odd if f(-x) = -f(x)
f(x,y,z) is odd if f(-x,-y,-z) = -f(x,y,z) ?

I'm thinking this might just mean it's odd across a plane.I just handed in my assignment and will have to wait 2 weeks before I know if what I did was right or wrong, so I'm asking here to speed things up!
I'm not sure about whether we refer to functions of more than one variable as even or odd in an overall sense, although I doubt that we do.

However, it's common to refer to a function of more than one variable as being even or odd with respect to an individual variable.

Factor z out of your function: f(x, y, z) = z(x2 + y2 + z2) .
 
PeteyCoco said:
f(x,y,z) is odd if f(-x,-y,-z) = -f(x,y,z) ?
The parity transformation takes (x,y,z) to (-x,-y,-z). It's common to say f has even or odd parity if f(-x,-y,-z) is equal to, respectively, f(x,y,z) or -f(x,y,z). But as SammyS noted, when you have more than one variable, it's ambiguous if all you say is "f is odd." You should specify under what type of transformation is the function odd.
 
Last edited:
Since the region of integration is symmetric in all three coordinates, the integral will be zero if the function is odd wrt any of the three.
 

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