Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

How to define the domain of a 2-variable function as this?

  1. Apr 8, 2012 #1
    Hi,

    I am wondering how to express the domain of a two-variable function f(x,y) as below.

    For any given y, f(x,y) is defined on [0, g(y)); for any given x, f(x,y) is defined on [0, h(x)). Then how should I specify the domain for the function f(x,y)?

    To be more specific, f(x,y)=x+y, and for any given y, it is defined on [0, y); for any given x, it is defined on [0, x). I want to know what is the domain of f(x,y).

    Thanks.
     
  2. jcsd
  3. Apr 8, 2012 #2
    the domain D is in general

    [itex] D=\{(x,y)\in R^2| x \in[0,g(y))[/itex] and [itex]y \in [0,h(x))\}[/itex]

    I don't think that without knowing more about h and g you can say more than this which is just a tidier formulation of what you say in your post.

    and in the specific case:

    [itex] D=\{(0,0)\}[/itex]

    maybe even this is excluded since does [0,0) include 0 or not...

    Anyway the example says in fact that f(x,y) is only defined for x and y greater or equal zero and x<y and y<x (with the exception above depending on [0,0) including 0 or not).

    although the function x+y is then quite arbitrarily restricted since we can easily define that on all of the real plane

    Actually now that I think about it the statement is rather contradictory since you say for any given y the function is defined on [0,y) so for instance f(3,4) should then be defined. but you say for any given x it is defined on [0,x) so f(3,4) should not be defined. unless you mean something like at least defined for this and that in which case the formulation of the problem is a bit weird and the answer would simply be:
    D is the real plane

    and in general

    [itex] D=\{(x,y)\in R^2| x \in[0,g(y))[/itex] or [itex]y \in [0,h(x))\}[/itex]

    I hope this helps!
     
  4. Apr 8, 2012 #3
    Thanks a lot for your reply.

    Yes, you are right. My initial statement is contradictory and not well-defined. Now I have cleared my mind with your comments. Thanks again.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: How to define the domain of a 2-variable function as this?
Loading...