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Domain of ln(x+y)

  1. Oct 12, 2016 #1
    1. The problem statement, all variables and given/known data
    in my notes , it's stated that the domain of ln(x+y) is x+y> 0 ...
    Can i write it as x ≠ 0 , y ≠ 0 ? Or should i write it as (x,y) ≠ (0,0) ?


    (x,y) ≠ (0,0) represent x , and y can be 0 at the same time ?

    Whereas x ≠ 0 , y ≠ 0 only represent x cant be equal to 0 ? for example, when x = 0 , y not = 0 , so , ln(x+y) is defined , when y = 0 , x not = 0 , so , ln(x+y) is defined .

    Correct me if i am wrong ?

    2. Relevant equations


    3. The attempt at a solution
     
  2. jcsd
  3. Oct 12, 2016 #2

    fresh_42

    Staff: Mentor

    What about the negative numbers?
     
  4. Oct 12, 2016 #3
    negative number are ok , as long as x+y >0 , so , Can i write the domain as (x,y) ≠ (0,0) ?
     
  5. Oct 12, 2016 #4
    Hi Chetzread,
    You may be a little confused because typically we see the equation y = f(x).
    My advice is to reword the equation to z = f(x + y) = f(u), where u = x + y

    So ask yourself what is the domain of f(u)? What values can u take? What values of x and y add to give u?

    You are correct, both x and y cannot be equal to 0 at the same time. However, you may be surprised to learn that either x and y can be negative (not both).

    Below is a plot of z = ln(x + y) (obtained from Wolfram), where x is the horizontal axis, y is the axis going out of the page, and z is the vertical axis. Note what we have discussed about the cases y<0, x<0, x=0, y=0.

    https://www4b.wolframalpha.com/Calculate/MSP/MSP37851g488iahb1eeddb60000639baefd52h542g2?MSPStoreType=image/gif&s=63 [Broken]

    EDIT: Sorry I have misunderstood the OP. x, y ≠ (0, 0) seems correct.
     
    Last edited by a moderator: May 8, 2017
  6. Oct 12, 2016 #5

    fresh_42

    Staff: Mentor

    ##(x,y) \neq (0,0)## doesn't exclude ##(-1,-1)##. You will have to add ##x+y>0## anyway, or an equivalent notation.
     
  7. Oct 12, 2016 #6
    Yes, I know what x and y can be negative, as long as x+y> 0 , I'm wondering is it wrong to write the domain as write it as (x,y) ≠ (0,0) ?
     
    Last edited by a moderator: May 8, 2017
  8. Oct 12, 2016 #7
    Look at Fresh_42's comment, he/she has corrected my edit.

    EDIT: You can see it in the plot btw, what Fresh_42 is saying :). Good luck.
     
  9. Oct 12, 2016 #8
    so , if i write the limit as ##(x,y) \neq (0,0)## and ##x+y>0## at the same time , then , my ans would be correct ?

    How if i write the domain as ##x+y>0## only , without ##(x,y) \neq (0,0)## , is it correct ?
     
  10. Oct 12, 2016 #9

    Mark44

    Staff: Mentor

    All you need is x + y > 0. That way, x can be 0, but y can't, and y can be 0, but x can't.

    Something you might be missing is that the inequality x + y > 0 is an entire half-plane: all the points in the plane that are above the line y = -x. The origin is not in this set, so you don't need to also specify that ##(x, y) \ne (0, 0)##.
     
  11. Oct 12, 2016 #10
    so , just leave the final ans as x + y >0 , will do ?
     
  12. Oct 12, 2016 #11

    fresh_42

    Staff: Mentor

    ##x+y>0## automatically implies ##(x,y) \neq (0,0)##. It doesn't need to be mentioned. However, ##x=0## or ##y=0## is still possible as long as the other one is positive. They even can be negative as you correctly have said (as long as the other one is positive and larger in its absolute value). You might write ##x > -y## if you like.

    What is the purpose of writing the domain other than ##\{(x,y) \in \mathbb{R}^2\; | \; x+y > 0\}##?
     
  13. Oct 12, 2016 #12
    ok , i have one more question here , for the domain of ln (x^2 + y^2 ) , it it given in my notes that the ans is x ≠ 0 , y ≠ 0

    IMO , there 's no need to give x ≠ 0 , y ≠ 0 , so just leave the ans (x^2 + y^2 ) > 0 , will do ?
     
  14. Oct 12, 2016 #13

    fresh_42

    Staff: Mentor

    In this case you don't have to deal with negative numbers, because the squares are always positive: ##(-1)^2=1##.
    Therefore only ##(x,y)=(0,0)## can violate this condition. So here ##x^2+y^2 > 0## is equivalent to ##(x,y) \neq (0,0)##. As soon as one is unequal zero, the point is with in the domain. Drawn as a picture it is the entire plane without the origin.
     
  15. Oct 12, 2016 #14
    so , just choose anyone of ##x^2+y^2 > 0## or ##(x,y) \neq (0,0)## , the ans would be correct ?
     
  16. Oct 12, 2016 #15
    how about x ≠ 0 , y ≠ 0 is it wrong to write the limit as x ≠ 0 , y ≠ 0 ?
     
  17. Oct 12, 2016 #16
    @fresh_42 correct?
     
  18. Oct 12, 2016 #17

    fresh_42

    Staff: Mentor

    I'm not sure I understand what you mean by limit. The difficulty with ##x \neq 0\, , \,y \neq 0## is, that it is not clear whether it has to hold for both or just for one of them. Therefore ##(x,y) \neq (0,0)## is clear: not both zero at the same time.
     
  19. Oct 12, 2016 #18

    fresh_42

    Staff: Mentor

  20. Oct 12, 2016 #19
    sorry , i mean domain as x ≠ 0 and y ≠ 0 ... Here's the original ans that the author gave ...
     
  21. Oct 12, 2016 #20

    fresh_42

    Staff: Mentor

    Even this is ambiguous: For ##(x,y)=(1,0)## there is ##[\;(x \neq 0) \text{ and } (y \neq 0) \;] = \text{ false }##, but the point is allowed!
    Language often isn't very clear, especially when it comes to negations and even more if two of them are combined by and or or.
    It definitely should be avoided. In general it is always better to express something positively, if possible, rather than using "not". E.g. it is usually harder to prove, that something cannot occur, than it is to prove that something is true. Think of how long it took to prove Fermat's last theorem: more than 300 years!

    So ##\{(x,y)\in \mathbb{R}^2 \,\vert \, x^2 +y^2 > 0\}## is better than something with ##x \neq 0 </whatever> y \neq 0##.
    The pairing ##(x,y) \neq (0,0)## avoids these troubles.
     
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