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Homework Help: Definition of continuity in math help

  1. Feb 27, 2009 #1
    1. The problem statement, all variables and given/known data
    given: w is any bounded 2pi periodic function of one variable. and u(x,y) is a function in cartesian coordinates.
    show that u(x,y)=r*w(theta) is continous at the origin.

    2. Relevant equations

    u(x,y)=r*w(theta) is equal to v(r,theta) where v is a function in polar coordinates

    3. The attempt at a solution

    I know that the definition of continuity is:
    f(x) is said to be continuous at x=x0 if the limit of f(x) as x tends to x0 is equal to f(x0).

    but i dont know how to apply it in this case just that our x0 is the origin but how do i apply this in terms of polar coordinates?
  2. jcsd
  3. Feb 28, 2009 #2
    Re: continuity

    You want to show that u(x,y)->0 as (x,y)->(0,0).
    Use the fact that r(x,y)->0 as (x,y)->(0,0) and w(theta(x,y)) is bounded.
  4. Feb 28, 2009 #3


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    Re: continuity

    A nice thing about polar coordinates is that the distance from a point to the origin is given by r alone. [itex]r w(\theta)[/itex] goes to 0 as r goes to 0 no matter what [itex]\theta[/itex] is so the limit as (x,y) goes to (0,0) is 0, the value of the function.
  5. Mar 3, 2009 #4
    Re: continuity

    ok, so clearly as (x,y) tends to zero, then r also ttends to zero since x=rcos(theta) and y=rsin(theta); but how do we use the fact that w(theta) is bounded?
  6. Mar 3, 2009 #5


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    Re: continuity

    because, if w tended to infinity as theta went to a particular value, r w(theta) would not necessarily go to 0 as r goest to 0. Knowing that w has a bound, say M, rw(theta)< rM and goes to 0 as r goes to 0.
    Last edited by a moderator: Mar 9, 2009
  7. Mar 8, 2009 #6
    Re: continuity

    Thank you, that makes a lot of sense.
    The question then says: give an example where u is not continuous outside of the origin.
    I take it i can choose any function as long as it's not continuous outside the origin.
    I thought tan(x) but tan(x) is not bounded. I need a bounded function.
    How about u=tanh(x) ? would this be a good example for u?
  8. Mar 9, 2009 #7
    Re: continuity

    There's another part to the question that says:
    give a condition on 'w' such that u is also continuous outside of the origin.
    Does anyone have any idea about this condition?
    Thank you
  9. Mar 9, 2009 #8
    Re: continuity

    Shouldn't u be of the same form as above, i.e. u=r*w(theta)? What discontinuous functions do you know? Remember that tanh is a nice infinitely differentiable function on R, so this will not work.

    You want u to be continuous, so what would be the obvious condition for w? Keep in mind that you can always divide u by r if r is not 0, and that the quotient of continuous functions is continuous where the denominator is not 0.
  10. Mar 9, 2009 #9
    Re: continuity

    I think you are right. u should be of the form r*w(theta).
    The discontinuos functions i know are:
    y=1/x; discontinuous at x=0
    y=sin(1/x); discontinuous at x=0
    but i need a function where it's discontinuous outside the origin. I cant think of one. How about: y=x (for integer values of x).

    For the second part:
    can the condition for w be: that it has to be differentiable? (becuase if it is differentiable, then it is continuous).
  11. Mar 9, 2009 #10
    Re: continuity

    You could, for example, take the so called http://en.wikipedia.org/wiki/Everywhere_discontinuous_function" [Broken]. It is defined as follows: If x is a rational number, then f(x)=1, otherwise (if x is irrational) f(x)=0. This function is nowhere continuous (and difficult to draw)! It is also obviously bounded, which is what you are looking for.
    You could also use the fractional part or "floor" function.

    It is even simpler than that, you only need continuity (do you see why this is sufficient?).
    Last edited by a moderator: May 4, 2017
  12. Mar 9, 2009 #11
    Re: continuity

    Thank you for the dirichlet function. It's first time i hear of it but it's interesting.

    For the second part,
    what do you mean in this case by sufficient?
    can i say that u has a limit at all points??
  13. Mar 9, 2009 #12
    Re: continuity

    Do you see why, if w is continuous, then u will also be continuous? (Hint: the product of continuous functions is continuous)

    One more thing: I am assuming that theta is restricted to some range like [0,2pi). Otherwise you will need to choose a [tex]2\pi[/tex]-periodic function for w, i.e [tex]w(\theta)=w(\theta+2\pi)[/tex]. The Dirichlet function D is not [tex]2\pi[/tex] periodic, but [tex]D(x/(2\pi))[/tex] is.
  14. Mar 9, 2009 #13
    Re: continuity

    ok, so a condition on w is that it is continuous because the product of two continuous functions is continuous (as you said).

    theta is between 0 and 2pi; so in that case, can i take the dirichlet function D or must i take D(x/(2pi)) ?
  15. Mar 9, 2009 #14
    Re: continuity

    If theta is in the interval [0,2pi), so 2pi is excluded, then it's no problem. But if theta can be both 0 and 2pi, then you need to choose w so that it gives the same value for 0 and 2pi (because that is the same angle). You probably defined polar coordinates with theta in [0,2pi). If you want to be on the safe side you can take D(x/(2pi)), which works in both cases.
  16. Mar 9, 2009 #15
    Re: continuity

    2pi is included so theta can be 0 and 2pi.
    so do i say that w=D(x/(2pi)) or is it u=D(x/(2pi))?
  17. Mar 9, 2009 #16
    Re: continuity

    We already agreed that u=r*w(theta), remember? So w(theta)=D(theta/(2pi)).
  18. Mar 9, 2009 #17
    Re: continuity

    ok that's good thanks a lot.
    so u=r*D(theta/(2pi))right?
  19. Mar 9, 2009 #18
    Re: continuity

    Yes, that is one possible example. Also, remember how D was defined, since the notation is not standard like sin,tanh.
  20. Mar 9, 2009 #19
    Re: continuity

    ok, thank you this helps a lot.
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