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Homework Help: Uniform continuity

  1. Jul 24, 2007 #1

    daniel_i_l

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    1. The problem statement, all variables and given/known data
    Prove that if y>=x>=0:
    a) [tex] y^2 arctan y - x^2 arctan x >= (y^2 - x^2) arctan x [/tex]

    b) [tex] \ | \ y^2 arctan y - x^2 arctan x \ | \ >= (y^2 - x^2) arctan x [/tex]

    c) use (b) to prove that x^2 arctan(x) isn't UC in R.

    2. Relevant equations



    3. The attempt at a solution

    a) We have to prove that [tex] y^2 ( arctan y - arctan x ) >= 0 [/tex]
    And since arctan(y) - arctan(x) >= 0 for all y>=x and y^2 > 0 this is true.
    b)Since [tex] y^2 arctan y - x^2 arctan x >= 0 [/tex] for all y>=x>=0 then this is obviously true from a.

    c)If we choose Epsilon (E) = 1/2 , Lambda (L) > 0 and y = x+L then
    (y^2 - x^2)arctan(x) = (2xL - L^2)arctanx and the limit of that at infinity is infinity. So we can find N>0 so that for every x>N
    (y^2 - x^2)arctan(x) = |y^2 arctan(y) - x^2 arctan(x)| > E

    Now my questions are:
    1) a & b seemed too trivial -are those the right are answers?
    2)Is (c) right?

    Thanks.
     
    Last edited: Jul 24, 2007
  2. jcsd
  3. Jul 24, 2007 #2
    How are a and b different?
     
  4. Jul 24, 2007 #3

    daniel_i_l

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    Oops, in b you need to prove that
    |y^2 arctan(y) - x^2 arctan(x)| >= (y^2 - x^2)arctanx
    the same as (a) just with the absolute value on the right.
    Thanks.
     
  5. Jul 24, 2007 #4
    These look suspiciously like mean value theorem applications.
     
  6. Jul 24, 2007 #5
    These look suspiciously like mean value theorem applications.

    Edit: Sorry for me being a dolt.

    a and b are both correct. The reasoning is thusly: Suppose y>x>=0

    We have [itex]arctan(y)\ge\arctan(x)[/itex] since [itex]f(x)=\arctan(x)[/itex] is monotonically increasing ([itex]f' \ge 0[/itex] on that particular domain). Therefore [itex]y^2\arctan(y)-x^2\arctan(x) \ge 0[/itex]. Since [itex]arctan(y)\ge\arctan(x)[/itex] we have [itex]y^2\arctan(y)-x^2arctan(x) \ge y^2\arctan(x)-x^2\arctan(x)=(y^2-x^2)\arctan(x)[/itex].

    b follows since both sides are positive, hence their absolute values are the same.

    you haven't properly finished c. "(y^2 - x^2)arctan(x) = |y^2 arctan(y) - x^2 arctan(x)| > E" is dead wrong, as the equality does not hold. You actually want to be doing a proof by contradiction. Suppose it was U.C. on the positive reals. Let E=1/2. Suppose L is the delta that works, ie for |y-x|<L then |f(y)-f(x)|<E=1/2. Since y>x then y=x+a for some constant a>0. Etc Etc and look for your contradiction. Your idea was right, that for large enough x, irregardless of distance between x and y, the difference between f(x) and f(y) exceed 1/2.

    I would do a finishing statement, saying something along the lines of "therefore, there does not exist any L>0 such that when |y-x|<L, |f(y)-f(x)|<1/2, ie, f is not UC on the positive reals.

    A cople of notes, use \ge and \le for your greater/less then or equal to signs. Classically we use the definitions of [itex]|x-y|<\delta, |f(x)-f(y)|<\epsilon[/itex], which is no big deal.
     
    Last edited: Jul 24, 2007
  7. Jul 24, 2007 #6

    daniel_i_l

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    Gold Member

    | "(y^2 - x^2)arctan(x) = |y^2 arctan(y) - x^2 arctan(x)| > E" is dead
    | wrong, as the equality does not hold.
    But isn't it enough to prove that there's some N>0 so that for all x>N it holds?
    Thanks.
     
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