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

Homework Help: Show x^2 is continuous.

  1. Jul 14, 2010 #1
    1. The problem statement, all variables and given/known data
    Show x^2 is continuous, on all reals, using a delta/epsilon argument.

    Let E>0. I want to find a D s.t. whenever d(x,y)<D d(f(x),f(y))<E.
    WLOG let x>y
    |x^2-y^2|=x^2-y^2=(x-y)(x+y)<D(x+y)

    I am trying to bound x+y, but can't figure out how.
     
  2. jcsd
  3. Jul 14, 2010 #2

    Hurkyl

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    One problem is that you have the epsilon-delta statement wrong. What you wrote is the definition of "uniformly continuous", and squaring is not uniformly continuous.

    Uniform continuity requires a delta that works for all values of y. Continuity only requires that, for each value of y, there exists a delta. If it helps, you should think of the delta you are choosing as a function of y.
     
  4. Jul 14, 2010 #3
    Maybe writing x+y=x-y+2y will help?
     
  5. Jul 15, 2010 #4

    HallsofIvy

    User Avatar
    Science Advisor

    You need to show: given any real number, a, then for any [itex]\epsilon> 0[/itex], there exist [itex]\delta> 0[/itex] such that if [itex]|x- a|< \delta[/itex] then [itex]|x^2- a^2|< \epsilon[/itex].

    You might start by factoring [itex]|x^2- a^2|< |x-a||x+ a|[/itex].

    Now, if x is close to a, so x- a is close to 0, how large can x+ a be?
     
    Last edited by a moderator: Jul 17, 2010
  6. Jul 16, 2010 #5

    then x+a approaches 2a...


    I feel like I should just take delta = sqrt(epsilon), and I'm fairly confident any delta less than that will suffice. I do not really understand how to show that though.
     
  7. Jul 17, 2010 #6

    HallsofIvy

    User Avatar
    Science Advisor

    Saying x+ a "approaches" 2a is not enough. If [itex]|x- a|< \delta[/itex] then [itex]-\delta< x- a< \delta[/itex] so, adding 2a to each part, [itex]2a- \delta< x+ a< 2a+ \delta[/itex].

    Now you can say [itex]|(x-a)(x+a)|= |x-a||x+a|< (a+\delta)|x- a|[/itex].
     
    Last edited by a moderator: Jul 17, 2010
  8. Jul 17, 2010 #7

    Mark44

    Staff: Mentor

    Fixed your LaTeX.
    Tip: Make sure that LaTeX expressions start and end with the same type of tag. [ tex] ... [ /tex] and [ itex] ... [ /itex]. You sometimes start the expression with [ itex] and end with [ /math]. I'm not sure that this Web site can render [ math] ... [ /math] expressions, but I am sure that you can't mix them.
     
  9. Jul 17, 2010 #8

    HallsofIvy

    User Avatar
    Science Advisor

    Thanks, Mark44. I just need to learn to check my responses before going on!
     
  10. Jul 17, 2010 #9

    Mark44

    Staff: Mentor

    That's advice I'm trying to give myself, too.:approve:
     
  11. Jul 17, 2010 #10
    How do this demonstrate the continuity? I mean, what you're saying is the same than [tex]\lim_{x \to c}{f(x)} = f(c)[/tex], but in delta epsilon notation? I don't get it :P
     
  12. Jul 18, 2010 #11
    shouldn't we conclude that

    [itex]
    |(x-a)(x+a)|= |x-a||x+a|< (2a+\delta)|x- a|
    [/itex]
    ?

    I still want to clearly define delta as a function of x, epsilon, and a (even though a is a constant)

    I see the trick that was used to bound |x^2-a^2| and that was neat, but now I am confused.

    If I solved
    [itex]
    (2a+\delta)|x- a|<\epsilon
    [/itex]

    for delta would that give me a correctly bounded delta?
     
  13. Jul 18, 2010 #12

    Hurkyl

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    I think you have a typo there....


    No, you want it as a function of epsilon and a. (Check your quantifiers -- delta is "chosen" before x enters the picture)
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook