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Continuity of x^n

  1. Apr 23, 2013 #1
    From Pugh's "Real Mathematical Analysis" Chapter 1
    In the case of n=1, ##\delta = \epsilon## satisfies the condition, i.e. ##|y-x| < \delta = \epsilon \implies |y - x| < \epsilon##.

    In the case of n=2, it needs to be shown that ##|y-x| < \delta \implies |y-x||y+x| < \epsilon##

    1. ##|y+x| < 1## implies that ##|y-x||y+x| < |y-x| < \delta##. If ##\delta = \epsilon##, then the condition is met.

    2. ##|y+x| \geq 1## implies that ##|y-x| \leq |y-x||y+x| < \delta \text{ } |y+x|##
    Here's my question: Is it valid to set ##\delta## to ##\frac{\epsilon}{|y+x|}##? In the case that ##|y+x| \geq 1##, there is no risk of division by zero or assigning ##\delta## to a negative value, so ##\delta## would exist and be positive. If so, then the condition is met.​

    [Note: This is only the first part of the problem; I do intend to solve the general case with induction.]


    Thanks!
     
  2. jcsd
  3. Apr 23, 2013 #2

    micromass

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    I'm afraid it is not valid. You want your ##\delta## to be independent of ##x##. This means that you want to find a ##\delta>0## such that your implication holds for all ##x##.
    If you choose ##\delta = \varepsilon / |x+y|##, then you will get a different value of ##\delta## for each ##x##. Clearly this is not what you want.
     
  4. Apr 25, 2013 #3
    That makes sense. Does this work?

    2. ##|y+x|≥1## implies that ##|y−x|≤|y−x||y+x|<ε.## Let ##δ=ε##, then the condition is met?

    If it does, I've been making it out to be more complicated than it is.

    Thanks!
     
  5. Apr 25, 2013 #4
    Is this alright? I notice that there is the possibility that ##\delta = \epsilon< |y-x|## for certain x, y, and ε, which would cause ##|y^2-x^2| < \epsilon## but ##|y-x| \geq \delta##; however, ##|y-x| < \delta \implies |y^2-x^2| < \epsilon## requires only that ##|y^2-x^2| < \epsilon## is true whenever ##|y-x| < \delta## and not the other way around, so it seems okay to me. Am I correct?

    Example: ##x=0.1, y=0.2, δ=ε=.04##
    ##.1 < .04 \implies .03<.04 \text{ is true.}##​


    Thanks!
     
  6. Apr 25, 2013 #5
    For n>2, the following formula is helpful: ##|y^n-x^n| = |y-x||y^{n-1} + xy^{n-2}+\cdots+x^{n-2}y+x^{n-1}|##

    It seems to me that, provided my methods for n=2 were valid, that the same arguments could be used for the cases ##|y^{n-1} + xy^{n-2}+\cdots+x^{n-2}y+x^{n-1}| < 1## and ##|y^{n-1} + xy^{n-2}+\cdots+x^{n-2}y+x^{n-1}| \geq 1## to show that ##\delta = \epsilon## satisfies all conditions for continuity.
     
  7. May 28, 2013 #6
    I think it is easier to do this problem if you prove [itex]{lim}_{h \rightarrow 0} |(x+h)^n -x^n| = 0[/itex].
     
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