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Continuity of y^n

  1. Oct 6, 2013 #1
    Hello. This is an improvement on a previous post, "Continuity of y^2". My original plan was to first prove that y and y^2 were continuous and then prove by induction that y^n was continuous; however, in the process of doing so I think I found a better way. This proof is for rudimentary practice of proof construction and should be taken in context of an undergraduate Real Analysis I course. How does this look?

    ---Start Proof---

    Define ##(f_n)_{n \in \mathbb{N}}## as a sequence of functions where ##f_n:\mathbb{R} \to \mathbb{R}## maps ##y \mapsto y^n##. I intend prove pointwise continuity for any ##f_n## by showing that:

    ##\forall n \in \mathbb{N} \;\; \forall e > 0 \;\; \forall u \;\; \exists \delta: \forall y (|y-u| < \delta \implies |f_n(y) - f_n(u)| = |y^n - u^n| < \epsilon)##​

    Choose any ##n \in N, e > 0,## and ##u##.

    Case 1: ##u = 0##
    Pick ##\delta = \epsilon^{\frac{1}{n}}## and then any ##y##. It is clear that ##|y-0
    | < \epsilon^{\frac{1}{n}} \implies |y^n - 0| < \epsilon##.​

    Case 2: ##u \neq 0##

    Choose a ##\delta < |y|## and ##\delta < \frac{\epsilon}{(2^{n}-1)|u|^{n-1}}.## A ##\delta## meeting this criteria can be found since both ##|y|## and ##\frac{\epsilon}{(2^{n}-1)|u|^{n-1}}## are positive, and there exists no smallest positive number.

    Note the following:
    1. ##|y^n-u^n| = |y-u| \left |\sum_{i=0}^{n-1} u^{n-1-i}y^i \right |##. This is obvious from evaluating ##\frac{y^n-u^n}{y-u}##.
    2. ##|y-u| < \delta < |y| \implies |u| < 2|y|##
      Proof of 2.
      ##|y-u| < |y|## so ##u - y < |y|## and ##u < |y| + y \leq 2|y|##. At the same time ##y-u < |y|##,which implies ##-u < |y| - y \leq 2|y|##. Both ##u## and ##-u## are ##<2|y|##; therefore ##|u|<2|y|##.​
    3. ##\left |\sum_{i=0}^{n-1} u^{n-1-i}y^i \right | < \sum_{i=0}^{n-1} 2^i|u|^{n-1} = (2^n-1)|u|^{n-1} ##. This follows from 2. and the evaluation of the geometric series.

    It follows that:
    ##\begin{array}
    &|y^n-u^n| &=& |y-u| | \sum_{i=0}^{n-1} u^{n-1-i}y^i | & \text{1.}\\
    &<& \delta (2^{n}-1)|u|^{n-1} & \text{By assumption that } |y-u| < \delta \text{ and 3.}\\
    &<& \frac{\epsilon}{(2^{n}-1)|u|^{n-1}}(2^{n}-1)|u|^{n-1} & \delta\text{ was chosen for this}\\
    &=& \epsilon
    \end{array}##​

    .
    QED​
    ---End Proof---

    Thanks in advance!
     
    Last edited: Oct 6, 2013
  2. jcsd
  3. Oct 7, 2013 #2
    If two functions are continuous, their product is as well.

    It's trivial to show identity continuous. Then you can multiply them together n times.
     
  4. Oct 7, 2013 #3
    This is true, and of course I knew that; my goal here is to prove the continuity of ##y_n## from the definition of continuity without assumptions concerning the properties of continuous function like such. With that said, I probably should have thought more about that when constructing this proof. Proving that the product of two continuous functions is continuous is probably simpler enough that I could have either 1) provided it as a lemma to the proof or 2) used the same methodology. For example

    Proof Sketch:
    Trivially, ##|y-u| < \delta = \epsilon \implies |y-u| < \epsilon## and therefore ##f(y) = y## is continuous. If ##y^n## is assumed to be continuous, then it can be shown that ##y^{n+1}## is also continuous by [insert a special case of the product of continuity functions here where ##f(y) = y## and ##g(y) = y^n##]. By induction, ##f(y) = y^n## is continuous for all ##n \in \mathbb{N}##.​

    Does this seem right?
     
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