Proving Continuity of y^n

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In summary, this is a proof of the continuity of the function ##y^n## that avoids assuming the properties of continuous functions. It uses the definition of continuity and induction to show that the function is continuous for all natural numbers. The author also acknowledges that it may have been simpler to prove the continuity of the function by showing that the product of two continuous functions is also continuous.
  • #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!
 
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  • #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.
 
  • #3
johnqwertyful said:
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.

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?
 

1. What is continuity?

Continuity refers to a mathematical concept that describes a function as being uninterrupted and without any gaps or discontinuities. In simpler terms, it means that the graph of a function can be drawn without lifting your pencil from the paper.

2. How is continuity related to y^n?

Y^n, or the nth derivative of a function, is a measure of the rate of change of that function. In order for a function to be continuous, all of its derivatives (including y^n) must also be continuous.

3. What is the process for proving continuity of y^n?

The process for proving continuity of y^n involves first showing that the function is continuous at a specific point, and then extending that proof to show that it is continuous for all values of x. This can be done using the definition of continuity, the limit definition of derivatives, and the properties of continuous functions.

4. What are the main challenges in proving continuity of y^n?

One of the main challenges in proving continuity of y^n is identifying the specific point at which the function is continuous, as well as determining the necessary conditions for continuity at that point. Additionally, the proof may become more complex as the order of the derivative increases.

5. Why is proving continuity of y^n important?

Proving continuity of y^n is important because it allows us to understand the behavior of a function and its derivatives at a specific point. It also helps us to better understand the overall behavior of a function and how it changes over its entire domain. This is crucial in many areas of mathematics, physics, and engineering where the concept of continuity is essential.

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