# Max's StatPage

Stat Student, Data Analysis Nerd, Chinese Speaker

# Epsilon Delta Proof (continuity)

### Max Lang / 2021-01-14

We are living in crazy times. Everything that was normal before has changed during the Covid-19 pandemic. That might make you wonder: “What did not change?”, “Are qudratic polynoms still continous?”. Ok, nobody did/would ever say this (LOL), but I really did not know how to introduce this topic.

In my Calculus class we introduced the Definition of conituity a couple of weeks ago which is the following

Let $$M \subseteq \mathbb{C}, \zeta \in M .$$ Fkt. $$f: M \longrightarrow \mathbb{K}$$ is called continous in $$\zeta$$ if and only if $\begin{array}{ccc} \forall & \exists & \forall \\ \epsilon>0 & \delta>0 & z \in M \end{array}(|z-\zeta|<\delta \Rightarrow|f(z)-f(\zeta)|<\epsilon)$

This basically just says that a small change on the X-Axis will result in a small change on the Y-Axis. To get a better visual perspective on this you can play a bit with this [GeoGebra illusstration]{https://www.geogebra.org/m/cxbjrvks}.

So our goal is to prove that the following function is continuous with an Epsilon Delta Argument. $f: \mathbb R^2 \rightarrow \mathbb R^2 \; x\mapsto(x^2+2x+3)$

The first step is always to figure out the delta. Therefore we start to simplify $$|f(z)-f(\zeta)|<\epsilon$$.

So I started with

(★) $\left\lvert f(x)-f(y) \right\rvert= \left\lvert (x^2+2x+3) - (y^2+2y+3) \right\rvert = \left\lvert (x^2+2x)-(y^2+2y) \right\rvert$

$\le 2\left\lvert \left\lvert \frac{x^2-y^2}{2} \right\rvert + \left\lvert x- y\right\rvert \right\rvert$ $\le 2\left\lvert \left\lvert \frac{(x+y)(x-y)}{2} \right\rvert + \delta \right\rvert$ $= 2\left\lvert \frac{\left\lvert(x+y)(x-y)\right\rvert} {2} + \delta \right\rvert$ $\le 2\left\lvert \frac{\left\lvert(x+y)\delta\right\rvert} {2} + \delta \right\rvert$ $= 2\left\lvert \frac{[\left\lvert(x-y)\right\rvert+\left\lvert 2y\right\rvert]\delta} {2} + \delta \right\rvert$

Because we can set any requirements for our Delta we let Delta be $$\delta \le1.$$ This means that our two points on the X-Axis can not be further apart than 1. Now we can substitute $$(x-y)$$ for 1.

$= 2\left\lvert \frac{(1+\left\lvert 2y \right\rvert)\delta} {2} + \delta \right\rvert$ $= \left\lvert(1+\left\lvert 2y \right\rvert)\delta + 2 \delta \right\rvert$ $= \delta \left\lvert(1+\left\lvert 2y \right\rvert) + 2 \right\rvert$ $= \delta \left\lvert \left\lvert 2y \right\rvert + 3 \right\rvert = \epsilon$ $$\delta = min\left(1, \frac{\epsilon}{\left\lvert\left\lvert 2y\right\rvert + 3\right\rvert} \right)$$

So that was a lot of Algebra, but we made it. Now we can write our formal proof.

Given $$\epsilon>0,$$ let $$\delta= min\left(1, \frac{\epsilon}{\left\lvert\left\lvert 2y \right\rvert +3 \right\rvert} \right)$$. If $$0<|x-y|<\delta$$ then (★), so $$|f(x)-L|<\epsilon$$ and $$f$$ is therefore continuous. So we could insert it into the Definition and it would look like this. $\begin{array}{ccc} \forall & \exists & \forall \\ \epsilon>0 & \delta>0 & x \in \mathbb R \end{array}(|x-y|<min\left(1, \frac{\epsilon}{\left\lvert\left\lvert 2y\right\rvert + 3 \right\rvert} \right) \Rightarrow|f(x)-f(y)|< \delta \left\lvert\left\lvert 2y \right\rvert + 3 \right\rvert)$

# Thoughts

So what can we take from this? Not everything has changed because of Covid-19, Polynoms are still continuous. Nice! If you take some time and reconsider the Definition of a continuous function, you find out it actually is a pretty genius way of describing continuity. However, nothing new, nothing special just describing obvious facts precisely.

Pure mathematics is, in its way, the poetry of logical ideas. - Albert Einstein