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. \[ \begin{array}{ccc} \forall & \exists & \forall \\ \epsilon>0 & \delta>0 & x \in \mathbb R \end{array}(|x-y|<min\left(1, \frac{\epsilon}{\left\lvert\left\lvert 2y\right\rvert + 3 \right\rvert} \right) \Rightarrow|f(x)-f(y)|< \delta \left\lvert\left\lvert 2y \right\rvert + 3 \right\rvert) \]