💡 Let $f\in {\mathbb{R}}^D$.

1. $f$ is bounded from above if $f(D)\subseteq \mathbb{R}$ is bounded from above.
2. $f$ is bounded from below if $f(D)\subseteq \mathbb{R}$ is bounded from below.
3. $f$ is bounded if $f(D)\subseteq \mathbb{R}$ is bounded.

💡 A function $f:D⟶\mathbb{R}$, where $D\subseteq \mathbb{R}$ is called

1. monotonically increasing if $\forall x,y\in D$

$$x\le y⟹f(x)\le f(y)$$

1. strictly monotonically increasing if $\forall x,y\in D$

$$x<y⟹f(x)<f(y)$$

1. monotonically decreasing if $\forall x,y\in D$

$$x\le y⟹f(x)\ge f(y)$$

1. strictly monotonically decreasing if $\forall x,y\in D$

$$x<y⟹f(x)>f(y)$$

1. monotonous if $f$ is monotonically increasing or monotonically decreasing
2. strictly monotonous if $f$ is strictly monotonically increasing or strictly monotonically decreasing.

💡 Let $D\subseteq \mathbb{R}$, $x_0\in D$. The function $f:D⟶\mathbb{R}$ is continuous in $x_0$ if for every $ϵ>0$ there exists a $\delta >0$ such that for every $x\in D$ the implication

$$|x-x_0|<\delta ⟹|f(x)-f(x_0)|<ϵ$$

holds.

💡 The function $f:D⟶\mathbb{R}$ is continuous if it is continuous in every point of $D$.

📖 Let $x_0\in D\subseteq \mathbb{R}$ and $f:D⟶\mathbb{R}$. The function $f$ is continuous in $x_0$ if and only if the following holds for every sequence $(a_n{)}_{n\ge 1}$ in $D$:

$$\underset{n\to \infty }{\lim }{a}_n=x_0⟹\underset{n\to \infty }{\lim }f(a_n)=f(x_0).$$

📎 Let $x_0\in D\subseteq \mathbb{R},\lambda \in \mathbb{R}$ and $f:D⟶\mathbb{R},g:D⟶\mathbb{R}$ both be continuous in $x_0$.

1. Then $f+g,\lambda \cdot f,f\cdot g$ are continuous in $x_0$.

2. If $g(x_0)\ne 0$ then

   $$\begin{array}{rl}\frac{f}{g}:D\cap \{x\in D|g(x)\ne 0\}& ⟶\mathbb{R}\\ x& ⟼\frac{f(x)}{g(x)}\end{array}$$

   is continuous in $x_0$.

💡 A polynomial function $P:\mathbb{R}⟶\mathbb{R}$ is a function of the form

$$P(x)=a_n{x}^n+\dots +a_0$$

where $a_n,\dots ,a_0\in \mathbb{R}$. If $a_n\ne 0$ then $n$ is the degree of $P$,

📎 Polynomial functions are continuous on all of $\mathbb{R}$.

📎 Let $P,Q$ be polynomial functions on $\mathbb{R}$ with $Q\ne 0$. Let $x_1,\dots ,x_m$ be the roots of $Q$. Then

$$\begin{array}{rl}\frac{P}{Q}:\mathbb{R}\setminus \{x_1,\dots ,x_m\}& ⟶\mathbb{R}\\ x& ⟼\frac{P(x)}{Q(x)}\end{array}$$

is continuous.

💡 Let $x_1,x_2\in \mathbb{R}$. Then $c$ lies between $x_1$ and $x_2$ if:

1. $\ast x_1\le x_{2}c\in [x_1,x_2]$*
2. $\ast x_2\le x_{1}c\in [x_2,x_1]$.*

📎 (Bolzano 1817).

Let $I\subseteq \mathbb{R}$ be an interval, $f:I⟶\mathbb{R}$ a continuous function and $a,b\in I.$ For every $c$ between $f(a)$ and $f(b)$ there exists a $z$ between $a$ and $b$ with $f(z)=c$.

📎 Let $P(x)=a_n{x}^n+a_{n-1}{x}^{n-1}+\dots +a_0$ be a polynomial with $a_n\ne 0$ and $n$ odd. Then $P$ has at least one root in $\mathbb{R}$.

💡 An interval $I\subseteq R$ is called compact if it is of the form

$$I=[a,b]a\le b.$$

💡 Let $D$ be a set and $f,g:D⟶\mathbb{R}$ be functions. We define

• $\ast |f|(x)=|f(x)|,\forall x\in D\ast$
• $\ast \max (f,g)(x)=\max (f(x),g(x)),\forall x\in D\ast$
• $\ast \min (f,g)(x)=\min (f(x),g(x)),\forall x\in D.\ast$

📌 Let $D\subseteq \mathbb{R},x_0\in D$ and $f,g:D⟶\mathbb{R}$ continuous in $x_0$. Then

$$|f|,\max (f,g),\min (f,g)$$

are continuous in $x_0$.

📌 Let $(x_n{)}_{n\ge 1}$ be a convergent sequence in $\mathbb{R}$ with limit

$$\underset{x\to \infty }{\lim }{x}_n\in \mathbb{R}.$$

Let $a\le b$. If $\{x_n|n\ge 1\}\subseteq [a,b]$ it follows that

$$\underset{n\to \infty }{\lim }{x}_n\in [a,b].$$

📖 Let $f:I=[a,b]⟶\mathbb{R}$ be continuous on a compact interval $I$. Then there exists $u\in I$ and $v\in I$ with

$$f(u)\le f(x)\le f(v)\forall x\in I.$$

In particular, $f$ is bounded.

📖 Let $D_1,D_2\subseteq \mathbb{R}$ be two subsets, $f:D_1⟶D_2,g:D_2⟶\mathbb{R}$ be two functions and $x_0\in D_1$. If $f$ is continuous in $x_0$ and $g$ is continuous in $f(x_0)$ then

$$g\circ f:D_1⟶\mathbb{R}$$

is continuous in $x_0$.

📎 If $f$ is continuous on $D_1$ and $g$ is continuous on $D_2$ then $g\circ f$ is continuous on $D_1$.

📖 Let $I\subseteq \mathbb{R}$ be an interval and $f:I⟶\mathbb{R}$ be continuous and strictly monotonous. Then $J=f(I)\subseteq \mathbb{R}$ is an interval and $f^{-1}:J⟶I$ is continuous and strictly monotonous.

📖 $\ast \exp :\mathbb{R}⟶(0,+\infty )$ is strictly monotonically increasing, continuous and surjective.*

📎 All of the following follow from the power series expansion of $\exp (x)$:

$$\exp (x)>0\forall x\in \mathbb{R}.$$

$$\exp (x)>1\forall x>0.$$

If now $y<z$, then it follows that:

$$\exp (z)=\exp (y+(z-y))=\exp (y)\exp (z-y)$$

and because $\exp (z-y)>1$ it follows that:

$$\exp (z)>\exp (y)\forall z>y.$$

$$\exp (x)\ge 1+x\forall x\in \mathbb{R}.$$

Let $x\in \mathbb{R}$ and $N$ such that

$$\frac{x}{N}>-1$$

from which it follows that:

$$\frac{x}{n}>-1\forall n\ge N$$

For all $n\ge N$ it follows through Bernoulli’s Inequality that:

$$(1+\frac{x}{n}{)}^n\ge 1+n(\frac{x}{n})=1+x$$

from which in turn it follows that:

$$\exp (x)=\underset{n\to \infty }{\lim }(1+{\frac{x}{n}}^n)\ge 1+x.$$

📎 The natural logarithm

$$\ln :(0,+\infty )⟶\mathbb{R}$$

is a strictly monotonically increasing, continuous, bijective function. Furthermore, it holds that

$$\ln (a\cdot b)=\ln a+\ln b\forall a,b,\in (0,+\infty ).$$

💡 We can now use the logarithm and exponential function to define general powers. For $x>0$ and $a\in \mathbb{R}$ arbitrary we define:

$$x^a=\exp (a\ln x).$$

In particular: $x^0=1\forall x>0.$

📎

1. For $a>0$

   $$\begin{array}{rl}(0,+\infty )& ⟶(0,+\infty )\\ x& ⟼x^a\end{array}$$

   is a continuous, strictly monotonically increasing bijection.

2. For $a<0$

   $$\begin{array}{rl}(0,+\infty )& ⟶(0,+\infty )\\ x& ⟼x^a\end{array}$$

   is a continuous, strictly monotonically decreasing bijection.

3. $\ast \ln (x^a)=a\ln (x)\forall a\in \mathbb{R},\forall x>0.\ast$
4. $\ast x^a\cdot x^b=x^{a+b}\forall a,b\in \mathbb{R},\forall x>0.\ast$
5. $\ast (x^a{)}^b=x^{a\cdot b}\forall a,b\in \mathbb{R},\forall x>0.\ast$

💡 Let $D$ be a set. Analogously to the definition of a sequence of real numbers a (real-valued) sequence of functions is a transformation

$$\begin{array}{rl}\mathbb{N}& ⟶{\mathbb{R}}^D\\ n& ⟼f(n).\end{array}$$

Just like we did in the case of sequences we will denote $f(n)$ by $f_n$ and the sequence of functions as $(f_n{)}_{n\ge 0}$. For every $x\in D$ we get a sequence $(f_n(x){)}_{n\ge 0}$ in $\mathbb{R}$.

💡 The sequence of functions $(f_n{)}_{n\ge 0}$ converges pointwise towards a function $f:D⟶\mathbb{R}$ if for all $x\in D$:

$$f(x)=\underset{n\to \infty }{\lim }{f}_n(x)$$

💡 (Weierstrass 1841)

The sequence

$$f_n:D⟶\mathbb{R}$$

converges uniformly in $D$ towards

$$f:D⟶\mathbb{R}$$

if it holds that: $\forall ϵ>0\exists N\ge 1$ such that:

$$\forall n\ge N,\forall x\in D|f_n(x)-f(x)|<ϵ.$$

📖 Let $D\subseteq \mathbb{R}$ and $f_n:D⟶\mathbb{R}$ be a sequence of functions consisting of continuous functions (in $D$) that converge uniformly (in $D$) towards a function $f:D⟶\mathbb{R}$. Then $f$ is continuous (in $D$).

💡 A sequence of functions

$$f_n:D⟶\mathbb{R}$$

is uniformly convergent if for every $x\in D$ the limit

$$f(x)=\underset{n\to \infty }{\lim }{f}_n(x)$$

exists and the sequence $(f_n{)}_{n\ge 0}$ converges uniformly towards $f$.

📎 The sequence of functions

$$f_n:D⟶\mathbb{R}$$

converges uniformly in $D$ if and only if:

$\ast \forall ϵ>0\exists N\ge 1$ such that $\forall n,m\ge N$ and $\forall x\in D$:*

$$|f_n(x)-f_m(x)|<ϵ.$$

📎 Let $D\subseteq \mathbb{R}$. If $f_n:D⟶\mathbb{R}$ is a uniformly convergent sequence of continuous functions, then the function

$$f(x)=\underset{n\to \infty }{\lim }{f}_n(x)$$

is continuous.

💡 Let $f_n:D⟶\mathbb{R}$ be a sequence of functions.

The series ${\sum }_{k=0}^{\infty }{f}_k(x)$ converges uniformly (in $D$), if the sequence of functions given by

$$S_n(x)=\sum _{k=0}^n{f}_k(x)$$

converges uniformly.

📖 Let $D\subseteq \mathbb{R}$ and

$$f_n:⟶\mathbb{R}$$

be a sequence of functions, We assume, that

$$|f_n(x)|\le c_n\forall x\in D$$

and that ${\sum }_{n=0}^{\infty }{c}_n$ converges. Then the series

$$\sum _{n=0}^{\infty }{f}_n(x)$$

converges uniformly in $D$ and its limit

$$f(x)=\sum _{n=0}^{\infty }{f}_n(x)$$

is a continuous function in $D$.

💡 The power series

$$\sum _{k=0}^{\infty }{c}_k{x}^k$$

has a positive radius of convergence if ${\mathrm{limsup}}_{k\to \infty }\sqrt[k]{|c_k|}$ exists.

The radius of convergence is then defines as:

$$\rho =\{\begin{array}{ll}+\infty & \text{if}{\mathrm{limsup}}_{k\to \infty }\sqrt[k]{|c_k|}=0,\\ \frac{1}{{\mathrm{limsup}}_{k\to \infty }\sqrt[k]{|c_k|}}& \text{if}{\mathrm{limsup}}_{k\to \infty }\sqrt[k]{|c_k|}>0.\end{array}$$

📖 Let ${\sum }_{k=0}^{\infty }{c}_k{x}^k$ be a power series with a positive radius of convergence $\rho >0$ and let

$$f(x)=\sum _{k=0}^{\infty }{c}_k{x}^k,|x|<\rho .$$

Then it holds that: $\forall 0\le r<\rho$ the series

$$\sum _{k=0}^{\infty }{c}_k{x}^k$$

converges uniformly on $[-r,r]$, in particular $f:(-\rho ,\rho )⟶\mathbb{R}$ is continuous.

💡 We define the sine function for $z\in \mathbb{C}$:

$$\sin z=z-\frac{z^3}{3!}+\frac{z^5}{5!}-\frac{z^7}{7!}+\dots =\sum _{n=0}^{\infty }\frac{(-1{)}^n{z}^{2n+1}}{(2n+1)!}$$

and the cosine function for $z\in \mathbb{C}$:

$$\cos z=z-\frac{z^2}{2!}+\frac{z^4}{4!}-\frac{z^6}{6!}+\dots =\sum _{n=0}^{\infty }\frac{(-1{)}^n{z}^{2n}}{(2n)!}$$

It follows by ratio test that both series converge absolutely for all $z\in \mathbb{C}$. Therefore, in both cases the radius of convergence is $+\infty$ and by the previous theorem the next theorem follows.

📖 $\ast \sin :\mathbb{R}⟶\mathbb{R}$ and $\cos :\mathbb{R}⟶\mathbb{R}$ are continuous functions.*

📖 Trigonometry equalities

1. $\exp (iz)=\cos (z)+i\sin (z)\forall z\in \mathbb{C}$
2. $\ast \cos (z)=\cos (-z)$ and $\sin (-z)=-\sin (z)\forall z\in \mathbb{C}$*
3. $\sin z=\frac{e^{iz}-e^{-iz}}{2i}$, $\cos z=\frac{e^{iz}+e^{-iz}}{2}$
4. $$\begin{gathered} \sin (z+w)=\sin (z)\cos (w)+\cos (z)\sin (w) \\ \cos (z+w)=\cos (z)\cos (w)-\sin (z)\sin (w) \end{gathered}$$
5. $\cos (z{)}^2+\sin (z{)}^2=1\forall z\in \mathbb{C}$

📎 More trigonometry equalities

$$\begin{array}{rl}\sin (2z)& =2\sin (z)\cos (z)\\ \cos (2z)& =\cos (z{)}^2-\sin (z{)}^2\end{array}$$

📖 The sine function has at least one root on $(0,+\infty )$.

Let

$$\pi =\inf \{t>0|\sin t=0\}.$$

Then the following hold:

1. $\sin \pi =0,\pi \in (2,4)$
2. $\sin x>0\forall x\in (0,\pi )$
3. $e^{\frac{i\pi }{2}}=i.$

💡 Let $x\ge 0$: the series

$$\sin z=z-\frac{z^3}{3!}+\frac{z^5}{5!}-\frac{z^7}{7!}+\dots$$

is alternating.

For $x\ge 0$, $\left(\frac{x^{2n+1}}{(2n+1)!}\right)$ is monotonically decreasing if and only if

$$\left(\frac{x^{2n+1}}{(2n+1)!}\right)\le \left(\frac{x^{2n-1}}{(2n-1)!}\right)\forall n\ge 1,$$

that means that

$$x^2\ge 2n\cdot (2n+1)\forall n\ge 1.$$

therefore

$$x\le \sqrt{6}.$$

📎 It follows that

$$x\ge \sin x\ge x-\frac{x^3}{3!}\forall 0\le x\le \sqrt{6}.$$

📎 More trigonometry equalities

1. $e^{i\pi }=-1,e^{2i\pi }=1$
2. $\sin (x+\frac{\pi }{2})=\cos (x),\cos (x+\frac{\pi }{2})=-\sin (x)\forall x\in \mathbb{R}$
3. $\sin (x+\pi )=-\sin (x),\sin (x+2\pi )=\sin (x)\forall x\in \mathbb{R}$
4. $\cos (x+\pi )=-\cos (x),\cos (x+2\pi )=\cos (x)\forall x\in \mathbb{R}$

5. Roots of the sine function $=\{k\cdot \pi |k\in \mathbb{Z}\}$

   $$\begin{gathered} \sin (x)>0\forall x\in (2k\pi ,(2k+1)\pi )k\in \mathbb{Z} \\ \sin (x)<0\forall x\in ((2k+1)\pi ,(2k+2)\pi )k\in \mathbb{Z} \end{gathered}$$

6. Roots of the cosine function $=\{\frac{\pi }{2}+k\cdot \pi |k\in \mathbb{Z}\}$

   $$\begin{gathered} \cos (x)>0\forall x\in (-\frac{\pi }{2}+2k\pi ,-\frac{\pi }{2}+(2k+1)\pi )k\in \mathbb{Z} \\ \cos (x)<0\forall x\in (-\frac{\pi }{2}+(2k+1)\pi ,-\frac{\pi }{2}+(2k+2)\pi )k\in \mathbb{Z} \end{gathered}$$

💡 We define the tangent function for $z\notin \frac{\pi }{2}+\pi \cdot \mathbb{Z}$:

$$\tan (z)=\frac{\sin (z)}{\cos (z)}$$

and the cotangent function for $z\notin \pi \cdot \mathbb{Z}$:

$$\cot (z)=\frac{\cos (z)}{\sin (z)}$$

💡 Let $f:D⟶\mathbb{R}$ where $D\subseteq \mathbb{R}$. $x_0\in \mathbb{R}$ is called a limit point (also accumulation point) of $D$ if $\forall \delta >0$:

$$((x_0-\delta ,x_0+\delta )\setminus \{x_0\})\cap D\ne \varnothing$$

💡 Let $f:D⟶\mathbb{R}$, $x_0\in \mathbb{R}$ be a limit point of $D$. Then $A\in \mathbb{R}$ is the limit of $f(x)$ for $x\to x_0$, denoted by

$$\underset{x\to x_0}{\lim }f(x)=A,$$

if $\forall ϵ>0\exists \delta >0$ such that

$$\forall x\in D\cap ((x_0-\delta ,x_0+\delta )\setminus \{x_0\})|f(x)-A|<ϵ.$$

💡 Note:

📖 Let $D,E\subseteq \mathbb{R}$, $x_0$ be an accumulation point of $D$, and let $f:D⟶E$ be a function. We assume that

$$y_0=\underset{x\to x_0}{\lim }f(x)$$

exists and $y_0\in E$. If $g:E⟶\mathbb{R}$ is continuous in $y_0$ it follows that:

$$\underset{x\to x_0}{\lim }g(f(x))=g(y_0).$$

💡 Let’s entertain the example

$$\begin{array}{rl}f:\mathbb{R}\setminus \{0\}& ⟶\mathbb{R}\\ x& ⟼\frac{1}{x}.\end{array}$$

Then for $x>0$, where $x$ is arbitrarily close to $0$, $\frac{1}{x}$ gets arbitrarily positively big and for $x<0$, where $x$ is arbitrarily close to 0, $\frac{1}{x}$ get arbitrarily negatively big. In both cases $\frac{1}{x}$ has a very simple behaviour.

In the case of $a\in \mathbb{R}$,

$$\begin{array}{rl}f:(0,\infty )& ⟶\mathbb{R}\\ x& ⟼x^a\end{array}$$

$f$ is defined on $(0,\infty )$. If $a>0$ we can see that

$$\underset{x\to 0}{\lim }f(x)=0.$$

Let $f:D⟶\mathbb{R}$ and $x_0\in \mathbb{R}$. We assume that $x_0$ is an accumulation point of $D\cap (x_0,+\infty )$; that is a right-side accumulation point.

If the limit of the restricted function

$$f{|}_{D\cap [x_0,+\infty )}$$

for $x\to x_0$exists it is denoted by

$$\underset{x\to x_0^{+}}{\lim }f(x)$$

and is called the right-side limit of $f$ at $x_0$.

We expand this definition to:

$$\underset{x\to x_0^{+}}{\lim }f(x)=+\infty$$

if it holds that:

$$\forall ϵ>0\exists \delta >0,\forall x\in D\cap (x_0,x_0+\delta )f(x)>\frac{1}{ϵ}$$

and analogously

$$\underset{x\to x_0^{+}}{\lim }f(x)=-\infty$$

if

$$\forall ϵ>0\exists \delta >0,\forall x\in D\cap (x_0,x_0+\delta )f(x)<-\frac{1}{ϵ}.$$

Left-side accumulation points and limits are defined analogously. With these definitions it holds that

$$\underset{x\to 0^{+}}{\lim }\frac{1}{x}=+\infty ,\underset{x\to 0^{-}}{\lim }\frac{1}{x}=-\infty .$$