The so-called quadratic complement is a method for solving quadratic equations of the form
\[ \begin{align} ax^2 + bx + c = 0 \end{align} \]
with $a \not= 0$. With the definitions
\[ \begin{align} p \coloneqq \frac{b}{a},& {} & q \coloneqq \frac{c}{a} \end{align} \]
you can do this in the form
\[ \begin{align} x^2 + px + q = 0 \end{align} \]
note down. The quadratic addition gets its name from the fact that a quadratic term is now added to this equation and then subtracted again:
\[ \begin{align} x^2 + px + \left(\frac{p}{2}\right)^2 - \left(\frac{p}{2}\right)^2 + q &= 0\nonumber\\ \Leftrightarrow\left(x + \frac{p}{2}\right)^2 - \left(\frac{p}{2}\right)^2 + q &= 0\nonumber\\ \Leftrightarrow\left(x + \frac{p}{2}\right)^2 &= \left(\frac{p}{2}\right)^2 - q \end{align} \]
From this it follows
\[ \begin{align} x_{1, 2} = -\frac{p}{2} \pm \sqrt{\frac{p^2}{4} - q}.\tag{A.5}\label{eq:pq} \end{align} \]
This formula is called pq formula.
First, the Gaussian empirical formula is viewed. It reads
\[ \begin{align} \sum_{i = 0}^{n}i = \frac{n\left(n + 1\right)}{2}\tag{A.6}\label{eq:kleiner_gauss}. \end{align} \]
This is proven by complete induction. For $n = 0$, $\sum_{i = 0}^{n = 0}i = 0 = \frac{1}{2}n\left(n + 1\right)$. The statement already applies to $n\in\mathbb{N}$. Then follows
\[ \begin{align} \sum_{i = 0}^{n + 1}i = \frac{n\left(n + 1\right)}{2} + n + 1 = \frac{n^2 + 3n + 2}{2} = \frac{\left(n + 1\right)\left(n + 2\right)}{2}. \end{align} \]
This proves the statement.
The Theorem about the geometric series is
\[ \begin{align} \sum_{n = 0}^{\infty}q^n = \frac{1}{1 - q}\tag{A.8}\label{eq:geometr_reihe} \end{align} \]
for $q\in \mathbb{C}$ with $\left|q\right| <1$.
First of all, for $N\in \mathbb {N}$
\[ \begin{align} \sum_{n = 0}^{N}q^n = \frac{1 - q^{N + 1}}{1 - q}. \end{align} \]
This can be shown by complete induction. For $N = 0$ this is clear, the statement applies to $N$. Then applies
\[ \begin{align} \sum_{n = 0}^{N + 1}q^n = \frac{1 - q^{N + 1}}{1 - q} + q^{N + 1} = \frac{1 - q^{N + 1}}{1 - q} + \frac{q^{N + 1} - q^{N + 2}}{1 - q} = \frac{1 - q^{N + 2}}{1 - q}. \end{align} \]
With the calculation rules for consequence limits and
\[ \begin{align} \sum_{n = 0}^{\infty}q^n = \lim\limits_{N\to\infty}\sum_{n = 0}^{N}q^n \end{align} \]
follows the statement. Furthermore, according to [11]
\[ \begin{align} \sum_{n = 1}^{\infty}\frac{1}{n^4} = \frac{\pi^4}{90}.\tag{A.12}\label{eq:reihe_1} \end{align} \]
One defines the binomial coefficient
\[ \begin{align} \left(\begin{array}{c} n\\ k \end{array}\right) \coloneqq \frac{n!}{k!\left(n - k\right)!} \end{align} \]
with the Faculty
\[ \begin{align} n! \coloneqq 1\cdot 2\cdot\dotsc\cdot n \end{align} \]
with $n, k\in \mathbb{N}$ and $n\geq k$. The number of ways to arrange $n\in \mathbb{N}$ elements is $n!$. For the first element you have $n$ possibilities. For the second only $n - 1$, etc., for the last element you only have exactly one way to arrange it. A Permutation is a bijetion on the set $\{1, 2, \dotsc, N\}$ with $1 \leq N \in \mathbb{N}$. The set of all these mappings is denoted by $S_N$. Based on the statement just proven, the number of all permutations is $\left|S_N\right|$
\[ \begin{align} \left|S_N\right| = N!. \end{align} \]
One further defines the sign of a permutation $\pi \in S_N$ by $\sign\left(\pi\right) \coloneqq \left(-1\right)^M$ with $M$ as the number of all pairwise swaps that have to be made in order to convert $\id \to \pi$. If $M$ is even, $\pi$ is called even permutation, otherwise as odd.
$\left(\begin{array}{c} n\\ k \end{array}\right)$ ist die Anzahl der $k-$elementigen Teilmengen einer $n-$elementigen Menge. Dies kann man sich leicht überlegen. Seien also eine $n-$elementige Menge $M$ gegeben sowie ein $k\in\mathbb{N}$ mit $k\leq n$. Für den Fall $k = 0$ ist $\left(\begin{array}{c} n\\ k \end{array}\right) = 1$, hier stimmt die Aussage also, weil die einzige in Frage kommende Menge die leere Menge $\emptyset$ ist, und diese ist Teilmenge jeder anderen Menge, auch von sich selbst. Im Fall $n, k>0$ hat man $n$ Möglichkeiten, das erste Element der Teilmenge auszuwählen. Für das zweite Element bleiben $n - 1$ Möglichkeiten und so weiter. Dies führt auf die Zahl $n\cdot\left(n - 1\right)\cdot\dotsc\cdot\left(n - k + 1\right) = \frac{n!}{\left(n - k\right)!}$. Hierbei wurden jedoch Mengen doppelt gezählt. Weil die Anzahl der Möglichkeiten, $k$ Elemente anzuordnen, $k!$ beträgt, ist die Anzahl der $k-$elementigen Teilmengen einer $n-$elementigen Menge gleich $\frac{n!}{k!\left(n - k\right)!} = \left(\begin{array}{c} n\\ k \end{array}\right)$.
Let $n, n_1, \dotsc, n_k\in \mathbb{N}$ with $\sum_{i = 1}^{k}n_i = n$. One defines the multinomial coefficient by
\[ \begin{align} \left(\begin{array}{c} n\\ n_1, \dotsc, n_k \end{array}\right) \coloneqq \frac{n!}{n_1!\dotsc n_k!}.\tag{A.16}\label{def:multinomialkoeffizient} \end{align} \]
Sei eine Menge mit $k$ Elementen gegeben. Aus diesen wählt man $n-$mal ein zufälliges aus, dabei kann $n\geq k$ oder $n\leq k$ sein. Als Ergebnis hält man die Zahlen $n_1, \dotsc, n_k$ fest, die angeben, wie oft die Elemente gezogen wurden, das Ergebnis ist also das Tupel $\left(n_1, \dotsc, n_k\right)$. Der Multinomialkoeffizient $\left(\begin{array}{c} n\\ n_1, \dotsc, n_k \end{array}\right)$ ist die Anzahl der Ziehungen mit Ergebnis $\left(n_1, \dotsc, n_k\right)$. $n!$ ist die Anzahl aller Möglichkeiten, die $n$ gezogenen Elemente anzuordnen. Für jede solche Anordnung kann man aber die gleichartigen Elemente untereinander vertauschen, ohne das Ergebnis zu verändern. Die Anzahl der Vertauschungen beträgt $n_1!\dotsc n_k!$.
The binomial theorem, also known as the general binomial formula, is
\[ \begin{align} \left(a + b\right)^n = \sum_{k = 0}^{n}\left(\begin{array}{c} n\\ k \end{array}\right)a^kb^{n - k}.\tag{A.17}\label{eq:gen_bin_formula} \end{align} \]
For $n = 0$ this is clear. Further applies
\[ \begin{align} \left(a + b\right)^{n + 1} &= \left(a + b\right)^n\left(a + b\right) = \left(a + b\right)\sum_{k = 0}^{n}\left(\begin{array}{c} n\\ k \end{array}\right)a^kb^{n - k}\nonumber\\ &= \sum_{k = 0}^{n}\frac{n!}{k!\left(n - k\right)!}a^{k + 1}b^{n - k} + \sum_{k = 0}^{n}\frac{n!}{k!\left(n - k\right)!}a^kb^{n - k + 1}\nonumber\\ &= \sum_{k = 1}^{n + 1}\frac{n!}{\left(k - 1\right)!\left(n + 1 - k\right)}a^kb^{n - k + 1} + \sum_{k = 0}^{n}\frac{n!}{k!\left(n - k\right)!}a^kb^{n - k + 1}\nonumber\\ &= \sum_{k = 1}^{n + 1}\frac{\left(n + 1\right)!}{k!\left(n + 1 - k\right)!}\frac{k}{n + 1}a^kb^{n - k + 1}\nonumber\\ & + \sum_{k = 0}^{n}\frac{\left(n + 1\right)!}{k!\left(n + 1 - k\right)!}\frac{n + 1 - k}{n + 1}a^kb^{n - k + 1}\nonumber\\ &= \sum_{k = 0}^{n + 1}\frac{\left(n + 1\right)!}{k!\left(n + 1 - k\right)!}a^kb^{n - k + 1}\left(\frac{k + n + 1 - k}{n + 1}\right)\nonumber\\ &= \sum_{k = 0}^{n + 1}\frac{\left(n + 1\right)!}{k!\left(n + 1 - k\right)!}a^kb^{n + 1 - k} = \sum_{k = 0}^{n + 1}\left(\begin{array}{c} n + 1\\ k \end{array}\right)a^kb^{n + 1 - k}. \end{align} \]
Let $\left(x_1, x_2, x_3\right)$ be three Cartesian coordinates and $\left(\mathbf{e}_1, \mathbf{e}_2, \mathbf{e}_3\right)$ be the corresponding basis. One defines the Levi-Civita symbol $\epsilon_{i, j, k}$ by
\[ \begin{align} \epsilon_{i, j, k} \coloneqq \begin{cases} 1, \:\left(i, j, k\right)\text{\:gerade\:Permutation\:von\:}\left(1, 2, 3\right),\\ - 1, \:\left(i, j, k\right)\text{\:ungerade\:Permutation\:von\:}\left(1, 2, 3\right),\\ 0, \:\text{sonst}. \end{cases} \end{align} \]
This means in particular
\[ \begin{align} \epsilon_{i, j, k} &= \epsilon_{k, i, j} = \epsilon_{j, k, i} = -\epsilon_{j, i, k} = -\epsilon_{k, j, i} = -\epsilon_{i, k, j},\\ \epsilon_{i, i, j} &= \epsilon_{i, j, i} = \epsilon_{j, i, i} = 0. \end{align} \]
It applies
\[ \begin{align} & \sum_{n = 1}^{3}\epsilon_{l, n, k}\epsilon_{n, o, j} - \epsilon_{l, n, j}\epsilon_{n, o, k} = \sum_{n = 1}^{3}\epsilon_{l, n, k}\epsilon_{j, n, o} + \epsilon_{l, n, j}\epsilon_{k, o, n} = \sum_{n = 1}^{3}\epsilon_{j, o, n}\epsilon_{l, k, n} + \epsilon_{j, l, n}\epsilon_{k, o, n}\nonumber\\ &= \sum_{n = 1}^{3}\epsilon_{j, o, n}\epsilon_{l, k, n} - \epsilon_{j, l, n}\epsilon_{o, k, n}.\tag{A.22}\label{eq:levi-civita_prop_1} \end{align} \]
Here you have to make a case distinction. For the left expression to be nonzero, either $j = l$ and $o = k$ or $j = k$ and $o = l$ must hold. For the left expression you can write $\delta _{j, l}\delta _{o, k} - \delta _{j,k}\delta _{o,l}$. For the right expression to be nonzero, either $j = o$ and $l = k$ or $j = k$ and $l = o$ must hold. For the right expression you can write $-\delta _{j, o}\delta _{l, k} + \delta _{j,k}\delta _{o,l}$. So you can continue
\[ \begin{align} \sum_{n = 1}^{3}\epsilon_{j, o, n}\epsilon_{l, k, n} - \epsilon_{j, l, n}\epsilon_{o, k, n} &= \delta_{j, l}\delta_{o, k} - \delta_{j, o}\delta_{l, k} = \sum_{n = 1}^{3}\epsilon_{j, k, n}\epsilon_{l, o, n}, \end{align} \]
which is easy to understand.
Let a square matrix $A \in \mathbb {C}^{N \times N}$ be given with $1 \leq N \in \mathbb {N}$. The determinant $\det\left(A\right)$ is defined by
\[ \begin{align} \det\left(A\right) \coloneqq \sum_{\pi \in S_N}\sign\left(\pi\right)A_{1, \pi\left(1\right)}\cdot\dotsc\cdot A_{N, \pi\left(N\right)}. \end{align} \]
Occasionally the spelling is also used
\[ \begin{align} \left|\begin{array}{ccc} A_{1, 1} & \dots & A_{1, N} \\ \vdots & \ddots & \vdots \\ A_{N, 1} & \dots & A_{N, N} \end{array}\right| \coloneqq \det\left(A\right). \end{align} \]
First of all
\[ \begin{align} \det\left(A^T\right) &= \sum_{\pi \in S_N}\sign\left(\pi\right)A^T_{1, \pi\left(1\right)}\cdot\dotsc\cdot A^T_{N, \pi\left(N\right)} = \sum_{\pi \in S_N}\sign\left(\pi\right)A_{\pi\left(1\right), 1}\cdot\dotsc\cdot A_{\pi\left(N\right), N}. \end{align} \]
By sorting each individual summand by ascending row indices, one obtains
\[ \begin{align} \det\left(A^T\right) &= \sum_{\pi \in S_N}\sign\left(\pi\right)A_{\pi\left(1\right), 1}\cdot\dotsc\cdot A_{\pi\left(N\right), N} = \sum_{\pi \in S_N}\sign\left(\pi\right)A_{1, \pi\left(1\right)}\cdot\dotsc\cdot A_{N, \pi\left(N\right)} = \det\left(A\right). \end{align} \]
All following statements are therefore only proven for rows; the statement for columns then follows analogously. If two lines $1 \leq i, j \leq N$ of $A$ are equal, then it is already
\[ \begin{align} \det\left(A\right) = 0. \end{align} \]
To do this, sort the elements $\pi \in S_N$ into $N!/2$ pairs $\left(\pi, \pi'\right)$ such that $\pi'$ emerges from $\pi$ by swapping $\pi\left(i\right)$ and $\pi\left(j\right)$. In this case, $\sign\left(\pi\right) = -\sign\left(\pi'\right)$, which shows the statement. If the $j-$th row emerges from the $i-$th row by multiplication with a complex constant $C \in \mathbb{C}$, then $A_{j, k} = CA_{i, k}$ for all $1 \leq k \leq N$, it also follows
\[ \begin{align} \det\left(A\right) = 0, \end{align} \]
by simply dragging $C$ in front of the sum defining the determinant. If the $j-$th row is a linear combination of the other $N - 1$ rows, there exist constants $C_l \in \mathbb{C}$ for $1 \leq l \leq N$ with $l \not= j$ with $A_{j, k} = \sum_{\substack{l = 1,\\l \not = j}}^NC_lA_{l, k}$ for all $1 \leq k \leq N$, also applies in this case
\[ \begin{align} \det\left(A\right) = 0. \end{align} \]
To do this you simply have to apply the previously proven statement $N - 1-$ times.
The concept of function is assumed. Let $K\in\{\mathbb{R}, \mathbb{C}\}$ and $a_n\in K$ for all $n\in\mathbb{N}$. A power series $p$ at the development point $x_0\in K$ is a series of the form
\[ \begin{align} p\left(x\right) = \sum_{n = 0}^{\infty}a_n\left(x - x_0\right)^n. \end{align} \]
For their derivatives the following applies (remember the sum rule)
\[ \begin{align} p'\left(x\right) &= \sum_{n = 0}^{\infty}a_nn\left(x - x_0\right)^{n - 1} = \sum_{n = 1}^{\infty}a_nn\left(x - x_0\right)^{n - 1}\sum_{n = 0}^{\infty}a_{n + 1}(n + 1)\left(x - x_0\right)^{n},\tag{A.32}\label{eq:1st_derivative_series}\\ \frac{d^2}{dx^2}\sum_{n = 0}^{N}a_ix^i &= \sum_{n = 0}^{\infty}\left(n + 2\right)\left(n + 1\right)a_{n + 2}\left(x - x_0\right)^n,\\ \frac{d^m}{dx^m}\sum_{n = 0}^{N}a_ix^i &= \sum_{n = 0}^{\infty}\frac{\left(n + m\right)!}{n!}a_{n + m}\left(x - x_0\right)^n\tag{A.34}\label{eq:nth_derivative_series} \end{align} \]
These formulas can also be used for terminating power series, so-called polynomials.
Let $K\in\{\mathbb{R}, \mathbb{C}\}$ and $f:K\to K$ be a function. The function $f$ is called locally analytic at a point $x_0\in K$ if there is a neighborhood M of $x_0$ and a power series $p$ at the development point $x_0$ such that for all $x\in M$ holds
\[ \begin{align} f\left(x\right) = p\left(x\right). \end{align} \]
A function is called analytic if it is locally analytic at every point. There are some standard power series called the basic functions:
\[ \begin{align} \exp\left(x\right) \coloneqq \sum_{n = 0}^{\infty}\frac{x^n}{n!}. \end{align} \]
\[ \begin{align} \ln \coloneqq \exp^{-1}. \end{align} \]
\[ \begin{align} \sinh\left(x\right) \coloneqq \frac{\exp\left(x\right) - \exp\left(-x\right)}{2} & {} & \cosh\left(x\right) \coloneqq \frac{\exp\left(x\right) + \exp\left(-x\right)}{2} & {} & \tanh \coloneqq \frac{\sinh}{\cosh}. \end{align} \]
\[ \begin{align} \arsinh \coloneqq \sinh^{-1}, & {} & \arcosh \coloneqq \cosh^{-1}, & {} & \artanh \coloneqq \tanh^{-1}. \end{align} \]
\[ \begin{align} \sin\left(x\right) \coloneqq -i\sinh\left(ix\right), & {} & \cos\left(x\right) \coloneqq \cosh\left(ix\right), & {} & \tan \coloneqq \frac{\sin}{\cos}. \end{align} \]
\[ \begin{align} \arcsin \coloneqq \sin^{-1}, & {} & \arccos \coloneqq \cos^{-1}, & {} & \arctan \coloneqq \tan^{-1} \end{align} \]
The exponential function fulfills the calculation rule
\[ \begin{align} \exp\left(x + y\right) & \stackrel{\href{#eq:gen_bin_formula}{\text{Glg. (A.17)}}}{=} \sum_{k = 0}^\infty\frac{1}{k!}\sum_{l = 0}^k\left(\begin{array}{c} k\\ l \end{array}\right)x^ly^{k - l} = \sum_{k = 0}^\infty\frac{1}{k!}\sum_{l = 0}^k\frac{k!}{l!\left(k - l\right)!}x^ly^{k - l} = \sum_{k = 0}^\infty\sum_{l = 0}^k\frac{1}{l!\left(k - l\right)!}x^ly^{k - l}\nonumber\\ &= \sum_{l = 0}^\infty\sum_{k = l}^\infty\frac{1}{l!\left(k - l\right)!}x^ly^{k - l} = \sum_{l = 0}^\infty\frac{x^l}{l!}\sum_{k = 0}^\infty\frac{y^k}{k!} = \exp\left(x\right)\exp\left(y\right).\tag{A.42}\label{eq:exp_prop_1} \end{align} \]
From this it follows
\[ \begin{align} 1 = \exp\left(x - x\right) = \exp\left(x\right)\exp\left(-x\right) \Rightarrow \exp\left(-x\right) = \frac{1}{\exp\left(x\right)}.\tag{A.43}\label{eq:exp_prop_2} \end{align} \]
It is
\[ \begin{align} \exp\left(x\right) > 0\tag{A.44}\label{eq:exp_prop_3} \end{align} \]
for all $x \in \mathbb{R}.$ For $x \geq 0$ this is clear. For $x < 0$ applies
\[ \begin{align} \exp\left(x\right) = \frac{1}{\exp\left(-x\right)} > 0. \end{align} \]
Therefore apply
\[ \begin{align} \cosh^2\left(x\right) &= \frac{1}{4}\left(e^{2x} + e^{-2x} + 2\right) = \frac{1}{4}\left(e^{2x} + e^{-2x} - 2\right) + 1 = \sinh^2\left(x\right) + 1,\\ \sin^2\left(x\right) + \cos^2\left(x\right) &= -\sinh^2\left(ix\right) + \cosh^2\left(ix\right) = 1,\\ \tanh'&= \frac{\cosh^2 - \sinh^2}{\cosh^2} = 1 - \tanh^2 = \frac{1}{\cosh^2},\\ \tan' &= \frac{\cos^2 + \sin^2}{\cos^2} = 1 + \tan^2 = \frac{1}{\cos^2}. \end{align} \]
The basic functions fulfill the following symmetry properties:
\[ \begin{align} \sinh\left(-x\right) &= \frac{e^{-x} - e^x}{2} = -\frac{e^{x} - e^{-x}}{2} = -\sinh\left(x\right)\\ \cosh\left(-x\right) &= \frac{e^{-x} + e^x}{2} = \frac{e^{x} + e^{-x}}{2} = \cosh\left(x\right)\\ \Rightarrow \tanh\left(-x\right) &= -\tanh\left(x\right)\\ \sin\left(-x\right) &= -i\sinh\left(-ix\right) = i\sinh\left(ix\right) = -\left(-i\sinh\left(ix\right)\right) = -\sin\left(x\right)\\ \cos\left(-x\right) &= \cosh\left(-ix\right) = \cosh\left(ix\right) = \cos\left(x\right)\\ \Rightarrow \tan\left(-x\right) &= -\tan\left(x\right) \end{align} \]
The so-called addition theorems apply.
\[ \begin{align} \sinh\left(x + y\right) &= \frac{e^xe^y - e^{-x}e^{-y}}{2} = \frac{1}{4}\left(2e^xe^y - 2e^{-x}e^{-y}\right) = \frac{1}{4}\left(\left(e^x - e^{-x}\right)\left(e^y + e^{-y}\right) + \left(e^y - e^{-y}\right)\left(e^x + e^{-x}\right)\right)\nonumber\\ &= \sinh\left(x\right)\cosh\left(y\right) + \cosh\left(x\right)\sinh\left(y\right), \tag{A.56}\label{eq:sinh_add_1}\\ \cosh\left(x + y\right) &= \frac{e^xe^y + e^{-x}e^{-y}}{2} = \frac{1}{4}\left(2e^xe^y + 2e^{-x}e^{-y}\right) = \frac{1}{4}\left(\left(e^x + e^{-x}\right)\left(e^y + e^{-y}\right) + \left(e^y - e^{-y}\right)\left(e^x - e^{-x}\right)\right)\nonumber\\ &= \cosh\left(x\right)\cosh\left(y\right) + \sinh\left(x\right)\sinh\left(y\right), \tag{A.57}\label{eq:cosh_add_1}\\ \tanh\left(x + y\right) &= \frac{\sinh\left(x + y\right)}{\cosh\left(x + y\right)} = \frac{\sinh\left(x\right)\cosh\left(y\right) + \cosh\left(x\right)\sinh\left(y\right)}{\cosh\left(x\right)\cosh\left(y\right) + \sinh\left(x\right)\sinh\left(y\right)} = \frac{\tanh\left(x\right)\cosh\left(y\right) + \sinh\left(y\right)}{\cosh\left(y\right) + \tanh\left(x\right)\sinh\left(y\right)}\nonumber\\ &= \frac{\tanh\left(x\right) + \tanh\left(y\right)}{1 + \tanh\left(x\right)\tanh\left(y\right)}.\tag{A.58}\label{eq:tanh_add_1} \end{align} \]
Applying this to the trigonometric functions gives
\[ \begin{align} \sin\left(x + y\right) &= -i\sinh\left(ix + iy\right) = -i\sinh\left(ix\right)\cosh\left(iy\right) - i\cosh\left(ix\right)\sinh\left(iy\right)\nonumber\\ &= \sin\left(x\right)\cos\left(y\right) + \cos\left(x\right)\sin\left(y\right),\\ \cos\left(x + y\right) &= \cosh\left(ix + iy\right) = \cosh\left(ix\right)\cosh\left(iy\right) + \sinh\left(ix\right)\sinh\left(iy\right)\nonumber\\ &= \cos\left(x\right)\cos\left(y\right) - \sin\left(x\right)\sin\left(y\right). \end{align} \]
From this it follows
\[ \begin{align} \cos\left(\alpha\right) = \cos^2\left(\frac{\alpha}{2}\right) - \sin^2\left(\frac{\alpha}{2}\right) = 1 - 2\sin^2\left(\frac{\alpha}{2}\right) = 2\cos^2\left(\frac{\alpha}{2}\right) - 1.\tag{A.61}\label{eq:trigo_add_lemma_1} \end{align} \]
The so-called Euler identity also applies
\[ \begin{align} \exp\left(ix\right) = \frac{\exp\left(ix\right) + \exp\left(-ix\right) + \exp\left(ix\right) - \exp\left(-ix\right)}{2} = \cosh\left(ix\right) + \sinh\left(ix\right) = \cos\left(x\right) + i\sin\left(x\right). \end{align} \]
The following applies to the derivations of the basic functions
\[ \begin{align} \exp'\left(x\right) & \stackrel{\href{#eq:1st_derivative_series}{\text{Glg. (A.32)}}}{=} \sum_{n = 0}^{\infty}\frac{n + 1}{\left(n + 1\right)!}x^n = \sum_{n = 0}^{\infty}\frac{x^n}{n!} = \exp\left(x\right),\\ \sinh'\left(x\right) &= \cosh\left(x\right),\\ \cosh'\left(x\right) &= \sinh\left(x\right),\\ \sin'\left(x\right) &= \cosh\left(ix\right) = \cos\left(x\right),\\ \cos'\left(x\right) &= -\sin\left(x\right). \end{align} \]
Because of Eq. (A.44) is $\exp$ strictly monotonically increasing on $\mathbb{R}$. Because $\cosh\left(x\right) > 0$ for all real $x$, $\sinh$ is strictly monotonically increasing, and we also have
\[ \begin{align} \lim\limits_{x\to\infty}\sinh\left(x\right) = \infty, & {} & \lim\limits_{x\to - \infty}\sinh\left(x\right) = -\infty. \end{align} \]
So is
\[ \begin{align} \arsinh:\mathbb{R}\to\mathbb{R} \end{align} \]
defined, still applies
\[ \begin{align} \arsinh'\left(x\right) = \frac{1}{\cosh\left(\arsinh'\left(x\right)\right)} = \frac{1}{\sqrt{1 + x^2}}. \end{align} \]
They still apply
\[ \begin{align} \lim\limits_{x\to\infty}\cosh\left(x\right) = \infty, & {} & \lim\limits_{x\to - \infty}\cosh\left(x\right) = \infty, \end{align} \]
the minimum is at $x = 0$, there $\cosh\left(0\right) = 1$, so define
\[ \begin{align} \arcosh\left[1, \infty\right)\to\left[0, \infty\right). \end{align} \]
For the derivation applies
\[ \begin{align} \arcosh'\left(x\right) = \frac{1}{\sinh\left(\arcosh\left(x\right)\right)} = \frac{1}{\sqrt{x^2 - 1}}. \end{align} \]
The arc functions have the definition and image sets
\[ \begin{align} \arcsin: & \left[-1, 1\right]\to\left[-\frac{\pi}{2}, \frac{\pi}{2}\right],\\ \arccos: & \left[-1, 1\right]\to\left[0, \pi\right],\\ \arctan: & \left(-\infty, \infty\right)\to\left(-\frac{\pi}{2}.\frac{\pi}{2}\right). \end{align} \]
Their derivatives are calculated as:
\[ \begin{align} \arcsin'\left(x\right) &= \frac{1}{\cos\left(\arcsin\left(x\right)\right)} = \frac{1}{\sqrt{1 - x^2}},\\ \arccos'\left(x\right) &= -\frac{1}{\sin\left(\arccos\left(x\right)\right)} = \frac{1}{\sqrt{1 - x^2}},\\ \arctan'\left(x\right) &= \frac{1}{\tan'\left(\arctan\left(x\right)\right)} = \frac{1}{1 + x^2}. \end{align} \]
For the first order Taylor expansions of the basic functions to the expansion point zero follows
\[ \begin{align} \exp\left(x\right) \approx 1 + x, & {} & \sinh\left(x\right) \approx x, & {} & \cosh\left(x\right) \approx 1, & {} & \tanh\left(x\right) \approx x,\\ \sin\left(x\right) \approx x, & {} & \cos\left(x\right) \approx 1, & {} & \tan\left(x\right) \approx x. \end{align} \]
Let $p\left(x\right) = \sum_{n = 0}^{\infty}a_nx^n$ be a power series $\mathbb{C}\to\mathbb{C}$ with $p\left(x\right) = 0$ for all $x\in \mathbb{C}$, then $a_n = 0$ already holds for all $n\in\mathbb{N}$. This should first be seen by proving by contradiction for a polynomial, i.e. for a terminating power series. So assume that $a_n = 0$ for all $N < n \in \mathbb{N}$ and further assume that $a_N \not = 0$. For $\left|x\right|\to\infty$ every polynomial diverges, since polynomials are continuous there is in particular a $b \in \mathbb{R}$ with $b > 0$ and $\left|p\left(x\right)\right| > 0$ for all $x \in \mathbb{C}$ with $\left|x\right| > b$ contradicts the assumption. Thus the statement for polynomials is shown.
At this point a note should be made about complex numbers. The obvious hurdle with complex numbers is to recognize the existence of the imaginary unit $i =\sqrt{-1}$. One cannot imagine a number $i$ that satisfies this equation. Nevertheless, it can be expected, so it is a theoretical aid. Before comparing with measurements, however, the result must be projected onto the real axis. Even with real numbers, one can ask oneself what one should do with a number that never ends, i.e. an irrational number like $e$. The clear limits have basically already been exceeded here.
The Delta distribution $\delta\left(x\right)$ is defined by the fact that for every continuous function $f = f\left(x\right)$
\[ \begin{align} \int_{a<0}^{b>0}\delta\left(x\right)f\left(x\right)dx = f\left(0\right) \end{align} \]
applies. It is according to [20]
\[ \begin{align} \lim_{t\to \infty}\frac{4\sin^2\left(\frac{\Omega t}{2}\right)}{\Omega ^2t} = 2\pi\delta\left(\Omega\right)\tag{A.83}\label{eq:delta_distribution_prop_1}. \end{align} \]
According to [19] the following applies
\[ \begin{align} \delta\left(x - x_0\right) = \frac{1}{2\pi}\int_{-\infty}^{\infty}\exp\left(ik\left(x - x_0\right)\right)dk\tag{A.84}\label{eq:delta_distribution_prop_2}. \end{align} \]
Let $f, g:\mathbb{R}\to\mathbb{R}$ be Riemann integrable. Then the product rule applies
\[ \begin{align} \left(fg\right)' &= f'g + fg'\Leftrightarrow fg = \int f'gdx + \int fg'dx\Leftrightarrow \int f'gdx = fg - \int fg'dx. \end{align} \]
This is called partial integration.
Let $a, b\in \mathbb {R}$ with $a \[
\begin{align}
\int_{a}^{b}f\left(x\right)dx = \left[F\left(x\right)\right]_a^b = \left[F\left(\varphi\left(x\right)\right)\right]_{\varphi^{-1}\left(a\right)}^{\varphi^{-1}\left(b\right)} = \int_{\varphi^{-1}\left(a\right)}^{\varphi^{-1}\left(b\right)}\frac{d}{dx}\left(F\left(\varphi\left(x\right)\right)\right)dx = \int_{\varphi^{-1}\left(a\right)}^{\varphi^{-1}\left(b\right)}f\left(\varphi\left(x\right)\right)\varphi'\left(x\right)dx.\tag{A.86}\label{eq:substitutionsregel}
\end{align}
\] This is called substitution rule. The method of separation of variables is derived. Let $a, b\in \mathbb {R}$ with $a \[
\begin{align}
\frac{df}{dx} = g\left(f\right)h\left(x\right)
\end{align}
\] solve on $\left[a, b\right]$. You do the math \[
\begin{align}
\frac{1}{g\left(f\right)}\frac{df}{dx} = h\left(x\right).
\end{align}
\] This is now integrated from $a$ to a desired $x\in\left[a, b\right]$: \[
\begin{align}
\int_{a}^x\frac{1}{g\left(f\left(x'\right)\right)}\frac{df}{dx'}dx' &= \int_{a}^{x}h\left(x'\right)dx'\Leftrightarrow \int_{f\left(a\right)}^{f\left(x\right)}\frac{1}{g\left(f\right)}df = \int_{a}^{x}h\left(x'\right)dx'
\end{align}
\] The substitution rule was used here. If these two integrals are solvable, one obtains an algebraic expression for $f\left(x\right)$. Let a continuously differentiable function $f: \mathbb{R}^2 \to \mathbb{R}$ be given. Furthermore, let two functions $a \leq b$ be given, which depend on a parameter $t$, $a = a\left(t\right)$, $b = b\left(t\right)$. Define the function $F = F\left(t\right)$ by \[
\begin{align}
F\left(t\right) \coloneqq \int_{a\left(t\right)}^{b\left(t\right)}f\left(x, t\right)dx.
\end{align}
\] The Leibniz rule is a statement about the derivative of $F$. To derive this, you first define an auxiliary function \[
\begin{align}
\newtilde{F}\left(a, b, t\right) \coloneqq \int_{a}^{b}f\left(x, t\right)dx.
\end{align}
\] Then the derivative of $\newtilde{F}$ holds \[
\begin{align}
\newtilde{F}'\left(a, b, t\right)&= \nabla\newtilde{F}\left(a, b, t\right) = \left(\begin{array}{c}
\frac{\partial\int_{a}^{b}f\left(x, t\right)dx}{\partial a}\\
\frac{\partial\int_{a}^{b}f\left(x, t\right)dx}{\partial b}\\
\frac{\partial\int_{a}^{b}f\left(x, t\right)dx}{\partial t}
\end{array}\right).
\end{align}
\] This can be done using the main theorem of analysis \[
\begin{align}
\left(\begin{array}{c}
\frac{\partial\int_{a}^{b}f\left(x, t\right)dx}{\partial a}\\
\frac{\partial\int_{a}^{b}f\left(x, t\right)dx}{\partial b}\\
\frac{\partial\int_{a}^{b}f\left(x, t\right)dx}{\partial t}
\end{array}\right) &= \left(\begin{array}{c}
-f\left(a, t\right)\\
f\left(b, t\right)\\
\int_{a}^{b}\frac{\partial f\left(x, t\right)}{\partial t}dx
\end{array}\right)
\end{align}
\] rephrase. With the multidimensional chain rule you get \[
\begin{align}
\frac{dF\left(t\right)}{dt} &= \left(\begin{array}{c}
\frac{da}{dt}\\
\frac{db}{dt}\\
1
\end{array}\right)\cdot\nabla\newtilde{F}\left(a, b, t\right)
\end{align}
\]
\[
\begin{align}
\Leftrightarrow\frac{dF\left(t\right)}{dt} = \int_{a\left(t\right)}^{b\left(t\right)}\frac{\partial f\left(x, t\right)}{\partial t}dx + f\left[b\left(t\right), t\right]\frac{db}{dt} - f\left[a\left(t\right), t\right]\frac{da}{dt}.\tag{A.95}\label{eq:leibniz-rule}
\end{align}
\] This is the Leibniz rule. Now should \[
\begin{align}
\int_{0}^{\infty}x^ne^{-\beta x}dx = \frac{n!}{\beta^{n + 1}}\tag{A.96}\label{eq:int_prop_1}
\end{align}
\] with $\beta\in \mathbb {R}$, $\beta>0$, $n\in \mathbb {N}$. This is best done via complete induction. For $n = 0$ applies \[
\begin{align}
\int_{0}^{\infty}e^{-\beta x}dx = \frac{1}{\beta},
\end{align}
\] so the statement is true for $n = 0$. Now the statement already applies to $n$. Then follows by partial integration \[
\begin{align}
\int_{0}^{\infty}x^{n + 1}e^{-\beta x} &= \left[-\frac{1}{\beta}x^{n + 1}e^{-\beta x}\right]_0^{\infty} + \frac{n + 1}{\beta}\int_{0}^{\infty}x^ne^{-\beta x}dx = \frac{n + 1}{\beta}\frac{n!}{\beta^{n + 1}} = \frac{\left(n + 1\right)!}{\beta^{n + 2}}.
\end{align}
\] The statement therefore applies to all $n\in\mathbb{N}$. Still applies \[
\begin{align}
\int_{0}^{\infty}\frac{x^3}{e^x - 1}dx = \frac{\pi^4}{15}.\tag{A.99}\label{eq:stefanboltzmann_help}
\end{align}
\] For this one first calculates with Eq. (A.8) \[
\begin{align}
\frac{1}{e^x - 1} = \frac{e^{-x}}{1 - e^{-x}} = \frac{e^{-x} + 1 - 1}{1 - e^{-x}} = \frac{1}{1 - e^{-x}} - 1 = \sum_{n = 0}^{\infty}e^{-nx} - 1 = \sum_{n = 1}^{\infty}e^{-nx}.
\end{align}
\] This means that with the equations (A.96) and (A.12) \[
\begin{align}
\int_{0}^\infty\frac{x^3}{e^x - 1}dx &= \sum_{n = 1}^{\infty}\int_{0}^{\infty}x^3e^{-nx}dx = 6\sum_{n = 1}^{\infty}\frac{1}{n^4} = \frac{\pi^4}{15}.\tag{A.101}\label{eq:int_prop_3}
\end{align}
\] The so-called Gamma function is by \[
\begin{align}
\Gamma\left(n+1\right)\coloneqq\int_{0}^{\infty}x^ne^{-x}dx
\end{align}
\] defined. For them it follows with Eq. (A.96) \[
\begin{align}
\Gamma\left(n\right) = \left(n - 1\right)!.
\end{align}
\] The error function $\text{erf}$ is defined by \[
\begin{align}
\text{erf}\left(x\right) \coloneqq \frac{2}{\sqrt{\pi}}\int_{0}^{x}e^{-t^2}dt.\tag{A.104}\label{eq:def_fehlerfunktion}
\end{align}
\] According to [11] \[
\begin{align}
\lim\limits_{x\to\infty}\text{erf}\left(x\right) = 1\tag{A.105}\label{eq:integral_fehlerfunktion}.
\end{align}
\] For $C>0$ follows \[
\begin{align}
\lim\limits_{x\to\infty}\int_{0}^{\infty}e^{-Cx^2}dx = \frac{1}{\sqrt{C}}\lim\limits_{x\to\infty}\int_{0}^{\infty}e^{-x^2}dx = \frac{1}{2}\sqrt{\frac{\pi}{C}}.
\end{align}
\] Follow it \[
\begin{align}
\int_{0}^{\infty}e^{-Cx^2}x^2dx &= \left[x\left(-\frac{1}{2C}e^{-Cx^2}\right)\right]_0^\infty + \frac{1}{2C}\int_{0}^{\infty}e^{-Cx^2}dx = \frac{1}{4}\sqrt{\frac{\pi}{C^3}}\tag{A.107}\label{eq:int_prop_5},\\
\int_{0}^{\infty}e^{-Cx^2}x^3dx &= \left[x^2\cdot\left(-\frac{1}{2C}e^{-Cx^2}\right)\right]_0^\infty + \int_{0}^{\infty}\frac{x}{C}e^{-Cx^2}dx\nonumber\\
&= \int_{0}^{\infty}\frac{x}{C}e^{-Cx^2}dx = -\frac{1}{2C^2}\left[e^{-Cx^2}\right]_0^\infty = \frac{1}{2C^2}.\tag{A.108}\label{eq:int_prop_4}
\end{align}
\] One defines the gamma function for $x>0$ by \[
\begin{align}
\Gamma\left(x\right) \coloneqq \int_{0}^{\infty}t^{x - 1}e^{-t}dt.
\end{align}
\] For $n\in\mathbb{N}$, because of Eq. (A.96) \[
\begin{align}
\Gamma\left(n + 1\right) = \int_{0}^{\infty}t^{n}e^{-t}dt = n!.\tag{A.110}\label{eq:gamma_prop_0}
\end{align}
\] This follows in particular \[
\begin{align}
\Gamma\left(n + 1\right) &= n! = n\left(n - 1\right)! = n\Gamma\left(n\right)\tag{A.111}\label{eq:gamma_prop_1}.
\end{align}
\] One can use this to derive the Stirling formula, which is an approximation of $n!$ for high $n$, which is very useful when, for example, the question is how much $\left(10^{23}\right)!$ is. This is what you first calculate \[
\begin{align}
n! = \Gamma\left(n + 1\right) = \int_{0}^{\infty}t^ne^{-t}dt = \int_{0}^{\infty}e^{-t + n\ln\left(t\right)}dt = \int_{0}^{\infty}e^{-n\varphi\left(t\right)}dt
\end{align}
\] with \[
\begin{align}
\varphi\left(t\right) \coloneqq \frac{t}{n} - \ln\left(t\right).
\end{align}
\] They apply \[
\begin{align}
\lim\limits_{t \to 0}\varphi\left(t\right) = \infty, & {} & \lim\limits_{t \to \infty}\varphi\left(t\right) = \infty.
\end{align}
\] For the first two derivatives applies \[
\begin{align}
\varphi'\left(t\right) = \frac{1}{n} - \frac{1}{t}, & {} & \varphi''\left(t\right) = \frac{1}{t^2}.
\end{align}
\] The unique minimum is therefore $t = n$. The Taylor expansion up to the second order to the point $x = n$ is therefore \[
\begin{align}
\varphi\left(t\right) = 1 - \ln\left(n\right) + \frac{1}{2}\frac{\left(t - n\right)^2}{n^2} + \mathcal{O}\left[\left(t - n\right)^3\right].
\end{align}
\] The expression $e^{-n\varphi}$ is only relevant for negative and small $\varphi$, so the Taylor expansion to the minimum is a good approximation in the integral of the gamma function: \[
\begin{align}
n!\approx \int_{0}^{\infty}e^{-n + n\ln\left(n\right) - \frac{1}{2}\frac{\left(t - n\right)^2}{n}}dt = n^ne^{-n}\int_{0}^{\infty}e^{-\frac{t^2}{2n}}dt = n^{n}e^{-n}\sqrt{2\pi n} = \sqrt{2\pi}n^{n + 1/2}e^{-n}\tag{A.117}\label{eq:stirling}
\end{align}
\] In the last step, Eq. (A.105) is used. Eq. (A.117) is the Stirling formula. Let $n\in \mathbb{N}$ with $n \geq 1$. Let $\Omega \subseteq \mathbb {R}^3$ be connected and $\psi: \Omega\to \mathbb {R}^n$ and $f: \mathbb {R}^n\to \mathbb {R}$ be continuously differentiable. Define the functional \[
\begin{align}
F\left[\psi\right] \coloneqq \int_\Omega f\left(\psi\left(\mathbf{r}\right)\right)d^3r \in \mathbb{R}.
\end{align}
\] Now define a set of helper functions \[
\begin{align}
D \coloneqq \lbrace h:\Omega\to\mathbb{R}^n \vline \text{stetig differenzierbar, }h\left(\partial\Omega\right) = 0\rbrace.
\end{align}
\] Based on this, a further set of difference quotients is created \[
\begin{align}
H: D \times \mathbb{R}_{> 0}: \left(h, \epsilon\right) \mapsto \frac{F\left[\psi + \epsilon h\right] - F\left[\psi\right]}{\epsilon}
\end{align}
\] defined. The functional derivative \[
\begin{align}
\frac{\delta F}{\delta\psi}: \Omega \to \mathbb{R}^n
\end{align}
\] is the function which \[
\begin{align}
\frac{dH\left(h, \epsilon\right)}{d\epsilon} = \lim_{\epsilon\to 0}\frac{F\left[\psi + \epsilon h\right] - F\left[\psi\right]}{\epsilon} = \lim_{\epsilon\to 0}\frac{1}{\epsilon}\int_\Omega f\left(\psi\left(\mathbf{r}\right) + \epsilon h\left(\mathbf{r}\right)\right) - f\left(\psi\left(\mathbf{r}\right)\right)d^3r = \int_\Omega\frac{\delta F}{\delta\psi}\cdot h\left(\mathbf{r}\right)d^3r\tag{A.122}\label{eq:def_functional_derivative}
\end{align}
\] satisfied for all $h \in D$. The functional derivative can be understood as the derivative of $F$ in the direction $\psi$. Let $K\in\{\mathbb{R}, \mathbb{C}\}$, then one defines a vector space $V$ over $K$ in the following way: There is an illustration \[
\begin{align}
+ :V\times V\to V;\left(\mathbf{x}, \mathbf{y}\right)\mapsto \mathbf{x} + \mathbf{y}
\end{align}
\] and another illustration \[
\begin{align}
\cdot :K\times V\to V;\left(\lambda, \mathbf{x}\right)\mapsto\lambda \mathbf{x}.
\end{align}
\] These two functions have the following properties: \[
\begin{align}
\left(\mathbf{x} + \mathbf{y}\right) + \mathbf{z} = \mathbf{x} + \left(\mathbf{y} + \mathbf{z}\right), & {} & \mathbf{x} + \mathbf{y} = \mathbf{y} + \mathbf{x}, & {} & \mathbf{0} + \mathbf{x} = \mathbf{x}.
\end{align}
\] For every $\mathbf{x}\in V$ there exists a $\left(-\mathbf{x}\right)\in V$ with \[
\begin{align}
\left(-\mathbf{x}\right) + \mathbf{x} = 0.
\end{align}
\] Should continue to apply \[
\begin{align}
\lambda_1\cdot\left(\lambda_2\mathbf{x}\right) = \left(\lambda_1\lambda_2\right)\cdot\mathbf{x}, & {} & 1\cdot\mathbf{x} = \mathbf{x},\\
\lambda\left(\mathbf{x} + \mathbf{y}\right) = \lambda\mathbf{x} + \lambda\mathbf{y}, & {} & \left(\lambda_1 + \lambda_2\right)\mathbf{x} = \lambda_1\mathbf{x} + \lambda_2\mathbf{x}.
\end{align}
\] Vectors that can be understood as arrows in space are noted in this book $\mathbf{bold italic}$. A matrix $A$ over $K\{\mathbb{R}, \mathbb{C}\}$ is a rectangular array of elements from $K$ with $m\geq 1$ rows and $n\geq 1$ columns. Write $A_{i, j}$ for the element in the $jth column of the $ith row. One defines the transpose $A^T$ of $A$ by $A_{j, i} = A_{i, j}$ for all $1\leq i\leq m$, $1\leq j\leq n$. The multiplication of such a matrix with a vector $\mathbf{b} = \left(b_1, \dotsc, b_n\right)^T$ is defined by, \[
\begin{align}
A\mathbf{b} = \sum_{i = 1}^{m}\sum_{j = 1}^{n}A_{i, j}b_j\mathbf{e}_i,
\end{align}
\] This creates an $m-$dimensional column vector. More generally, given a matrix $B$ over $K$ with $n$ rows and $l\geq 1$ columns, then the matrix product $A\cdot B$ is defined by \[
\begin{align}
\left(A\cdot B\right)_{i, j} \coloneqq\sum_{\alpha = 1}^{n}A_{i, \alpha}B_{\alpha, j}.
\end{align}
\] It applies \[
\begin{align}
\left(A\cdot B\right)^T_{i, j}&= \left(A\cdot B\right)_{j, i} = \sum_{\alpha = 1}^{n}A_{j, \alpha}B_{\alpha, i} = \sum_{\alpha = 1}^{n}B_{\alpha, i}A_{j, \alpha} = \sum_{\alpha = 1}^{n}B^T_{i, \alpha}A^T_{\alpha, j} = \left(B^T\cdot A^T\right)_{i, j}
\end{align}
\] and thus is in this case \[
\begin{align}
\left(A\cdot B\right)^T = B^T \cdot A^T.
\end{align}
\] One further defines the dot product of two vectors $\mathbf{a}, \mathbf{b}\in \mathbb{R}^n$ by \[
\begin{align}
\mathbf{a}\cdot\mathbf{b} = \left\langle\mathbf{a}|\mathbf{b}\right\rangle \coloneqq \mathbf{a}^T\mathbf{b}.
\end{align}
\] If $A$ is a real matrix, then it follows \[
\begin{align}
\left\langle\mathbf{a}|A\mathbf{b}\right\rangle = \mathbf{a}^TA\mathbf{b} = \left(A^T\mathbf{a}\right)^T\mathbf{b} = \left\langle A^T\mathbf{a}\big|\mathbf{b}\right\rangle.
\end{align}
\] Matrices with as many rows as columns are called square. Operators are functions between functions. In meteorology only two function spaces are relevant: \[
\begin{align}
F_1 &= \{A\times\mathbb{R}\to\mathbb{R}, \:\text{unendlich oft stetig-differenzierbar}\}\\
F_2&= \{A\times\mathbb{R}\to\mathbb{R}^3, \:\text{unendlich oft stetig-differenzierbar}\}
\end{align}
\] Here $A$ is the atmosphere. Operators are marked with an umbrella symbol. You define \[
\begin{align}
& \newhat{A}\:\text{linearer}\:\text{Operator}:\nonumber\\
&\Leftrightarrow&\newhat{A}\left(\lambda\psi\right) = \lambda\newhat{A}\left(\psi\right)\land\newhat{A}\left(\psi_1 + \psi_2\right) = \newhat{A}\left(\psi_1\right) + \newhat{A}\left(\psi_2\right)
\end{align}
\] for each $\lambda\in \mathbb{C}$. Let $\newhat{A}$ and $\newhat{B}$ be operators, then define \[
\begin{align}
\newhat{A}\newhat{B} \coloneqq \newhat{A}\circ\newhat{B}
\end{align}
\] If $\newhat{A}$ and $\newhat{B}$ are linear, then the following applies \[
\begin{align}
\newhat{A}\left(\newhat{B}\left(\lambda\psi\right)\right) = \newhat{A}\left(\lambda\newhat{B}\left(\psi\right)\right) = \lambda\newhat{A}\left(\newhat{B}\left(\psi\right)\right)\Leftrightarrow \newhat{A}\newhat{B}\left(\lambda\psi\right) = \lambda\newhat{A}\newhat{B}\left(\psi\right)
\end{align}
\] Still applies \[
\begin{align}
\newhat{A}\newhat{B}\left(\psi_1 + \psi_2\right) = \newhat{A}\left(\newhat{B}\left(\psi_1\right) + \newhat{B}\left(\psi_2\right)\right) = \newhat{A}\newhat{B}\left(\psi_1\right) + \newhat{A}\newhat{B}\left(\psi_2\right).
\end{align}
\] Successive executions of linear operators are therefore linear. Furthermore, the brackets around the argument for operators are often left out: \[
\begin{align}
\newhat{A}\psi \coloneqq \newhat{A}\left(\psi\right)
\end{align}
\] Also applies \[
\begin{align}
\left(\lambda_1\newhat{A} + \lambda_2\newhat{B}\right)\lambda\psi &= \lambda_1\newhat{A}\lambda\psi + \lambda_2\newhat{B}\lambda\psi = \lambda_1\lambda\newhat{A}\psi + \lambda_2\lambda\newhat{B}\psi = \lambda\left(\lambda_1\newhat{A}\psi + \lambda_2\newhat{B}\psi\right)\nonumber\\
&= \lambda\left(\lambda_1\newhat{A} + \lambda_2\newhat{B}\right)\psi
\end{align}
\] as well as \[
\begin{align}
\left(\lambda_1\newhat{A} + \lambda_2\newhat{B}\right)\left(\psi_1 + \psi_2\right) &= \lambda_1\newhat{A}\left(\psi_1 + \psi_2\right) + \lambda_2\newhat{B}\left(\psi_1 + \psi_2\right)\nonumber\\
&= \lambda_1\newhat{A}\psi_1 + \lambda_1\newhat{A}\psi_2 + \lambda_2\newhat{B}\psi_1 + \lambda_2\newhat{B}\psi_2\nonumber\\
&= \left(\lambda_1\newhat{A} + \lambda_2\newhat{B}\right)\psi_1 + \left(\lambda_1\newhat{A} + \lambda_2\newhat{B}\right)\psi_2.
\end{align}
\] So linear combinations of linear operators are linear. Since the normal derivation rules apply to partial derivatives, we get \[
\begin{align}
\frac{\partial}{\partial x_i}\lambda\psi &= \lambda\frac{\partial}{\partial x_i}\psi\\
\frac{\partial}{\partial x_i}\left(\psi_1 + \psi_2\right) &= \frac{\partial}{\partial x_i}\psi_1 + \frac{\partial}{\partial x_i}\psi_2,
\end{align}
\] where the time is covered by one of the $x_i$. Partial derivatives and linear combinations of partial derivatives (also with basis vectors) are therefore linear. Let $\left(x_1, x_2, x_3\right)$ be three Cartesian coordinates and $\left(\mathbf{e}_1, \mathbf{e}_2, \mathbf{e}_3\right)$ be the corresponding basis. The vector product $\mathbf{A}\times\mathbf{B}$ of two vectors $\mathbf{A} = \left(A_1, A_2, A_3\right)^T$, $B = \left(B_1, B_2, B_3\right)^T$ is defined by \[
\begin{align}
\mathbf{A}\times\mathbf{B} \coloneqq \sum_{i, j, k = 1}^{3}\epsilon_{i, j, k}A_iB_j\mathbf{e}_k.
\end{align}
\] This is linear because for $\lambda\in \mathbb{R}$ and $\mathbf{C}\in \mathbb{R}^3$ \[
\begin{align}
\mathbf{A}\times\left(\mathbf{B} + \mathbf{C}\right) &= \sum_{i, j, k = 1}^{3}\epsilon_{i, j, k}A_i\left(B_j + C_j\right)\mathbf{e}_k = \sum_{i, j, k = 1}^{3}\epsilon_{i, j, k}A_iB_j\mathbf{e}_k + \sum_{i, j, k = 1}^{3}\epsilon_{i, j, k}A_iC_j\mathbf{e}_k\nonumber\\
&= \mathbf{A}\times\mathbf{B} + \mathbf{A}\times\mathbf{C}, \tag{A.147}\label{eq:vektorprodukt_linear_summe}\\
\mathbf{A}\times\left(\lambda\mathbf{B}\right) &= \sum_{i, j, k = 1}^{3}\epsilon_{i, j, k}A_i\lambda B_j\mathbf{e}_k = \lambda\mathbf{A}\times\mathbf{B}.\tag{A.148}\label{eq:vektorprodukt_linear_faktor}
\end{align}
\] Furthermore, it is anticommutative, \[
\begin{align}
\mathbf{A}\times\mathbf{B} = \sum_{i, j, k = 1}^{3}\epsilon_{i, j, k}A_iB_j\mathbf{e}_k = \sum_{i, j, k = 1}^{3} - \epsilon_{j,i,k}B_jA_i\mathbf{e}_k = -\mathbf{B}\times\mathbf{A}.\tag{A.149}\label{eq:cross_product_anticommutative}
\end{align}
\] For the $l$th component, $1 \leq l \leq 3$, of the vector product one finds \[
\begin{align}
\left(\mathbf{A}\times\mathbf{B}\right)_l &= \left(\sum_{i, j, k = 1}^{3}\epsilon_{i, j, k}A_iB_j\mathbf{e}_k\right)\cdot\mathbf{e}_l = \sum_{i, j, k = 1}^{3}\epsilon_{i, j, k}A_iB_j\delta_{k,l} = \sum_{i, j = 1}^{3}\epsilon_{i, j, l}A_iB_j.
\end{align}
\] The vector product has the orthogonality properties \[
\begin{align}
\mathbf{A}\cdot\left(\mathbf{A}\times\mathbf{B}\right) &= \sum_{i = 1}^3A_i\left(\mathbf{A}\times\mathbf{B}\right)_i = \sum_{i = 1}^3A_i\sum_{j, k = 1}^{3}\epsilon_{j, k, i}A_jB_k = \sum_{k = 1}^3B_k\sum_{i, j = 1}^3\epsilon_{k, i, j}A_iA_j\nonumber\\
&= \sum_{k = 1}^3B_k\sum_{i, j = 1}^3A_iA_j - A_jA_i = 0
\end{align}
\] as well as \[
\begin{align}
\mathbf{B}\cdot\left(\mathbf{A}\times\mathbf{B}\right) & \stackrel{\href{#eq:cross_product_anticommutative}{\text{Glg. (A.149)}}}{=} -\mathbf{B}\cdot\left(\mathbf{B}\times\mathbf{A}\right) = -0 = 0.
\end{align}
\] Furthermore, the so-called Graßmann identity applies: \[
\begin{align}
\mathbf{A}\times\left(\mathbf{B}\times\mathbf{C}\right) &= \sum_{i, j, k = 1}^{3}\epsilon_{i, j, k}A_i\left(\mathbf{B}\times\mathbf{C}\right)_j\mathbf{e}_k = \sum_{i, j, k = 1}^{3}\epsilon_{i, j, k}A_i\left(\sum_{l, m, n = 1}^3\epsilon_{l, m, n}B_lC_m\mathbf{e}_n\right)_j\mathbf{e}_k\nonumber\\
&= \sum_{i, j, k = 1}^{3}\epsilon_{i, j, k}A_i\sum_{l, m = 1}^3\epsilon_{l, m, j}B_lC_m\mathbf{e}_k = \sum_{k = 1}^3\mathbf{e}_k\sum_{i = 1}^3A_i\sum_{j, l, m = 1}^{3}\epsilon_{j, k, i}\epsilon_{j, l, m}B_lC_m\nonumber\\
&= \sum_{k = 1}^3\mathbf{e}_k\sum_{i = 1}^{3}A_i\left(B_kC_i - B_iC_k\right) = \mathbf{B}\left(\mathbf{A}\cdot\mathbf{C}\right) - \mathbf{C}\left(\mathbf{A}\cdot\mathbf{B}\right)\tag{A.153}\label{eq:bac-cab}
\end{align}
\] This is also known as BAC-CAB rule. You can still write \[
\begin{align}
\mathbf{B} \equiv \mathbf{B}_\parallel + \mathbf{B}_\perp,
\end{align}
\] where $\mathbf{B}_\parallel$ should denote the part of $\mathbf{B}$ parallel to $\mathbf{A}$ and \[
\begin{align}
\mathbf{B}_\perp \coloneqq \mathbf{B} - \mathbf{B}_\parallel
\end{align}
\] the remaining portion of $\mathbf{B}$ that is perpendicular to $\mathbf{A}$, as will be shown. It applies \[
\begin{align}
\mathbf{B}_\parallel = \left(\frac{1}{\left|\mathbf{A}\right|^2}\mathbf{A}\cdot\mathbf{B}\right)\mathbf{A}.
\end{align}
\] From this it follows \[
\begin{align}
\mathbf{A}\cdot\mathbf{B}_\perp = \mathbf{A}\cdot\left(\mathbf{B} - \mathbf{B}_\parallel\right) = \mathbf{A}\cdot\mathbf{B} - \mathbf{A}\cdot\left(\frac{1}{\left|\mathbf{A}\right|^2}\mathbf{A}\cdot\mathbf{B}\right)\mathbf{A} = 0,
\end{align}
\] so $\mathbf{B}_\perp$ is indeed perpendicular to $\mathbf{A}$. With Eq. (A.153) can also be noted \[
\begin{align}
\mathbf{A}\times\left(\mathbf{B}\times\mathbf{A}\right) &= \mathbf{B}A^2 - \mathbf{A}\left(\mathbf{A}\cdot\mathbf{B}\right) = A^2\mathbf{B} - A^2\mathbf{B}_\parallel = A^2\mathbf{B}_\perp\nonumber\\
\Rightarrow \mathbf{B}_\perp &= \frac{1}{A^2}\mathbf{A}\times\left(\mathbf{B}\times\mathbf{A}\right).
\end{align}
\] It applies \[
\begin{align}
\mathbf{A}\cdot\left(\mathbf{B}\times\mathbf{C}\right) &= \sum_{i = 1}^3A_i\left(\mathbf{B}\times\mathbf{C}\right)_i = \sum_{i = 1}^3A_i\sum_{j, k = 1}^{3}\epsilon_{j, k, i}B_jC_k = \sum_{i, j, k = 1}^{3}\epsilon_{j, k, i}A_iB_jC_k = \sum_{i, j, k = 1}^{3}\epsilon_{i, j, k}A_iB_jC_k\nonumber\\
&= \sum_{k = 1}^{3}C_k\sum_{i, j = 1}^{3}\epsilon_{i, j, k}A_iB_j = \sum_{i = 1}^3C_k\left(\mathbf{A}\times\mathbf{B}\right)_k = \mathbf{C}\cdot\left(\mathbf{A}\times\mathbf{B}\right).\tag{A.159}\label{eq:spat_prod_prop_0}
\end{align}
\]A.7.3 Separation of variables
A.7.4 Leibniz rule
A.7.5 More formulas
A.7.6 Functional derivatives
A.8 Vector spaces
A.8.1 Vector product