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        <dc:date>2025-12-23T03:33:58+0200</dc:date>
        <dc:creator>nikolaj (nikolaj@undisclosed.example.com)</dc:creator>
        <title>Nikolaj-K's notebook</title>
        <link>https://axiomsofchoice.org/nikolajs_notebook?rev=1766457238&amp;do=diff</link>
        <description>Nikolaj-K's notebook



This wiki is inactive since 2016 or so, but here I kept track of some of the vast amount of mathematical objects and learn about their relationships. 
This is the credible and neatly interlinked (interlinked) notebook of a physicist and where the content leans towards applications, it's with an eye on stochastics, statistical physics and their computational implementation.
It also contains content for a book and ideas for a typed programming language for the formal scienc…</description>
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        <dc:creator>nikolaj (nikolaj@undisclosed.example.com)</dc:creator>
        <title>Hermitian matrix</title>
        <link>https://axiomsofchoice.org/hermitian_matrix?rev=1598193206&amp;do=diff</link>
        <description>Hermitian matrix

Set
  context       $n\in\mathbb N$   definiendum   $ A \in \mathrm{HermitianMatrix}(n) $   postulate     $ A \in \mathrm{SquareMatrix}(n,\mathbb C) $   postulate     $ A^*=A $ 
Discussion

Reference

Wikipedia: Hermitian matrix

Parents

Subset of

Square matrix

Context

Matrix conjugate transpose

Related

Symmetric matrix</description>
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        <dc:creator>nikolaj (nikolaj@undisclosed.example.com)</dc:creator>
        <title>First infinite von Neumann ordinal</title>
        <link>https://axiomsofchoice.org/first_infinite_von_neumann_ordinal?rev=1589977428&amp;do=diff</link>
        <description>First infinite von Neumann ordinal

Set
  definiendum   $ \omega_{\mathcal N}$ 
...

----------

As is common, I'll also use the symbol $\mathbb N$ to denote the set theoretic object $\omega_{\mathcal N}$.



Idea

This is probably the most straightforward way to set up a countably infinite set.

Elaboration
$\omega_{\mathcal N}$$\emptyset$$\omega_{\mathcal N}$$\omega_{\mathcal N}$$\omega_{\mathcal N}$$\emptyset$$m$${\mathrm{succ}}\ m\equiv m\cup\{m\}$$0\equiv \emptyset$$1\equiv {\mathrm{succ}}\…</description>
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        <dc:date>2020-01-10T23:08:53+0200</dc:date>
        <dc:creator>nikolaj (nikolaj@undisclosed.example.com)</dc:creator>
        <title>x^x</title>
        <link>https://axiomsofchoice.org/x_x?rev=1578694133&amp;do=diff</link>
        <description>x^x

Function
  definiendum   $ \zeta: \mathbb{C}\setminus\{???\} \to \mathbb C$   definiendum   $ x\mapsto x^x$ 
----------

Note

Representations
 $x^x={\mathrm e}^{x\log(x)}=\left({\mathrm e}^x\right)^{\log(x)}$ 
	&quot;todo: write down the above with an expanded $\log$ to third order&quot;

Because of this, the local minimum of $x^x$ is that of $x\log(x)$, namely $\frac{1}{\mathrm e}\approx 0.37$, and then see
Secretary problem (Wikipedia)

Furthermore
 $x^x = \sum_{n=0}^\infty \prod_{k=1}^n (1-x)\lef…</description>
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        <title>Holomorphic function - [Discussion] </title>
        <link>https://axiomsofchoice.org/holomorphic_function?rev=1570397829&amp;do=diff</link>
        <description>Holomorphic function

Set
  context       $\mathcal O\subset \mathbb C$   definiendum   $f\in \mathrm{it}$   inclusion     $f:\mathcal O\to\mathbb C$   for all       $z_0\in\mathcal O$   postulate     $\left(\lim_{z \to z_0} {f(z) - f(z_0) \over z - z_0 }\right)\in\mathbb C $ 
Discussion

The following discussion is an elaboration/derivation on holomorphic functions from the viewpoint of analysis on $\mathbb R^n$. The article $\mathbb R^n, n&gt;1$$a+i\,b\in\mathbb C$$\langle a,b\rangle\in\mathbb R^…</description>
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        <title>Exponential function - typo</title>
        <link>https://axiomsofchoice.org/exponential_function?rev=1570375564&amp;do=diff</link>
        <description>Exponential function

Function
  definition    $\exp: \mathbb C\to\mathbb C$   definition    $\exp(z):=\sum_{k=0}^\infty \frac{1}{k!} z^k $ 
----------

Discussion

Theorems
 $\mathrm{e}^z = \exp(z) $ 
Because per definition $\mathrm{e}^z:=\exp(z\cdot \mathrm{ln}(\mathrm{e}))$.
 $\mathrm{e}^z \neq 0 $  $\frac{\mathrm d}{\mathrm d z}\mathrm{e}^{f(z)} = \frac{\mathrm d}{\mathrm dz}f(z)\cdot \mathrm{e}^{f(z)} $ 
$a,b,r,\theta\in\mathbb R$
 $\exp(i\theta)=\cos(\theta)+i\sin(\theta)$  $\forall a,b.\ …</description>
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        <dc:date>2019-09-28T18:11:56+0200</dc:date>
        <dc:creator>nikolaj (nikolaj@undisclosed.example.com)</dc:creator>
        <title>Dependent product functor</title>
        <link>https://axiomsofchoice.org/dependent_product_functor?rev=1569687116&amp;do=diff</link>
        <description>Dependent product functor

Functor
  context       ${\bf C}$ ... Cartesian closed category with all limits   let           $\prod_f$ ... right adjoint to the pullback functor $f^*$, where $f$ is an arrow in ${\bf C}$   context       $!_X:X\to *$ ... terminal morphisms for $X\in{\bf C}$   definition    $\prod_X p := {\mathrm{dom}}\prod_{!_X} p$ 
	&quot;todo: fmap of $\prod_X$$f^*$${\bf C}/Y$${\bf C}/X$$Y=*$$\prod_{!_X} p$${{\bf C}/*}$$\prod_X p$${\bf C}$${{\bf C}/*}$${\bf C}$$!_B:B\to{*}$$B$${\mathrm{…</description>
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        <dc:creator>nikolaj (nikolaj@undisclosed.example.com)</dc:creator>
        <title>Magic Gaussian integral</title>
        <link>https://axiomsofchoice.org/magic_gaussian_integral?rev=1569536340&amp;do=diff</link>
        <description>Magic Gaussian integral

Partial function

Discussion

The $\varepsilon$ prescription in the definition is just there so that one can evaluate the integral for certain complex matrices $A$ where it wouldn't exist otherwise. For example if $A$ has imaginary eigenvalues, then the naive integral will not be finite, while if we use $A_\varepsilon:=A-\varepsilon\,\mathrm{1}$$\mathrm{e}^{-\varepsilon\,\left\langle\phi\left|\right.\phi\right\rangle}$$I_a:=\int_{-\infty}^\infty{\mathrm e}^{-\tfrac{1}{2}…</description>
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        <dc:date>2019-09-23T21:19:31+0200</dc:date>
        <dc:creator>nikolaj (nikolaj@undisclosed.example.com)</dc:creator>
        <title>Infinite geometric series</title>
        <link>https://axiomsofchoice.org/infinite_geometric_series?rev=1569266371&amp;do=diff</link>
        <description>Infinite geometric series

Function
  definition    $Q_\infty: \{z\in{\mathbb C}\mid \vert{z}\vert&lt;1\}\to\mathbb C$   definition    $Q_\infty(z):=\sum_{k=0}^\infty z^k $ 
----------

$Q_\infty(z)=\dfrac{1}{1-z}$

This can also be written as

$\sum_{k=0}^\infty\left(\dfrac{1}{1+z}\right)^k = 1+\dfrac{1}{z}$

and

$\sum_{k=0}^\infty\left(1-\dfrac{1}{z}\right)^k = z$

or, for $z&gt;0$ and $X&lt;1+z$ resp. $X&lt;z/(z-1)$

$\sum_{k=0}^\infty\left(\dfrac{1}{1+z}\right)^kX^k = 1+\dfrac{1}{z}+(X-1)(z-1) \,z\dfra…</description>
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        <dc:date>2019-09-03T15:33:33+0200</dc:date>
        <dc:creator>nikolaj (nikolaj@undisclosed.example.com)</dc:creator>
        <title>Natural logarithm of real numbers</title>
        <link>https://axiomsofchoice.org/natural_logarithm_of_real_numbers?rev=1567517613&amp;do=diff</link>
        <description>Natural logarithm of real numbers

Function
  definiendum   $\mathrm{ln}:\mathbb R_+^*\to \mathbb R$   postulate     $\mathrm{ln}=\mathrm{exp}^{-1}$ 
----------

$\int_1^y \frac {1 } {x} {\mathrm d}x = \ln(y) $

$\int_0^{y} \frac {1 } {1+x } {\mathrm d}x = \ln(1+y) $


Log[a] == Log[b] + Integrate[1/(t+b)-1/(t+a),{t,0,Infinity}]


The function $x\mapsto\frac{x}{x-1}\log(x)$ is one without bad behaviours (singularities) on $[0,\infty)$.

----------

Subset of

Real logarithm,</description>
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        <title>Diagonal construction - fix $</title>
        <link>https://axiomsofchoice.org/diagonal_construction?rev=1566733719&amp;do=diff</link>
        <description>Diagonal construction

Set
  context       $ f:C\to \mathcal {\mathcal P}(C) $   definiendum   $ x\in D_f $   inclusion     $ D_f \subseteq C $   postulate     $ x \notin f(x) $ 
Discussion

We take an arbitrary set $C$ and argue about all the functions $f:C\to \mathcal {\mathcal P}(C)$ from $C$ to the powerset ${\mathcal P}(C)$. 
For any such $f$, we define $ D_f$ as the subset of $C$ containing the elements $x\in C$$ x \notin f(x) $$C=\{0,1\}$${\mathcal P}(C)=\{\{\},\{0\},\{1\},\{0,1\}\}$$f$$0…</description>
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