$$\mathfrak{a}$$
Test avec un seul signe dollar comme délimiteur MathJax : \(\newcommand{\schroederimp}{{\mathbin{=\!\!\!\!\!(\ }}}\) $$a \schroederimp b$$
$ \begin{array}{c} \mbox{{Primary} divisibilities:} \\ \begin{array}{ c | c |} d''''<b''', c''' & d_1> c_0, a' \\ \hline b'''<a'', c' & c_0>a'', c'\\ c'''<a'' & a'> a'' \\ \hline a''<a', c_0 & a''> c''', b'''\\ c'<c_0& c'> b'''\\ \hline a'< d_1 c='''' > d''''\\ c_{0}< d_1 b='''' > d'''' \\ \hline \end{array}\end{array} $
Les trucs qui ne fonctionnent pas sur Philiumm :
$$\frac{\sqrt[{}^{\textrm{\fbox{$\frac{z}{1+z}$}}}]{ar}}{\frac{z}{1+z}}$$
-
$$\int,dx \left(1-x^{\dotuline{2:p}}\right)^{\dotuline{n=m:2}}$$
$$\sqrt{\frac{d}{2}, \frac{e}{2} , \frac{f}{2} , \frac{d+e+f}{2}} \ \textrm{æqu.} \ \begin{tiny}\frac{\sqrt{\begin{matrix} def & \cap & d \\ &&e \\ &&f \end{matrix}}}{4}\end{tiny}$$
$$\sqrt{\frac{d}{2}, \frac{e}{2} , \frac{f}{2} , \frac{d+e+f}{2}} \ \textrm{aequ.} \ \frac{\sqrt{\begin{matrix} def & \cap & d \\ &&e \\ &&f \end{matrix}}}{4}$$
$$\sqrt{\frac{d}{2}, \frac{e}{2} , \frac{f}{2} , \frac{d+e+f}{2}} \ {\ae qu.} \ \frac{\sqrt{\begin{matrix} def & \cap & d \\ &&e \\ &&f \end{matrix}}}{4}$$
æ \(æ œ\)
-
$$y^3 = {\substack{y^{(3)} \\ ~}} {\substack{-2 y^{(2)} \\ -1 \ \ \ }} {\substack{~ \\ +y}} = y^{(3)} - 3y^{(2)} + y^{(1)}$$
$$y^{(3)} = y^3 + 3y^2 + 2y = y^3 + 1.2.3 y^{(2)} \Large{\substack{\ - \ 3y\\ + \ 2y}}$$
$$y^{(3)} = y^3 + 3y^2 + 2y = y^3 + 1.2.3 y^{(2)} {\substack{\ - \ 3y\\ + \ 2y}}$$
$$
\begin{array}{lll|llllllllll} \int y^1 & = & \frac{1}{2}y^{(2)}& dy & = & 1 & dy & & & & & & \\ \int y^2 & = & \frac{1}{3}y^{(3)}-\frac{1}{2}y^{(2)} & dy^2 & = & 1.2 & y^{(1)}dy & - & 1. & dy && & \\ \int y^3 & = & \frac{1}{4}y^{(4)}-\frac{3}{3}y^{(3)}+\frac{1}{2}y^{(2)} & d y^3 & = & 1.3 & y^{(2)}dy & - & 3.2 & y^{(1)}dy & + & 1 & dy\\ \int y^4 & = & \frac{1}{5}y^{(5)}-\frac{6}{4}y^{(4)}+\frac{7}{3}y^{(3)}-\frac{1}{2}y^{(2)} & d y^4 & = & 1.4 & y^{(3)}ddy & - & 6.3 & y^{(2)}dy & + & 7 & y^{(1)}dy \\ & \&c. & \&c. & & \&c. & & & \&c. & & & & & \end{array}
$$
--
Un autre test plus vieux :
$$ f(q)=2\sum_1^{\frac{k}{2}-1}+2\sum_{\frac{k}{2}-1}^{k-1}+2\sum_{k+1}^{\frac{n?k}{2}-1}+\ldots+\log4\frac{\pi^2}{3n^2}+\lim\limits_{q=q_0}\frac{(-1)^{\frac{n}{2}}}{n^2}16\sum_{1,\infty}\frac{\log(1+q^{p\frac{n}{2}}}{p^2} $$ pars priori hujus expressionis fit? ,? supra: \( \frac{\pi^2}{n^2}\sum_{1,\frac{n}{2}-1}^2 \frac{2(-1)^s\log4\cos\frac{sm\pi^2}{n}}{\sin \pi \frac{s^2}{n}}+\frac{\pi^2}{3?n^2}\log4\), beneficio formulae notae $$\frac{1}{\sin^2z}=\frac{1}{z^2}+\frac{1}{(\pi+z)^2}+\frac{1}{(\pi-z)^2}++\frac{1}{(2\pi+z)^2}++\frac{1}{(2\pi-z)^2}+\ldots$$
\(a < b \)
Lines antem ita determinari potesti ponamus \(
$$\lim\frac{(-1)^{\frac{n}{2}16}{n^2}}\sum_{1,\infty}\frac{\log(1-\varrho^{p\frac{n}{2}})}{p^2}$$
$\log\frac{q_0-q}{q_0+q}=\log\frac{1-\varrho}{1+\varrho}$ vel [rature illisible] in limite\(=\log \frac{1-p}{2}\), porro habemus
$$\lim \frac{1-\varrho^{p\frac{n}{2}}}{1-\varrho}=\frac{pn}{2} $$
unde ?faillime? formula nostra demana adjumenta formulae : \(1+\frac{1}{9}+\frac{1}{25}+\frac{1}{49}+\ldots=\frac{\pi^2}{8}\)
Perinde obtinetur
\begin{align*}
&63) \, \int_0^{e^{xi}}-\log k'\frac{dq}{q}=-2\log\tg~\frac{x^2}{2}-\frac{2}{9}\log\tg~\frac{3x^2}{2}-\frac{2}{25}\log\tg~\frac{5x^2}{2}-\ldots\\
&+4\pi i\left\{\left(\frac{x}{2\pi}\right)-\left(\frac{x}{2\pi}+\frac{1}{2}\right)\right\}+\frac{4\pi i}{9}\left\{\left(\frac{3x}{2\pi}\right)-\left(\frac{3x}{2\pi}+\frac{1}{2}\right)\right\}+\frac{4\pi i}{25}\left\{\left(\frac{5x}{2\pi}\right)-\left(\frac{5x}{2\pi}+\frac{1}{2}\right)\right.\\
\end{align*}
\begin{align*}
& 55)\, \int_0^q (\log k - \log 4\sqrt q)\frac{dq}{q}=-4\sum_p \frac{\varphi (p)}{p^2}(q^p-\frac{3q^{2p}}{4}-\frac{3}{16}q^{4p}-\frac{3}{64}q^{8p}-\frac{3}{256}q^{16p}-\ldots)\\
& 56)\, \int_0^q -\log k' \frac{dq}{q}=8\sum_p \frac{\varphi (p)}{p^2}q^p\\
& 57) \, \int_0^q\log\frac{2K}{\pi} \frac{dq}{q}=4\sum_p\frac{\varphi (p)}{p^2}(q^p-\frac{1}{2}q^{2p}-\frac{1}{4}q^{4p}-\frac{1}{8}q^{8p}-\frac{1}{16}q^{16p}-\ldots)\\
\end{align*}
\begin{align*} & 55)\, \int_0^q (\log k - \log 4\sqrt q)\frac{dq}{q}=-4\sum_p \frac{\varphi (p)}{p^2}(q^p-\frac{3q^{2p}}{4}-\frac{3}{16}q^{4p}-\frac{3}{64}q^{8p}-\frac{3}{256}q^{16p}-\ldots)\\ & 56)\, \int_0^q -\log k' \frac{dq}{q}=8\sum_p \frac{\varphi (p)}{p^2}q^p\\ & 57) \, \int_0^q\log\frac{2K}{\pi} \frac{dq}{q}=4\sum_p\frac{\varphi (p)}{p^2}(q^p-\frac{1}{2}q^{2p}-\frac{1}{4}q^{4p}-\frac{1}{8}q^{8p}-\frac{1}{16}q^{16p}-\ldots)\\ \end{align*} \begin{align*} & 58) \, \int_0^q\left(\frac{2K}{\pi}-1\right)\frac{dq}{q}=4\sum\frac{\psi(n)q^{2^l(4m-1)^2n}}{2^l(4m-1)^2n}\\ & 59)\, \int_0^q\left(\frac{2kK}{\pi}\right)\frac{dq}{q}=8\sum\frac{\psi(n)q^{\frac{(4m-1)^2n}{2}}}{(4m-1)^2n}\\ & 60)\, \int_0^q\left(\frac{2k'K}{\pi}-1\right)\frac{dq}{q}=-4\sum\frac{\psi(n)q^{(4m-1)^2n}}{(4m-1)^2n}+4\sum\frac{\psi(n)q^{2^{l+1}(4m-1)^2n}}{2^{l+1}(4m-1)^2n}\\ & 61)\, \int_0^q\left(\frac{2\sqrt k'K}{\pi}-1\right)\frac{dq}{q}=-4\sum\frac{\psi(n)q^{2(4m-1)^2n}}{2(4m-1)^2n}+4\sum\frac{\psi(n)q^{2^{l+2}(4m-1)^2n}}{2^{l+2}(4m-1)^2n}\ \end{align*}
\(\begin{array}{c c c c} \mbox{ All } & \frak a - \frak b & \mbox{ are themselves } & \frak c \\ \mbox{ " } & \frak b - \frak c & \mbox{ " ~ "} & \frak c \\ \mbox{ " } & \frak a + \frak d & \mbox{ " ~ "} & \frak d \\ \mbox{ " } & \frak b + \frak d & \mbox{ " ~ "} & \frak d \\ \mbox{ " } & \frak a \pm \frak a' & \mbox{ " ~ "} & \frak a \\ \mbox{ " } & \frak b \pm \frak b' & \mbox{ " ~ "} & \frak b \\ \mbox{ " } & \frak c - \frak c' & \mbox{ " ~ "} & \frak c \\ \mbox{ " } & \frak d + \frak d' & \mbox{ " ~ "} & \frak d \\ \end{array} \)
$$ \begin{array}{c} \mbox{{Primary} divisibilities:} \\ \begin{array}{ c | c |} d''''<b''', c''' & d_1> c_0, a' \\ \hline b'''<a'', c' & c_0>a'', c'\\ c'''<a'' & a'> a'' \\ \hline a''<a', c_0 & a''> c''', b'''\\ c'<c_0& c'> b'''\\ \hline a'< d_1 c='''' > d''''\\ c_{0}< d_1 b='''' > d'''' \\ \hline \end{array}\end{array} $$
\begin{eqnarray*} \frak{a'}_n=\frak{a+b}, \, \frak{a'}_{n-1} ... \frak{a'}_1, \frak{a'}_0=\frak{a'} \end{eqnarray*}
eine vollständige Kette der Gruppen \(\frak{M_{a+b,a}}\) und setzt man
\begin{eqnarray*}\frak{b^{(r)}_1=b-a'_r} \end{eqnarray*}
so bilden die \(n+1\) Moduln
\begin{eqnarray*}\frak{b}^{(n)}_1=b, \, \frak{b}^{(n-1)}_1 ..., \frak{b^1, b^{(0)}_1=a-b} \end{eqnarray*}
ein vollständige Kette der Gruppe \(\frak{M_{b,a-b}}\).