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diff --git a/2025math.tex b/2025math.tex
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--- a/2025math.tex
+++ b/2025math.tex
@@ -49,6 +49,7 @@
% \everymath{\displaystyle}
\DeclareMathOperator{\sgn}{sgn}
\DeclareMathOperator{\arctg}{arctg}
+%\DeclareMathOperator{\arcsin}{arcsin}
\title {Экзамен по матеше, 2025, I семестр}
\author {\texttt{justanothercatgirl}}
@@ -1104,13 +1105,32 @@ $f^{(n)} = -2(n-1)f^{n-2}(0)$
\item 99: Сами решите)))
-
-
-
-
-
-
-
+\item 114: найти $\arcsin^{(n)}(0)$
+\[\vsystem . {
+& f(x) = \arcsin x,\;
+ f'(x) = \frac 1 {\left.f^{-1}\right.'(y)} = \frac 1 {\cos y} = \frac 1 {\sqrt{1 - x^2}},\;
+ f'(x)\sqrt{1-x^2} = 1,\;
+ \left(f'(x)\right)^2(1 - x^2) = 1\\
+& f''(x) = 2f'f''(1-x^2) - 2f'f'x \overset{\all x\;f' \neq 0}= f''(1-x^2) - 2f'x = 0. \text{ Берём производную $n$ раз...} \\
+& \sum_{k=0}^nC_n^k(1-x^2)^{(n-k)}f^{(n+2)} - \sum_{k=0}^nC_n^k (x)^{(n-k)}f^{(n+1)} = \\
+& = 0 + \dots + 0 + \frac{n!}{(n-2)!(2)!}(1-x^2)''f^{(n)} + \frac{n!}{(n-1)!1!}(1-x^2)'f^{(n+1)} + (1-x^2)f^{(n+2)} - \\
+& \hspace*{1cm} - 0 - \dots - 0 - \frac{n!}{(n-1)!1!}x'f^{(n)} - xf^{(n+1)} = \\
+& = -n(n-1)f^{(n)} - 2xnf^{(n+1)} + (1-x^2)f^{(n+2)} - nf^{(n)} - xf^{(n+1)} = \\
+& = -n^2f^{(n)} - x(2n-1)f^{(n+1)} + (1-x^2)f^{(n+2)} = 0\quad\text{ Для }x=0: \\
+& f^{(n+2)} = n^2f^{(n)};\;\system{f'(0) = 1\\ f''(0) = 0} \implies \system{
+ &f^{n}(0) = 0,&n \bmod 2 = 0 \\
+ &f^{n}(0) = (n-2)^2\cdot(n-4)^2\dots3^2\cdot1^2 = (n-2)!!^2,&n \bmod 2 = 1
+ }\\
+&\text{Ответ: } \arcsin^{(n)}(0) = (n \bmod 2)\cdot \left((n-2)!!\right)^2
+}\]
+
+Проверка: \\$\dst
+ \arcsin x = \sum_{k=0}^\infty \frac{\arcsin^{(k)}(0)x^k}{k!} = x + \sum_{k=3}^\infty \frac{(k \bmod 2)((k-2)!!)^2x^k}{k!}
+ \overset{n := 2k}= x+\sum_{n=1}^\infty \frac{x^{2n+1}(2n-1)!!(2n-1)!!}{(2n+1)!} = x+ \sum_{n=1}^\infty \frac{(2n-1)!!}{(2n)!!}\frac{x^{2n+1}}{2n+1}
+$
+
+Сверяясь с википедией, вроде оно
\end{itemize}
+
\end{document}