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Diffstat (limited to '2025math.tex')
-rw-r--r-- | 2025math.tex | 34 |
1 files changed, 27 insertions, 7 deletions
diff --git a/2025math.tex b/2025math.tex index e10f50d..181877b 100644 --- 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} |