diff --git a/advanced-math/exercise/4-integal-of-functions-of-single-variable/integal-of-functions-of-single-variable.pdf b/advanced-math/exercise/4-integal-of-functions-of-single-variable/integal-of-functions-of-single-variable.pdf index a1bc565..da55b42 100644 Binary files a/advanced-math/exercise/4-integal-of-functions-of-single-variable/integal-of-functions-of-single-variable.pdf and b/advanced-math/exercise/4-integal-of-functions-of-single-variable/integal-of-functions-of-single-variable.pdf differ diff --git a/advanced-math/exercise/4-integal-of-functions-of-single-variable/integal-of-functions-of-single-variable.tex b/advanced-math/exercise/4-integal-of-functions-of-single-variable/integal-of-functions-of-single-variable.tex index de1194b..57bb555 100644 --- a/advanced-math/exercise/4-integal-of-functions-of-single-variable/integal-of-functions-of-single-variable.tex +++ b/advanced-math/exercise/4-integal-of-functions-of-single-variable/integal-of-functions-of-single-variable.tex @@ -179,7 +179,7 @@ \subsubsection{第一类换元} \hline \end{tabular} \medskip -$\therefore=t^2\sin t-2t\cos t-\sin t+C=(\arcsin x)^2x+2\arcsin x\sqrt{1-x^2}-2x+C$。 +$\therefore=t^2\sin t+2t\cos t-2\sin t+C=(\arcsin x)^2x+2\arcsin x\sqrt{1-x^2}-2x+C$。 \paragraph{指对换元} \leavevmode \medskip diff --git a/advanced-math/knowledge/0-perpare/perpare.pdf b/advanced-math/knowledge/0-perpare/perpare.pdf index 1ea5869..34b99b0 100644 Binary files a/advanced-math/knowledge/0-perpare/perpare.pdf and b/advanced-math/knowledge/0-perpare/perpare.pdf differ diff --git a/advanced-math/knowledge/0-perpare/perpare.tex b/advanced-math/knowledge/0-perpare/perpare.tex index f012aff..f07e69e 100644 --- a/advanced-math/knowledge/0-perpare/perpare.tex +++ b/advanced-math/knowledge/0-perpare/perpare.tex @@ -1004,7 +1004,7 @@ \subsubsection{倍角公式} $1+\sin2\alpha=(\sin\alpha+\cos\alpha)^2$，$1-\sin2\alpha=(\sin\alpha-\cos\alpha)^2$。 -$\sin 3\alpha=-4\sin^3\alpha_3\sin\alpha,\cos 3\alpha=4\cos^3\alpha-3\cos\alpha$。 +$\sin 3\alpha=-4\sin^3\alpha+3\sin\alpha,\cos 3\alpha=4\cos^3\alpha-3\cos\alpha$。 $\tan 2\alpha=\dfrac{2\tan\alpha}{1-\tan^2\alpha},\cot 2\alpha=\dfrac{\cot^2\alpha-1}{2\cot\alpha}$。 diff --git a/advanced-math/knowledge/1-function-and-limit/function-and-limit.pdf b/advanced-math/knowledge/1-function-and-limit/function-and-limit.pdf index 870e8f8..6df9b18 100644 Binary files a/advanced-math/knowledge/1-function-and-limit/function-and-limit.pdf and b/advanced-math/knowledge/1-function-and-limit/function-and-limit.pdf differ diff --git a/advanced-math/knowledge/1-function-and-limit/function-and-limit.tex b/advanced-math/knowledge/1-function-and-limit/function-and-limit.tex index c39a2da..d8c7970 100644 --- a/advanced-math/knowledge/1-function-and-limit/function-and-limit.tex +++ b/advanced-math/knowledge/1-function-and-limit/function-and-limit.tex @@ -990,7 +990,7 @@ \subsubsection{麦克劳林公式} \item $e^x=\sum\limits_{i=0}^n\dfrac{1}{i!}x^i$，$=1+\dfrac{1}{1!}x+\dfrac{1}{2!}x^2+\dfrac{1}{3!}x^3+o(x^3)$。 \item $\ln(1+x)=\sum\limits_{i=1}^n(-1)^{i+1}\dfrac{1}{i}x^i$，$=x-\dfrac{1}{2}x^2+\dfrac{1}{3}x^3+o(x^3)$。 \item $\sin x=\sum\limits_{i=1}^{2i-1}(-1)^{2i-1}\dfrac{1}{(2i-1)!}x^{2i-1}$，$=x-\dfrac{1}{3!}x^3+\dfrac{1}{5!}x^5+o(x^5)$。 - \item $\cos x=\sum\limits_{i=1}^{2i}(-1)^{2i-1}\dfrac{1}{(2i-2)!}x^{2i-2}$，$=x-\dfrac{1}{2!}x^2+\dfrac{1}{4!}x^4+o(x^4)$。 + \item $\cos x=\sum\limits_{i=1}^{2i}(-1)^{2i-1}\dfrac{1}{(2i-2)!}x^{2i-2}$，$=1-\dfrac{1}{2!}x^2+\dfrac{1}{4!}x^4+o(x^4)$。 \item $\arcsin x=\sum\limits_{i=1}^{2i-1}\dfrac{(2i-3)!!}{(2i-2)!!}\dfrac{x^{2i-1}}{2i-1}$，$=x+\dfrac{1}{2}\dfrac{x^3}{3}+\dfrac{1\times3}{2\times4}\dfrac{x^5}{5}+\dfrac{1\times3\time5}{2\times4\times6}\dfrac{x^7}{7}+o(x^7)$。（假定$-1!=0!$） \item $\dfrac{1}{1-x}=\sum\limits_{i=0}^nx^i$，$=1+x+x^2+x^3+o(x^3)$。 \item $(1+x)^a=1+\sum\limits_{i=1}^n\dfrac{\prod_{j=1}^i(a-j+1)}{i!}x^i$，$=1+\dfrac{a}{1!}x+\dfrac{a(a-1)}{2!}x^2\\+\dfrac{a(a-1)(a-2)}{3!}x^3+o(x^3)$。 diff --git a/probability-theory-and-mathematical-statistics/knowledge/2-random-variables-and-distribution/random-variables-and-distribution.pdf b/probability-theory-and-mathematical-statistics/knowledge/2-random-variables-and-distribution/random-variables-and-distribution.pdf index 90dc040..32ca862 100644 Binary files a/probability-theory-and-mathematical-statistics/knowledge/2-random-variables-and-distribution/random-variables-and-distribution.pdf and b/probability-theory-and-mathematical-statistics/knowledge/2-random-variables-and-distribution/random-variables-and-distribution.pdf differ diff --git a/probability-theory-and-mathematical-statistics/knowledge/2-random-variables-and-distribution/random-variables-and-distribution.tex b/probability-theory-and-mathematical-statistics/knowledge/2-random-variables-and-distribution/random-variables-and-distribution.tex index c23419d..a1022e9 100644 --- a/probability-theory-and-mathematical-statistics/knowledge/2-random-variables-and-distribution/random-variables-and-distribution.tex +++ b/probability-theory-and-mathematical-statistics/knowledge/2-random-variables-and-distribution/random-variables-and-distribution.tex @@ -379,7 +379,7 @@ \subsection{离散型} \hline \end{tabular} -\subsection{连续性} +\subsection{连续型} 设$X$为离散型随机变量，其分布函数、概率密度为$F_X(x)$与$f_X(x)$，随机变量$Y=g(X)$也是$X$的函数，则$Y$的分布函数或概率密度可用分布函数法得到$F_Y(y)=P\{Y\leqslant y\}=P\{g(X)\leqslant y\}=\int_{g(X)\leqslant y}f_X(x)\,\textrm{d}x$。 @@ -387,7 +387,7 @@ \subsection{连续性} 首先已知$X$的概率密度函数为$f_X(x)$，分布函数为$F_X(x)$，已知$Y=g(X)$，即$Y$对$X$的映射关系。现在要求$Y$的概率规律，即要求$Y$的概率密度$f_Y(y)$与分布函数$F_Y(y)$。 -先求分布函数$F_Y(y)=P\{Y\leqslant y\}=P\{g(X)\leqslant y\}$，然后用$y$来表示$X$，这是连续性随机变量函数分布的重点。 +先求分布函数$F_Y(y)=P\{Y\leqslant y\}=P\{g(X)\leqslant y\}$，然后用$y$来表示$X$，这是连续型随机变量函数分布的重点。 即$X$在以$y$表示的一个区间上，$X\in I_y$，所以解得$Y$分布函数$\int_{I_y}f_X(x)\,\textrm{d}x$。 diff --git a/probability-theory-and-mathematical-statistics/knowledge/4-law-of-large-numbers-and-central-limit-theorem/law-of-large-numbers-and-central-limit-theorem.pdf b/probability-theory-and-mathematical-statistics/knowledge/4-law-of-large-numbers-and-central-limit-theorem/law-of-large-numbers-and-central-limit-theorem.pdf index b8fbb74..004639e 100644 Binary files a/probability-theory-and-mathematical-statistics/knowledge/4-law-of-large-numbers-and-central-limit-theorem/law-of-large-numbers-and-central-limit-theorem.pdf and b/probability-theory-and-mathematical-statistics/knowledge/4-law-of-large-numbers-and-central-limit-theorem/law-of-large-numbers-and-central-limit-theorem.pdf differ diff --git a/probability-theory-and-mathematical-statistics/knowledge/4-law-of-large-numbers-and-central-limit-theorem/law-of-large-numbers-and-central-limit-theorem.tex b/probability-theory-and-mathematical-statistics/knowledge/4-law-of-large-numbers-and-central-limit-theorem/law-of-large-numbers-and-central-limit-theorem.tex index 3d6ad21..02fb31a 100644 --- a/probability-theory-and-mathematical-statistics/knowledge/4-law-of-large-numbers-and-central-limit-theorem/law-of-large-numbers-and-central-limit-theorem.tex +++ b/probability-theory-and-mathematical-statistics/knowledge/4-law-of-large-numbers-and-central-limit-theorem/law-of-large-numbers-and-central-limit-theorem.tex @@ -72,7 +72,7 @@ \subsection{切比雪夫大数定律} 根据夹逼定理$\lim\limits_{n\to\infty}P\left\{\left\vert\dfrac{1}{n}\sum\limits_{i=1}^nX_i-\dfrac{1}{n}\sum\limits_{i=1}^nEX_i\right\vert<\epsilon\right\}$=1。 -\textbf{例题：}设$X_1,X_2,\cdots,X_n,\cdots$为相互独立的随机变量序列，$X_n$服从参数为$n$的指数分布（$n\leqslant1$），则下列随机变量序列中不服从切比雪夫大数定律的是()。 +\textbf{例题：}设$X_1,X_2,\cdots,X_n,\cdots$为相互独立的随机变量序列，$X_n$服从参数为$n$的指数分布（$n\geqslant1$），则下列随机变量序列中不服从切比雪夫大数定律的是()。 $A.X_1,\dfrac{1}{2}X_2,\cdots,\dfrac{1}{n}X_n,\cdots$\qquad$B.X_1,X_2,\cdots,X_n,\cdots$