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第一讲完成
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Didnelpsun
@@ -25,9 +25,13 @@
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% 为了实现相对位置的设定
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\usepackage{xcolor}
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% 为了实现不同的颜色
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\usepackage{array}
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% 设置表格行距
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\author{Didnelpsun}
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\title{考研数学准备}
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\begin{document}
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\renewcommand\arraystretch{1.5}
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% 表格长1.5倍
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\maketitle
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\thispagestyle{empty}
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\tableofcontents
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@@ -60,7 +64,7 @@ $y=f(x)$，定义域为$D$，值域为$R$，若对于每一个$y\in R$，必然
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$\therefore x\in[-1,3]$
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$\therefore\frac{\rm{d}\psi(x)}{\rm{d}x}=(-x^2+2x+3)'=-2x+2=0$
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$\therefore\dfrac{\rm{d}\psi(x)}{\rm{d}x}=(-x^2+2x+3)'=-2x+2=0$
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$\therefore x=1$，驻点为1
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@@ -82,13 +86,13 @@ $x+\sqrt{x^2+1}>x+\vert x\vert \geqslant 0$，所以$x\in R$。
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而研究其奇偶性：
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$f(-x)=\ln(-x+\sqrt{x^2+1})=\ln(\frac{1}{\sqrt{x^2+1}+x})=-\ln(x+\sqrt{x^2+1})=-f(x)$
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$f(-x)=\ln(-x+\sqrt{x^2+1})=\ln(\dfrac{1}{\sqrt{x^2+1}+x})=-\ln(x+\sqrt{x^2+1})=-f(x)$
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所以该函数为奇函数。
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对其求单调性，即通过链式法则求导：
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$\frac{\rm{d}y}{\rm{d}x}=\frac{1}{x+\sqrt{x^2+1}}\cdot (1+\frac{2x}{2\sqrt{x^2+1}})=\frac{1}{\sqrt{x^2+1}}>0$
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$\dfrac{\rm{d}y}{\rm{d}x}=\dfrac{1}{x+\sqrt{x^2+1}}\cdot (1+\dfrac{2x}{2\sqrt{x^2+1}})=\dfrac{1}{\sqrt{x^2+1}}>0$
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所以该函数严格单调增。
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@@ -105,7 +109,7 @@ $$
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$$
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\begin{aligned}
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\because -y & =-\ln(x+\sqrt{x^2+1}) \\
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& =\ln(\frac{1}{x+\sqrt{x^2+1}}) \\
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& =\ln(\dfrac{1}{x+\sqrt{x^2+1}}) \\
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& =\ln(\sqrt{x^2+1}-x) \\
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e^{-y} & =\sqrt{x^2+1}-x
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\end{aligned}
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@@ -114,17 +118,17 @@ $$
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$$
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\begin{aligned}
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\therefore e^y-e^{-y} & =2x \\
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x & =\frac{e^y-e^{-y}}{2}
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x & =\dfrac{e^y-e^{-y}}{2}
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\end{aligned}
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$$
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解出了用x表示y的函数表达$x=f^{-1}(y)$，即反函数，则$f^{-1}(x)=\frac{e^x-e^{-x}}{2}$
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解出了用x表示y的函数表达$x=f^{-1}(y)$，即反函数，则$f^{-1}(x)=\dfrac{e^x-e^{-x}}{2}$
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这种曲线为一种常见曲线：
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\begin{itemize}
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\item $\frac{e^x-e^{-x}}{2}$：双曲正弦。
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\item $\frac{e^x+e^{-x}}{2}$：双曲余弦。（为一种悬链线）
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\item $\dfrac{e^x-e^{-x}}{2}$：双曲正弦。
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\item $\dfrac{e^x+e^{-x}}{2}$：双曲余弦。（为一种悬链线）
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\item $\ln(x+\sqrt{x^2+1})$：反双曲正弦。
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\item $\ln(x+\sqrt{x^2-1})$：反双曲余弦。
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\end{itemize}
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@@ -184,8 +188,8 @@ $$
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\subsection{单调性}
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$\begin{matrix}
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\frac{\rm{d}y}{\rm{d}x}>0 & \Rightarrow & (x_1-x_2)[f(x_1)-f(x_2)]>0 & \Rightarrow & f(x)\nearrow \\
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\frac{\rm{d}y}{\rm{d}x}<0 & \Rightarrow & (x_1-x_2)[f(x_1)-f(x_2)]<0 & \Rightarrow & f(x)\searrow
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\dfrac{\rm{d}y}{\rm{d}x}>0 & \Rightarrow & (x_1-x_2)[f(x_1)-f(x_2)]>0 & \Rightarrow & f(x)\nearrow \\
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\dfrac{\rm{d}y}{\rm{d}x}<0 & \Rightarrow & (x_1-x_2)[f(x_1)-f(x_2)]<0 & \Rightarrow & f(x)\searrow
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\end{matrix}
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$
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@@ -204,7 +208,7 @@ $
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如当a为0时：$f(b)-f(a)=f'(\xi )(b-a)=f(b)=bf'(\xi)$
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如$f(x)>xf(1)$变形为$\frac{f(x)}{x}>f(1)$，辅助函数$F(x)=\frac{f(x)}{x}$
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如$f(x)>xf(1)$变形为$\dfrac{f(x)}{x}>f(1)$，辅助函数$F(x)=\dfrac{f(x)}{x}$
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所以加减法警惕0，乘除法警惕1。
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@@ -252,7 +256,7 @@ $y=x^{\mu}$，$\mu$为实数，当$x>0$，$y=x^{\mu}$都有定义：
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\draw[-latex](0,-2) -- (0,4) node[above]{$y$};
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\draw[black, thick, smooth, domain=0.3:2] plot (\x,1/\x) node[right]{$\mu =-1$};
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\draw[black, thick, smooth, domain=-2:-0.5] plot (\x,1/\x) node[right]{$\mu =-1$};
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\draw[black, thick, smooth, domain=0.01:2] plot (\x, {sqrt(\x)}) node[right]{$\mu =\frac{1}{2}$};
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\draw[black, thick, smooth, domain=0.01:2] plot (\x, {sqrt(\x)}) node[right]{$\mu =\dfrac{1}{2}$};
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\draw[black, thick, smooth, domain=-2:2] plot (\x,\x) node[right]{$\mu =1$};
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\draw[black, thick, smooth, domain=-2:2] plot (\x, {\x*\x}) node[right]{$\mu =2$};
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\filldraw[black] (0,0) node[below]{$O$};
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@@ -264,7 +268,7 @@ $y=x^{\mu}$，$\mu$为实数，当$x>0$，$y=x^{\mu}$都有定义：
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\begin{enumerate}
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\item $\sqrt{u}，\sqrt[3]{u}$可以使用$u$来研究。
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\item $\vert u\vert$可以使用$u^2$来研究。
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\item $\frac{1}{u},u>0$可以使用$u$来研究，但是最值相反。
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\item $\dfrac{1}{u},u>0$可以使用$u$来研究，但是最值相反。
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\item $u_1u_2...u_n$可以使用$\sum_{i=1}^{n}\ln u_i$来研究。
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\end{enumerate}
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@@ -323,9 +327,9 @@ $y=log_ax(a>0,a\neq 1)$为$y=a^x$的反函数：
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\draw[black, thick, smooth, domain=-5:5] plot (\x,{sin(\x r)}) node at (0,1.5){$\sin(x)$};
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\draw[black, densely dashed](-5,1) -- (5,1) node[right]{$x=1$};
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\draw[black, densely dashed](-5,-1) -- (5,-1) node[right]{$x=-1$};
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\draw[black, densely dashed](-pi/2*3,1) -- (-pi/2*3,0) node[below]{$-\frac{3\pi}{2}$};
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\draw[black, densely dashed](-pi/2,-1) -- (-pi/2,0) node[above]{$-\frac{\pi}{2}$};
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\draw[black, densely dashed](pi/2,1) -- (pi/2,0) node[below]{$\frac{\pi}{2}$};
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\draw[black, densely dashed](-pi/2*3,1) -- (-pi/2*3,0) node[below]{$-\dfrac{3\pi}{2}$};
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\draw[black, densely dashed](-pi/2,-1) -- (-pi/2,0) node[above]{$-\dfrac{\pi}{2}$};
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\draw[black, densely dashed](pi/2,1) -- (pi/2,0) node[below]{$\dfrac{\pi}{2}$};
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\draw[black](0,0) -- (0,0) node[above]{$O$};
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\filldraw[black] (-pi-0.1,0) node[below]{$-\pi$};
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\filldraw[black] (pi,0) node[below]{$\pi$};
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@@ -342,15 +346,15 @@ $y=log_ax(a>0,a\neq 1)$为$y=a^x$的反函数：
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\draw[black, densely dashed](-pi,-1) -- (-pi,0) node[above]{$-\pi$};
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\draw[black, densely dashed](pi,-1) -- (pi,0) node[above]{$\pi$};
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\filldraw[black] (0,0) node[below]{$O$};
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\filldraw[black] (-pi/2*3-0.25,0) node[below]{$-\frac{3\pi}{2}$};
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\filldraw[black] (-pi/2,0) node[below]{$-\frac{\pi}{2}$};
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\filldraw[black] (pi/2,0) node[below]{$\frac{\pi}{2}$};
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\filldraw[black] (-pi/2*3-0.25,0) node[below]{$-\dfrac{3\pi}{2}$};
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\filldraw[black] (-pi/2,0) node[below]{$-\dfrac{\pi}{2}$};
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\filldraw[black] (pi/2,0) node[below]{$\dfrac{\pi}{2}$};
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\end{tikzpicture}
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弦函数有如下特征：
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\begin{enumerate}
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\item 特殊函数值：$\sin 0=0$，$\sin\frac{\pi}{6}=\frac{1}{2}$，$\sin\frac{\pi}{4}=\frac{\sqrt{2}}{2}$，$\sin\frac{\pi}{3}=\frac{\sqrt{3}}{2}$，$\sin\frac{\pi}{2}=1$，$\sin\pi=0$，$\sin\frac{3\pi}{2}=-1$，$\sin 2\pi=0$，$\cos 0=1$，$\cos\frac{\pi}{6}=\frac{\sqrt{3}}{2}$，$\cos\frac{\pi}{4}=\frac{\sqrt{2}}{2}$，$\cos\frac{\pi}{3}=\frac{1}{2}$，$\cos\frac{\pi}{2}=0$，$\cos\pi=-1$，$\cos\frac{3\pi}{2}=0$，$\cos 2\pi=1$。
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\item 特殊函数值：$\sin 0=0$，$\sin\dfrac{\pi}{6}=\dfrac{1}{2}$，$\sin\dfrac{\pi}{4}=\dfrac{\sqrt{2}}{2}$，$\sin\dfrac{\pi}{3}=\dfrac{\sqrt{3}}{2}$，$\sin\dfrac{\pi}{2}=1$，$\sin\pi=0$，$\sin\dfrac{3\pi}{2}=-1$，$\sin 2\pi=0$，$\cos 0=1$，$\cos\dfrac{\pi}{6}=\dfrac{\sqrt{3}}{2}$，$\cos\dfrac{\pi}{4}=\dfrac{\sqrt{2}}{2}$，$\cos\dfrac{\pi}{3}=\dfrac{1}{2}$，$\cos\dfrac{\pi}{2}=0$，$\cos\pi=-1$，$\cos\dfrac{3\pi}{2}=0$，$\cos 2\pi=1$。
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\item 定义域：$(-\infty, +\infty)$，值域：$[-1,+1]$。
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\item 奇偶性：$y=\sin x$为奇函数，$y=\cos x$为偶函数。
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\item 周期性：最小正周期为$2\pi$。
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@@ -370,10 +374,10 @@ $y=log_ax(a>0,a\neq 1)$为$y=a^x$的反函数：
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\draw[black, thick, domain=pi/2+0.5:pi/2*3-0.5] plot (\x,{tan(\x r)}) node[above]{$\tan(x)$};
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\draw[black, densely dashed](-pi/2*3,2) -- (-pi/2*3,-2);
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\filldraw[black] (0,0) node[below]{$O$};
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\filldraw[black] (pi/2+0.5,-0.5) node{$\frac{\pi}{2}$};
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\filldraw[black] (-pi/2-0.75,-0.5) node{$-\frac{\pi}{2}$};
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\filldraw[black] (pi/2*3+0.5,-0.5) node{$\frac{3\pi}{2}$};
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\filldraw[black] (-pi/2*3-0.75,-0.5) node{$-\frac{3\pi}{2}$};
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\filldraw[black] (pi/2+0.5,-0.75) node{$\dfrac{\pi}{2}$};
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\filldraw[black] (-pi/2-0.75,-0.75) node{$-\dfrac{\pi}{2}$};
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\filldraw[black] (pi/2*3+0.5,-0.75) node{$\dfrac{3\pi}{2}$};
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\filldraw[black] (-pi/2*3-0.75,-0.75) node{$-\dfrac{3\pi}{2}$};
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\end{tikzpicture}
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余切函数：
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@@ -386,23 +390,23 @@ $y=log_ax(a>0,a\neq 1)$为$y=a^x$的反函数：
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\draw[black, thick, domain=-0.5:-pi+0.5] plot (\x,{cot(\x r)}) node at(-1,2){$\cot(x)$};
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\draw[black, densely dashed](-pi,2) -- (-pi,-2);
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\filldraw[black] (0,0) node[below]{$O$};
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\filldraw[black] (pi/2,0) node[below]{$\frac{\pi}{2}$};
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\filldraw[black] (pi/2,0) node[below]{$\dfrac{\pi}{2}$};
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\filldraw[black] (pi+0.5,-0.5) node{$\pi$};
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\filldraw[black] (-pi/2-0.25,0) node[below]{$-\frac{\pi}{2}$};
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\filldraw[black] (-pi/2-0.25,0) node[below]{$-\dfrac{\pi}{2}$};
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\filldraw[black] (-pi-0.5,-0.5) node{$-\pi$};
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\end{tikzpicture}
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切函数有如下特征：
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\begin{enumerate}
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\item 特殊函数值：$\tan 0=0$，$\tan\frac{\pi}{6}=\frac{\sqrt{3}}{3}$，$\tan\frac{\pi}{4}=1$，$\tan\frac{\pi}{3}=\sqrt{3}$，$\lim_{x\to\frac{\pi}{2}}\tan x=\infty$，$\tan\pi=0$，$\lim_{x\to\frac{3\pi}{2}}\tan x=\infty$，$\tan 2\pi=0$，$\lim_{x\to 0}\cot x=\infty$，$\cot\frac{\pi}{6}=\sqrt{3}$，$\cot\frac{\pi}{4}=1$，$\cot\frac{\pi}{3}=\frac{\sqrt{3}}{3}$，$\cot\frac{\pi}{2}=0$，$\lim_{x\to\pi}\cot x=\infty$，$\cot\frac{3\pi}{2}=0$，$\lim_{x\to 2\pi}\cot x=\infty$。
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\item 定义域：$\tan x:x\neq k\pi+\frac{\pi}{2}(k\in Z)$，$\cot x:x\neq k\pi(k\in Z)$，值域：$(-\infty,+\infty)$。
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\item 特殊函数值：$\tan 0=0$，$\tan\frac{\pi}{6}=\frac{\sqrt{3}}{3}$，$\tan\frac{\pi}{4}=1$，$\tan\frac{\pi}{3}=\sqrt{3}$，$\lim_{x\to\frac{\pi}{2}}\tan x=\infty$，$\tan\pi=0$，$\lim_{x\to\frac{3\pi}{2}}\tan x=\infty$，$\tan 2\pi=0$，$\lim_{x\to 0}\cot x=\infty$，$\cot\dfrac{\pi}{6}=\sqrt{3}$，$\cot\dfrac{\pi}{4}=1$，$\cot\dfrac{\pi}{3}=\dfrac{\sqrt{3}}{3}$，$\cot\dfrac{\pi}{2}=0$，$\lim_{x\to\pi}\cot x=\infty$，$\cot\dfrac{3\pi}{2}=0$，$\lim_{x\to 2\pi}\cot x=\infty$。
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\item 定义域：$\tan x:x\neq k\pi+\dfrac{\pi}{2}(k\in Z)$，$\cot x:x\neq k\pi(k\in Z)$，值域：$(-\infty,+\infty)$。
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\item 奇偶性：定义域内均为奇函数。
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\item 周期性：最小正周期为$\pi$。
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\end{enumerate}
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$$
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\sec x=\frac{1}{\cos x},\csc x=\frac{1}{\sin x}
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\sec x=\dfrac{1}{\cos x},\csc x=\dfrac{1}{\sin x}
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$$
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正割函数：
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@@ -424,10 +428,10 @@ $$
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\filldraw[black] (0,0) node[below]{$O$};
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\filldraw[black] (0.5,0.5) node{$1$};
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\filldraw[black] (0.5,-1.5) node{$-1$};
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\filldraw[black] (-pi/2*3-0.75,-0.5) node{$-\frac{3\pi}{2}$};
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\filldraw[black] (-pi/2-0.75,-0.5) node{$-\frac{\pi}{2}$};
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\filldraw[black] (pi/2+0.5,-0.5) node{$\frac{\pi}{2}$};
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\filldraw[black] (pi/2*3+0.5,-0.5) node{$\frac{3\pi}{2}$};
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\filldraw[black] (-pi/2*3-0.75,-0.5) node{$-\dfrac{3\pi}{2}$};
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\filldraw[black] (-pi/2-0.75,-0.5) node{$-\dfrac{\pi}{2}$};
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\filldraw[black] (pi/2+0.5,-0.5) node{$\dfrac{\pi}{2}$};
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\filldraw[black] (pi/2*3+0.5,-0.5) node{$\dfrac{3\pi}{2}$};
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\end{tikzpicture}
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余割函数：
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@@ -457,7 +461,7 @@ $$
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割函数有如下特征：
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\begin{enumerate}
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\item 定义域：$\sec x:x\neq k\pi+\frac{\pi}{2}(k\in Z)$，$\csc x:x\neq k\pi(k\in Z)$，值域：$(-\infty,-1]\cup [1,+\infty)$。
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\item 定义域：$\sec x:x\neq k\pi+\dfrac{\pi}{2}(k\in Z)$，$\csc x:x\neq k\pi(k\in Z)$，值域：$(-\infty,-1]\cup [1,+\infty)$。
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\item 奇偶性：$y=\sec x$为偶函数，$y=\csc x$为奇函数。
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\item 周期性：最小正周期为$2\pi$。
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\end{enumerate}
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@@ -470,10 +474,10 @@ $$
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\draw[-latex](-1.5,0) -- (1.5,0) node[below]{$x$};
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\draw[-latex](0,-2) -- (0,2) node[above]{$y$};
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\draw[black, thick, domain=-1:1] plot (\x,{rad(asin(\x))}) node[right]{$\arcsin(x)$};
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\draw[black, densely dashed](1,pi/2) -- (0,pi/2) node[left]{$\frac{\pi}{2}$};
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\draw[black, densely dashed](1,pi/2) -- (0,pi/2) node[left]{$\dfrac{\pi}{2}$};
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\filldraw[black] (0,0) node[below]{$O$};
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\draw[black, densely dashed](1,pi/2) -- (1,0) node[below]{$1$};
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\draw[black, densely dashed](-1,-pi/2) -- (0,-pi/2) node[right]{$-\frac{\pi}{2}$};
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\draw[black, densely dashed](-1,-pi/2) -- (0,-pi/2) node[right]{$-\dfrac{\pi}{2}$};
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\draw[black, densely dashed](-1,-pi/2) -- (-1,0) node[above]{$-1$};
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\end{tikzpicture}
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@@ -483,7 +487,7 @@ $$
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\draw[-latex](-1.5,0) -- (1.5,0) node[below]{$x$};
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\draw[-latex](0,-0.5) -- (0,4) node[above]{$y$};
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\draw[black, thick, domain=-1:1] plot (\x,{rad(acos(\x)}) node at (-2, pi){$\arccos(x)$};
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\filldraw[black] (0,pi/2) node[right]{$\frac{\pi}{2}$};
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\filldraw[black] (0,pi/2+0.5) node[right]{$\dfrac{\pi}{2}$};
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\draw[black](1,0) -- (1,0) node[below]{$1$};
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\filldraw[black] (0,0) node[below]{$O$};
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\draw[black, densely dashed](-1,pi) -- (0,pi) node[right]{$\pi$};
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@@ -493,19 +497,19 @@ $$
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反弦函数有如下特征：
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\begin{enumerate}
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\item 特殊函数值：$\arcsin 0=0$，$\arcsin\frac{1}{2}=\frac{\pi}{6}$，$\arcsin\frac{\sqrt{2}}{2}=\frac{\pi}{4}$，$\arcsin\frac{\sqrt{3}}{2}=\frac{\pi}{3}$，$\arcsin 1=\frac{\pi}{2}$，$\arccos 1=0$，$\arccos\frac{\sqrt{3}}{2}=\frac{\pi}{6}$，$\arccos\frac{\sqrt{2}}{2}=\frac{\pi}{4}$，$\arccos\frac{1}{2}=\frac{\pi}{3}$，$\arccos 0=\frac{\pi}{2}$。
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\item 定义域：$(-1, +1)$，值域：$\arcsin x:[-\frac{\pi}{2},+\frac{\pi}{2}]$，$\arccos x:[0,\pi]$。
|
||||
\item 特殊函数值：$\arcsin 0=0$，$\arcsin\dfrac{1}{2}=\dfrac{\pi}{6}$，$\arcsin\dfrac{\sqrt{2}}{2}=\dfrac{\pi}{4}$，$\arcsin\dfrac{\sqrt{3}}{2}=\dfrac{\pi}{3}$，$\arcsin 1=\dfrac{\pi}{2}$，$\arccos 1=0$，$\arccos\dfrac{\sqrt{3}}{2}=\dfrac{\pi}{6}$，$\arccos\dfrac{\sqrt{2}}{2}=\dfrac{\pi}{4}$，$\arccos\dfrac{1}{2}=\dfrac{\pi}{3}$，$\arccos 0=\dfrac{\pi}{2}$。
|
||||
\item 定义域：$(-1, +1)$，值域：$\arcsin x:[-\dfrac{\pi}{2},+\dfrac{\pi}{2}]$，$\arccos x:[0,\pi]$。
|
||||
\item 单调性：$y=\arcsin x$单调增，$y=\arccos x$单调减。
|
||||
\item 奇偶性：$y=\arcsin x$为奇函数。
|
||||
\item 有界性：$\vert\arcsin x\vert\leqslant\frac{\pi}{2}$，$0\leqslant\arccos x\leqslant\pi$。
|
||||
\item 性质：$\arcsin x+\arccos x=\frac{\pi}{2}(-1\leqslant x\leqslant 1)$
|
||||
\item 有界性：$\vert\arcsin x\vert\leqslant\dfrac{\pi}{2}$，$0\leqslant\arccos x\leqslant\pi$。
|
||||
\item 性质：$\arcsin x+\arccos x=\dfrac{\pi}{2}(-1\leqslant x\leqslant 1)$
|
||||
\end{enumerate}
|
||||
|
||||
对反弦函数性质进行证明：
|
||||
|
||||
令$f(x)=\arcsin x+\arccos x$，对其求导得：$f'(x)=\frac{1}{\sqrt{1-x^2}}-\frac{1}{1-x^2}=0$，所以$f(x)$是个常数函数。
|
||||
令$f(x)=\arcsin x+\arccos x$，对其求导得：$f'(x)=\dfrac{1}{\sqrt{1-x^2}}-\dfrac{1}{1-x^2}=0$，所以$f(x)$是个常数函数。
|
||||
|
||||
又$f(0)=\frac{\pi}{2}$，所以该函数等于$\frac{\pi}{2}$。
|
||||
又$f(0)=\dfrac{\pi}{2}$，所以该函数等于$\dfrac{\pi}{2}$。
|
||||
|
||||
反正切函数：
|
||||
|
||||
@@ -516,8 +520,8 @@ $$
|
||||
\filldraw[black] (0,0) node[below]{$O$};
|
||||
\draw[black, densely dashed](-3,pi/2) -- (3,pi/2);
|
||||
\draw[black, densely dashed](-3,-pi/2) -- (3,-pi/2);
|
||||
\filldraw[black] (0.5,pi/2-0.5) node{$\frac{\pi}{2}$};
|
||||
\filldraw[black] (0.5,-pi/2-0.5) node{$-\frac{\pi}{2}$};
|
||||
\filldraw[black] (0.5,pi/2-0.5) node{$\dfrac{\pi}{2}$};
|
||||
\filldraw[black] (0.5,-pi/2-0.5) node{$-\dfrac{\pi}{2}$};
|
||||
\end{tikzpicture}
|
||||
|
||||
反余切函数：
|
||||
@@ -528,18 +532,18 @@ $$
|
||||
\draw[black, thick, domain=-3:3] plot (\x,{pi/2-rad(atan(\x))}) node[right]{$\rm{arccot}(x)$};
|
||||
\filldraw[black] (0,0) node[below]{$O$};
|
||||
\draw[black, densely dashed](-3,pi) -- (3,pi);
|
||||
\filldraw[black] (-0.5,pi/2-0.5) node{$\frac{\pi}{2}$};
|
||||
\filldraw[black] (-0.5,pi/2-0.5) node{$\dfrac{\pi}{2}$};
|
||||
\end{tikzpicture}
|
||||
|
||||
反切函数有如下特征：
|
||||
|
||||
\begin{enumerate}
|
||||
\item 特殊函数值：$\arctan 0=0$，$\arctan\frac{\pi}{6}=\frac{\sqrt{3}}{3}=$，$\arctan 1=\frac{\pi}{4}$，$\arctan\sqrt{3}=\frac{\pi}{3}$，$\rm{arccot}0=\frac{\pi}{2}$，$\rm{arccot}\sqrt{3}=\frac{\pi}{6}$，$\rm{arccot}1=\frac{\pi}{4}$，$\rm{arccot}\frac{\sqrt{3}}{3}=\frac{\pi}{3}$。
|
||||
\item 定义域：$(-\infty, +\infty)$，值域：$\arctan x:[-\frac{\pi}{2},+\frac{\pi}{2}]$，$\rm{arccot}x:[0,\pi]$。
|
||||
\item 特殊函数值：$\arctan 0=0$，$\arctan\dfrac{\pi}{6}=\dfrac{\sqrt{3}}{3}=$，$\arctan 1=\dfrac{\pi}{4}$，$\arctan\sqrt{3}=\dfrac{\pi}{3}$，$\rm{arccot}0=\dfrac{\pi}{2}$，$\rm{arccot}\sqrt{3}=\dfrac{\pi}{6}$，$\rm{arccot}1=\dfrac{\pi}{4}$，$\rm{arccot}\dfrac{\sqrt{3}}{3}=\dfrac{\pi}{3}$。
|
||||
\item 定义域：$(-\infty, +\infty)$，值域：$\arctan x:[-\dfrac{\pi}{2},+\dfrac{\pi}{2}]$，$\rm{arccot}x:[0,\pi]$。
|
||||
\item 单调性：$y=\arctan x$单调增，$y=\rm{arccot}x$单调减。
|
||||
\item 奇偶性：$y=\arctan x$为奇函数。
|
||||
\item 有界性：$\vert\arctan x\vert\leqslant\frac{\pi}{2}$，$0\leqslant\rm{arccot}x\leqslant\pi$。
|
||||
\item 性质：$\arctan x+\rm{arccot}x=\frac{\pi}{2}(-\infty<x<+\infty)$
|
||||
\item 有界性：$\vert\arctan x\vert\leqslant\dfrac{\pi}{2}$，$0\leqslant\rm{arccot}x\leqslant\pi$。
|
||||
\item 性质：$\arctan x+\rm{arccot}x=\dfrac{\pi}{2}(-\infty<x<+\infty)$
|
||||
\end{enumerate}
|
||||
|
||||
\subparagraph{初等函数} \leavevmode \bigskip
|
||||
@@ -780,7 +784,7 @@ $f(x)$沿$y$轴上移$y_0$个单位长度得到$f(x)+y_0$，向下移动$y_0$个
|
||||
\draw[-latex](-5,0) -- (5,0) node[below]{$x$};
|
||||
\draw[-latex](0,-1.5) -- (0,1.5) node[above]{$y$};
|
||||
\draw[black, thick, smooth, domain=-5:5] plot (\x,{sin(\x r)}) node[right]{$\sin(x)$};
|
||||
\draw[blue, thick, smooth, domain=-5:5] plot (\x,{sin(\x/2 r)}) node[right]{$\sin(\frac{x}{2})$};
|
||||
\draw[blue, thick, smooth, domain=-5:5] plot (\x,{sin(\x/2 r)}) node[right]{$\sin(\dfrac{x}{2})$};
|
||||
\draw[brown, thick, smooth, domain=-5:5] plot (\x,{sin(\x*2 r)}) node[right]{$\sin(2x)$};
|
||||
\filldraw[black] (0,0) node[below]{$O$};
|
||||
\end{tikzpicture}
|
||||
@@ -794,7 +798,7 @@ $f(x)$沿$y$轴上移$y_0$个单位长度得到$f(x)+y_0$，向下移动$y_0$个
|
||||
\draw[-latex](-5,0) -- (5,0) node[below]{$x$};
|
||||
\draw[-latex](0,-1.5) -- (0,1.5) node[above]{$y$};
|
||||
\draw[black, thick, smooth, domain=-5:5] plot (\x,{sin(\x r)}) node[right]{$\sin(x)$};
|
||||
\draw[blue, thick, smooth, domain=-5:5] plot (\x,{sin(\x r)/2}) node[right]{$\frac{sin(x)}{2}$};
|
||||
\draw[blue, thick, smooth, domain=-5:5] plot (\x,{sin(\x r)/2}) node at(5.5,1){$\dfrac{sin(x)}{2}$};
|
||||
\filldraw[black] (0,0) node[below]{$O$};
|
||||
\end{tikzpicture}
|
||||
|
||||
@@ -808,7 +812,7 @@ $f(x)$沿$y$轴上移$y_0$个单位长度得到$f(x)+y_0$，向下移动$y_0$个
|
||||
|
||||
其中$r$为线的极径，$\theta$为极角，$a$为形状参数且$a>0$，周期为$2\pi$。
|
||||
|
||||
在直角坐标系下表达式：$x^2+y^2+a*x=a\cdot\sqrt{x^2+y^2}$和$x^2+y^2-a\cdot x=a\cdot\sqrt{x^2+y^2}$。
|
||||
在直角坐标系下表达式：$x^2+y^2+a\cdot x=a\cdot\sqrt{x^2+y^2}$和$x^2+y^2-a\cdot x=a\cdot\sqrt{x^2+y^2}$。
|
||||
|
||||
参数方程：$x=a\cdot(2\cdot\cos(t)-cos(2\cdot t))$与$y=a\cdot(2\cdot\sin(t)-sin(2\cdot t))$
|
||||
|
||||
@@ -823,16 +827,16 @@ $f(x)$沿$y$轴上移$y_0$个单位长度得到$f(x)+y_0$，向下移动$y_0$个
|
||||
|
||||
\begin{tabular}{|c|c|c|c|c|c|c|c|c|c|}
|
||||
\hline
|
||||
$\theta$ & $0$ & $\frac{\pi}{6}$ & $\frac{\pi}{4}$ & $\frac{\pi}{3}$ & $\frac{\pi}{2}$ & $\frac{2\pi}{3}$ & $\frac{3\pi}{4}$ & $\frac{5\pi}{6}$ & $\pi$ \\ \hline
|
||||
$r$ & $0$ & $\frac{2-\sqrt{3}}{2}a$ & $\frac{2-\sqrt{2}}{2}a$ & $\frac{1}{2}a$ & $a$ & $\frac{3}{2}a$ & $\frac{2+\sqrt{2}}{2}a$ & $\frac{2+\sqrt{3}}{2}a$ & $2a$ \\
|
||||
$\theta$ & $0$ & $\dfrac{\pi}{6}$ & $\dfrac{\pi}{4}$ & $\dfrac{\pi}{3}$ & $\dfrac{\pi}{2}$ & $\dfrac{2\pi}{3}$ & $\dfrac{3\pi}{4}$ & $\dfrac{5\pi}{6}$ & $\pi$ \\ \hline
|
||||
$r$ & $0$ & $\dfrac{2-\sqrt{3}}{2}a$ & $\dfrac{2-\sqrt{2}}{2}a$ & $\dfrac{1}{2}a$ & $a$ & $\dfrac{3}{2}a$ & $\dfrac{2+\sqrt{2}}{2}a$ & $\dfrac{2+\sqrt{3}}{2}a$ & $2a$ \\
|
||||
\hline
|
||||
\end{tabular}
|
||||
|
||||
\paragraph{玫瑰线} \leavevmode \bigskip
|
||||
|
||||
表达式：$r=a\sin(n\theta)$，周期为$\frac{2\pi}{n}$。
|
||||
表达式：$r=a\sin(n\theta)$，周期为$\dfrac{2\pi}{n}$。
|
||||
|
||||
当$n$为3时为三叶，2时为四叶，$\frac{3}{2}$为六叶。三叶时周期为$\frac{2\pi}{3}$。
|
||||
当$n$为3时为三叶，2时为四叶，$\dfrac{3}{2}$为六叶。三叶时周期为$\dfrac{2\pi}{3}$。
|
||||
|
||||
直角坐标系下表达式：$x=a\cdot\sin(n\cdot\theta)\cdot\cos(\theta)$与$y=a\cdot\sin(n\cdot)\cdot\sin(\theta)$
|
||||
|
||||
@@ -847,8 +851,8 @@ $f(x)$沿$y$轴上移$y_0$个单位长度得到$f(x)+y_0$，向下移动$y_0$个
|
||||
|
||||
\begin{tabular}{|c|c|c|c|c|c|c|c|c|c|}
|
||||
\hline
|
||||
$\theta$ & $0$ & $\frac{\pi}{12}$ & $\frac{\pi}{6}$ & $\frac{\pi}{4}$ & $\frac{\pi}{3}$ & $\frac{5\pi}{12}$ & $\frac{\pi}{2}$ & $\frac{7\pi}{12}$ & $\frac{3\pi}{2}$ \\ \hline
|
||||
$r$ & $0$ & $\frac{\sqrt{2}}{2}a$ & $a$ & $\frac{\sqrt{2}}{2}a$ & $0$ & $-frac{\sqrt{2}}{2}a$ & $-a$ & $-frac{\sqrt{2}}{2}a$ & $0$ \\
|
||||
$\theta$ & $0$ & $\dfrac{\pi}{12}$ & $\dfrac{\pi}{6}$ & $\dfrac{\pi}{4}$ & $\dfrac{\pi}{3}$ & $\dfrac{5\pi}{12}$ & $\dfrac{\pi}{2}$ & $\dfrac{7\pi}{12}$ & $\dfrac{3\pi}{2}$ \\ \hline
|
||||
$r$ & $0$ & $\dfrac{\sqrt{2}}{2}a$ & $a$ & $\dfrac{\sqrt{2}}{2}a$ & $0$ & $-frac{\sqrt{2}}{2}a$ & $-a$ & $-frac{\sqrt{2}}{2}a$ & $0$ \\
|
||||
\hline
|
||||
\end{tabular}
|
||||
|
||||
@@ -930,7 +934,7 @@ $$
|
||||
|
||||
\subsubsection{星形线（内摆线）}
|
||||
|
||||
与半径为$r$的定圆内切的半径为$\frac{r}{4}$的动圆沿定圆无滑动地滚动，动圆上一点的轨迹称为星形线。
|
||||
与半径为$r$的定圆内切的半径为$\dfrac{r}{4}$的动圆沿定圆无滑动地滚动，动圆上一点的轨迹称为星形线。
|
||||
|
||||
令$t$表示摆点与圆心的连线所构成夹角的弧度，其中$t\in[0,2\pi]$，得对应参数方程：
|
||||
|
||||
@@ -953,7 +957,7 @@ $$
|
||||
|
||||
通项公式：$a_n=a_1+(n-1)d$。
|
||||
|
||||
前$n$项和：$S_n=\frac{n}{2}[2a_1+(n-1)d]=\frac{n}{2}(a_1+a_n)$
|
||||
前$n$项和：$S_n=\dfrac{n}{2}[2a_1+(n-1)d]=\dfrac{n}{2}(a_1+a_n)$
|
||||
|
||||
\subsubsection{等比数列}
|
||||
|
||||
@@ -965,27 +969,27 @@ $$
|
||||
\left\{
|
||||
\begin{array}{lcl}
|
||||
na_1, & & r=1 \\
|
||||
\frac{a_1(1-r^n)}{1-r}, & & r\neq 1
|
||||
\dfrac{a_1(1-r^n)}{1-r}, & & r\neq 1
|
||||
\end{array}
|
||||
\right.$
|
||||
\right.$
|
||||
|
||||
若首项为1，则$1+r+r^2+\cdots+r^{n-1}=\frac{1-r^n}{1-r}(r\neq 1)$。
|
||||
若首项为1，则$1+r+r^2+\cdots+r^{n-1}=\dfrac{1-r^n}{1-r}(r\neq 1)$。
|
||||
|
||||
则对无穷的极限为$\frac{1}{1-r}$。
|
||||
则对无穷的极限为$\dfrac{1}{1-r}$。
|
||||
|
||||
\subsubsection{常见数列前$n$项和}
|
||||
|
||||
\begin{enumerate}
|
||||
\item $\sum_{k=1}^nk=1+2+\cdots+n=\frac{n(n+1)}{2}$。
|
||||
\item $\sum_{k=1}^nk^2=1^2+2^2+\cdots+n^2=\frac{n(n+1)(2n+1)}{6}$。
|
||||
\item $\sum_{k=1}^n\frac{1}{k(k+1)}=\frac{1}{1\times 2}+\frac{1}{2\times 3}+\cdots+\frac{1}{n(n+1)}=\frac{n}{n+1}$。
|
||||
\item $\sum_{k=1}^nk=1+2+\cdots+n=\dfrac{n(n+1)}{2}$。
|
||||
\item $\sum_{k=1}^nk^2=1^2+2^2+\cdots+n^2=\dfrac{n(n+1)(2n+1)}{6}$。
|
||||
\item $\sum_{k=1}^n\dfrac{1}{k(k+1)}=\dfrac{1}{1\times 2}+\dfrac{1}{2\times 3}+\cdots+\dfrac{1}{n(n+1)}=\dfrac{n}{n+1}$。
|
||||
\end{enumerate}
|
||||
|
||||
\subsection{三角函数}
|
||||
|
||||
\subsubsection{基本关系}
|
||||
|
||||
$\csc\alpha=\frac{1}{\sin\alpha},\sec\alpha=\frac{1}{\cos\alpha},\cot\alpha=\frac{1}{\tan\alpha},\tan\alpha=\frac{\sin\alpha}{\cos\alpha},\cot\alpha=\frac{\cos\alpha}{\sin\alpha}$。
|
||||
$\csc\alpha=\dfrac{1}{\sin\alpha},\sec\alpha=\dfrac{1}{\cos\alpha},\cot\alpha=\dfrac{1}{\tan\alpha},\tan\alpha=\dfrac{\sin\alpha}{\cos\alpha},\cot\alpha=\dfrac{\cos\alpha}{\sin\alpha}$。
|
||||
|
||||
$\sin^2\alpha+\cos^2\alpha=1,1+\tan^2\alpha=\sec^2\alpha,1+\cot^2\alpha=\csc^2\alpha$。
|
||||
|
||||
@@ -993,13 +997,12 @@ $\sin^2\alpha+\cos^2\alpha=1,1+\tan^2\alpha=\sec^2\alpha,1+\cot^2\alpha=\csc^2\a
|
||||
|
||||
奇变偶不变，符号看象限。奇指前面添加的常数项是否为$\pi$的整数倍，是就需要改变函数，看象限指添加了常数后整体的符号看函数所在象限的符号。
|
||||
|
||||
$\sin(\frac{\pi}{2}\pm\alpha)=\cos\alpha$
|
||||
|
||||
$\cos(\frac{\pi}{2}\pm\alpha)=\mp\sin\alpha$
|
||||
|
||||
$\sin(\pi\pm\alpha)=\mp\sin\alpha$
|
||||
|
||||
$\cos(\pi\pm\alpha)=-\cos\alpha$
|
||||
\begin{enumerate}
|
||||
\item $\sin(\dfrac{\pi}{2}\pm\alpha)=\cos\alpha$
|
||||
\item $\cos(\dfrac{\pi}{2}\pm\alpha)=\mp\sin\alpha$
|
||||
\item $\sin(\pi\pm\alpha)=\mp\sin\alpha$
|
||||
\item $\cos(\pi\pm\alpha)=-\cos\alpha$
|
||||
\end{enumerate}
|
||||
|
||||
\subsubsection{倍角公式}
|
||||
|
||||
@@ -1007,12 +1010,250 @@ $\sin 2\alpha=2\sin\alpha\cos\alpha,\cos 2\alpha=\cos^2\alpha-\sin^2\alpha=1-2\s
|
||||
|
||||
$\sin 3\alpha=-4\sin^3\alpha_3\sin\alpha,\cos 3\alpha=4\cos^3\alpha-3\cos\alpha$。
|
||||
|
||||
$\tan 2\alpha=\frac{2\tan\alpha}{1-\tan^2\alpha},\cot 2\alpha=\frac{\cot^2\alpha-1}{2\cot\alpha}$。
|
||||
$\tan 2\alpha=\dfrac{2\tan\alpha}{1-\tan^2\alpha},\cot 2\alpha=\dfrac{\cot^2\alpha-1}{2\cot\alpha}$。
|
||||
|
||||
\subsubsection{半角公式}
|
||||
|
||||
$\sin^2\dfrac{\alpha}{2}=\dfrac{1}{2}(1-\cos\alpha),\cos^2\dfrac{\alpha}{2}=\dfrac{1}{2}(1+\cos\alpha)\text{（降幂公式）}$。
|
||||
|
||||
$\sin\dfrac{\alpha}{2}=\pm\sqrt{\dfrac{1-\cos\alpha}{2}},\cos\dfrac{\alpha}{2}=\pm\sqrt{\dfrac{1+\cos\alpha}{2}}$。
|
||||
|
||||
$\tan\dfrac{\alpha}{2}=\dfrac{1-\cos\alpha}{\sin\alpha}=\dfrac{\sin\alpha}{1+\cos\alpha}=\pm\sqrt{\dfrac{1-\cos\alpha}{1+\cos\alpha}}=\dfrac{1}{\cot\dfrac{\alpha}{2}}$。
|
||||
|
||||
\subsubsection{和差公式}
|
||||
|
||||
$\sin$和$\tan$的和差公式更容易考到。
|
||||
|
||||
$\sin(\alpha\pm\beta)=\sin\alpha\cos\beta\pm\cos\alpha\sin\beta,\cos(\alpha\pm\beta)=\cos\alpha\cos\beta\mp\sin\alpha\sin\beta$。
|
||||
|
||||
$\tan(\alpha\pm\beta)=\dfrac{\tan\alpha\pm\tan\beta}{1\mp\tan\alpha\tan\beta},\cot(\alpha\pm\beta)=\dfrac{\cot\alpha\cot\beta\mp 1}{\cot\beta\pm\cot\alpha}$。
|
||||
|
||||
\subsubsection{积化和差公式}
|
||||
|
||||
和差化积与积化和差不需要背。
|
||||
|
||||
$\sin\alpha\cos\beta=\dfrac{1}{2}[\sin(\alpha+\beta)+\sin(\alpha-\beta)],\cos\alpha\sin\beta=\dfrac{1}{2}[\sin(\alpha+\beta)-\sin(\alpha-\beta)]$。
|
||||
|
||||
$\cos\alpha\cos\beta=\dfrac{1}{2}[\cos(\alpha+\beta)+\cos(\alpha-\beta)],\sin\alpha\sin\beta=\dfrac{1}{2}[\cos(\alpha-\beta)-\cos(\alpha+\beta)]$。
|
||||
|
||||
\subsubsection{和差化积公式}
|
||||
|
||||
$\sin\alpha+\sin\beta=2\sin\dfrac{\alpha+\beta}{2}\cos\dfrac{\alpha-\beta}{2},\sin\alpha-\sin\beta=2\sin\dfrac{\alpha-\beta}{2}\cos\dfrac{\alpha+\beta}{2}$。
|
||||
|
||||
$\cos\alpha+\cos\beta=2\cos\dfrac{\alpha+\beta}{2}\cos\dfrac{\alpha-\beta}{2},\cos\alpha-\cos\beta=-2\sin\dfrac{\alpha+\beta}{2}\sin\dfrac{\alpha-\beta}{2}$。
|
||||
|
||||
\subsubsection{万能公式}
|
||||
|
||||
一般不会用到。
|
||||
|
||||
若$u=\tan\dfrac{x}{2}(-\pi<x<\pi)$，则$\sin x=\dfrac{2u}{1+u^2},\cos x=\dfrac{1-u^2}{1+u^2}$。
|
||||
|
||||
\subsection{指数运算法则}
|
||||
|
||||
$a^\alpha\cdot a^\beta=a^{\alpha+\beta},\dfrac{a^\alpha}{a^\beta}=a^{\alpha-\beta},(a^\alpha)^\beta=a^{\alpha\beta},(ab)^\alpha=a^\alpha b^\alpha,(\dfrac{a}{b})^\alpha=\dfrac{a^\alpha}{b^\alpha}$。
|
||||
|
||||
其中$a$，$b$为正实数，$\alpha$，$\beta$为任意实数。
|
||||
|
||||
\subsection{对数运算法则}
|
||||
|
||||
重点：
|
||||
|
||||
\begin{enumerate}
|
||||
\item $\log_a(MN)=\log_aM+\log_aN$（积的对数=对数的和）。
|
||||
\item $\log_a(\dfrac{M}{N})=\log_aM-\log_aN$（商的对数=对数的差）。
|
||||
\item $\log_aM^n=n\log_aM$（幂的对数=对数的倍数）。
|
||||
\item $\log_a\sqrt[n]{M}=\dfrac{1}{n}\log_aM$。
|
||||
\end{enumerate}
|
||||
|
||||
所以以后多项相乘相除乘方开放都\textcolor{orange}{取对数}进行化简。
|
||||
|
||||
对于分数相减的对数先\textcolor{orange}{通分}再进行对数减法。
|
||||
|
||||
如下面这个题（先不要求能直接证明）：
|
||||
|
||||
\textbf{例题4：}证明$\dfrac{1}{x+1}<\ln(1+\dfrac{1}{x})<\dfrac{1}{x}$，其中$x>0$。
|
||||
|
||||
首先因为证明中间项无法进行直接处理，又看到是一个对数，所以进行通分：$\ln(1+\dfrac{1}{x})=\ln\dfrac{x+1}{x}=\ln(x+1)-\ln x$。
|
||||
|
||||
又因为是证明该中间式在一个区间，所以很明显会想到拉格朗日中值定理：$f(b)-f(a)=f'(\xi)(b-a)$。
|
||||
|
||||
得到原式$=f'(\xi)=(\ln\xi)'=\dfrac{1}{\xi}$，又中值定理下$a<\xi<b$且$x>0$，所以$\dfrac{1}{b}<\dfrac{1}{\xi}<\dfrac{1}{a}$，得到$0<\dfrac{1}{x+1}<\dfrac{1}{\xi}<\dfrac{1}{x}$。
|
||||
|
||||
所以原式$\dfrac{1}{x+1}<\ln(1+\dfrac{1}{x})<\dfrac{1}{x}$成立。
|
||||
|
||||
\subsection{一元二次方程基础}
|
||||
|
||||
\begin{enumerate}
|
||||
\item 基本格式为$ax^2+bx+c=0(a\neq 0)$。
|
||||
\item 如果$\Delta=\sqrt{b^2-4ac}\geqslant 0$，根的公式为$x_{1,2}=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$，其中如果等于0为唯一实根，如果大于0为二重实根，如果$\Delta<0$则得到共轭复数根$-\dfrac{b}{2a}\pm\dfrac{\sqrt{4ac-b^2}}{2a}i$。
|
||||
\item 根与系数的关系（韦达定理）为$x_1+x_2=-\dfrac{b}{a},x_1x_2=\dfrac{c}{a}$。
|
||||
\item 抛物线顶点为$(-\dfrac{b}{2a},c-\dfrac{b^2}{4a})$。
|
||||
\end{enumerate}
|
||||
|
||||
\subsection{因式分解公式}
|
||||
|
||||
重点为3、4、7和11的公式。
|
||||
|
||||
\begin{enumerate}
|
||||
\item $(a+b)^2=a^2+2ab+b^2$。
|
||||
\item $(a-b)^2=a^2-2ab+b^2$。
|
||||
\item $(a+b)^3=a^3+3a^2b+3ab^2+b^3$。
|
||||
\item $(a-b)^3=a^3-3a^2b+3ab^2-b^3$。
|
||||
\item $a^2-b^2=(a+b)(a-b)$。
|
||||
\item $a^3-b^3=(a-b)(a^2+ab+b^2)$。
|
||||
\item $a^3+b^3=(a+b)(a^2-ab+b^2)$。
|
||||
\item $n$为正整数时，$a^n-b^n=(a-b)(a^{n-1}+a^{n-2}b+\cdots+ab^{n-2}+b^{n-1})$。
|
||||
\item $n$为正偶数时，$a^n-b^n=(a+b)(a^{n-1}-a^{n-2}b+\cdots+ab^{n-2}-b^{n-1})$。
|
||||
\item $n$为正奇数时，$a^n+b^n=(a+b)(a^{n-1}-a^{n-2}b+\cdots-ab^{n-2}+b^{n-1})$。
|
||||
\item 二项式定理$(a+b)^n=\sum_{k=0}^nC_n^ka^{n-k}b^k=a^n+na^{n-1}b+\dfrac{n(n-1)}{2!}a^{n-2}b^2+\cdots+\dfrac{n(n-1)\cdots(n-k+1)}{k!}a^{n-k}b^k+\cdots+nab^{n-1}+b^n$。
|
||||
\end{enumerate}
|
||||
|
||||
对于二项式定理需要记忆，后面的幂比较简单，而前面的系数比较困难，可以使用杨辉三角形来记忆：
|
||||
|
||||
|
||||
\begin{tikzpicture}[scale=0.9]
|
||||
\node[black] at (0,0) {$C_0^0$};
|
||||
\node[black] at (-1,-1) {$C_1^0$};
|
||||
\node[black] at (0,-1) {$C_1^1$};
|
||||
\node[black] at (-2,-2) {$C_2^0$};
|
||||
\node[black] at (-1,-2) {$C_2^1$};
|
||||
\node[black] at (-0,-2) {$C_2^2$};
|
||||
\node[black] at (-3,-3) {$C_3^0$};
|
||||
\node[black] at (-2,-3) {$C_3^1$};
|
||||
\node[black] at (-1,-3) {$C_3^2$};
|
||||
\node[black] at (-0,-3) {$C_3^3$};
|
||||
\node[black] at (-4,-4) {$C_4^0$};
|
||||
\node[black] at (-3,-4) {$C_4^1$};
|
||||
\node[black] at (-2,-4) {$C_4^2$};
|
||||
\node[black] at (-1,-4) {$C_4^3$};
|
||||
\node[black] at (-0,-4) {$C_4^4$};
|
||||
\end{tikzpicture}
|
||||
\hspace{2.5em}
|
||||
\begin{tikzpicture}[scale=0.9]
|
||||
\node[black] (0) at (0,0) {1};
|
||||
\node[black] (1) at (-1,-1) {1};
|
||||
\node[black] (2) at (1,-1) {1};
|
||||
\node[black] (3) at (-2,-2) {1};
|
||||
\node[black] (4) at (0,-2) {2};
|
||||
\node[black] (5) at (2,-2) {1};
|
||||
\node[black] (6) at (-3,-3) {1};
|
||||
\node[black] (7) at (-1,-3) {3};
|
||||
\node[black] (8) at (1,-3) {3};
|
||||
\node[black] (9) at (3,-3) {1};
|
||||
\node[black] (10) at (-4,-4) {1};
|
||||
\node[black] (11) at (-2,-4) {4};
|
||||
\node[black] (12) at (0,-4) {6};
|
||||
\node[black] (13) at (2,-4) {4};
|
||||
\node[black] (14) at (4,-4) {1};
|
||||
\draw[-,thick] (0) to (1);
|
||||
\draw[-,thick] (0) to (2);
|
||||
\draw[-,thick] (1) to (3);
|
||||
\draw[-,thick] (1) to (4);
|
||||
\draw[-,thick] (2) to (4);
|
||||
\draw[-,thick] (2) to (5);
|
||||
\draw[-,thick] (3) to (6);
|
||||
\draw[-,thick] (3) to (7);
|
||||
\draw[-,thick] (4) to (7);
|
||||
\draw[-,thick] (4) to (8);
|
||||
\draw[-,thick] (5) to (8);
|
||||
\draw[-,thick] (5) to (9);
|
||||
\draw[-,thick] (6) to (10);
|
||||
\draw[-,thick] (6) to (11);
|
||||
\draw[-,thick] (7) to (11);
|
||||
\draw[-,thick] (7) to (12);
|
||||
\draw[-,thick] (8) to (12);
|
||||
\draw[-,thick] (8) to (13);
|
||||
\draw[-,thick] (9) to (13);
|
||||
\draw[-,thick] (9) to (14);
|
||||
\end{tikzpicture}
|
||||
|
||||
\subsection{阶乘与双阶乘}
|
||||
|
||||
\begin{enumerate}
|
||||
\item $n!=1\times 2\times 3\times\cdots\times n$，其中$0!=1$。
|
||||
\item $(2n)!!=2\times 4\times 6\times\cdots\times(2n)=2^n\cdot n!$。
|
||||
\item $(2n-1)!!=1\times 3\times 5\times\cdots\times(2n-1)$。
|
||||
\end{enumerate}
|
||||
|
||||
以后的华里士公式（点火公式）会使用到，如下面的题目：
|
||||
|
||||
\textbf{例题5：}计算$\int_0^{\frac{\pi}{2}}\sin^{10}x\rm{d}x$与$\int_0^{\frac{\pi}{2}}\cos^9x\rm{d}x$。
|
||||
|
||||
原式1为偶数次幂，所以$=\dfrac{9}{10}\cdot\dfrac{7}{8}\cdot\dfrac{5}{6}\cdot\dfrac{3}{4}\cdot\dfrac{1}{2}\cdot\dfrac{\pi}{2}=\dfrac{\pi}{2}\cdot\dfrac{9!!}{10!!}$
|
||||
|
||||
原式2为奇数次幂，所以$=\dfrac{8}{9}\cdot\dfrac{6}{7}\cdot\dfrac{4}{5}\cdot\dfrac{2}{3}=\dfrac{8!!}{9!!}$
|
||||
|
||||
\subsection{常用不等式}
|
||||
|
||||
非常重要。
|
||||
|
||||
\subsubsection{绝对值不等式}
|
||||
|
||||
若$a$，$b$为实数，则：
|
||||
|
||||
\begin{enumerate}
|
||||
\item $\vert a\pm b\vert\leqslant\vert a\vert+\vert b\vert$。
|
||||
\item 推广公式一到离散区间：$\vert a_1\pm a_2\pm a_3\pm\cdots\pm a_n\vert\leqslant\vert a_1\vert+\vert a_2\vert+\cdots+\vert a_n\vert$。
|
||||
\item 推广公式一到连续区间且$f(x)$在$[a,b](a<b)$上可积：$\vert\int_a^bf(x)\rm{d}x\vert\leqslant\int_a^b\vert f(x)\vert\rm{d}x$。因为符号不一定相同的面积代数和一定小于同为正的面积代数和。
|
||||
\item $\vert\vert a\vert-\vert b\vert\vert\leqslant\vert a-b\vert$。
|
||||
\end{enumerate}
|
||||
|
||||
\subsubsection{根号不等式}
|
||||
|
||||
公式一非常重要，即二分之一的算数平均大几何平均。
|
||||
|
||||
$a,b,c>0$：
|
||||
|
||||
\begin{enumerate}
|
||||
\item $\sqrt{ab}\leqslant\dfrac{a+b}{2}\leqslant\sqrt{\dfrac{a^2+b^2}{2}}$。
|
||||
\item $\sqrt[3]{abc}\leqslant\dfrac{a+b+c}{3}\leqslant\dfrac{a^2+b^2+c^2}{3}$。
|
||||
\end{enumerate}
|
||||
|
||||
\textbf{例题6：}证明函数$f(x)=\dfrac{x}{1+x^2}$在$(-\infty,+\infty)$内有界。
|
||||
|
||||
可以使用极限，若极限存在则函数有界，这里使用有界性定义与不等式来完成。
|
||||
|
||||
\textcircled{1}当$x=0$时，$f(0)=\dfrac{0}{1}$，有界。
|
||||
|
||||
\textcircled{2}当$x\neq 0$时，原式分式上下都有$x$，所以简化公式：$f(x)=\dfrac{\dfrac{x}{x}}{\dfrac{1+x^2}{x}}=\dfrac{1}{\dfrac{1}{x}+x}$。
|
||||
|
||||
$\because$需要证明有界性，以及根号不等式下需要参数大于0，所以需要证明$\vert f(x)\vert=\dfrac{1}{\dfrac{1}{\vert x\vert}+\vert x\vert}\leqslant M$
|
||||
|
||||
又$\because\dfrac{a+b}{2}\geqslant\sqrt{ab}$，$\therefore \dfrac{\dfrac{1}{\vert x\vert}+\vert x\vert}{2}\geqslant\sqrt{\dfrac{1}{\vert x\vert}\cdot\vert x\vert}=1$
|
||||
|
||||
$\therefore\vert f(x)\vert=\dfrac{1}{\dfrac{1}{\vert x\vert}+\vert x\vert}\leqslant\dfrac{1}{2}$。
|
||||
|
||||
故整个函数在$R$上有界。
|
||||
|
||||
\subsubsection{指数不等式}
|
||||
|
||||
设$a>b>0$，则$
|
||||
\left\{
|
||||
\begin{array}{lcl}
|
||||
a^n>b^n, & & \text{当}n>0\text{时} \\
|
||||
a^n<b^n, & & \text{当}n<0\text{时}
|
||||
\end{array}
|
||||
\right.$。
|
||||
|
||||
$e^x\geqslant x+1(\forall x)$。
|
||||
|
||||
\subsubsection{分数不等式}
|
||||
|
||||
若$0<a<x<b,0<c<y<d$，则$\dfrac{c}{b}<\dfrac{y}{x}<\dfrac{d}{a}$。
|
||||
|
||||
\subsubsection{三角不等式}
|
||||
|
||||
\begin{enumerate}
|
||||
\item $\sin x<x<\tan x(0<x<\dfrac{\pi}{2})$。
|
||||
\item $\sin x<x(x>0)$。
|
||||
\item $\arctan x\leqslant x\leqslant\arcsin x(0\leqslant x\leqslant 1)$。
|
||||
\end{enumerate}
|
||||
|
||||
\subsubsection{对数不等式}
|
||||
|
||||
\begin{enumerate}
|
||||
\item $x-1\geqslant\ln x(x>0)$。
|
||||
\item $\dfrac{1}{1+x}<\ln(1+\dfrac{1}{x})<\dfrac{1}{x}(x>0)$。
|
||||
\end{enumerate}
|
||||
|
||||
\end{document}
|
||||
Reference in New Issue
Block a user