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[TOC]

预备知识

三角恒等式

\[sin^2 + cos^2 = 1\]

证明:

​ 在三角形中斜边为1,直边分别为sin(θ) 和 cos(θ)

​ 由勾股定理得此公式

\[1 + tan^2 = sec^2\]

证明: 公式(1)等式两边同时除以cos

\[1 + cot^2 = csc^2\]

证明:

​ 公式(1)等式两边同时除以sin

和角公式

\[cos(A+B) = cosAcosB-sinAsinB\\ sin(A+B)=sinAcosB+cosAsinB\]

Proof:

preview

​ The figure shows

​ \(b=\frac{sin(β)}{cos(\alpha)}\)

​ \(a=sin(\alpha)\times(cos(\beta)-b\times sin(\alpha))\)

​ \(sin(\alpha+\beta)=a+b\)

​ Integrate the reletionship and get formula(4)

Proof:

​ The figure shows

​ \(b=\frac{sin(β)}{cos(\alpha)}\)

​ \(a=sin(\alpha)\times(cos(\beta)-b\times sin(\alpha))\)

​ \(cos(\alpha)=\frac{a}{tan(\alpha)}\)

​ Integate the relationship and get formula(5)

余弦定理

\[cos(A) = \frac{b^2+c^2-a^2}{2bc}\]

1.1

Proof:

​ \(\vec{a}^2=(\vec{b}-\vec{c})^2\\\quad=b^2+c^2-2bccos(A)\)

​ Integrate and get formula(6)

极限与连续

极限的定义

定义 极限的正式定义

f(x)定义在可能不包含\(x_0\)的开区间上, 当x趋于\(x_0\)时f(x)趋于极限L,记为
\(\lim_{x \to x_0}{f(x)}=L\) 如果,对任何数\(\epsilon>0\),存在相应的数\(\delta>0\)使得对所有满足\(0<|x-x_0|<\delta\)的x,有 \(|f(x)-L|<\epsilon\)

定义 右侧极限和左侧极限

设f(x)定义在(a,b)上, a<b,如果在区间(a,b)内趋于a时f(x)任意接近地趋近于L,f在a有右侧极限,记作
\(\lim_{x \to a^+}f(x)=L\) 设f(x)定义在(c,a)上, c<a,如果在区间(c,a)内趋于a时f(x)任意接近地趋近于L,f在a有左侧极限,记作
\(\lim_{x \to a^-}f(x)=L\)

定理 单侧极限和双侧极限的关系

当\(x \to c\)时函数f(x)有极限当且仅当f的左侧极限和右侧极限存在且相等:
\(\lim_{x \to c}=L \Leftrightarrow \lim_{x \to c-}=L 且\lim_{x \to c+}=L\)

定义 无穷极限

x趋于\(x_0\)时f(x)趋于无穷,记作
\(\lim_{x \to c}=\infty\) 如果对于任何正实数B存在相应的\(\delta>0\),使得对一起满足\(0<|x-x_0|<\delta\)的x,有f(x)>B

x趋于\(x_0\)时f(x)趋于负无穷,记作
\(\lim_{x \to c}=-\infty\) 如果对于任何正实数B存在相应的\(\delta>0\),使得对一起满足\(0<|x-x_0|<\delta\)的x,有f(x)>-B

定义 水平渐近线和垂直渐近线

直线\(y=b\)是函数\(y=f(x)\)图形的水平渐近线,如果有
\(\lim_{x \to \infty}f(x)=b或\lim_{x \to -\infty}f(x)=b\) 直线\(x=a\)是函数\(y=f(x)\)图形的水平渐近线,如果有
\(\lim_{x \to a^+}f(x)=\pm\infty或\lim_{x \to a^-}f(x)=\pm\infty\)

极限的性质

加减乘除

定理 极限法则

如果L,M,c,k为实数,且 \(\lim_{x \to x_0}f(x)=L\)和\(\lim_{x \to x_0}g(x)=M\)
\(\begin{align} &\lim_{x \to c}(f(x)+g(x))=L+M\\ &\lim_{x \to c}(f(x)-g(x))=L-M\\ &\lim_{x \to c}(f(x) \times g(x))=L \times M\\ &\lim_{x \to c}(k \cdot f(x))=k \cdot f(x)\\ &\lim_{x \to c}(\frac{f(x)}{g(x)})=\frac{L}{M},M \neq 0\\ &\lim_{x \to c}(f(x))^\frac{r}{s}=L^\frac{r}{s} \end{align}\)

求极限

多项式

定理 使用代入法求多项式极限

如果\(P(x)=a_nx^n+a_{n-1}x^{n-1}+...+a_0\),那么 \(\lim_{x \to c}P(x)=P(c)\)

有理函数

定理 代入法求有理函数的极限

如果P(x)和Q(x)都是多项式且Q(c)\(\neq\)0,那么 \(\lim_{x \to c}\frac{P(x)}{Q(x)}=\frac{P(c)}{Q(c)}\)

夹逼定理

定理 夹逼定理

如果在包含c的某个开区间中\(x=c\)处除外的所有x,有\(g(x) \leq f(x) \leq h(x)\). 又假设 \(\lim_{x \to c}g(x)=\lim_{x \to c}h(x)=L\) 那么 \(\lim_{x \to c}f(x)=L\)

三角函数

定理 sin的极限 \(\lim_{\theta \to 0}\frac{sin \theta}{\theta}=1\)

Proof:

1.2

​ The figure shows \(\frac{1}{2}sin\theta < \frac{1}{2}\theta < \frac{1}{2}tan\theta\) ​ Sort the expression and get \(1>\frac{sin\theta}{\theta}>cos\theta\) ​ Also because \(\lim_{x \to 0}cos\theta=1\) ​ Thus \(\lim_{x \to 0}{\frac{sin\theta}{\theta}}=1\)

定理 cos的极限 \(\lim_{x \to 0}\frac{1-cos (x)}{x}=0\)

Proof:
\(\begin{align} \lim_{x \to 0}\frac{cos (x)-1}{x}&=\lim_{x \to 0}\frac{1-cos^2(x)}{x}\times\frac{1}{1+cos(x)}\\ &=\lim_{x \to 0}sin(x)\times\frac{sin(x)}{x}\times \frac{1}{1+cos(x)}\\ &=0 \times1\times\frac{1}{1+1}\\ &=0 \end{align}\)

不定式

定理 L’Hopital法则

假定\(f(x_0)=g(x_0)=0或\pm\infty\) \(\lim_{x \to x_0}\frac{f(x)}{g(x)}=\lim_{x \to x_0}\frac{f'(x)}{g'(x)}\)

Proof:

Let \(F(x)=f(x)-f(a)-\frac{f(b)-f(a)}{g(b)-g(a)}[g(x)-g(a)]\) That \(F(a)=F(b)=0\) According to Median Value Theorem,there is figure ‘c’ let \(F'(c)=0\) Which is \(\frac{f'(c)}{g'(c)}=\frac{f(b)-f(a)}{g(b)-g(a)}\) Then compelet the proof of Cauthy Median Value Theorem

Let x right of \(x_0\)

Accourding to CMVT, there is c between x and \(x_o\) let \(\frac{f'(c)}{g'(c)}=\frac{f(x)-f(x_0)}{g(x)-g(x_0)}\) because \(f(x_0)=g(x_0)=0\) \(\frac{f'(c)}{g'(c)}=\frac{f(x)}{g(x)}\) because \(x_0<c<x\) \(x_0 \to x \Rightarrow c \to x\) Thus \(\lim_{x \to x_0+}\frac{f(x)}{g(x)}=\lim_{x \to x_0+}\frac{f'(x)}{g'(x)}\) The same \(\lim_{x \to x_0-}\frac{f(x)}{g(x)}=\lim_{x \to x_0-}\frac{f'(x)}{g'(x)}\) Thus \(\lim_{x \to x_0}\frac{f(x)}{g(x)}=\lim_{x \to x_0}\frac{f'(x)}{g'(x)}\) then complete proof of the \(\frac{0}{0}\)form of L’Hopital Theorem

Let \(F(x)=\frac{1}{f(x)},G(x)=\frac{1}{g(x)}\)

That \(\lim_{x \to x_0}F(x)=\infty,\lim_{x \to x_0}G(x)=\infty\)

Suppose \(\lim_{x \to x_0}\frac{F(x)}{G(x)}=\lim_{x \to x_0}\frac{F'(x)}{G'(x)}\)

\[\lim_{x \to x_0}{\frac{g(x)}{f(x)}}=\lim_{x \to x_0}{\frac{-\frac{1}{f^2(x)}f'(x)}{-\frac{1}{g^2(x)}g'(x)}}=\lim_{x \to x_0}\frac{g^2(x)f'(x)}{f^2(x)g'(x)}\] \[\lim_{x \to x_0}{\frac{g(x)}{f(x)}}=\lim_{x \to x_0}{\frac{g'(x)}{f'(x)}}\]

Which is the \(\frac{0}{0}\)form of L’hopital Theorem

Then we have proof the suppose, which is \(\lim_{x \to x_0}\frac{F(x)}{G(x)}=\lim_{x \to x_0}\frac{F'(x)}{G'(x)}\) The \(\frac{\infty}{\infty}\)form of L’Hopital Theorem

连续性的定义

定义 在一点的连续性

内点: 函数f(x)在定义域的内点c处是连续的,如果 \(\lim_{x \to c}f(x)=f(x)\) 端点: 函数载器定义域的左端点a或右端点b是连续的,如果 \(\lim_{x \to a^+}f(x)=f(a)\quad\lim_{x \to b^-}f(x)=f(b)\)

连续性的性质

加减乘除

定理 连续函数的性质

如果函数f和g在x=c连续,下列函数在x=c连续 \(\begin{align} &f + g \\ &f - g \\ &f \cdot g \\ &\frac{f}{g},倘若g(c)\neq0\\ &f \circ g,倘若f在g(c)连续 \end{align}\)

连续函数的中值定理

定理 连续函数的中值定理

在闭区间[a,b]上连续的函数一定取到f(a)和f(b)之间的每一个值

连续性和可导性

定理 可导性蕴含着连续性

如果f在x=c有导数,那么f在x=c连续

导数

定义

1.3

定义 导函数

函数f(x)关于变量x的导数是函数f’,它在x处的值为 \(f'(x)=\lim_{h\to0}{\frac{f(x+h)-f(x)}{h}}\) 如果该极限存在

导数的性质

常数倍

\[\frac{\mathrm{d}}{\mathrm{d}x}(cu)=c\frac{\mathrm{d}}{\mathrm{d}x}\]

\[\frac{\mathrm{d}}{\mathrm{d}x}(u+v)=\frac{\mathrm{d}u}{\mathrm{d}x}+\frac{\mathrm{d}v}{\mathrm{d}x}\]

\[\frac{\mathrm{d}}{\mathrm{d}x}(uv)=u\frac{\mathrm{d}v}{\mathrm{d}x}+v\frac{\mathrm{d}u}{\mathrm{d}x}\]

Proof:
\(\begin{align} \frac{\mathrm{d}}{\mathrm{d}x}(uv)&=\lim_{\Delta x\to 0}\frac{\Delta (uv)}{\Delta x}\\ &=\lim_{\Delta x\to 0}(u\frac{\Delta v}{\Delta x}+v\frac{\Delta u}{\Delta x}+\frac{\Delta u \Delta v}{\Delta x})\\ &=u\frac{\mathrm{d}v}{\mathrm{d}x}+v\frac{\mathrm{d}u}{\mathrm{d}x} \end{align}\)

\[\frac{\mathrm{d}}{\mathrm{d}x}(\frac{u}{v})=\frac{v\frac{\mathrm{d}u}{\mathrm{d}x}-u\frac{\mathrm{d}v}{\mathrm{d}x}}{v^2}\]

Proof:

Regard \(\frac{u}{v}\) as \(u \times \frac{1}{v}\)

According to the Product Rule of the Derivative
\(\begin{align} \frac{\mathrm{d}}{\mathrm{d}x}(u \times \frac{1}{v})&=u(-\frac{1}{v^2})(\frac{\mathrm{d}v}{\mathrm{d}x})+\frac{1}{v}\frac{\mathrm{d}u}{\mathrm{d}x}\\ &=\frac{v\frac{\mathrm{d}u}{\mathrm{d}x}-u\frac{\mathrm{d}v}{\mathrm{d}x}}{v^2} \end{align}\)

复合

\[\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{\mathrm{d}y}{\mathrm{d}u}\cdot\frac{\mathrm{d}u}{\mathrm{d}x}\]

Proof:

\(\frac{\mathrm{d}}{\mathrm{d}x}f(g(x))=f'(g(x))g'(x)\) is another form of \(\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{\mathrm{d}y}{\mathrm{d}u}\cdot\frac{\mathrm{d}u}{\mathrm{d}x}\)

Let \(h(x)=f(g(x))\) Let \(\epsilon=g(x+\Delta x)-g(x)\) That \(x \to 0 \Rightarrow \epsilon \to 0\) Then
\(\begin{align} h'(x)&=\lim_{\Delta x \to 0}\frac{f(g(x + \Delta x))-f(g(x))}{\Delta x}\\ &=\lim_{\Delta x \to 0}\frac{f(g(x + \Delta x))-f(g(x))}{g(x + \Delta x)-g(x)}\times\frac{g(x + \Delta x)-g(x)}{\Delta x}\\ &=\lim_{\Delta x \to 0}\frac{f(g(x) + \epsilon)-f(g(x))}{\epsilon}\times\frac{g(x + \Delta x)-g(x)}{\Delta x}\\ &=f'(g(x))g'(x) \end{align}\)

求导数

常函数

\[\frac{\mathrm{d}f}{\mathrm{d}x}=\frac{\mathrm{d}}{\mathrm{d}x}(c)=0\]

幂函数

\[\frac{\mathrm{d}}{\mathrm{d}x}(x^n)=nx^{n-1}\]

Proof: \(\frac{\mathrm{d}}{\mathrm{d}x}(x^n)=\frac{\mathrm{d}}{\mathrm{d}x}\mathrm{e}^{\ln{x}\,n}=nx^{n-1}\)

自然对数

\[\mathrm{e}=\lim_{h \to 0^+}(1+h)^{1/h}\]

Proof:

let \(F(x)=\int_{1}^{x}\frac{1}{t}\mathrm(d)t\)

that
\(F(1)=0\\ \lim_{x \to \infty}F(x)=\infty\\ \frac{\mathrm{d}}{\mathrm{d}x}F(x)=\frac{1}{x}\) According to Intermediate Value Theorem

There is \(\mathrm{}e\) let \(F(e)=1\)

let \(G(x)=F(x^a)\)

that
\(G'(x)=a\frac{1}{x}=aF'(x)\\ G(x)=aF(x)+C\) let \(x=1 \Rightarrow C=0\) \(aF(x)=F(x^a)\) let \(x=\mathrm{e},a=x\)

Thus \(F(\mathrm{e})=1\)
\(F(\mathrm{e}^x)=x\\ F(x)=\log_{\mathrm{e}}x=\ln(x)\\ \frac{\mathrm{d}}{\mathrm{d}x}ln(x)=\frac{1}{x}\\ \frac{\mathrm{d}}{\mathrm{d}x}(\mathrm{e}^x)=\mathrm{e}^x\) let \(y=(1+h)^{1/h}\)

According to the L’Hopital Theorem
\(\lim_{h \to 0^+}\ln(y)=\lim_{h \to 0^+}\frac{\ln(1+h)}{h}=1\) Thus \(\ln(\mathrm{e})=1\)
\(\mathrm{e}=\lim_{h \to 0^+}(1+h)^{1/h}\)

对数

\[\frac{\mathrm{d}}{\mathrm{d}x}\log_{b}(x)=\frac{1}{x\ln(b)}\\ \frac{\mathrm{d}}{\mathrm{d}x}\ln(x)=\frac{1}{x}\]

Proof:

According to the definition of derivative
\(\begin{align} \frac{\mathrm{d}}{\mathrm{d}x}\log_b(x)=&\lim_{h \to 0}\frac{\log_b(x+h)-\log_b(x)}{h}\\ =&\lim_{h \to 0}\log_b(\frac{x+h}{x})^{1/h}\\ =&\frac{1}{x}\log_b\mathrm{e}\\ =&\frac{1}{x\ln(b)} \end{align}\) Then
\(\begin{align} \frac{\mathrm{d}}{\mathrm{d}x}\ln(x)&=\frac{\mathrm{d}}{\mathrm{d}x}\log_{\mathrm{e}}(x)\\ &=\frac{1}{ln(\mathrm{e})x}\\ &=\frac{1}{x} \end{align}\)

自然指数

\[\frac{\mathrm{d}}{\mathrm{d}x}(b^x)=b^x\ln(b)\\ \frac{\mathrm{d}}{\mathrm{d}x}(\mathrm{e}^x)=\mathrm{e}^x\]

Proof:
\(\begin{align} y&=b^x\\ \log_by&=x\\ \frac{\mathrm{d}}{\mathrm{d}x}log_by&=1\\ \frac{\mathrm{d}y}{\mathrm{d}x}\frac{1}{\ln(b)y}&=1\\ \frac{\mathrm{d}}{\mathrm{d}x}(b^x)&=b^x\ln(b) \end{align}\)

Then

\[\begin{align} \frac{\mathrm{d}}{\mathrm{d}x}(\mathrm{e}^x)&=e^x\ln(\mathrm{e})\\ &=\mathrm{e}^x \end{align}\]

三角函数

\[\frac{\mathrm{d}}{\mathrm{d}x}sin(x)=cos(x)\]

Proof:

According to the definition of the derivative
\(\begin{align} \frac{\mathrm{d}}{\mathrm{d}x}sin(x)&=\lim_{h \to 0}\frac{sin(x+h)-sin(x)}{h}\\ &=\lim_{h \to 0}\frac{sin(x)cos(h)+cos(x)sin(h)-sin(x)}{h}\\ &=\lim_{h \to 0}sin(x)\frac{cos(h)-1}{h}+cos(x)\frac{sin(h)}{h}\\ &=sin(x) \times 0+cos(x) \times 1\\ &=cos(x) \end{align}\)

\[\frac{\mathrm{d}}{\mathrm{d}x}cos(x)=-sin(x)\]

Proof:

According to the definition of the derivative
\(\begin{align} \frac{\mathrm{d}}{\mathrm{d}x}cos(x)&=\lim_{h \to 0}\frac{cos(x+h)-cos(x)}{h}\\ &=\lim_{h \to 0}\frac{cos(x)cos(h)-sin(x)sin(h)-cos(x)}{h}\\ &=\lim_{h \to 0}cos(x)\frac{cos(h)-1}{h}-sin(x)\frac{sin(h)}{h}\\ &=cos(x) \times 0-sin(x) \times 1\\ &=-sin(x) \end{align}\)

\[\begin{align} &\frac{\mathrm{d}}{\mathrm{d}x}tan(x)=sec^2(x)\\ &\frac{\mathrm{d}}{\mathrm{d}x}sec(x)=sec(x)tan(x)\\ &\frac{\mathrm{d}}{\mathrm{d}x}csc(x)=-csc(x)cot(x)\\ &\frac{\mathrm{d}}{\mathrm{d}x}cot(x)=-csc^2(x) \end{align}\]

反三角函数

\[\begin{align} &\frac{\mathrm{d}}{\mathrm{d}x}sin^{-1}(x)=\frac{1}{\sqrt{1-x^2}}\\ &\frac{\mathrm{d}}{\mathrm{d}x}cos^{-1}(x)=-\frac{1}{\sqrt{1-x^2}}\\ &\frac{\mathrm{d}}{\mathrm{d}x}tan^{-1}(x)=\frac{1}{1+x^2}\\ &\frac{\mathrm{d}}{\mathrm{d}x}csc^{-1}(x)=-\frac{1}{|x|\sqrt{1-x^2}}\\ &\frac{\mathrm{d}}{\mathrm{d}x}sec^{-1}(x)=\frac{1}{|x|\sqrt{1-x^2}}\\ &\frac{\mathrm{d}}{\mathrm{d}x}cot^{-1}(x)=-\frac{1}{1+x^2} \end{align}\]

Proof:
\(\begin{align} y&=sin^{-1}(x)\\ sin(y)&=x\\ \frac{\mathrm{d}y}{\mathrm{d}x}cos(y)&=1\\ \frac{\mathrm{d}y}{\mathrm{d}x}&=\frac{1}{\sqrt{1-sin(y)^2}}\\ \frac{\mathrm{d}}{\mathrm{d}x}sin^{-1}(x)&=\frac{1}{\sqrt{1-x^2}} \end{align}\) As the same
\(\begin{align} &\frac{\mathrm{d}}{\mathrm{d}x}cos(x)=-sin(x) &\qquad &-\frac{1}{sin(y)}=-\frac{1}{\sqrt{1-cos(y)^2}} \\ &\frac{\mathrm{d}}{\mathrm{d}x}tan(x)=sec^2(x) &\qquad &-\frac{1}{sec^2(y)}=\frac{1}{1+tan(y)^2} \\ &\frac{\mathrm{d}}{\mathrm{d}x}sec(x)=sec(x)tan(x) &\qquad &\frac{1}{sec(y)tan(y)}=\frac{1}{|sec(y)|\sqrt{1-sec(y)^2}} \\ &\frac{\mathrm{d}}{\mathrm{d}x}csc(x)=-csc(x)cot(x) &\qquad &-\frac{1}{csc(y)cot(y)}=-\frac{1}{|csc(y)|\sqrt{1-csc(y)^2}} \\ &\frac{\mathrm{d}}{\mathrm{d}x}cot(x)=-csc^2(x) &\qquad &-\frac{1}{csc^2(y)}=-\frac{1}{1+cot(y)^2} \\ \end{align}\)

双曲函数

定义 双曲函数
\(\begin{aligned} &\cosh x=\frac{e^{x}+e^{-x}}{2}\\ &\sinh x=\frac{e^{x}-e^{-x}}{2}\\ &\tanh x=\frac{\sinh x}{\cosh x}\\ &\operatorname{coth} x=\frac{\cosh x}{\sinh x}\\ &\operatorname{sech} x=\frac{1}{\cosh x}\\ &\operatorname{csch} x=\frac{1}{\sin x}\\ \end{aligned}\)

性质
\(\begin{align} &\sinh 2x =\sinh x \cosh x\\ &\cosh 2 x=\cosh ^{2} x+\sinh ^{2} x\\ &\cosh ^{2} x=\frac{\cosh 2 x+1}{2}\\ &\sinh ^{2} x=\frac{\cos h 2 x-1}{2}\\ &\cosh ^{2} x-\sinh ^{2} x=1\\ &\tanh ^{2} \alpha=1-\operatorname{sech}^{2} x\\ &\operatorname{coth}^{2} x=1+\csc h^{2} x\\ &\operatorname{sech} ^{-1} x=\cosh ^{-1} \frac{1}{x}\\ &\operatorname{csch} ^{-1} x=\sinh ^{-1} \frac{1}{x}\\ &\coth ^{-1} x=\tanh ^{-1} \frac{1}{x} \end{align}\)

导数
\(\frac{d}{d x} \sinh x=\operatorname{cosh} x \\ \frac{d}{d x} \cosh x=\sinh x \\ \frac{d}{d x} \operatorname{tanh} x=\operatorname{sech}^{2} x \\ \frac{d}{d x} \operatorname{coth} x=-\operatorname{csch}^{2} x \\ \frac{d}{d x} \operatorname{sech} x=-\operatorname{sech} x \tan x \\ \frac{d}{d x} \operatorname{csch} x=-\operatorname{csch} x \operatorname{coth} x\\ \frac{d}{d x} \sinh ^{-1} x =\frac{1}{\sqrt{1+x^{2}}} \\ \frac{d}{d x} \cosh ^{-1} x =\frac{1}{\sqrt{x^{2}-1}},x>1 \\ \frac{d}{a x} \tanh ^{-1} x =\frac{1}{1-x^{2}},|x|<1 \\ \frac{d}{d x} \coth^{-1} x =\frac{1}{1-x^{2}},|x|>1 \\ \frac{d}{d x} \operatorname{sech} ^{-1} x =-\frac{1}{x \sqrt{1-x^{2}}},0<x<1 \\ \frac{d}{d x} \operatorname{csch}^{-1} x =-\frac{-1}{|x|\sqrt{1+x^{2}}},x\ne0\)

隐函数

解题 隐函数求导

  1. 处理方程 y=f(x) 方程两边对x求导数
  2. 并项\(\frac{\mathrm{d}y}{\mathrm{d}x}\)到一边
  3. 提出因子\(\frac{\mathrm{d}y}{\mathrm{d}x}\)
  4. 解\(\frac{\mathrm{d}y}{\mathrm{d}x}\)

相关变化率

定义 相关变化率

如果Q为某个量,那么Q的变化率为\(\frac{\mathrm{d}Q}{\mathrm{d}t}\)

解题 隐函数求导

  1. 对涉及变化率的问题进行建模
  2. 变量对\(t\)进行求导,\(\frac{\mathrm{d}Q}{\mathrm{d}t}\)使用已知条件v(变化率)进行替换
  3. 对方程进行求解变量之间的相关变化率

导数的应用

物理中的应用

速度 \(v=\frac{\mathrm{d}s}{\mathrm{d}t}\)

加速度 \(a=\frac{\mathrm{d}v}{\mathrm{d}t}\)

急推 \(z=\frac{\mathrm{d}a}{\mathrm{d}t}\)

函数的极值

函数的极值

如果函数\(f\)在定义域\(c\)点取到局部最小值或局部最大值,那么 \(f'(c)=0 或 f'(c)不存在\)

中值定理

罗尔定理

假设函数\(f\)在闭区间\([a,b]\)内连续,在开区间\((a,b)\)可导,如果\(f(a)=f(b)\),那么在开区间\((a,b)\)内至少存在一点\(c\),使得\(f'(c)=0\).

中值定理

假设函数\(f\)在闭区间\([a,b]\)内连续,在开区间\((a,b)\)可导,如果\(f(a)=f(b)\),那么在开区间\((a,b)\)内至少存在一点\(c\),使得 \(f'(c)=\frac{f(b)-f(a)}{b-a}\)

Proof:

let \(g(x)=f(x)-\frac{f(b)-f(a)}{b-a}(x-a)\)

that \(g(a)=g(b)=f(a)\) According to Rolle Theorem

There is c let \(g'(x)=0\) which is \(g'(x)=f'(x)-\frac{f(b)-f(a)}{b-a}\\ f'(c)=\frac{f(b)-f(a)}{b-a}\)

最优化

解决最优化问题

  1. 分析问题,数学建模
  2. 求导使用函数极值解决最优化问题

线性化

使用线性化进行估值 \(f(a+\Delta x) \approx f(a)+f'(a)\Delta x\)

1.4

Newton法

Newton法

\(a\)为\(f(x)=0\)的近似解 \(b=a-\frac{f(a)}{f'(a)}\) \(b\)为更优解

1.5

积分

定义

定义 定积分作为黎曼和的极限

设f时定义在区间[a,b]的一个函数, 对于[a,b]的任意划分P,设\(c_k\)是在子区间\([x_{k-1},x_k]\)上任意选取的数.

如果存在一个数I,使得不论划分P怎样和\(c_k\)如何选取,都有 \(\lim_{||P|| \to 0}\sum_{k=1}^{n}f(c_k)\Delta x_k = I\) 则称f在[a,b]上是可积的,而I称为f在区间[a,b]上的定积分 \(I=\int_a^bf(x)\mathrm{d}x\)

images

性质

定积分性质
\(\begin{align} &\int_a^bf(x)\mathrm{d}x=-\int_b^af(x)\mathrm{d}x\\ &\int_a^af(x)\mathrm{d}x=0\\ &\int_a^bf(x)\mathrm{d}x=\int_a^cf(x)\mathrm{d}x+\int_c^bf(x)\mathrm{d}x\\ &\int_a^bcf(x)\mathrm{d}x=c\int_a^bf(x)\mathrm{d}x\\ &\int_a^b(f(x)+g(x))\mathrm{d}x=\int_a^bf(x)\mathrm{d}x+\int_a^bg(x)\mathrm{d}x \end{align}\)

微积分基本定理

微积分第一基本定理(不定积分为反导数) \(\frac{d}{d x}\int_a^xf(t)\mathrm{d}t=f(x)\)

Proof:
\(\begin{align} \frac{d}{d x}\int_a^xf(t)\mathrm{d}t&=\lim_{h \to 0}\frac{\int_a^{x+h}f(t)\mathrm{d}t-\int_a^{x}f(t)\mathrm{d}t}{h}\\ &=\lim_{h \to 0}\frac{\int_h^{x+h}f(t)\mathrm{d}t}{h}\\ &=\lim_{h \to 0}\frac{f(x)h}{h}\\ &=f(x) \end{align}\)

微积分第二基本定理 \(\int_a^bf(x)\mathrm{d}x=F(b)-F(a)\)

Proof:

let \(G(x)=\int_a^xf(t)\mathrm{d}t\), G is one of anti-derivative of f
let \(F(x)=G(x)+C\), F is any of anti-derivative of f
\(\begin{align} F(b)-F(a)&=[G(b)+C]-[G(a)+C]\\ &=G(b)-G(a)\\ &=\int_a^bf(t)dt-\int_a^af(t)dt\\ &=\int_a^bf(t)dt \end{align}\)

求积分方法

积分公式

\[\begin{aligned} &\int d u=u+c\\ &\int k d u=k u+c\\ &\int du + dv=\int d u+\int d v\\ &\int u^{n} d u=\frac{u^{n+1}}{n+1}+C\\ &\int \frac{1}{u} d u=\ln |u|+c\\ &\int \sin u d u=-\cos u+c\\ &\int \cos u d u=\sin u+c\\ &\int \sec u d u=\ln |\sec x+\tan x|+c\\ &\int \csc u d u=-\ln |\csc x+\cot x|+c\\ &\int \sec ^{2} u d u=\tan u+c\\ &\int \csc ^{2} u d u=-\cot u+c\\ &\int \sec u \tan u d u=\sec u+c\\ &\int \csc u \cot u d u=-\csc u+c\\ &\int \tan u d u=\ln |\sec u|+C\\ &\int \cot u d u=\ln |\sin u|+C\\ &\int e^{u} d u=e^{u}+c \\ &\int a^{u} d u=\frac{a^{u}}{\ln a}+c\\ &\int \sinh u d u=\cosh u d x+c\\ &\int \cosh u d u=\sinh u+c\\ &\int \frac{d u}{\sqrt{a^{2}-u^{2}}}=\sin ^{-1}\left(\frac{u}{a}\right)+c\\ &\int \frac{d u}{\sqrt{u^{2}-a^{2}}}=\cos h^{-1}\left(\frac{u}{a}\right)+c \\ &\int \frac{d u}{\sqrt{a^{2}+u^{2}}}=\sinh ^{-1}\left(\frac{u}{a}\right)+c \\ &\int \frac{d u}{a^{2}+u^{2}}=\frac{1}{a} \tan ^{-1}\left(\frac{u}{a}\right)+c\\ &\int \frac{d u}{u \sqrt{u^{2}-a^{2}}}=\frac{1}{a} \sec ^{-1}\left(\frac{u}{a}\right)+C \end{aligned}\]

变量替换

替换积分法

当f和g’是连续函数时,为求积分\(\int f(g(x))g'(x)dx\)

  1. 做替换\(u=g(x)\),则\(du=g'(x)dx\),得到积分\(\int f(u)du\)
  2. 对u积分
  3. 使用g(x)替代u

定积分的变量替换 \(\int _a^bf(g(x))\cdot g'(x)dx=\int_{g(a)}^{g(b)}f(u)du\)

分部积分

分部积分公式 \(\int u\mathrm{d}v=uv-\int v\mathrm{d}u\)

Proof:
\(\begin{align} \frac{\mathrm{d}}{\mathrm{d}x}(uv)&=u\frac{\mathrm{d}v}{\mathrm{d}x}+v\frac{\mathrm{d}u}{\mathrm{d}x}\\ \int(\frac{\mathrm{d}}{\mathrm{d}x}(uv)\mathrm{d}x&=\int(u\frac{\mathrm{d}v}{\mathrm{d}x})+\int (v\frac{\mathrm{d}u}{\mathrm{d}x})\\\int u\mathrm{d}v&=uv-\int v\mathrm{d}u \end{align}\)

列表积分

对于形如\(\int f(x)g(x)dx\)的积分, 其中f(x)可以反复求导得到0, g(x)可以简单地重复积分


1.15

列表积分2

1.7

部分分式

形如\(\int \frac{dx}{x^2+px+q}\)

\(\Delta>0\):
\(\frac{1}{x^2+px+q}=\frac{A_1}{x-a_1}+\frac{A_2}{x-a_2}\\\) 其中
\(A_1(x-a_2)+A_2(x-a_1)=1\) \(\Delta=0\):
\(\int \frac{dx}{x^2+px+q}=\int \frac{dx}{(x-a)^2}=-\frac{1}{x-a}+C\) \(\Delta<0\):

配方 \(x^2+px+q=(x+\frac{p}{2})^2-\frac{\Delta}{4}\) 化简
\(\begin{align} \int \frac{dx}{x^2+px+q}=&-\frac{4}{\Delta}\int \frac{dx}{1+(\frac{2x+p}{\sqrt{-\Delta}})^2}\\ =&\frac{2}{\sqrt{-\Delta}}\int\frac{du}{u^2+1}\\ =&\frac{2}{\sqrt{-\Delta}}tan^{-1}(\frac{2x+p}{\sqrt{-\Delta}})+C \end{align}\)

形如\(\int\frac{mx+n}{x^2+px+q}dx\)

\(\Delta>0\):
\(\frac{mx+n}{x^2+px+q}=\frac{A_1}{x-a_1}+\frac{A_2}{x-a_2}\) \(\Delta=0\):
\(\frac{mx+n}{x^2+px+q}=\frac{A}{x-a}+\frac{B}{(x-a)^2}\) \(\Delta<0\): 换元后长除
\(u=x^2+px+q\\ du=2x+p\\ \int\frac{mx+n}{x^2+px+q}dx=\int (r\frac{2x+p}{x^2+px+q}+s\frac{1}{x^2+px+q})dx\\ mx+n=r(2x+q)+s\)

三角换元

三角替换

  1. \(x=atan\theta\) \(a^2+x^2=a^2sec^2\theta\)
  2. \(x=asin\theta\) \(a^2-x^2=a^2cos^2\theta\)
  3. \(x=asec\theta\) \(x^2-a^2=a^2tan^2\theta\)

估算积分

  1. 黎曼和

    在区间[a,b],均分为n等分 \(h=\frac{b-a}{n}\\ \int_a^bf(x)dx\approx h\sum_{j=1}^{n}f(c_j)\)

  2. 梯形法则

    在区间[a,b],均分为n等分 \(h=\frac{b-a}{n}\\ \int_a^bf(x)dx\approx \frac{h}{2}\sum_{j=1}^{n}(f(x_{j-1})+f(x_j))\)

    1.8

  3. 辛普森法则(二次曲线逼近)

    在区间[a,b],均分为n等分 \(h=\frac{b-a}{n}\\ \int_a^bf(x)dx\approx \frac{h}{3}\sum_{j=1}^{\frac{1}{2}n}(f(x_{2j-2})+4f(x_{2j-1})+f(x_{2j}))\)

    Proof: \(A=\frac{h}{3}(y_0+4y_1+y_2)\) 1.9

积分的应用

计算体积

切片法

旋转轴为y=h \(V=\int_a^b\pi (y-h)^2 dx\)

1.11

壳法

旋转轴为x=h \(V=\int_a^b2\pi(x-h)ydx\)

1.12

椎体体积

椎体体积 \(V=\int_0^hA(x)dx=\frac{1}{3}Ah\)

1.10

Proof:
\(\begin{align} \frac{x}{l}=\frac{h}{L} \Rightarrow \frac{A}{A(x)}=&(\frac{L}{l})^2 \Rightarrow A(x)=\frac{Ax^2}{h^2}\\ V=&\int _0^hA(x)dx\\ =&\int _0^h\frac{A}{h^2}x^2dx\\ =&\frac{1}{3}Ah \end{align}\)

计算弧长

\[L=\int_a^b \sqrt{1+(\frac{dy}{dx})^2}dx\\ L=\int_{t_0}^{t_1} \sqrt{(\frac{dy}{dt})^2+(\frac{dx}{dt})^2}dx\\ L=\int_{\theta_0}^{\theta_1} \sqrt{f(\theta)^2+f'(\theta)^2}dx\]

1.14

计算旋转体表面积

\[S=\int_a^b2\pi y\sqrt{1+(\frac{dy}{dx})^2}dx\\ S=\int_{t_0}^{t_1}2\pi y\sqrt{(\frac{dy}{dt})^2+(\frac{dx}{dt})^2}dt\]

1.13

微分方程

可分离变量的微分方程

指数变化率

如果y以正比与当前数量的速率变化(\(\frac{\mathrm{d}y}{\mathrm{d}t}=ky\))并且当t=0时\(y=y_0\),则 \(y=y_0\mathrm{e}^{kt}\) k>0表示增长, k<0表示衰减,k为速率常数

Proof:

Solve differential equations with separable varaibles

  1. separate varaibles
    \(\begin{align} \frac{\mathrm{d}y}{\mathrm{d}t}&=ky\\ \frac{1}{y}\mathrm{dy}&=k\mathrm{d}t \end{align}\)

  2. Intefral at both ends of the equation
    \(\begin{align} \int \frac{1}{y}\mathrm{dy}&=\int k\mathrm{d}t\\ ln|y|&=kt+C \end{align}\)

  3. Simplify the equation \(y=y_0\mathrm{e}^{kt}\)

线性一阶微分方程

线性一阶微分方程的解

线性方程 \(\frac{\mathrm{d}y}{\mathrm{d}x}+P(x)y=Q(x)\) 的解为 \(y=\frac{1}{v(x)}\int v(x)Q(x)\mathrm{d}x\) 其中 \(v(x)=\mathrm{e}^{\int P(x)\mathrm{d}x}\)

Proof:

Inspired by the Deravitive Product Rule

\[\frac{\mathrm{d}y}{\mathrm{d}x}(vy)=y'v+yv'\]

Product v(x) to the both ends of the equation, turn it to the form of DPR \(\begin{align} v(x)\frac{\mathrm{d}y}{\mathrm{d}x}+v(x)P(x)y&=v(x)Q(x)\\ \frac{\mathrm{d}}{\mathrm{d}x}(v(x)y)&=v(x)Q(x)\\ y&=\frac{1}{v(x)}\int v(x)Q(x)\mathrm{d}x \end{align}\) For v(x)
\(\frac{\mathrm{dv}}{\mathrm{d}x}=v(x)P(x)\\ v(x)=\mathrm{e}^{\int P(x)\mathrm{d}x}\)

高阶微分方程

暂无

反常积分

定义

定义 有穷积分限的反常积分

有穷积分限的积分是反常积分

  1. 如果f(x)在\([a,\infty)\)是连续的,则 \(\int_a^{\infty}f(x)\mathrm{d}x=\lim_{N \to \infty}\int_a^{N}f(x)\mathrm{d}x\)

  2. 如果f(x)在\((-\infty,b]\)是连续的,则 \(\int_{\infty}^bf(x)\mathrm{d}x=\lim_{N \to -\infty}\int_N^bf(x)\mathrm{d}x\)

  3. 如果f(x)在\((-\infty,\infty)\)是连续的,则 \(\int_{-\infty}^{\infty}f(x)\mathrm{d}x=\int_{-\infty}^{c}f(x)\mathrm{d}x+\int_{c}^{\infty}f(x)\mathrm{d}x\)

定义 无界不连续函数的反常积分

  1. 如果f(x)在\([a,b)\)是连续的,则 \(\int_a^bf(x)\mathrm{d}x=\lim_{N \to b^-}\int_a^Nf(x)\mathrm{d}x\)

  2. 如果f(x)在\((a,b]\)是连续的,则 \(\int_{a}^bf(x)\mathrm{d}x=\lim_{N \to a^+}\int_N^bf(x)\mathrm{d}x\)

  3. 如果f(x)在\((a,b)\)是连续的,则 \(\int_{a}^{b}f(x)\mathrm{d}x=\int_{a}^{c}f(x)\mathrm{d}x+\int_{c}^{b}f(x)\mathrm{d}x\)

无穷级数

数列的极限

定义

定义 收敛,发散,极限

极限序列{\(a_n\)}收敛到数L,如果每个正数\(\varepsilon\),都对应一个整数,使得对所有n: \(n>N\Rightarrow|a_n-L|<\varepsilon\) 如果这样的数L不存在,我们说{\(a_n\)}发散.

若{\(a_n\)}收敛到数L,我们记成\(\lim_{n \to \infty}=L\),或简单记成\(a_n\to L\),并称L是序列{\(a_n\)}的极限

无穷级数

发散级数

发散级数的第n项判别法

若\(\lim_{n \to \infty}a_n\)不存在或异于零,则级数\(\sum_{n=1}^\infty a_n\)发散

收敛级数

收敛级数的第n项极限

若\(\sum_{n=1}^\infty a_n\)收敛,则\(a_n \to 0\)

几何级数

几何级数 \(a_n=ar^{n-1}\\ S_n=\frac{a_1(1-r^n)}{1-r}\\\)

Proof:
\(\begin{align} s_n&=a+ar+ar^2+...+ar^{n-1}\\ rs_n&=ar +ar^2+...+ar^{n-1}+ar^{n}\\ s_n-rs_n&=a-ar^n\\ s_n&=\frac{a_1(1-r^n)}{1-r}\\ \end{align}\)

几何级数的收敛和发散 \(\sum_{n=1}^{\infty}ar^{n-1}=\frac{a_1}{1-r}\) 当|r|<1是收敛的上式,当|r|\(\geq\)1时发散

p级数

P级数 \(\sum_{n=1}^{\infty}\frac{1}{n^p}\) 当p>1时收敛,当p\(\leq\)1时发散

Proof:

According to the Integral Method
\(\sum_{n=1}^{\infty}\frac{1}{n^p}\sim\int_{n=1}^{\infty}\frac{1}{x^p}\mathrm{d}x\\ \begin{align} \int_{n=1}^{\infty}\frac{1}{x^p}\mathrm{d}x&=\lim_{N \to \infty}\int_{n=1}^{N}\frac{1}{x^p}\mathrm{d}x\\ &=\lim_{N \to \infty}[\frac{x^{1-p}}{1-p}]^{N}_{1}\\ &=\lim_{N \to \infty}(\frac{N^{1-p}}{1-p}-\frac{1}{1-p}) \end{align}\)

非负项级数

  1. 积分判别法

    \[\{a_n\}是一个正数序列,f(x)是x的连续函数,a_n=f(n).则级数\sum_{n=N}^{\infty}a_n和\int_{n=N}^{\infty}f(x)\mathrm{d}x同时收敛或发散\]

    Proof:
    \(\int_1^{n+1}f(x)dx\leq a_1+a_2+\dots+a_n,(估算积分取上和)\\ \int_1^{n+1}f(x)dx发散\Rightarrow\sum_{n=N}^{\infty}a_n发散\\ \int_1^{n+1}f(x)dx+a_1\geq a_1+a_2+\dots+a_n,(估算积分取下和)\\ \int_1^{n+1}f(x)dx收敛\Rightarrow\sum_{n=N}^{\infty}a_n收敛\)

  2. 比较判别法

    设\(\sum a_n\)是非负项级数

    1. 如果存在收敛级数\(\sum c_n\)和整数N,使得n>N有\(a_n\leq c_n\),则\(\sum a_n\)收敛
    2. 如果存在非负项发散级数\(\sum b_n\)和整数N,使得n>N有\(a_n\geq c_n\),则\(\sum a_n\)发散
  3. 极限比较判别法

    1. 若\(\lim_{n\to \infty}\frac{a_n}{b_n}=c\),0<c<\(\infty\),则\(\sum a_n\)和\(\sum b_n\)同时收敛或发散
    2. 若\(\lim_{n\to \infty}\frac{a_n}{b_n}=0\),\(\sum b_n\)收敛,则\(\sum a_n\)收敛
    3. 若\(\lim_{n\to \infty}\frac{a_n}{b_n}=\infty\),\(\sum b_n\)发散,则\(\sum a_n\)发散

    Proof:
    \(\lim_{n \to \infty}\frac{f(x)}{g(x)}=1\Rightarrow\ \frac{1}{2}\leq\frac{f(x)}{g(x)}\leq2\\ \frac{1}{2}g(x)\leq f(x)\leq2g(x)\) According to the Comparative discrimination, we can proof item1
    \(\lim_{n \to \infty}\frac{f(x)}{g(x)}=0\Rightarrow\ \frac{f(x)}{g(x)}\leq\frac{1}{2}\\ f(x)\leq\frac{1}{2}g(x)\) According to the Comparative discrimination, we can proof item2
    \(\lim_{n \to \infty}\frac{f(x)}{g(x)}=0\Rightarrow\ \frac{f(x)}{g(x)}\geq2\\f(x)\geq2g(x)\) According to the Comparative discrimination, we can proof item3

  4. 比式判别法

    \[\lim_{n \to \infty}\frac{a_{n+1}}{a_n}=p\]
    1. p<1:收敛
    2. p>1:发散
    3. p=1:无定论

    Proof:
    let r that p<r<1

    there is m big enough that
    \(\sum_{n=m}^{\infty}a_n<\sum_{n=m}^{\infty}Ar^n\) According to the Comparative discrimination, we can proof item1
    \(p>1\Rightarrow \lim_{n \to \infty}a_n\neq0\) According to the N-th discrimination method, we can proof item2

    when p=1 , we can know nothing about \(\lim_{n\to\infty}a_n\)

  5. n次根式判别法

    \[\lim_{n \to \infty}\sqrt[n]a_n=p\]
    1. p<1:收敛
    2. p>1:发散
    3. p=1:无定论

    let r that p<r<1

    there is m big enough that
    \(\sum_{n=m}^{\infty}a_n<\sum_{n=m}^{\infty}r^n\) According to the Comparative discrimination, we can proof item1
    \(p>1\Rightarrow \lim_{n \to \infty}a_n\neq0\) According to the N-th discrimination method, we can proof item2

    when p=1 , we can know nothing about \(\lim_{n\to\infty}a_n\)

    As a example , we discuss the function \(\sum\frac{1}{p^n}\)
    \(\lim_{n \to \infty}(\frac{1}{p^n})^{\frac{1}{n}}=\lim_{n \to \infty}n^{-p/n}=1\)

交错级数

交错级数判别法

级数 \(\sum_{n=1}^{\infty}(-1)^{n+1}u_n\) 收敛,如果一下条件满足:

  1. \(u_n\)全是正的
  2. \(u_n\)递减
  3. \(u_n \to 0\)并且\(\lvert L-s_n \lvert< \lvert u_{n+1}\lvert\)并且\(L-s_n\)和\(u_{n+1}\)同号

Proof:

\(\begin{align} s_{2m} &= (u_1 - u_2)+ \dots +(u_{2m-1} - u_{2m}) \\ &= u_1 -(u_2-u_3)-\dots-(u_{2m-2}-u_{2m-1})-u_{2m} \end{align}\) According to the first line and condition(1)(2), \(s_{2m}\)is increase

According to the second line and condition(1)(2), \(s_{2m}<u_1\)
\(\lim_{m\to\infty}s_{2m}=L\) According to the condition(3)
\(\lim_{m\to\infty}u_{2m+1}=0\\ \lim_{m\to\infty}s_{2m+1}=L+0=L\)

绝对收敛 \(\sum_{n=1}^{\infty}|a_n|收敛\Rightarrow \sum_{n=1}^{\infty}a_n收敛\)

Proof:
\(a_n\leq|a_n|\\ \sum_{n=0}^{\infty}a_n\leq\sum_{n=0}^{\infty}|a_n|\\ \sum_{n=0}^{\infty}|a_n|收敛\Rightarrow\sum_{n=0}^{\infty}收敛\)

幂级数

定义

中心在a的幂级数: \(\sum_{n=0}^{\infty}c_n(x-a)^n\)

性质

逐项求导定理
\(\frac{\mathrm{d}}{\mathrm{d}x}f(x)=\frac{\mathrm{d}}{\mathrm{d}x}\sum_{n=0}^{\infty}c_n(x-a)^n,|x-a|<R\) f(x)在收敛区间的所有阶导数, 可以逐项求导原级数得到

逐项积分定理
\(\int f(x)=\int\sum_{n=0}^{\infty}c_n(x-a)^n,|x-a|<R\) f(x)在收敛区间的积分, 可以逐项积分原级数得到

收敛性

幂级数收敛定理

对于\(\sum_{n=0}^{\infty}c_n(x-a)^n\)的收敛有三种可能.

  1. \(\lvert x-a \lvert >R\)发散,\(\lvert x-a\lvert<R\)收敛,在a+R和a-R,可能收敛也可能发散
  2. 级数对每个x收敛
  3. 级数在x=a收敛,在其余的点发散

求收敛区间

  1. 使用比值判别法n次判别法求使函数绝对收敛的区间
  2. 带入端点值进行检验(使用非负项级数判别法)

Taylor级数

2.1

  1. 假设光滑函数可以用多项式函数逼近
    \(f(x)=c_{0}+c_{1}(x-a)+c_{2}(x-a)^{2}+\cdots+c_{n}(x-a)^{n}\)

  2. 使用麦克劳林待定系数法\(f^{(n)}(x)=P^{(n)}(x)\)
    \(\begin{align} &f^{(1)}(x)=c_1+2c_2(x-a)+ \dots +nc_n(x-a)^{n-1}\\ &令x=a得c_1=f^{(1)}(a)\\ &f^{(2)}(x)=2c_2+2\times 3c_3(x-a)+ \dots +n(n-1)c_n(x-a)^{n-2}\\ &令x=a得c_2=\frac{f^{(2)}(a)}{2!}\\ &\dots\\ &f^{(n)}(x)=c_Nc!\\ &令x=a得c_n=\frac{f^{(n)}(a)}{n!}\\ \\ &P_N(x)=\sum_{n=0}^{N}\frac{f^{(n)}(a)}{n!}(x-a)^{n} \end{align}\)

  3. 讨论taylor级数和原函数的差值

\[\begin{aligned} R_{n} &=f(x)-P_{N}(x) \\ &=f(x)-\left(f(a)+\frac{f'(a)}{1 !}(x-a)+\cdots+\frac{f^{(n)}(a)}{n !}(x-a)^{n}\right) \end{aligned}\]

​ 做辅助函数\(g(t)=f(x)-(f(t)+\frac{f'(t)}{1!}(x-t)+\cdots+\frac{f^{(n)}(x)}{n!}(x-t)^{n}),t\in[a,x]\)
\(g'(t)=-\frac{f^{n+1}(t)}{n!}(x-t)^{n}\\ g(x)=0\\ g(a)=R_N(x)\) ​ 做辅助函数\(h(t)=(x-t)^{(n+1)}\)
\(\begin{align} &h^{\prime}(t)=-(n+1)(x-t)^{n} \\ &h(x)=0 \\ &h(a)=(x-a)^{(n+1)} \end{align}\) ​ 由Cauthy MVT得
\(\frac{g(x)-g(a)}{h(x)-h(a)}=\frac{g^{\prime}(c)}{h^{\prime}(c)} \quad c \in[a, x]\) ​ 解的
\(R_{N}(x)=\frac{f^{(n+1)}(C)}{(n+1) !}(x-a)^{(n+1)}\)

  1. 在所有N次或N次一下的多项式中,P是在a附近最佳近似

    Q为级数不超过N的多项式
    \(|f(x)-P_N(x)|<|f(x)-Q_N(x)|\\ |R_N(x)|<|S(x)+R_N(x)|,S(x)=P_N(x)-R_N(x)\)

    1. 当\(x \to a时,c \to a\)
      \(\begin{aligned} \left|R_{N}(x)\right| &=\left|\frac{f^{(N+1)}(x)}{(N+1) !}(x-a)^{N+1}\right| \sim \frac{f^{(x+1)}(a)}{(N+1) !}(x-a)^{N+1} \end{aligned}\)

    2. \(S(x)=a_m(x-a)^m+\dots\)其中\(a_m(x-a)^m\)为最低项,0<m<n
      \(\begin{aligned} &S(x) \sim a m(x-a)^{m}\\ &R_{N}(x) \sim C(x-a)^{N+1},C=\frac{f^{(N+1)}(a)}{(N+1) !}\\ &\because m<M+1\\ &\therefore S(x)+R_N(x)\sim a_m(x-a)^m\\ \end{aligned}\)


    \(\begin{aligned} &|f(x)-P_{N}(x)| \sim |c||x-a|^{N+1}\\ &|f(x)-Q_N(x)| \sim |a_{m}||x-a|^{m}\\ \\ &\frac{|c||x-a|^{N+1}}{|a_{m}||x-a|^{m}}=C_1|x-a|^{-m+N-1}\\ &当x \to a时,C_1|x-a|^{-m+N-1} \to 0\\ &|c||x-a|^{N+1}<|a_{m}||x-a|^{m}\\ &|f(x)-P_N(x)|<|f(x)-Q_N(x)| \end{aligned}\)

  2. 得出结论
    \(f(x)=\sum_{n=0}^{\infty}\frac{f^{(n)}(a)}{n!}(x-a)^{n}\)

Fourier级数

2.2

  1. 假设周期函数f(x)能表示为三角函数之和
    \(f(x)=A_{0}+\sum_{i=h}^{\infty}A_n \frac{n \pi x}{L}+\varphi_{n},T(周期)=2L\)

  2. 通过和角公式合并常数
    \(\begin{aligned} &A_{n} \sin \left(\frac{n \pi x}{L}+\varphi_{n}\right)=A_{n} \sin \psi_{n} \cos \left(\frac{n \pi x}{L}\right)+A_{n} \cos \psi_{n} \cos \left(\frac{n \pi x}{L}\right)\\ &f(x)=A_{0}+\sum_{n=1}^{\infty}[ a_{n} \cos (\frac{n \pi x}{L})+b_{n} \sin (\frac{n \pi x}{L}) ] \end{aligned}\)

  3. 通过三角积分球未知数\(\int_{-L}^Lf(x)\mathrm{d}x=\int_{-L}^LF(x)\mathrm{d}x\)

    三角积分
    \(\begin{aligned} &\int_{-L}^{L} \cos \left(\frac{n \pi x}{L}\right) d x=0 \\ &\int_{-L}^{L} \sin \left(\frac{1 \pi x}{L}\right) d x=0 \\ &\int_{-L}^{L} \cos \left(\frac{n \pi x}{L}\right) \cos \left(\frac{m \pi x}{L}\right) d x=\left\{\begin{array}{l} 0 . m \neq n \\ L, m=n \end{array}\right.\\ &\int_{-L}^{L} \sin \left(\frac{n \pi x}{L}\right) \cos \left(\frac{m \pi x}{L}\right) d x=0\\ &\int_{-L}^{L} \sin \left(\frac{n \pi x}{L}\right) \sin \left(\frac{m \pi x}{L}\right) d x=\left\{\begin{array}{l} 0, m \neq n \\ L, m=n \end{array}\right. \end{aligned}\) Proof:
    \(\begin{align} \cos (\alpha) \cos (\beta)&=\frac{1}{2}[\cos (\alpha+\beta)-\cos (\alpha-\beta)] \\ \sin (\alpha) \sin (\beta)&=\frac{1}{2}[\cos (\alpha-\beta)-\cos (\alpha+\beta)] \\ \sin (\alpha) \cos (\beta)&=\frac{1}{2}[\sin (\alpha+\beta)-\sin (\alpha-\beta)] \\ \end{align}\) (3)
    \(\begin{align} \int^{L}_{-L}cos(\frac{n \pi x}{L})cos(\frac{m \pi x}{L})\mathrm{d}x&=\frac{1}{2}[\int_{-L}^{L}cos(\frac{n \pi x}{L}+\frac{m \pi x}{L})\mathrm{d}x+\int_{-L}^{L}cos(\frac{n \pi x}{L}-\frac{m \pi x}{L})\mathrm{d}x]\\ m &\neq n\Rightarrow0,(根据公式1)\\ m&=n\Rightarrow\frac{1}{2}[0+\int_{-L}^{L}1\mathrm{d}x]=L \end{align}\) (4)
    \(\begin{align} \int^{L}_{-L}sin(\frac{n \pi x}{L})cos(\frac{m \pi x}{L})\mathrm{d}x&=\frac{1}{2}[\int_{-L}^{L}sin(\frac{n \pi x}{L}+\frac{m \pi x}{L})\mathrm{d}x-\int_{-L}^{L}sin(\frac{n \pi x}{L}-\frac{m \pi x}{L})\mathrm{d}x]\\ m &\neq n\Rightarrow0,(根据公式2)\\ m&=n\Rightarrow\frac{1}{2}[0-\int_{-L}^{L}0\mathrm{d}x]=0 \end{align}\) (5)
    \(\begin{align} \int^{L}_{-L}sin(\frac{n \pi x}{L})sin(\frac{m \pi x}{L})\mathrm{d}x&=\frac{1}{2}[\int_{-L}^{L}cos(\frac{n \pi x}{L}-\frac{m \pi x}{L})\mathrm{d}x-\int_{-L}^{L}cos(\frac{n \pi x}{L}+\frac{m \pi x}{L})\mathrm{d}x]\\ m &\neq n\Rightarrow0,(根据公式1)\\ m&=n\Rightarrow\frac{1}{2}[\int_{-L}^{L}1\mathrm{d}x-0]=L \end{align}\)

\(a_0的计算\) \(\begin{align} \\ \int_{-L}^{L} f(x) d x &=\int_{-L}^{L} A_{0} d x+\int_{-L}^{L} \sum_{h=1}^{\infty}\left[a_{n} \cos \left(\frac{n\pi x}{L}\right)+b_{n} \sin \left(\frac{n\pi x}{L}\right)\right] d x \\ &=2 L A_{0},(根据公式1)\\ A_0=\frac{1}{2L}&\int_{-L}^{L}f(x)\mathrm{d}x=\frac{1}{2}a_0 \end{align}\)

\(a_n的计算\) \(\begin{align} 两边同乘cos(\frac{m\pi x}{L}),m>0\\ \int_{-L}^{L} f(x)cos(\frac{m\pi x}{L}) d x &=\int_{-L}^{L} A_{0}cos(\frac{m\pi x}{L}) d x+\int_{-L}^{L} \sum_{h=1}^{\infty}\left[a_{n} cos(\frac{m\pi x}{L})\cos \left(\frac{n\pi x}{L}\right)+b_{n} cos(\frac{m\pi x}{L})\sin \left(\frac{n\pi x}{L}\right)\right] d x \\ &=a_nL,(m=n)\\ a_n&=\frac{1}{L}\int_{-L}^{L}f(x)cos(\frac{n\pi x}{L}) d x \end{align}\)

\(b_n的计算\) \(\begin{align} 两边同乘sin(\frac{m\pi x}{L}),m>0\\ \int_{-L}^{L} f(x)sin(\frac{m\pi x}{L}) d x &=\int_{-L}^{L} A_{0}sin(\frac{m\pi x}{L}) d x+\int_{-L}^{L} \sum_{h=1}^{\infty}\left[a_{n} sin(\frac{m\pi x}{L})\cos \left(\frac{n\pi x}{L}\right)+b_{n} sin(\frac{m\pi x}{L})\sin \left(\frac{n\pi x}{L}\right)\right] d x \\ &=b_nL,(m=n)\\ b_n&=\frac{1}{L}\int_{-L}^{L}f(x)sin(\frac{n\pi x}{L}) d x \end{align}\)

得出结论 \(\begin{align} 定义&在-L<x<L的f(x)的Fourier级数是\\ &f(x)=A_{0}+\sum_{n=1}^{\infty}[ a_{n} \cos (\frac{n \pi x}{L})+b_{n} \sin (\frac{n \pi x}{L}) ]\\ 其中&\\ &a_0=\frac{1}{L}\int_{-L}^{L}f(x)\mathrm{d}x\\ &a_n=\frac{1}{L}\int_{-L}^{L}f(x)cos(\frac{n\pi x}{L}) d x\\ &b_n=\frac{1}{L}\int_{-L}^{L}f(x)sin(\frac{n\pi x}{L}) d x \end{align}\)

偶函数延拓 \(\begin{aligned} &f(x)=\frac{a_{0}}{2}+\sum_{n=1}^{\infty} a_{n} \cos \frac{n n x}{L}\\ &\alpha_{0}=\frac{2}{L} \int_{0}^{L} f(x) d x\\ &a_{n}=\frac{2}{L} \int_{0}^{L} f(x) \cos \frac{a \pi x}{L} d x\\ \end{aligned}\) Proof: 如果g(x)是奇函数, f(x)是偶函数
\(\int_{-L}^{L} g(x) d x=0\\ \int_{-L}^{L} f(x) d x=2 \int_{0}^{L} f(x) d x\\\) Proof:
\(a_{0}=\frac{1}{L} \int_{-L}^{L} f(x) d x\\ a_{n}=\frac{1}{2} \int_{-L}^{L} f(x) \cos \frac{m x}{L} d x=\frac{2}{L} \int_{0}^{2} f(x) \cos \frac{n \pi x}{L} d x\\ b_{n}=\frac{1}{L} \int_{-L}^{L} f(x) \sin \frac{m n}{L} d x=0\\\) 奇函数延拓
\(\begin{align} &f(x)=\sum_{n=1}^{\infty} b_{n} \sin \frac{n \pi x}{L}\\ &b_{n}=\frac{2}{L} \int_{0}^{L} f(x) \sin \frac{n \pi x}{L} d x\\ \end{align}\) Proof:
\(a_{0}=\frac{1}{L} \int_{-L}^{L} f(N) d x=0\\ a_{n}=\frac{1}{2} \int_{-L}^{L} f(x) \cos \frac{n \pi x}{L} d x=0 \\ b_{n}=\frac{1}{L} \int_{-L}^{L} f(x) \sin \frac{m x}{L} d x=\frac{2}{2} \int_{0}^{1} f(x) \sin \frac{n \pi x}{L} d x\)

极坐标

定义及其图形

极坐标
\(P(r,\theta)\) r:从O到P的有向距离

\(\theta\):从初始射线到射线OP的有向角

r=a 中心为O半径为r的圆周

\(\theta\)=a 过O且与初始线段成角a的一条直线

对称性

  1. 关于x轴对称 \((r,\theta) \Rightarrow (r,-\theta)(r,-\theta+\pi)\)
  2. 关于y轴对称 \((r,\theta) \Rightarrow (-r,-\theta)(-r,-\theta+\pi)\)
  3. 关于O对称 \((r,\theta) \Rightarrow (-r,\theta)\)

笛卡尔坐标系和极坐标系的转换
\(x=r\cos(\theta),\quad y=r\sin(\theta),\quad x^2+y^2=r^2,\quad \frac{y}{x}=\tan(\theta)\)

极坐标的微积分

\[\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{f'(\theta)\sin\theta+f(\theta)\cos\theta}{f'(\theta)\cos\theta+f(\theta)\sin\theta}\]
\[A=\int_{\alpha}^{\beta}\frac{1}{2}r^2\mathrm{d}\theta\]

2.3

\[L=\int_{\theta_0}^{\theta_1} \sqrt{f(\theta)^2+f'(\theta)^2}dx\]

通过把极坐标写为参数形式导出

平面向量

计算

位置向量等同于 \(v=<x_2-x_1,y_2-y_1>\)

向量长度,方向,夹角

向量长度 \(\lvert v\lvert =\sqrt{v_1^2+v_2^2}=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\)

向量方向 \(单位向量\frac{\lvert v\lvert }{v}与v同方向\)

分解向量为方向和大小表示 \(v=\lvert v\lvert \frac{v}{\lvert v\lvert }\)

向量夹角 \(cos(\theta)=\frac{u_1v_1+u_2v_2}{\lvert u \lvert \lvert v\lvert }=\frac{u\cdot v}{\lvert u\lvert \lvert v\lvert }\)

内积(点积)

点积的含义 \(u\cdot v=\lvert u\lvert \lvert v\lvert cos(\theta)\)

点积计算公式 \(u\cdot v=u_1v_1+u_2v_2\)

Proof:

设\(w=u-v\)

根据余弦定理\({\lvert w\lvert}^2={\lvert u\lvert }^2+\lvert v\lvert ^2-2\lvert u\lvert \lvert v\lvert cos(\theta)\)

其中: \(\lvert u\lvert ^2=u_1^2+u_2^2\\ \lvert v\lvert ^2=v_1^2+v_2^2\\ \lvert w\lvert ^2=w_1^2+w_2^2\\\) 代入得\(cos(\theta)=\frac{u_1v_1+u_2v_2}{\lvert u\lvert \lvert v\lvert }=\frac{u\cdot v}{\lvert u\lvert \lvert v\lvert }\)

向量投影 \(\mathrm{proj}_vu=(\frac{u\cdot v}{\lvert v\lvert ^2})v\)

Proof:
\(\begin{align} \mathrm{proj}_vu&=(\lvert u\lvert cos\theta)\frac{v}{\lvert v\lvert }\\ &=(\frac{u\cdot v\cdot\lvert u\lvert }{\lvert u\lvert \lvert v\lvert })\frac{v}{\lvert v\lvert }\\ &=(\frac{u\cdot v}{\lvert v\lvert ^2})v \end{align}\)

分解向量为正交向量的和 \(u=\mathrm{proj}_vu+(u-\mathrm{proj_vu})\)

向量值函数

极限和连续

极限

设\(r(t)=f(t)i+g(t)j\),若 \(\lim_{t\to c}f(t)=L_1 \quad \& \quad \lim_{t \to c}g(t)=L_2\) 则\(r(t)\)当t趋于c时的极限是 \(\lim_{t\to c}r(t)=L=L_1i+L_2j\)

连续

一个向量函数\(r(t)\)在定义域内的一点t=c是连续的, 如果 \(\lim_{t \to c}r(t)=r(c)\)

导数和积分

导数 \(r'(t)=\frac{\mathrm{d}r}{\mathrm{d}t}=\lim_{\Delta t\to 0}\frac{r(t+\Delta t)-r(t)}{\Delta t}=\frac{\mathrm{d}f}{\mathrm{d}t}i+\frac{\mathrm{d}g}{\mathrm{d}t}j\)

求导法则
\(\begin{aligned} &\frac{d}{d t} C=0\\ &\frac{\mathrm{d}}{\mathrm{d} t}[\mathrm{u}(t)]=c \mathbf{u}^{\prime}(t)\\ &\frac{d}{d t}[\mathbf{u}(t)+\mathbf{v}(t)]=\mathbf{u}^{\prime}(t)+\mathbf{v}^{\prime}(t)\\ &\frac{d}{d t}[\mathbf{u}(t)-\mathbf{v}(t)]=\mathbf{u}^{\prime}(t)-\mathbf{v}^{\prime}(t)\\ &\frac{d}{d t}[\mathbf{u}(t) \cdot \mathbf{v}(t)]=\mathbf{u}^{\prime}(t) \cdot \mathbf{v}(t)+\mathbf{u}(t) \cdot \mathbf{v}^{\prime}(t)\\ &\frac{\mathrm{d}}{\mathrm{d} t}[\mathbf{u}(f(t))]=f^{\prime}(t) \mathbf{u}^{\prime}(f(t)) \end{aligned}\) Proof:

向量点积的求导法则

设\(u=u_1(t)i+u_2(t)j\),\(v=v_1(t)i+v_2(t)j\)
\(\begin{aligned} \frac{\mathrm{d}}{\mathrm{d} t}(\mathbf{u} \cdot \mathbf{v}) &=\frac{\mathrm{d}}{\mathrm{d} t}\left(u_{1} v_{1}+u_{2} v_{2}\right) \\ &=u_{1}^{\prime} v_{1}+u_{2}^{\prime} v_{2}+u_{1} v_{1}^{\prime}+u_{2} v_{2}^{\prime} \\ &=\mathbf{u}^{\prime} \cdot \mathbf{v}+\mathbf{u} \cdot \mathbf{v}^{\prime} \end{aligned}\)

不定积分 \(\int \mathbf{r}(t) \mathrm{d} t=\mathbf{R}(t)+\mathbf{C}\) 定积分 \(\int_{a}^{b} \mathbf{r}(t) \mathrm{d} t=\left(\int_{a}^{b} f(t) \mathrm{d} t\right) \mathbf{i}+\left(\int_{a}^{b} g(t) \mathrm{d} t\right) \mathbf{j}\)

抛射物体建模

抛物线

微分方程:\(\frac{d^2\vec{r}}{dt^2}=-g\vec{j}\)

初值条件:\(\vec{r}(0)=\vec{r_0}\) , \(\frac{d\vec{r}}{dt}=v_0\)

解得\(\vec{r}=(v_0cos\theta)t\vec{i}+((v_0sin\theta )t-\frac12gt^2)\vec{j}\)

带有线性阻力(空气阻力)的抛射运动

微分方程:\(\frac{d^2\vec{r}}{dt^2}=-g\vec{j}-k\frac{d\vec{r}}{dt}\)

初值条件:\(\vec{r}(0)=0\) , \(\frac{dr}{dt}\lvert _{t=0}=\vec{v_0}=(v_0cos\theta)\vec{i}+(v_0sin\theta)\vec{j}\)

导出方程:(解微分方程中,根据初值确定C分解运动为水平方向和竖直方向,分别解两个微分方程) \(x=\frac{v_0}{k}(1-\mathrm{e}^{-kt})cos\theta\\ y=\frac{v_0}{k}(1-\mathrm{e}^{-kt})sin\theta+\frac{g}{k^2}(1-kt-\mathrm{e}^{-kt})\)

以下暂无

空间向量

计算

向量长度,方向, 夹角

内积(点积)

向量积(叉积)

数量积(箱积)

空间中的直线和面

直线

平面

二次曲面