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微积分(2):微分方程->极坐标的微积分

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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

  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

  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

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

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

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

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

通过把极坐标写为参数形式导出