数学分析基础知识整理

基本求导公式
莱布尼茨求导公式
高阶导数
复合函数求导法则
隐函数求导法则
基本积分公式
- 万能代换
- 分部积分
级数定积分转换
变限积分函数求导
二重积分
三重积分
常用泰勒公式
重要幂级数展开式
无穷级数的和函数
傅里叶级数

基本求导公式

(C)′=0(C)^{\prime}=0(C)′=0
(xμ)′=μxμ−1\left(x^{\mu}\right)^{\prime}=\mu x^{\mu-1}(xμ)′=μxμ−1

(sin⁡x)′=cos⁡x(\sin x)^{\prime}=\cos x(sinx)′=cosx
(cos⁡x)′=−sin⁡x(\cos x)^{\prime}=-\sin x(cosx)′=−sinx
(tan⁡x)′=sec⁡2x(\tan x)^{\prime}=\sec ^{2} x(tanx)′=sec2x
(cot⁡x)′=−csc⁡2x(\cot x)^{\prime}=-\csc ^{2} x(cotx)′=−csc2x
(sec⁡x)′=sec⁡xtan⁡x(\sec x)^{\prime}=\sec x \tan x(secx)′=secxtanx
(csc⁡x)′=−csc⁡xcot⁡x(\csc x)^{\prime}=-\csc x \cot x(cscx)′=−cscxcotx

(arcsin⁡x)′=11−x2(arccos⁡x)′=−11−x2(arctan⁡x)′=11+x2(arccot⁡x)′=−11+x2\begin{aligned}(\arcsin x)^{\prime} &=\frac{1}{\sqrt{1-x^{2}}} \\(\arccos x)^{\prime} &=\frac{-1}{\sqrt{1-x^{2}}} \\(\arctan x)^{\prime} &=\frac{1}{1+x^{2}} \\(\operatorname{arccot} x)^{\prime} &=\frac{-1}{1+x^{2}} \end{aligned}(arcsinx)′(arccosx)′(arctanx)′(arccotx)′=1−x21=1−x2−1=1+x21=1+x2−1

(ax)′=axln⁡a\left(a^{x}\right)^{\prime}=a^{x} \ln a(ax)′=axlna
(ex)′=ex\left(e^{x}\right)^{\prime}=e^{x}(ex)′=ex
(log⁡ax)′=1xln⁡a\left(\log _{a} x\right)^{\prime}=\frac{1}{x \ln a}(logax)′=xlna1
(ln⁡x)′=1x(\ln x)^{\prime}=\frac{1}{x}(lnx)′=x1

(u±v)′=u′±v′(u \pm v)^{\prime}=u^{\prime} \pm v^{\prime}(u±v)′=u′±v′
(Cu)′=Cu′(C u)^{\prime}=C u^{\prime}(Cu)′=Cu′
(uv)′=u′v+uv′(u v)^{\prime}=u^{\prime} v+u v^{\prime}(uv)′=u′v+uv′
(uv)′=u′v−uv′v2\left(\frac{u}{v}\right)^{\prime}=\frac{u^{\prime} v-u v^{\prime}}{v^{2}}(vu)′=v2u′v−uv′

莱布尼茨求导公式

(u⋅v)′′=u′′⋅v+2u′⋅v′+u⋅v′′(u \cdot v)^{\prime \prime}=u^{\prime \prime} \cdot v+2 u^{\prime} \cdot v^{\prime}+u \cdot v^{\prime \prime}(u⋅v)′′=u′′⋅v+2u′⋅v′+u⋅v′′
(u⋅v)(n)=∑k=0nCnku(n−k)v(k)(u \cdot v)^{(n)}=\sum_{k=0}^{n} C_{n}^{k} u^{(n-k)} v^{(k)}(u⋅v)(n)=∑k=0nCnku(n−k)v(k)

高阶导数

(ax)(n)=(ln⁡a)n⋅ax\left(a^{x}\right)^{(n)}=(\ln a)^{n} \cdot a^{x}(ax)(n)=(lna)n⋅ax
(ex)(n)=ex\left(e^{x}\right)^{(n)}=e^{x}(ex)(n)=ex
(sin⁡kx)(n)=knsin⁡(nπ2+kx)(\sin k x)^{(n)}=k^{n} \sin \left(\frac{n \pi}{2}+k x\right)(sinkx)(n)=knsin(2nπ+kx)
(cos⁡kx)(n)=kncos⁡(nπ2+kx)(\cos k x)^{(n)}=k^{n} \cos \left(\frac{n \pi}{2}+k x\right)(coskx)(n)=kncos(2nπ+kx)
(ln⁡x)(n)=(−1)n−1(n−1)!xn(\ln x)^{(n)}=(-1)^{n-1} \frac{(n-1) !}{x^{n}}(lnx)(n)=(−1)n−1xn(n−1)!
(1x)(n)=(−1)nn!xn+1\left(\frac{1}{x}\right)^{(n)}=(-1)^{n} \frac{n !}{x^{n+1}}(x1)(n)=(−1)nxn+1n!
(1x+a)(n)=(−1)nn!(x+a)n+1\left(\frac{1}{x+a}\right)^{(n)}=(-1)^{n} \frac{n !}{(x+a)^{n+1}}(x+a1)(n)=(−1)n(x+a)n+1n!
(xm)(n)=m(m−1)⋯(m−n+1)xm−n\left(x^{m}\right)^{(n)}=m(m-1) \cdots(m-n+1) x^{m-n}(xm)(n)=m(m−1)⋯(m−n+1)xm−n
(u±v)(n)=u(n)±v(n)(u \pm v)^{(n)}=u^{(n)} \pm v^{(n)}(u±v)(n)=u(n)±v(n)

复合函数求导法则

设函数z=f(u,v)z=f(u, v)z=f(u,v)可微u=u(x,y),v=v(x,y)u=u(x, y), \quad v=v(x, y)u=u(x,y),v=v(x,y)具有一阶偏导数，并且它们可以构成，zzz关于(x,y)(x,y)(x,y)在某区域D内的复合函数，则在D内有复合函数求导法则：
∂z∂x=∂z∂u⋅∂u∂x+∂z∂v⋅∂v∂x\frac{\partial z}{\partial x}=\frac{\partial z}{\partial u} \cdot \frac{\partial u}{\partial x}+\frac{\partial z}{\partial v} \cdot \frac{\partial v}{\partial x}∂x∂z=∂u∂z⋅∂x∂u+∂v∂z⋅∂x∂v

∂z∂y=∂z∂u⋅∂u∂y+∂z∂v⋅∂v∂y\frac{\partial z}{\partial y}=\frac{\partial z}{\partial u} \cdot \frac{\partial u}{\partial y}+\frac{\partial z}{\partial v} \cdot \frac{\partial v}{\partial y}∂y∂z=∂u∂z⋅∂y∂u+∂v∂z⋅∂y∂v

隐函数求导法则

设函数 F(x,y,z)F(x, y, z)F(x,y,z) 在点 P0(x0,y0,z0)P_{0}\left(x_{0}, y_{0}, z_{0}\right)P0(x0,y0,z0) 的某邻域内有连续偏导数, 并且 F(x0,y0,z0)=0F\left(x_{0}, y_{0}, z_{0}\right)=0F(x0,y0,z0)=0, Fz′(x0,y0,z0)≠0,F_{z}^{\prime}\left(x_{0}, y_{0}, z_{0}\right) \neq 0,Fz′(x0,y0,z0)=0, 则方程 F(x0,y0,z0)=0F\left(x_{0}, y_{0}, z_{0}\right)=0F(x0,y0,z0)=0 在点 P0P_{0}P0 的某一邻域内恒能确定唯一的连续函数 z=f(x,y)z=f(x, y)z=f(x,y)满足:

z0=f(x0,y0)z_{0}=f\left(x_{0}, y_{0}\right)z0=f(x0,y0)
F(x,y,f(x,y))≡0F(x, y, f(x, y)) \equiv 0F(x,y,f(x,y))≡0
z=f(x,y)z=f(x, y)z=f(x,y) 具有连续偏导数
则:
∂z∂x=−Fx′(x,y,z)Fz′(x,y,z)\frac{\partial z}{\partial x}=-\frac{F_{x}^{\prime}(x, y, z)}{F_{z}^{\prime}(x, y, z)}∂x∂z=−Fz′(x,y,z)Fx′(x,y,z)
∂z∂y=−Fy′(x,y,z)Fz′(x,y,z)\frac{\partial z}{\partial y}=-\frac{F_{y}^{\prime}(x, y, z)}{F_{z}^{\prime}(x, y, z)}∂y∂z=−Fz′(x,y,z)Fy′(x,y,z)

基本积分公式

∫kdx=kx+C\int k d x=k x+C∫kdx=kx+C
∫xμdx=1μ+1⋅xμ+1+C(μ≠−1)\int x^{\mu} d x=\frac{1}{\mu+1} \cdot x^{\mu+1}+C(\mu \neq-1)∫xμdx=μ+11⋅xμ+1+C(μ=−1)
∫dxx=ln⁡∣x∣+C\int \frac{d x}{x}=\ln |x|+C∫xdx=ln∣x∣+C

∫cos⁡xdx=sin⁡x+C\int \cos x d x=\sin x+C∫cosxdx=sinx+C
∫sin⁡xdx=−cos⁡x+C\int \sin x d x=-\cos x+C∫sinxdx=−cosx+C
∫dxcos⁡2x=∫sec⁡2xdx=tan⁡x+C\int \frac{d x}{\cos ^{2} x}=\int \sec ^{2} x d x=\tan x+C∫cos2xdx=∫sec2xdx=tanx+C
∫dxsin⁡2x=∫csc⁡2xdx=−cot⁡x+C\int \frac{d x}{\sin ^{2} x}=\int \csc ^{2} x d x=-\cot x+C∫sin2xdx=∫csc2xdx=−cotx+C
∫sec⁡xtan⁡xdx=sec⁡x+C\int \sec x \tan x d x=\sec x+C∫secxtanxdx=secx+C
∫csc⁡xcot⁡xdx=−csc⁡x+C\int \csc x \cot x d x=-\csc x+C∫cscxcotxdx=−cscx+C

∫dx1+x2=arctan⁡x+C\int \frac{d x}{1+x^{2}}=\arctan x+C∫1+x2dx=arctanx+C
∫dx1−x2=arcsin⁡x+C\int \frac{d x}{\sqrt{1-x^{2}}}=\arcsin x+C∫1−x2dx=arcsinx+C
∫dxa2+x2=1aarctan⁡xa+C\int \frac{d x}{a^{2}+x^{2}}=\frac{1}{a} \arctan \frac{x}{a}+C∫a2+x2dx=a1arctanax+C
∫dxa2−x2=arcsin⁡xa+C\int \frac{d x}{\sqrt{a^{2}-x^{2}}}=\arcsin \frac{x}{a}+C∫a2−x2dx=arcsinax+C
∫dxx2−a2=12aln⁡∣x−ax+a∣+C\int \frac{d x}{x^{2}-a^{2}}=\frac{1}{2 a} \ln \left|\frac{x-a}{x+a}\right|+C∫x2−a2dx=2a1ln∣∣x+ax−a∣∣+C
∫dxa2−x2=12aln⁡∣x+ax−a∣+C\int \frac{d x}{a^{2}-x^{2}}=\frac{1}{2 a} \ln \left|\frac{x+a}{x-a}\right|+C∫a2−x2dx=2a1ln∣∣x−ax+a∣∣+C
∫dxx2+a2=ln⁡(x+x2+a2)+C\int \frac{d x}{\sqrt{x^{2}+a^{2}}}=\ln \left(x+\sqrt{x^{2}+a^{2}}\right)+C∫x2+a2dx=ln(x+x2+a2)+C
∫dxx2−a2=ln⁡∣x+x2−a2∣+C\int \frac{d x}{\sqrt{x^{2}-a^{2}}}=\ln \left|x+\sqrt{x^{2}-a^{2}}\right|+C∫x2−a2dx=ln∣∣x+x2−a2∣∣+C
∫x2−x2dx=−x2a2−x2+a22arcsin⁡xa+C\int \sqrt{x^{2}-x^{2}} d x=-\frac{x}{2} \sqrt{a^{2}-x^{2}}+\frac{a^{2}}{2} \arcsin \frac{x}{a}+C∫x2−x2dx=−2xa2−x2+2a2arcsinax+C

∫tan⁡xdx=∫sin⁡xcos⁡xdx=−∫dcos⁡xcos⁡x=−ln⁡∣cos⁡x∣+C\int \tan x d x=\int \frac{\sin x}{\cos x} d x=-\int \frac{d \cos x}{\cos x}=-\ln |\cos x|+C∫tanxdx=∫cosxsinxdx=−∫cosxdcosx=−ln∣cosx∣+C
∫cot⁡xdx=ln⁡∣sin⁡x∣+C\int \cot x d x=\ln |\sin x|+C∫cotxdx=ln∣sinx∣+C
∫csc⁡xdx=ln⁡∣tan⁡x2∣+C=ln⁡∣csc⁡x−cot⁡x∣+C\int \csc x d x=\ln \left|\tan \frac{x}{2}\right|+C=\ln |\csc x-\cot x|+C∫cscxdx=ln∣∣tan2x∣∣+C=ln∣cscx−cotx∣+C
∫sec⁡xdx=ln⁡∣sec⁡x+tan⁡x∣+C\int \sec x d x=\ln |\sec x+\tan x|+C∫secxdx=ln∣secx+tanx∣+C

∫exdx=ex+C\int e^{x} d x=e^{x}+C∫exdx=ex+C
∫axdx=1ln⁡aax+C\int a^{x} d x=\frac{1}{\ln a} a^{x}+C∫axdx=lna1ax+C

万能代换

u=tan⁡x2sin⁡x=2u1+u2cos⁡x=1−u21+u2dx=21+u2du\begin{aligned} u &=\tan \frac{x}{2} \\ \sin x &=\frac{2 u}{1+u^{2}} \\ \cos x &=\frac{1-u^{2}}{1+u^{2}} \\ d x &=\frac{2}{1+u^{2}} d u \end{aligned}usinxcosxdx=tan2x=1+u22u=1+u21−u2=1+u22du

万能代换可以将三角函数换成有理分式的形式来进行积分。

{a2−x2→x=asin⁡ta2+x2→x=atan⁡tx2−a2→x=asec⁡t\left\{\begin{array}{l}\sqrt{a^{2}-x^{2}} \rightarrow x=a \sin t \\ \sqrt{a^{2}+x^{2}} \rightarrow x=\operatorname{atan} t \\ \sqrt{x^{2}-a^{2}} \rightarrow x=\operatorname{asec} t\end{array}\right.⎩⎨⎧a2−x2→x=asinta2+x2→x=atantx2−a2→x=asect

分部积分

∫udv=uv−∫vdu\int u d v=u v-\int v d u∫udv=uv−∫vdu

级数定积分转换

{∑i=1∞f(in)1n=∫01f(t)dt∑i=1∞f(x⋅in)xn=∫0xf(t)dt∑i=1∞f(a+(b−a)⋅in)b−an=∫abf(t)dt\left\{\begin{array}{l}\sum_{i=1}^{\infty} f\left(\frac{i}{n}\right) \frac{1}{n}=\int_{0}^{1} f(t) d t \\ \sum_{i=1}^{\infty} f\left(\frac{x \cdot i}{n}\right) \frac{x}{n}=\int_{0}^{x} f(t) d t \\ \sum_{i=1}^{\infty} f\left(a+\frac{(b-a) \cdot i}{n}\right) \frac{b-a}{n}=\int_{a}^{b} f(t) d t\end{array}\right.⎩⎪⎨⎪⎧∑i=1∞f(ni)n1=∫01f(t)dt∑i=1∞f(nx⋅i)nx=∫0xf(t)dt∑i=1∞f(a+n(b−a)⋅i)nb−a=∫abf(t)dt

变限积分函数求导

F(x)=∫φ1(x)φ2(x)f(t)dtF(x)=\int_{\varphi_{1}(x)}^{\varphi_{2}(x)} f(t) d tF(x)=∫φ1(x)φ2(x)f(t)dt
F′(x)=ddx[∫φ1(x)φ2(x)f(t)dt]=f[φ2(x)]φ2′(x)−f[φ1(x)]φ1′(x)F^{\prime}(x)=\frac{d}{d x}\left[\int_{\varphi_{1}(x)}^{\varphi_{2}(x)} f(t) d t\right]=f\left[\varphi_{2}(x)\right] \varphi_{2}^{\prime}(x)-f\left[\varphi_{1}(x)\right] \varphi_{1}^{\prime}(x)F′(x)=dxd[∫φ1(x)φ2(x)f(t)dt]=f[φ2(x)]φ2′(x)−f[φ1(x)]φ1′(x)

二重积分

二重积分可以化为累次积分的定积分，表示曲顶柱体的体积。

∬Df(x,y)dσ\iint_{D} f(x, y) d \sigma∬Df(x,y)dσ

=∫axbxdx∫aybyf(x,y)dy\stackrel{}{=} \int_{a_{x}}^{b_{x}} d x \int_{a_{y}}^{b_{y}} f(x, y) d y=∫axbxdx∫aybyf(x,y)dy

=∫aθbθdθ∫arbrf(rcos⁡θ,rsin⁡θ)rdr=\int_{a_{\theta}}^{b_{\theta}} d \theta \int_{a_{r}}^{b_{r}} f(r \cos \theta, r \sin \theta) r d r=∫aθbθdθ∫arbrf(rcosθ,rsinθ)rdr
∑i=1∞∑j=1∞f(in,jn)1n2=∫01∫01f(x,y)dxdy\sum_{i=1}^{\infty} \sum_{j=1}^{\infty} f\left(\frac{i}{n}, \frac{j}{n}\right) \frac{1}{n^{2}}=\int_{0}^{1} \int_{0}^{1} f(x, y) d x d y∑i=1∞∑j=1∞f(ni,nj)n21=∫01∫01f(x,y)dxdy

三重积分

三重积分可以化为累次积分的定积分，表示空间区域以密度ρ=f(x,y,z)\rho=f(x, y, z)ρ=f(x,y,z)的质量。
∭Ωf(x,y,z)dv=∫axbxdx∫aybydy∫azbzf(x,y,z)dz=∫azbzdz∬Df(x,y,z)dσ=∭Ωf(rcos⁡θ,rsin⁡θ,z)rdrdθ=∭Ωf(rsin⁡φcos⁡θ,rsin⁡φsin⁡θ,rcos⁡φ)⋅r2sin⁡φdrdφdθ\begin{aligned} & \iiint_{\Omega} f(x, y, z) d v \\=& \int_{a_{x}}^{b_{x}} d x \int_{a_{y}}^{b_{y}} d y \int_{a_{z}}^{b_{z}} f(x, y, z) d z \\=& \int_{a_{z}}^{b_{z}} d z \iint_{D} f(x, y, z) d \sigma \\=& \iiint_{\Omega} f(r \cos \theta, r \sin \theta, z) r d r d \theta \\=& \iiint_{\Omega} f(r \sin \varphi \cos \theta, r \sin \varphi \sin \theta, r \cos \varphi) \cdot r^{2} \sin \varphi d r d \varphi d \theta \end{aligned}====∭Ωf(x,y,z)dv∫axbxdx∫aybydy∫azbzf(x,y,z)dz∫azbzdz∬Df(x,y,z)dσ∭Ωf(rcosθ,rsinθ,z)rdrdθ∭Ωf(rsinφcosθ,rsinφsinθ,rcosφ)⋅r2sinφdrdφdθ

常用泰勒公式

f(x)=f(x0)+f′(x0)(x−x0)+f′′(x0)2!(x−x0)2+⋯+f(n)(x0)n!(x−x0)n+Rn(x)f(x)=f\left(x_{0}\right)+f^{\prime}\left(x_{0}\right)\left(x-x_{0}\right)+\frac{f^{\prime \prime}\left(x_{0}\right)}{2 !}\left(x-x_{0}\right)^{2}+\cdots+\frac{f^{(n)}\left(x_{0}\right)}{n !}\left(x-x_{0}\right)^{n}+R_{n}(x)f(x)=f(x0)+f′(x0)(x−x0)+2!f′′(x0)(x−x0)2+⋯+n!f(n)(x0)(x−x0)n+Rn(x)
Rn(x)=f(n+1)(ξ)(n+1)!(x−x0)(n+1)ξ∈(x,x0)R_{n}(x)=\frac{f^{(n+1)}(\xi)}{(n+1) !}\left(x-x_{0}\right)^{(n+1)} \xi \in\left(x, x_{0}\right)Rn(x)=(n+1)!f(n+1)(ξ)(x−x0)(n+1)ξ∈(x,x0)
Rn(x)=o[(x−x0)n]R_{n}(x)=o\left[\left(x-x_{0}\right)^{n}\right]Rn(x)=o[(x−x0)n]
sin⁡x=x−x33!+x55!+⋯+(−1)nx2n+1(2n+1)!+o(x2n+1)\sin x=x-\frac{x^{3}}{3 !}+\frac{x^{5}}{5 !}+\cdots+(-1)^{n} \frac{x^{2 n+1}}{(2 n+1) !}+o\left(x^{2 n+1}\right)sinx=x−3!x3+5!x5+⋯+(−1)n(2n+1)!x2n+1+o(x2n+1)
cos⁡x=1−x22!+x44!+⋯+(−1)nx2n(2n)!+o(x2n)\cos x=1-\frac{x^{2}}{2 !}+\frac{x^{4}}{4 !}+\cdots+(-1)^{n} \frac{x^{2 n}}{(2 n) !}+o\left(x^{2 n}\right)cosx=1−2!x2+4!x4+⋯+(−1)n(2n)!x2n+o(x2n)
ex=1+x+x22!+x33!+⋯+xnn!+o(xn)e^{x}=1+x+\frac{x^{2}}{2 !}+\frac{x^{3}}{3 !}+\cdots+\frac{x^{n}}{n !}+o\left(x^{n}\right)ex=1+x+2!x2+3!x3+⋯+n!xn+o(xn)
ln⁡(1+x)=x−x22+x33+⋯+(−1)n−1xnn+o(xn)\ln (1+x)=x-\frac{x^{2}}{2}+\frac{x^{3}}{3}+\cdots+(-1)^{n-1} \frac{x^{n}}{n}+o\left(x^{n}\right)ln(1+x)=x−2x2+3x3+⋯+(−1)n−1nxn+o(xn)
tan⁡x=x+x33+o(x3)\tan x=x+\frac{x^{3}}{3}+o\left(x^{3}\right)tanx=x+3x3+o(x3)
arcsin⁡x=x+x33!+o(x3)\arcsin x=x+\frac{x^{3}}{3 !}+o\left(x^{3}\right)arcsinx=x+3!x3+o(x3)
arctan⁡x=x−x33+o(x3)\arctan x=x-\frac{x^{3}}{3}+o\left(x^{3}\right)arctanx=x−3x3+o(x3)
(1+x)a=1+ax+a(a−1)2!x2+o(x2)(1+x)^{a}=1+a x+\frac{a(a-1)}{2 !} x^{2}+o\left(x^{2}\right)(1+x)a=1+ax+2!a(a−1)x2+o(x2)

重要幂级数展开式

ex=∑n=0∞xnn!=1+x+x22!+x33!+⋯+xnn!+⋯,−∞<x<+∞e^{x}=\sum_{n=0}^{\infty} \frac{x^{n}}{n !}=1+x+\frac{x^{2}}{2 !}+\frac{x^{3}}{3 !}+\cdots+\frac{x^{n}}{n !}+\cdots,-\infty<x<+\inftyex=∑n=0∞n!xn=1+x+2!x2+3!x3+⋯+n!xn+⋯,−∞<x<+∞
11+x=∑n=0∞(−1)nxn=1−x+x2−⋯+(−1)nxn+⋯,−1<x<1\frac{1}{1+x}=\sum_{n=0}^{\infty}(-1)^{n} x^{n}=1-x+x^{2}-\cdots+(-1)^{n} x^{n}+\cdots,-1<x<11+x1=∑n=0∞(−1)nxn=1−x+x2−⋯+(−1)nxn+⋯,−1<x<1
11−x=∑n=0∞xn=1+x+x2+⋯+xn+⋯,−1<x<1\frac{1}{1-x}=\sum_{n=0}^{\infty} x^{n}=1+x+x^{2}+\cdots+x^{n}+\cdots,-1<x<11−x1=∑n=0∞xn=1+x+x2+⋯+xn+⋯,−1<x<1
ln⁡(1+x)=∑n=1∞(−1)n−1xnn=x−x22+⋯+(−1)n−1xnn+⋯,−1<x⩽1\ln (1+x)=\sum_{n=1}^{\infty}(-1)^{n-1} \frac{x^{n}}{n}=x-\frac{x^{2}}{2}+\cdots+(-1)^{n-1} \frac{x^{n}}{n}+\cdots,-1<x \leqslant 1ln(1+x)=∑n=1∞(−1)n−1nxn=x−2x2+⋯+(−1)n−1nxn+⋯,−1<x⩽1
sin⁡x=∑n=0∞(−1)nx2n+1(2n+1)!=x−x33!+⋯+(−1)nx2n+1(2n+1)!+⋯,−∞<x<+∞\sin x=\sum_{n=0}^{\infty}(-1)^{n} \frac{x^{2 n+1}}{(2 n+1) !}=x-\frac{x^{3}}{3 !}+\cdots+(-1)^{n} \frac{x^{2 n+1}}{(2 n+1) !}+\cdots,-\infty<x<+\inftysinx=∑n=0∞(−1)n(2n+1)!x2n+1=x−3!x3+⋯+(−1)n(2n+1)!x2n+1+⋯,−∞<x<+∞
cos⁡x=∑n=0∞(−1)nx2n(2n)!=1−x22!+⋯+(−1)nx2n(2n)!+⋯,−∞<x<+∞\cos x=\sum_{n=0}^{\infty}(-1)^{n} \frac{x^{2 n}}{(2 n) !}=1-\frac{x^{2}}{2 !}+\cdots+(-1)^{n} \frac{x^{2 n}}{(2 n) !}+\cdots,-\infty<x<+\inftycosx=∑n=0∞(−1)n(2n)!x2n=1−2!x2+⋯+(−1)n(2n)!x2n+⋯,−∞<x<+∞

无穷级数的和函数

{∑n=1∞nxn−1=1(1−x)2(−1<x<1)∑n=1∞1nxn=−ln⁡(1−x)(−1⩽x<1)\left\{\begin{array}{ll}\sum_{n=1}^{\infty} n x^{n-1}=\frac{1}{(1-x)^{2}} & (-1<x<1) \\ \sum_{n=1}^{\infty} \frac{1}{n} x^{n}=-\ln (1-x) & (-1 \leqslant x<1)\end{array}\right.{∑n=1∞nxn−1=(1−x)21∑n=1∞n1xn=−ln(1−x)(−1<x<1)(−1⩽x<1)

傅里叶级数

f(x)=a02+∑n=1∞(ancos⁡nπxl+bnsin⁡nπxl)a0=1l∫−llf(x)dxan=1l∫−llf(x)cos⁡nπxldxbn=1l∫−llf(x)sin⁡nπxldx\begin{aligned} f(x) &=\frac{a_{0}}{2}+\sum_{n=1}^{\infty}\left(a_{n} \cos \frac{n \pi x}{l}+b_{n} \sin \frac{n \pi x}{l}\right) \\ a_{0} &=\frac{1}{l} \int_{-l}^{l} f(x) 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{aligned}f(x)a0anbn=2a0+n=1∑∞(ancoslnπx+bnsinlnπx)=l1∫−llf(x)dx=l1∫−llf(x)coslnπxdx=l1∫−llf(x)sinlnπxdx