文章目录
AM@常见函数的幂级数(series)展开@泰勒级数TaylorSeriesref几何级数🎈二项式级数🎈指数函数和自然对数🎈三角函数🎈常用三角双曲函数🎈朗伯W函数🎈多元函数的展开🎈幂级数小结特点AM@常见函数的幂级数(series)展开@泰勒级数TaylorSeries
泰勒级数是一种用无限项连加式来表示一个函数的方法，这些相加的项由函数在某一点的导数求得。泰勒级数可以用多项式来近似函数，使得多项式的表达比函数的形式更加友好ref
Power series - Wikipedia幂级数 - 维基百科，自由的百科全书 ()Taylor series - Wikipedia泰勒级数 - 维基百科，自由的百科全书 ()几何级数🎈
11−x=∑n=0∞xn=1+x+x2+⋯+xn+⋯∀x:∣x∣<1{\displaystyle {\frac {1}{1-x}}=\sum _{n=0}^{\infty }x^{n}=1+x+x^{2}+\cdots +x^{n}+\cdots \quad \forall x:\left|x\right|<1}1−x1=n=0∑∞xn=1+x+x2+⋯+xn+⋯∀x:∣x∣<1二项式级数🎈
(1+x)α=∑n=0∞(αn)xn=1+αx+α(α−1)2!x2+⋯+α(α−1)⋯(α−n+1)n!xn+⋯{\displaystyle (1+x)^{\alpha } =\sum _{n=0}^{\infty }{\binom {\alpha }{n}}x^{n} =1+\alpha x+{\frac {\alpha (\alpha -1)}{2!}}x^{2}+\cdots +{\frac {\alpha (\alpha -1)\cdots (\alpha -n+1)}{n!}}x^{n}+\cdots } (1+x)α=n=0∑∞(nα)xn=1+αx+2!α(α−1)x2+⋯+n!α(α−1)⋯(α−n+1)xn+⋯
∀x:∣x∣<1,∀α∈C{\displaystyle \forall x:\left|x\right|<1,\forall \alpha \in \mathbb {C} }∀x:∣x∣<1,∀α∈C二项式系数(αn)=∏k=1nα−k+1k=α(α−1)⋯(α−n+1)n!{\displaystyle {\binom {\alpha }{n}}=\prod _{k=1}^{n}{\frac {\alpha -k+1}{k}}={\frac {\alpha (\alpha -1)\cdots (\alpha -n+1)}{n!}}}(nα)=k=1∏nkα−k+1=n!α(α−1)⋯(α−n+1)
指数函数和自然对数🎈
以eee为底数的指数函数的麦克劳林序列是ex=∑n=0∞xnn!=1+x+x22!+x33!+⋯+xnn!+⋯∀x{\displaystyle 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 \quad \forall x}ex=n=0∑∞n!xn=1+x+2!x2+3!x3+⋯+n!xn+⋯∀x （对所有X都成立）ln(1−x)=−∑n=1∞xnn=−x−x22−x33−⋯−xnn−⋯∀x∈[−1,1){\displaystyle \ln(1-x)=-\sum _{n=1}^{\infty }{\frac {x^{n}}{n}}=-x-{\frac {x^{2}}{2}}-{\frac {x^{3}}{3}}-\cdots -{\frac {x^{n}}{n}}-\cdots \quad \forall x\in [-1,1)}ln(1−x)=−n=1∑∞nxn=−x−2x2−3x3−⋯−nxn−⋯∀x∈[−1,1) 对于在区间[-1,1)内所有的X都成立 ln(1+x)=∑n=1∞(−1)n+1nxn=x−x22+x33−x44+x55−⋯+(−1)n+1nxn+⋯∀x∈(−1,1]{\displaystyle \ln(1+x)=\sum _{n=1}^{\infty }{\frac {(-1)^{n+1}}{n}}x^{n}=x-{\frac {x^{2}}{2}}+{\frac {x^{3}}{3}}-\frac{x^4}{4}+\frac{x^5}{5}-\cdots +{\frac {(-1)^{n+1}}{n}}x^{n}+\cdots \quad \forall x\in (-1,1]}ln(1+x)=n=1∑∞n(−1)n+1xn=x−2x2+3x3−4x4+5x5−⋯+n(−1)n+1xn+⋯∀x∈(−1,1] 对于在区间(-1,1]内所有的X都成立三角函数🎈
常用的三角函数可以被展开为以下的麦克劳林序列：
sinx=∑n=0∞(−1)n(2n+1)!x2n+1=x−x33!+x55!−⋯∀xcosx=∑n=0∞(−1)n(2n)!x2n=1−x22!+x44!−⋯∀xtanx=∑n=1∞B2n(−4)n(1−4n)(2n)!x2n−1=x+x33+2x515+⋯∀x:∣x∣<π2secx=∑n=0∞(−1)nE2n(2n)!x2n=1+x22+5x424+⋯∀x:∣x∣<π2arcsinx=∑n=0∞(2n)!4n(n!)2(2n+1)x2n+1=x+x36+3x540+⋯∀x:∣x∣≤1arccosx=π2−arcsinx=π2−∑n=0∞(2n)!4n(n!)2(2n+1)x2n+1=π2−x−x36−3x540+⋯∀x:∣x∣≤1arctanx=∑n=0∞(−1)n2n+1x2n+1=x−x33+x55−⋯∀x:∣x∣≤1,x≠±i{\displaystyle {\begin{aligned}\sin x&=\sum _{n=0}^{\infty }{\frac {(-1)^{n}}{(2n+1)!}}x^{2n+1}&&=x-{\frac {x^{3}}{3!}}+{\frac {x^{5}}{5!}}-\cdots &&\forall x\\[6pt] \cos x&=\sum _{n=0}^{\infty }{\frac {(-1)^{n}}{(2n)!}}x^{2n}&&=1-{\frac {x^{2}}{2!}}+{\frac {x^{4}}{4!}}-\cdots &&\forall x\\[6pt] \tan x&=\sum _{n=1}^{\infty }{\frac {B_{2n}(-4)^{n}\left(1-4^{n}\right)}{(2n)!}}x^{2n-1}&&=x+{\frac {x^{3}}{3}}+{\frac {2x^{5}}{15}}+\cdots &&\forall x:|x|<{\frac {\pi }{2}}\\[6pt] \sec x&=\sum _{n=0}^{\infty }{\frac {(-1)^{n}E_{2n}}{(2n)!}}x^{2n}&&=1+{\frac {x^{2}}{2}}+{\frac {5x^{4}}{24}}+\cdots &&\forall x:|x|<{\frac {\pi }{2}}\\[6pt] \arcsin x&=\sum _{n=0}^{\infty }{\frac {(2n)!}{4^{n}(n!)^{2}(2n+1)}}x^{2n+1}&&=x+{\frac {x^{3}}{6}}+{\frac {3x^{5}}{40}}+\cdots &&\forall x:|x|\leq 1\\[6pt] \arccos x&={\frac {\pi }{2}}-\arcsin x\\ &={\frac {\pi }{2}}-\sum _{n=0}^{\infty }{\frac {(2n)!}{4^{n}(n!)^{2}(2n+1)}}x^{2n+1}&&={\frac {\pi }{2}}-x-{\frac {x^{3}}{6}}-{\frac {3x^{5}}{40}}+\cdots &&\forall x:|x|\leq 1\\[6pt] \arctan x&=\sum _{n=0}^{\infty }{\frac {(-1)^{n}}{2n+1}}x^{2n+1}&&=x-{\frac {x^{3}}{3}}+{\frac {x^{5}}{5}}-\cdots &&\forall x:|x|\leq 1,\ x\neq \pm i\end{aligned}} } sinxcosxtanxsecxarcsinxarccosxarctanx=n=0∑∞(2n+1)!(−1)nx2n+1=n=0∑∞(2n)!(−1)nx2n=n=1∑∞(2n)!B2n(−4)n(1−4n)x2n−1=n=0∑∞(2n)!(−1)nE2nx2n=n=0∑∞4n(n!)2(2n+1)(2n)!x2n+1=2π−arcsinx=2π−n=0∑∞4n(n!)2(2n+1)(2n)!x2n+1=n=0∑∞2n+1(−1)nx2n+1=x−3!x3+5!x5−⋯=1−2!x2+4!x4−⋯=x+3x3+152x5+⋯=1+2x2+245x4+⋯=x+6x3+403x5+⋯=2π−x−6x3−403x5+⋯=x−3x3+5x5−⋯∀x∀x∀x:∣x∣<2π∀x:∣x∣<2π∀x:∣x∣≤1∀x:∣x∣≤1∀x:∣x∣≤1,x=±i
在tan(x){\displaystyle \tan(x)}tan(x)展开式中的BkB_kBk是伯努利数。
在sec(x){\displaystyle \sec(x)}sec(x)展开式中的EkE_kEk是欧拉数。
常用三角
sinx=∑n=0∞(−1)nx2n+12n+1=∑n=0∞(−1)nxtt=x−x33!+x55!−x77!+x99!⋯t=1,3,5,7,9⋯\sin{x}=\sum\limits_{n=0}^{\infin}(-1)^n\frac{x^{2n+1}}{2n+1} =\sum\limits_{n=0}^{\infin}(-1)^n\frac{x^t}{t} =x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\frac{x^9}{9!}\cdots \\ \\ t=1,3,5,7,9\cdots sinx=n=0∑∞(−1)n2n+1x2n+1=n=0∑∞(−1)ntxt=x−3!x3+5!x5−7!x7+9!x9⋯t=1,3,5,7,9⋯
cosx=∑n=0∞(−1)nx2n2n=∑n=0∞(−1)nxtt=1−x22!+x44!−x66!+x88!⋯t=0,2,4,6,8⋯\cos{x}=\sum\limits_{n=0}^{\infin}(-1)^n\frac{x^{2n}}{2n} =\sum\limits_{n=0}^{\infin}(-1)^n\frac{x^t}{t} =1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+\frac{x^8}{8!}\cdots \\ \\ t=0,2,4,6,8\cdots cosx=n=0∑∞(−1)n2nx2n=n=0∑∞(−1)ntxt=1−2!x2+4!x4−6!x6+8!x8⋯t=0,2,4,6,8⋯
双曲函数🎈
sinhx=∑n=0∞1(2n+1)!x2n+1∀x\sinh x=\sum _{n=0}^{\infty }{\frac {1}{(2n+1)!}}x^{2n+1}\quad \forall xsinhx=∑n=0∞(2n+1)!1x2n+1∀x
coshx=∑n=0∞1(2n)!x2n∀x\cosh x=\sum _{n=0}^{\infty }{\frac {1}{(2n)!}}x^{2n}\quad \forall xcoshx=∑n=0∞(2n)!1x2n∀x
tanhx=∑n=1∞B2n4n(4n−1)(2n)!x2n−1∀x:∣x∣<π2\tanh x=\sum _{n=1}^{\infty }{\frac {B_{2n}4^{n}(4^{n}-1)}{(2n)!}}x^{2n-1}\quad \forall x:\left|x\right|<{\frac {\pi }{2}}tanhx=∑n=1∞(2n)!B2n4n(4n−1)x2n−1∀x:∣x∣<2π
sinh−1x=∑n=0∞(−1)n(2n)!4n(n!)2(2n+1)x2n+1∀x:∣x∣<1\sinh ^{-1}x=\sum _{n=0}^{\infty }{\frac {(-1)^{n}(2n)!}{4^{n}(n!)^{2}(2n+1)}}x^{2n+1}\quad \forall x:\left|x\right|<1sinh−1x=∑n=0∞4n(n!)2(2n+1)(−1)n(2n)!x2n+1∀x:∣x∣<1
tanh−1x=∑n=0∞12n+1x2n+1∀x:∣x∣<1\tanh ^{-1}x=\sum _{n=0}^{\infty }{\frac {1}{2n+1}}x^{2n+1}\quad \forall x:\left|x\right|<1tanh−1x=∑n=0∞2n+11x2n+1∀x:∣x∣<1
tanh(x){\displaystyle \tanh(x)}tanh(x)展开式中的BkB_kBk是伯努利数。
朗伯W函数🎈
W0(x)=∑n=1∞(−n)n−1n!xn∀x:∣x∣<1eW_{0}(x)=\sum _{n=1}^{\infty }{\frac {(-n)^{n-1}}{n!}}x^{n}\quad \forall x:\left|x\right|<{\frac {1}{e}}W0(x)=∑n=1∞n!(−n)n−1xn∀x:∣x∣<e1多元函数的展开🎈
泰勒级数可以推广到有多个变量的函数：
∑n1=0∞⋯∑nd=0∞∂n1+⋯+nd∂x1n1⋯∂xdndf(a1,⋯,ad)n1!⋯nd!(x1−a1)n1⋯(xd−ad)nd\displaystyle\Large\sum _{n_{1}=0}^{\infty }\cdots \sum _{n_{d}=0}^{\infty }{\frac {\partial ^{n_{1}+\cdots +n_{d}}}{\partial x_{1}^{n_{1}}\cdots \partial x_{d}^{n_{d}}}}{\frac {f(a_{1},\cdots ,a_{d})}{n_{1}!\cdots n_{d}!}}(x_{1}-a_{1})^{n_{1}}\cdots (x_{d}-a_{d})^{n_{d}} n1=0∑∞⋯nd=0∑∞∂x1n1⋯∂xdnd∂n1+⋯+ndn1!⋯nd!f(a1,⋯,ad)(x1−a1)n1⋯(xd−ad)nd
幂级数小结
常见函数的幂级数展开🎈运用这些展开可以得到一些重要的恒等式。 ∀x∈C,ex=∑n=0+∞xnn!.\forall x\in {\mathbb {C}},\,e^{x}=\sum _{{n=0}}^{{+{\infty }}}{{\frac {x^{n}}{n!}}}.∀x∈C,ex=∑n=0+∞n!xn.∀x∈R,cosx=∑n=0+∞(−1)nx2n(2n)!.\forall x\in {\mathbb {R}},\,\cos x=\sum _{{n=0}}^{{+{\infty }}}(-1)^{n}\,{{\frac {x^{{2\,n}}}{(2\,n)!}}}.∀x∈R,cosx=∑n=0+∞(−1)n(2n)!x2n.∀x∈R,sinx=∑n=0+∞(−1)nx2n+1(2n+1)!.\forall x\in {\mathbb {R}},\,\sin x=\sum _{{n=0}}^{{+{\infty }}}(-1)^{n}\,{{\frac {x^{{2\,n+1}}}{(2\,n+1)!}}}.∀x∈R,sinx=∑n=0+∞(−1)n(2n+1)!x2n+1.∀x∈R,chx=∑n=0+∞x2n(2n)!.\forall x\in {\mathbb {R}},\,\operatorname {ch}\,x=\sum _{{n=0}}^{{+{\infty }}}{{\frac {x^{{2\,n}}}{(2\,n)!}}}.∀x∈R,chx=∑n=0+∞(2n)!x2n.∀x∈R,shx=∑n=0+∞x2n+1(2n+1)!.\forall x\in {\mathbb {R}},\,\operatorname {sh}\,x=\sum _{{n=0}}^{{+{\infty }}}{{\frac {x^{{2\,n+1}}}{(2\,n+1)!}}}.∀x∈R,shx=∑n=0+∞(2n+1)!x2n+1.∀x∈D(0,1),11−x=∑n=0+∞xn.\forall x\in D(0,1),\,{1 \over {1-x}}=\sum _{{n=0}}^{{+{\infty }}}{x^{n}}.∀x∈D(0,1),1−x1=∑n=0+∞xn.∀x∈(−1,1],ln(1+x)=∑n=1+∞(−1)n+1xnn.)\forall x\in (-1,1],\,\ln(1+x)=\sum _{{n=1}}^{{+{\infty }}}(-1)^{{n+1}}{x^{{n}} \over {n}}.)∀x∈(−1,1],ln(1+x)=∑n=1+∞(−1)n+1nxn.)∀x∈[−1,1],arctanx=∑n=0+∞(−1)nx2n+12n+1\forall x\in [-1,1],\,\arctan \,x=\sum _{{n=0}}^{{+{\infty }}}(-1)^{n}\,{{\frac {x^{{2\,n+1}}}{2\,n+1}}}\;∀x∈[−1,1],arctanx=∑n=0+∞(−1)n2n+1x2n+1，特别地，π=4∑n=0+∞(−1)n2n+1\pi =4\,\sum _{{n=0}}^{{+{\infty }}}{{\frac {(-1)^{{n}}}{2\,n+1}}}π=4∑n=0+∞2n+1(−1)n。∀x∈(−1,1),∀α∉N,(1+x)α=1+∑n=1+∞α(α−1)⋯(α−n+1)n!xn.\forall x\in \,(-1,1),\ \forall \alpha \,\not \in \,{\mathbb {N}},\,(1+x)^{\alpha }\,=1\;+\;\sum _{{n=1}}^{{+{\infty }}}{{\frac {\alpha \,(\alpha -1)\cdots (\alpha -n+1)}{n!}}\,x^{n}}.∀x∈(−1,1),∀α∈N,(1+x)α=1+∑n=1+∞n!α(α−1)⋯(α−n+1)xn.∀x∈R,∀α∈N,(1+x)α=1+∑n=1+∞α(α−1)⋯(α−n+1)n!xn=∑n=0α(αn)xn.\forall x\in {\mathbb {R}},\,\forall \alpha \,\in \,{\mathbb {N}},\,(1+x)^{\alpha }\,=1\;+\;\sum _{{n=1}}^{{+{\infty }}}{{\frac {\alpha \,(\alpha -1)\cdots (\alpha -n+1)}{n!}}\,x^{n}}=\sum _{{n=0}}^{{\alpha }}{{\alpha \choose n}\,x^{n}}.∀x∈R,∀α∈N,(1+x)α=1+∑n=1+∞n!α(α−1)⋯(α−n+1)xn=∑n=0α(nα)xn.∀x∈(−1,1),artanhx=∑n=0+∞x2n+12n+1.\forall x\in (-1,1),\,\operatorname {artanh}\,x=\sum _{{n=0}}^{{+{\infty }}}\,{{\frac {x^{{2\,n+1}}}{2\,n+1}}}.∀x∈(−1,1),artanhx=∑n=0+∞2n+1x2n+1.∀x∈(−1,1),arcsinx=x+∑n=1+∞(∏k=1n(2k−1)∏k=1n2k)x2n+12n+1\forall x\in (-1,1),\,\arcsin \,x=x+\sum _{{n=1}}^{{+{\infty }}}\,\left({{\frac {\prod _{{k=1}}^{{n}}\,(2\,k-1)}{\prod _{{k=1}}^{{n}}\,2\,k}}}\right){{\frac {x^{{2\,n+1}}}{2\,n+1}}}∀x∈(−1,1),arcsinx=x+∑n=1+∞(∏k=1n2k∏k=1n(2k−1))2n+1x2n+1∀x∈(−1,1),arsinhx=x+∑n=0+∞(−1)n(∏k=1n(2k−1)∏k=1n2k)x2n+12n+1\forall x\in (-1,1),\,\operatorname {arsinh}\,x=x+\sum _{{n=0}}^{{+{\infty }}}\,(-1)^{n}\,\left({{\frac {\prod _{{k=1}}^{{n}}\,(2\,k-1)}{\prod _{{k=1}}^{{n}}\,2\,k}}}\right){{\frac {x^{{2\,n+1}}}{2\,n+1}}}∀x∈(−1,1),arsinhx=x+∑n=0+∞(−1)n(∏k=1n2k∏k=1n(2k−1))2n+1x2n+1∀x∈(−π2,π2),tanx=2π∑n=0+∞(xπ)2n+1(22n+2−1)ζ(2n+2)\forall x\in \,\left(-{\frac {\pi }{2}},{\frac {\pi }{2}}\right),\ \tan x={\frac {2}{\pi }}\,\sum _{{n=0}}^{{+{\infty }}}\,{\left({{\frac {x}{\pi }}}\right)}^{{2\,n+1}}(2^{{2\,n+2}}-1)\;\zeta (2\,n+2)∀x∈(−2π,2π),tanx=π2∑n=0+∞(πx)2n+1(22n+2−1)ζ(2n+2)，其中∀p>1,ζ(p)=∑n=1+∞1np\forall p>1,\,\zeta (p)=\sum _{{n=1}}^{{+{\infty }}}\,{\frac {1}{n^{p}}}∀p>1,ζ(p)=∑n=1+∞np1特点
三角函数的幂级数展开公式的累加下限大多从n=0n=0n=0开始计算
注意到两个公差d=2d=2d=2的数列:(n=0,1,2,...n=0,1,2,...n=0,1,2,...)借助这几个序列,我们可以快速地准确地流水地写出幂级数展开式😁😎☆*: .｡. o(≧▽≦)o .｡.:*☆
{p=2n=0,2,4,6,...q=2n+1=1,3,5,7,...\begin{cases} {p=2n}=0,2,4,6,... \\ {q=2n+1}=1,3,5,7,... \end{cases} {p=2n=0,2,4,6,...q=2n+1=1,3,5,7,...
交错符号sg(n);
sg=(−1)n=1,−1,1,−1,...sg=(-1)^n=1,-1,1,-1,...sg=(−1)n=1,−1,1,−1,...
两个交错级数可以写成
cosx=∑n=0∞(−1)np!⋅xpcosx=\sum\limits_{n=0}^{\infin}\frac{(-1)^{n}}{p!}\cdot x^{p}cosx=n=0∑∞p!(−1)n⋅xp
sinx=∑n=0∞(−1)nq!⋅xqsinx=\sum\limits_{n=0}^{\infin}\frac{(-1)^{n}}{q!}\cdot x^{q}sinx=n=0∑∞q!(−1)n⋅xq
进一步,可以抽象出T(t)=∑n=0∞(−1)nt!⋅xt,cosx=T(p)=T(2n),sinx=T(q)=T(2n+1)进一步,可以抽象出T(t)=\sum\limits_{n=0}^{\infin}\frac{(-1)^{n}}{t!}\cdot x^{t} ,cosx=T(p)=T(2n),sinx=T(q)=T(2n+1)进一步,可以抽象出T(t)=n=0∑∞t!(−1)n⋅xt,cosx=T(p)=T(2n),sinx=T(q)=T(2n+1)
最后,流水的写出展开式的各项的因子:最后,流水的写出展开式的各项的因子:最后,流水的写出展开式的各项的因子:
符号sg,系数绝对值1t!,x的幂xt;符号sg,系数绝对值\frac{1}{t!},x的幂x^t;符号sg,系数绝对值t!1,x的幂xt;
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