文章目录

• AM@常见函数的幂级数(series)展开@泰勒级数TaylorSeries
• • ref
  • 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 (wikipedia.org)
• Taylor series – Wikipedia
• 泰勒级数 – 维基百科，自由的百科全书 (wikipedia.org)

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+⋯

  • $\forall x:\left|x\right|<1,\forall \alpha \in \mathbb{C}$
  • 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∏n​kα−k+1​=n!α(α−1)⋯(α−n+1)​

指数函数和自然对数🎈

• 以$e$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+1​xn=x−2x2​+3x3​−4x4​+5x5​−⋯+n(−1)n+1​xn+⋯∀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：

• sin⁡x=∑n=0∞(−1)n(2n+1)!x2n+1=x−x33!+x55!−⋯∀xcos⁡x=∑n=0∞(−1)n(2n)!x2n=1−x22!+x44!−⋯∀xtan⁡x=∑n=1∞B2n(−4)n(1−4n)(2n)!x2n−1=x+x33+2×515+⋯∀x:∣x∣<π2sec⁡x=∑n=0∞(−1)nE2n(2n)!x2n=1+x22+5×424+⋯∀x:∣x∣<π2arcsin⁡x=∑n=0∞(2n)!4n(n!)2(2n+1)x2n+1=x+x36+3×540+⋯∀x:∣x∣≤1arccos⁡x=π2−arcsin⁡x=π2−∑n=0∞(2n)!4n(n!)2(2n+1)x2n+1=π2−x−x36−3×540+⋯∀x:∣x∣≤1arctan⁡x=∑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)n​x2n+1=n=0∑∞​(2n)!(−1)n​x2n=n=1∑∞​(2n)!B2n​(−4)n(1−4n)​x2n−1=n=0∑∞​(2n)!(−1)nE2n​​x2n=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)n​x2n+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$\tan (x)$展开式中的BkB_kBk​是伯努利数。

• 在$\sec (x)$展开式中的EkE_kEk​是欧拉数。

常用三角

• sin⁡x=∑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⋯

• cos⁡x=∑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🎈

• sinh⁡x=∑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)!1​x2n+1∀x

• cosh⁡x=∑n=0∞1(2n)!x2n∀x\cosh x=\sum _{n=0}^{\infty }{\frac {1}{(2n)!}}x^{2n}\quad \forall xcoshx=∑n=0∞​(2n)!1​x2n∀x

• tanh⁡x=∑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)!B2n​4n(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+11​x2n+1∀x:∣x∣<1

• $\tanh (x)$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−1​xn∀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​+⋯+nd​​n1​!⋯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。

1. ∀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​.
2. ∀x∈R,cos⁡x=∑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​.
3. ∀x∈R,sin⁡x=∑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​.
4. ∀x∈R,ch⁡x=∑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​.
5. ∀x∈R,sh⁡x=∑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​.
6. ∀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.
7. ∀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​.)
8. ∀x∈[−1,1],arctan⁡x=∑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​。
9. ∀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.
10. ∀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.
11. ∀x∈(−1,1),artanh⁡x=∑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​.
12. ∀x∈(−1,1),arcsin⁡x=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=1n​2k∏k=1n​(2k−1)​)2n+1x2n+1​
13. ∀x∈(−1,1),arsinh⁡x=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=1n​2k∏k=1n​(2k−1)​)2n+1x2n+1​
14. ∀x∈(−π2,π2),tan⁡x=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$n=0$开始计算

  • 注意到两个公差$d=2$的数列:($n=0,1,2,\dots$)
  • 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)

• $最后,流水的写出展开式的各项的因子:$

  • 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;