Formulaire de développements en séries

• ${\displaystyle \forall x\in D(0,1),\,{1 \over {1-x}}=\sum _{n=0}^{+{\infty }}{x^{n}}.}$
• ${\displaystyle \forall x\in \,]-1,1[,\ \forall \alpha \in \mathbb {R} ,\,(1+x)^{\alpha }\,=1\;+\;\sum _{n=1}^{+{\infty }}{{\frac {\alpha \,(\alpha -1)\ldots (\alpha -n+1)}{n!}}\,x^{n}}.}$^[1]

En particulier :

• ${\displaystyle \forall x\in \mathbb {R} ,\,\forall m\in \mathbb {N} ,\,(1+x)^{m}\,=1\;+\;\sum _{n=1}^{m}{{\frac {m\,(m-1)\ldots (m-n+1)}{n!}}\,x^{n}}=\sum _{n=0}^{m}{{m \choose n}\,x^{n}}.}$

• ${\displaystyle \forall x\in \,[-1,1],\ {\sqrt {1+x}}\,=\sum _{n=0}^{+{\infty }}{{\frac {(-1)^{n+1}}{2n-1}}{\frac {\binom {2n}{n}}{2^{2n}}}\,x^{n}}.}$
• ${\displaystyle \forall x\in \,]-1,1[,\ \forall \alpha \in \mathbb {R} ,\,{\frac {1}{(1-x)^{\alpha }}}\,=1\;+\;\sum _{n=1}^{+{\infty }}{{\frac {\alpha \,(\alpha +1)\ldots (\alpha +n-1)}{n!}}\,x^{n}}.}$
• ${\displaystyle \forall x\in \,]-1,1[,\ \forall m\in \mathbb {N} ^{*},\,{\frac {1}{(1-x)^{m}}}\,=\sum _{n=0}^{+{\infty }}{{\binom {m+n-1}{n}}\,x^{n}}}$ (formule du binôme négatif).
• ${\displaystyle \forall x\in \,[-1,1[,\ {\frac {1}{\sqrt {1-x}}}\,=\sum _{n=0}^{+{\infty }}{{\frac {\binom {2n}{n}}{2^{2n}}}\,x^{n}}.}$

Pour tout nombre complexe z et tout réel a > 0 :

• ${\displaystyle {\rm {e}}^{z}=\sum _{n=0}^{+\infty }{\frac {z^{n}}{n!}}=1+{\frac {z}{1!}}+{\frac {z^{2}}{2!}}+{\frac {z^{3}}{3!}}+{\frac {z^{4}}{4!}}+\cdots }$
• ${\displaystyle a^{z}={\rm {e}}^{z\ln a}=\sum _{n=0}^{+\infty }{\frac {(\ln a)^{n}}{n!}}{z^{n}}=1+{\frac {\ln a}{1!}}z+{\frac {(\ln a)^{2}}{2!}}z^{2}+{\frac {(\ln a)^{3}}{3!}}z^{3}+{\frac {(\ln a)^{4}}{4!}}z^{4}+\cdots }$
• ${\displaystyle |z|\leq 1{\text{ et }}z\neq -1\Rightarrow \ln(1+z)=\sum _{n=1}^{+\infty }{(-1)^{n+1}{\frac {z^{n}}{n}}}=z-{\frac {z^{2}}{2}}+{\frac {z^{3}}{3}}-{\frac {z^{4}}{4}}+\cdots }$^[1]

• ${\displaystyle \forall x\in \mathbb {C} ,\,\sin x=\sum _{n=0}^{+{\infty }}(-1)^{n}\,{\frac {x^{2n+1}}{(2\,n+1)!}}=x-{\frac {x^{3}}{3!}}+{\frac {x^{5}}{5!}}-{\frac {x^{7}}{7!}}+\cdots }$
• ${\displaystyle \forall x\in \mathbb {C} ,\,\cos x=\sum _{n=0}^{+{\infty }}(-1)^{n}\,{\frac {x^{2n}}{(2\,n)!}}=1-{\frac {x^{2}}{2!}}+{\frac {x^{4}}{4!}}-{\frac {x^{6}}{6!}}+\cdots }$
• ${\displaystyle \forall x\in D(0,{\frac {\pi }{2}}),\tan x=\sum _{n=1}^{+\infty }(-1)^{n-1}\,{\frac {2^{2n}(2^{2n}-1)}{(2n)!}}B_{2n}x^{2n-1}=x+{\frac {1}{3}}x^{3}+{\frac {2}{15}}x^{5}+{\frac {17}{315}}x^{7}+{\frac {62}{2835}}x^{9}+\cdots }$
• ${\displaystyle \forall x\in \,\left]-{\frac {\pi }{2}},{\frac {\pi }{2}}\right[,\,\tan x={\frac {2}{\pi }}\,\sum _{n=0}^{+{\infty }}\,{\left({\frac {x}{\pi }}\right)}^{2\,n+1}(2^{2\,n+2}-1)\;\zeta (2\,n+2)\quad {\text{avec}}\;\forall p>1,\,\zeta (2p)=\sum _{n=1}^{+{\infty }}\,{\frac {1}{n^{2p}}}={\frac {|B_{2p}|(2\pi )^{2p}}{2(2p)!}}}$ où ${\displaystyle \zeta }$ est la fonction zêta de Riemann et les ${\displaystyle B_{k}}$ sont les nombres de Bernoulli.
• ${\displaystyle \forall x\in D(0,\pi )\setminus \lbrace 0\rbrace ,\cot x={\frac {1}{x}}-\sum _{n=1}^{+\infty }{\frac {2^{2n}}{(2n)!}}B_{2n}x^{2n-1}={\frac {1}{x}}-{\frac {x}{3}}-{\frac {x^{3}}{45}}-{\frac {2x^{5}}{945}}\cdots }$
• ${\displaystyle \forall x\in [-1,1],\,\arcsin x=\sum _{n=0}^{+\infty }{\frac {\binom {2n}{n}}{2^{2n}}}\times {\frac {x^{2n+1}}{2n+1}}=x+{\frac {1}{2\cdot 3}}x^{3}+{\frac {1\cdot 3}{2\cdot 4\cdot 5}}x^{5}+{\frac {1\cdot 3\cdot 5}{2\cdot 4\cdot 6\cdot 7}}x^{7}+...}$^[1]
• ${\displaystyle \forall x\in [-1,1],\,\arccos x={\frac {\pi }{2}}-\arcsin x={\frac {\pi }{2}}-\sum _{n=0}^{+\infty }{\frac {\binom {2n}{n}}{2^{2n}}}\times {\frac {x^{2n+1}}{2n+1}}={\frac {\pi }{2}}-\left(x+{\frac {1}{2\cdot 3}}x^{3}+{\frac {1\cdot 3}{2\cdot 4\cdot 5}}x^{5}+{\frac {1\cdot 3\cdot 5}{2\cdot 4\cdot 6\cdot 7}}x^{7}+\cdots \right)}$^[1]
• ${\displaystyle \forall x\in [-1,1],\arctan x=\sum _{n=0}^{+{\infty }}(-1)^{n}\,{\frac {x^{2\,n+1}}{2\,n+1}}=x-{\frac {x^{3}}{3}}+{\frac {x^{5}}{5}}-{\frac {x^{7}}{7}}+\cdots }$^[1] et en particulier, pour ${\displaystyle x=1}$, ${\displaystyle \pi =4\,\sum _{n=0}^{+{\infty }}{\frac {(-1)^{n}}{2\,n+1}}}$.
• ${\displaystyle \forall x\in [-1,1],\operatorname {arccot} x={\frac {\pi }{2}}-\arctan x={\frac {\pi }{2}}-\sum _{n=0}^{+{\infty }}(-1)^{n}\,{\frac {x^{2\,n+1}}{2\,n+1}}={\frac {\pi }{2}}-\left(x-{\frac {x^{3}}{3}}+{\frac {x^{5}}{5}}-{\frac {x^{7}}{7}}+\cdots \right)}$
• ${\displaystyle \forall x\in [-1,1],\arcsin ^{2}x={\frac {1}{2}}\sum _{n=1}^{\infty }{\frac {(2x)^{2n}}{n^{2}{\binom {2n}{n}}}}=x^{2}+{\frac {x^{4}}{3}}+{\frac {8x^{6}}{45}}+\cdots }$^[2],[3]

• ${\displaystyle \forall x\in \mathbb {C} ,\,\sinh x=\sum _{n=0}^{+\infty }{\frac {x^{2n+1}}{(2\,n+1)!}}=x+{\frac {x^{3}}{3!}}+{\frac {x^{5}}{5!}}+{\frac {x^{7}}{7!}}+\cdots }$
• ${\displaystyle \forall x\in \mathbb {C} ,\,\cosh x=\sum _{n=0}^{+{\infty }}{\frac {x^{2n}}{(2\,n)!}}=1+{\frac {x^{2}}{2!}}+{\frac {x^{4}}{4!}}+{\frac {x^{6}}{6!}}+\cdots }$
• ${\displaystyle \forall x\in \left]-{\frac {\pi }{2}},{\frac {\pi }{2}}\right[,\operatorname {tanh} \,x=\sum _{n=1}^{+\infty }{\frac {2^{2n}(2^{2n}-1)}{(2n)!}}B_{2n}x^{2n-1}=x-{\frac {1}{3}}x^{3}+{\frac {2}{15}}x^{5}-{\frac {17}{315}}x^{7}+\cdots }$
• ${\displaystyle \forall x\in ]0,\pi [,\coth x={\frac {1}{x}}+\sum _{n=1}^{+\infty }{\frac {2^{2n}}{(2n)!}}B_{2n}x^{2n-1}={\frac {1}{x}}+{\frac {1}{3}}x-{\frac {1}{45}}x^{3}+{\frac {2}{945}}x^{5}-\cdots }$
• ${\displaystyle \forall x\in [-1,1],\,\operatorname {arsinh} \,x=\ln \left(x+{\sqrt {x^{2}+1}}\right)=x+\sum _{n=1}^{+{\infty }}\,(-1)^{n}{\frac {\binom {2n}{n}}{2^{2n}}}\times {\frac {x^{2n+1}}{2\,n+1}}=x-{\frac {1}{2\cdot 3}}x^{3}+{\frac {1\cdot 3}{2\cdot 4\cdot 5}}x^{5}-{\frac {1\cdot 3\cdot 5}{2\cdot 4\cdot 6\cdot 7}}x^{7}+\cdots }$
• ${\displaystyle \forall x\in [1,+\infty [,\,\operatorname {arcosh} \,x=\ln \left(x+{\sqrt {x^{2}-1}}\right)=\ln(2x)-\sum _{n=1}^{+\infty }\,{\frac {\binom {2n}{n}}{2^{2n}}}\times {\frac {1}{2n\times x^{2n}}}=\ln(2x)-{\frac {1}{2\cdot 2x^{2}}}-{\frac {1\cdot 3}{2\cdot 4\cdot 4x^{4}}}-{\frac {1\cdot 3\cdot 5}{2\cdot 4\cdot 6\cdot 6x^{6}}}-\cdots }$
• ${\displaystyle \forall x\in ]-1,1[,\,\operatorname {artanh} \,x={\frac {1}{2}}\ln \left({\frac {1+x}{1-x}}\right)=\sum _{n=0}^{+{\infty }}\,{\frac {x^{2n+1}}{2n+1}}=x+{\frac {x^{3}}{3}}+{\frac {x^{5}}{5}}+{\frac {x^{7}}{7}}+\cdots }$
• ${\displaystyle \forall x\in ]-\infty ,1[\cup ]1,+\infty [,\,\operatorname {arcoth} \,x={\frac {1}{2}}\ln \left({\frac {x+1}{x-1}}\right)=\sum _{n=0}^{+{\infty }}\,{\frac {1}{(2n+1)\times x^{2n+1}}}={\frac {1}{x}}+{\frac {1}{3x^{3}}}+{\frac {1}{5x^{5}}}+{\frac {1}{7x^{7}}}+\cdots }$
• ${\displaystyle \forall x\in [-1,1],{\text{arsinh}}^{2}x=\sum _{n=1}^{\infty }(-1)^{n+1}{\frac {(2x)^{2n}}{n^{2}{\binom {2n}{n}}}}=x^{2}-{\frac {x^{4}}{3}}+{\frac {8x^{6}}{45}}+\cdots }$