1.3 KiB
up::développement limité #s/maths/analyse
Ici, on pose \displaystyle\forall i\in\mathbb N, \lim_{x\rightarrow0}\varepsilon_i(x) = 0
(1+x)^\alpha = 1+\alpha x + \dfrac{\alpha(\alpha - 1)x^2}{2!} + \cdots + \dfrac{\alpha(\alpha -1)\cdots (\alpha-n+1)x^n}{n!} + x^n\varepsilon_1(x), \alpha\in\mathbb R
\dfrac1{1-x} = 1+x+x^2+x^3+ \cdots +x^n+x^n\varepsilon_2(x)
\dfrac1{1+x} = 1-x+x^2+ \cdots +(-1)^nx^n + x^n\varepsilon_3(x) soit (1+x)^\alpha avec \alpha = -1
\ln(1+x) = x-\dfrac{x^2}2+\dfrac{x^3}3+\cdots+(-1)^{n-1}\dfrac{x^n}n+x^n\varepsilon_4(x) penser que \displaystyle\left(x\mapsto\ln(1+x)\right)' = \left(x\mapsto \dfrac1{1+x}\right)
e^x = 1+\dfrac{x}{1!}+\dfrac{x^2}{2!}+\dfrac{x^3}{3!}+\cdots+\dfrac{x^n}{n!} + x^n\varepsilon_5(x)
\cos x = 1-\dfrac{x^2}{2!}+\dfrac{x^4}{4!}+\cdots+(-1)^n\dfrac{x^{2n}}{(2n)!}+x^{2n+1}\varepsilon_6(x)
\sin x = x-\dfrac{x^3}{3!}+\dfrac{x^5}{5!}+\cdots+(-1)^n\dfrac{x^{2n+1}}{(2n+1)!}+x^{2n+2}\varepsilon_7(x)
\text{sh}(x) = x + \frac{x^{3}}{3!} + \frac{x^{5}}{5!} + \cdots + \frac{x^{2n+1}}{(2n+1)!}+o(x^{2n+2}) = \sum\limits_{k=0} ^{n} \frac{x^{k}}{k!}[2\nmid k]
\text{ch}(x) = 1 + \frac{x^{2}}{2!} + \frac{x^{4}}{4!} + \cdots + \frac{x^{2n}}{(2n)!}+o(x^{2n+1}) = \sum\limits_{k=0} ^{n} \frac{x^{k}}{k!}[2\mid k]