在此篇文章中有一些泰勒展开式,主要用在极限求法上,可以稍微记忆
ex=∞∑n=0xnn!=1+x+x22!+x33!+⋯+xnn!+⋯
ln(1+x)=∞∑n=1(−1)n−1xnn=x−x22+x33−x44+⋯+(−1)n−1xnn+⋯(−1<x≤1)
sinx=∞∑k=0(−1)kx2k+1(2k+1)!=x−x33!+x55!−x77!+⋯+(−1)kx2k+1(2k+1)!+⋯
cosx=∞∑k=0(−1)kx2k(2k)!=1−x22!+x44!−x66!+⋯+(−1)kx2k(2k)!+⋯
tanx=x+13x3+215x5+⋯(|x|<π2)
cotx=1x−x3−x345−⋯(0<|x|<π)
secx=1+x22+5x424+61x6720+⋯=∞∑n=0E2n(2n)!x2n(|x|<π2)
cscx=1x+x6+7x3360+31x515120+⋯(0<|x|<π)
(1+x)α=∞∑n=0α(α−1)⋯(α−n+1)n!xn=1+αx+α(α−1)2!x2+⋯+α(α−1)⋯(α−n+1)n!xn+⋯(收敛区间依α而定)
arctanx=∞∑n=0(−1)n2n+1x2n+1=x−x33+x55−x77+⋯+(−1)n2n+1x2n+1+⋯(|x|<1)
arcsinx=∞∑n=0(2n)!4n(n!)2(2n+1)x2n+1=x+16x3+340x5+⋯+(2n)!4n(n!)2(2n+1)x2n+1+⋯(|x|<1)
arccosx=π2−∞∑n=0(2n)!(2n+1)4n(n!)2x2n+1=π2−x−x36−3x540−⋯(−1≤x≤1)
sinhx=∞∑n=0x2n+1(2n+1)!=x+x33!+x55!+⋯+x2n+1(2n+1)!+⋯
coshx=∞∑n=0x2n(2n)!=1+x22!+x44!+⋯+x2n(2n)!+⋯
erf(x)=2√π(x−x33+x510−x742+⋯)
Si(x)=x−x33×3!+x55×5!−x77×7!+⋯
Ci(x)=γ+lnx−x22×2!+x44×4!−x66×6!+⋯