泰勒 1/2 (e^{(1-x)})

解答

泰勒

\frac{1}{2} (e^{(1 - x)})

\frac{e}{2} - \frac{e}{2} x + \frac{e}{4} x^{2} - \frac{e}{12} x^{3} + \frac{e}{48} x^{4} + \dots

求解步骤

应用泰勒公式

= \frac{e}{2} + \frac{\frac{d}{d x} ( \frac{1}{2} e ^{1 - x} ) ( 0 )}{1 !} x + \frac{\frac{d ^{2}}{d x ^{2}} ( \frac{1}{2} e ^{1 - x} ) ( 0 )}{2 !} x^{2} + \frac{\frac{d ^{3}}{d x ^{3}} ( \frac{1}{2} e ^{1 - x} ) ( 0 )}{3 !} x^{3} + \dots

求导数

= \frac{e}{2} + \frac{- \frac{e}{2}}{1 !} x + \frac{\frac{e}{2}}{2 !} x^{2} + \frac{- \frac{e}{2}}{3 !} x^{3} + \frac{\frac{e}{2}}{4 !} x^{4} + \dots

整理后得

= \frac{e}{2} - \frac{e}{2} x + \frac{e}{4} x^{2} - \frac{e}{12} x^{3} + \frac{e}{48} x^{4} + \dots

s l o p e y = - \frac{12}{7} x = + b

\frac{d}{d x} y = tan^{2} (x)

\int e^{- x} sin (x) d x

n = 1 \sum \infty \frac{5 n}{4 n + 15}

x \to - 1 - lim (\frac{x + 1}{x ^{4} - 1})