泰勒 x^2e^{x-1}1

解答

泰勒

x^{2} e^{x - 1} 1

\frac{1}{e} x^{2} + \frac{1}{e} x^{3} + \frac{1}{2 e} x^{4} + \frac{1}{6 e} x^{5} + \frac{1}{24 e} x^{6} + \dots

求解步骤

应用泰勒公式

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

求导数

= 0 + \frac{0}{1 !} x + \frac{\frac{2}{e}}{2 !} x^{2} + \frac{\frac{6}{e}}{3 !} x^{3} + \frac{\frac{12}{e}}{4 !} x^{4} + \frac{\frac{20}{e}}{5 !} x^{5} + \frac{\frac{30}{e}}{6 !} x^{6} + \dots

整理后得

= \frac{1}{e} x^{2} + \frac{1}{e} x^{3} + \frac{1}{2 e} x^{4} + \frac{1}{6 e} x^{5} + \frac{1}{24 e} x^{6} + \dots

l a pl a ce t r an s f or m e^{- 2 t}

\frac{d}{d x} (x^{4} - 4 x^{3} + 6)

x^{^{''}} - 5 x^{^{'}} + 6 x = cos (t) e^{- t}

t an g e n t f (x) = \frac{2}{x + 3}, at a = - 4

n = 1 \sum \infty sin (n + 1)