麦克劳林 xe^{-x^2}

解答

麦克劳林

x e^{- x^{2}}

x - x^{3} + \frac{1}{2} x^{5} - \frac{1}{6} x^{7} + \frac{1}{24} x^{9} + \dots

求解步骤

应用麦克劳林公式

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

求导数

= 0 + \frac{1}{1 !} x + \frac{0}{2 !} x^{2} + \frac{- 6}{3 !} x^{3} + \frac{0}{4 !} x^{4} + \frac{60}{5 !} x^{5} + \frac{0}{6 !} x^{6} + \frac{- 840}{7 !} x^{7} + \frac{0}{8 !} x^{8} + \frac{15120}{9 !} x^{9} + \dots

整理后得

= x - x^{3} + \frac{1}{2} x^{5} - \frac{1}{6} x^{7} + \frac{1}{24} x^{9} + \dots

\int \frac{3 x}{( 2 x ^{2} + 1 ) ^{3}} d x

k = 0 \sum \infty x^{k}

t a y l or x^{2} - 4, 0

\frac{d}{d x} (\frac{3 e ^{x}}{5 e ^{x} + 4})

t an g e n t f (x) = - \frac{3}{x ^{2} - 4}, (1, 1)