麦克劳林 (4x^2)(e^{-4x})

解答

麦克劳林

(4 x^{2}) (e^{- 4 x})

4 x^{2} - 16 x^{3} + 32 x^{4} - \frac{128}{3} x^{5} + \frac{128}{3} x^{6} + \dots

求解步骤

应用麦克劳林公式

= 0 + \frac{\frac{d}{d x} ( 4 x ^{2} e ^{- 4 x} ) ( 0 )}{1 !} x + \frac{\frac{d ^{2}}{d x ^{2}} ( 4 x ^{2} e ^{- 4 x} ) ( 0 )}{2 !} x^{2} + \frac{\frac{d ^{3}}{d x ^{3}} ( 4 x ^{2} e ^{- 4 x} ) ( 0 )}{3 !} x^{3} + \dots

求导数

= 0 + \frac{0}{1 !} x + \frac{8}{2 !} x^{2} + \frac{- 96}{3 !} x^{3} + \frac{768}{4 !} x^{4} + \frac{- 5120}{5 !} x^{5} + \frac{30720}{6 !} x^{6} + \dots

整理后得

= 4 x^{2} - 16 x^{3} + 32 x^{4} - \frac{128}{3} x^{5} + \frac{128}{3} x^{6} + \dots

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