テイラー xe^{-x+1},\at 1

解

テイラー

x e^{- x + 1}, at 1

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

解答ステップ

テイラーの公式を適用する

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

導関数を評価する

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

改良

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