テイラー x^2e^{x-1},\at 1

解

テイラー

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

\frac{2 at 1 \cdot e ^{at 1 - 1} + e ^{at 1 - 1} ( at 1 ) ^{2}}{1 !} (x - at 1) + \frac{e ^{at 1 - 1} ( at 1 ) ^{2} + 4 e ^{at 1 - 1} at 1 + 2 e ^{at 1 - 1}}{2 !} (x - at 1)^{2} + \frac{e ^{at 1 - 1} ( at 1 ) ^{2} + 6 e ^{at 1 - 1} at 1 + 6 e ^{at 1 - 1}}{3 !} (x - at 1)^{3} + \frac{e ^{at 1 - 1} ( at 1 ) ^{2} + 8 e ^{at 1 - 1} at 1 + 12 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)^{2} e^{at 1 - 1} + \frac{\frac{d}{d x} ( x ^{2} e ^{x - 1} ) ( at 1 )}{1 !} (x - at 1) + \frac{\frac{d ^{2}}{d x ^{2}} ( x ^{2} e ^{x - 1} ) ( at 1 )}{2 !} (x - at 1)^{2} + \frac{\frac{d ^{3}}{d x ^{3}} ( x ^{2} e ^{x - 1} ) ( at 1 )}{3 !} (x - at 1)^{3} + \dots

導関数を評価する

= (at 1)^{2} e^{at 1 - 1} + \frac{2 at 1 \cdot e ^{at 1 - 1} + e ^{at 1 - 1} ( at 1 ) ^{2}}{1 !} (x - at 1) + \frac{e ^{at 1 - 1} ( at 1 ) ^{2} + 4 e ^{at 1 - 1} at 1 + 2 e ^{at 1 - 1}}{2 !} (x - at 1)^{2} + \frac{e ^{at 1 - 1} ( at 1 ) ^{2} + 6 e ^{at 1 - 1} at 1 + 6 e ^{at 1 - 1}}{3 !} (x - at 1)^{3} + \frac{e ^{at 1 - 1} ( at 1 ) ^{2} + 8 e ^{at 1 - 1} at 1 + 12 e ^{at 1 - 1}}{4 !} (x - at 1)^{4} + \dots

改良

= \frac{2 at 1 \cdot e ^{at 1 - 1} + e ^{at 1 - 1} ( at 1 ) ^{2}}{1 !} (x - at 1) + \frac{e ^{at 1 - 1} ( at 1 ) ^{2} + 4 e ^{at 1 - 1} at 1 + 2 e ^{at 1 - 1}}{2 !} (x - at 1)^{2} + \frac{e ^{at 1 - 1} ( at 1 ) ^{2} + 6 e ^{at 1 - 1} at 1 + 6 e ^{at 1 - 1}}{3 !} (x - at 1)^{3} + \frac{e ^{at 1 - 1} ( at 1 ) ^{2} + 8 e ^{at 1 - 1} at 1 + 12 e ^{at 1 - 1}}{4 !} (x - at 1)^{4} + \dots