taylor xe^{2x}

Lösung

taylor

x e^{2 x}

x + 2 x^{2} + 2 x^{3} + \frac{4}{3} x^{4} + \frac{2}{3} x^{5} + \dots

Schritte zur Lösung

Apply the Taylor Formula

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

Evaluate Derivatives

= 0 + \frac{1}{1 !} x + \frac{4}{2 !} x^{2} + \frac{12}{3 !} x^{3} + \frac{32}{4 !} x^{4} + \frac{80}{5 !} x^{5} + \dots

Fasse zusammen

= x + 2 x^{2} + 2 x^{3} + \frac{4}{3} x^{4} + \frac{2}{3} x^{5} + \dots

\frac{d}{d t} (cos (6 t))

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