taylor xe^x

Lösung

taylor

x e^{x}

x + x^{2} + \frac{1}{2} x^{3} + \frac{1}{6} x^{4} + \frac{1}{24} x^{5} + \dots

Schritte zur Lösung

Apply the Taylor Formula

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

Evaluate Derivatives

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

Fasse zusammen

= x + x^{2} + \frac{1}{2} x^{3} + \frac{1}{6} x^{4} + \frac{1}{24} x^{5} + \dots

d er i v a t i v e f (x) = \frac{20}{3} x^{3} + 380 x^{2}

\frac{\partial}{\partial x} (3 ln (x) + ln (y))

\frac{\partial}{\partial x} (400 e^{x y})

n = 1 \sum \infty n sin (n)

n = 1 \sum \infty \frac{5}{3 ^{2 n}}