maclaurin ln(1/3 (2+e^{-3x}))

Lösung

maclaurin

ln (\frac{1}{3} (2 + e^{- 3 x}))

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

Schritte zur Lösung

Apply the Maclaurin Formula

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

Evaluate Derivatives

= 0 + \frac{- 1}{1 !} x + \frac{2}{2 !} x^{2} + \frac{- 2}{3 !} x^{3} + \frac{- 6}{4 !} x^{4} + \frac{30}{5 !} x^{5} + \dots

Fasse zusammen

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

y^{^{''}} - 4 y + 4 y = e^{- t}, y (0) = 0, y^{^{'}} (0) = 0

x \to + 0 lim (3^{x} + 7)

d er i v a t i v e (\frac{t ^{3}}{t ^{8} + 9})^{2}

\int_{0.5}^{1} 1.5 x^{2} d x

\int (7 x + 4 y) d y