taylor ln(1-4x)

Lösung

taylor

ln (1 - 4 x)

- 4 x - 8 x^{2} - \frac{64}{3} x^{3} - 64 x^{4} - \frac{1024}{5} x^{5} + \dots

Schritte zur Lösung

Apply the Taylor Formula

= 0 + \frac{\frac{d}{d x} ( ln ( 1 - 4 x ) ) ( 0 )}{1 !} x + \frac{\frac{d ^{2}}{d x ^{2}} ( ln ( 1 - 4 x ) ) ( 0 )}{2 !} x^{2} + \frac{\frac{d ^{3}}{d x ^{3}} ( ln ( 1 - 4 x ) ) ( 0 )}{3 !} x^{3} + \dots

Evaluate Derivatives

= 0 + \frac{- 4}{1 !} x + \frac{- 16}{2 !} x^{2} + \frac{- 128}{3 !} x^{3} + \frac{- 1536}{4 !} x^{4} + \frac{- 24576}{5 !} x^{5} + \dots

Fasse zusammen

= - 4 x - 8 x^{2} - \frac{64}{3} x^{3} - 64 x^{4} - \frac{1024}{5} x^{5} + \dots

\int e^{t^{2} - t} d t

\frac{\partial}{\partial x} (ln (5 x + 3 y))

f (z) = e^{z}

\frac{d}{d x} (1 - cos (6 x))

n = 4 \sum \infty 1