taylor ln(1-z)

Lösung

taylor

ln (1 - z)

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

Schritte zur Lösung

Apply the Taylor Formula

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

Evaluate Derivatives

= 0 + \frac{- 1}{1 !} z + \frac{- 1}{2 !} z^{2} + \frac{- 2}{3 !} z^{3} + \frac{- 6}{4 !} z^{4} + \frac{- 24}{5 !} z^{5} + \dots

Fasse zusammen

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

\int 3 x^{2} - 24 x + 48 d x

\frac{d}{d x} (3 x^{4} - 8 x^{3} + 2)

d er i v a t i v e lo g_{6} (x)

y^{^{''}} + 2 t = 0

\int_{- 1}^{2} - x^{2} + x + 2 d x