maclaurin 1-(1-x)^{sqrt(2)-1}

Lösung

maclaurin

1 - (1 - x)^{2 - 1}

(2 - 1) x + \frac{- 4 + 3 2}{2} x^{2} + \frac{13 2 - 18}{6} x^{3} + \frac{7 ( 5 2 - 7 )}{12} x^{4} + \frac{7 ( 32 2 - 45 )}{60} x^{5} + \dots

Schritte zur Lösung

Apply the Maclaurin Formula

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

Evaluate Derivatives

= 0 + \frac{2 - 1}{1 !} x + \frac{- 4 + 3 2}{2 !} x^{2} + \frac{13 2 - 18}{3 !} x^{3} + \frac{- 98 + 70 2}{4 !} x^{4} + \frac{448 2 - 630}{5 !} x^{5} + \dots

Fasse zusammen

= (2 - 1) x + \frac{- 4 + 3 2}{2} x^{2} + \frac{13 2 - 18}{6} x^{3} + \frac{7 ( 5 2 - 7 )}{12} x^{4} + \frac{7 ( 32 2 - 45 )}{60} x^{5} + \dots

t a y l or ln (1 - 4 x)

\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))