taylor 1/2 arctan(2x)

Lösung

taylor

\frac{1}{2} arctan (2 x)

x - \frac{4}{3} x^{3} + \frac{16}{5} x^{5} - \frac{64}{7} x^{7} + \frac{256}{9} x^{9} + \dots

Schritte zur Lösung

Apply the Taylor Formula

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

Evaluate Derivatives

= 0 + \frac{1}{1 !} x + \frac{0}{2 !} x^{2} + \frac{- 8}{3 !} x^{3} + \frac{0}{4 !} x^{4} + \frac{384}{5 !} x^{5} + \frac{0}{6 !} x^{6} + \frac{- 46080}{7 !} x^{7} + \frac{0}{8 !} x^{8} + \frac{10321920}{9 !} x^{9} + \dots

Fasse zusammen

= x - \frac{4}{3} x^{3} + \frac{16}{5} x^{5} - \frac{64}{7} x^{7} + \frac{256}{9} x^{9} + \dots

n = 1 \sum \infty 3 e^{- 4 n}

t an g e n t f (x) = \frac{1}{x ^{2} + 4}, (1, \frac{1}{5})

\int 3 e^{x} - 3 x e^{x^{2}} d x

\frac{d}{d x} (k x + d)