taylor 1/((2+x)^3)

Lösung

taylor

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

\frac{1}{8} - \frac{3}{16} x + \frac{3}{16} x^{2} - \frac{5}{32} x^{3} + \frac{15}{128} x^{4} + \dots

Schritte zur Lösung

Apply the Taylor Formula

= \frac{1}{8} + \frac{\frac{d}{d x} ( \frac{1}{( 2 + x ) ^{3}} ) ( 0 )}{1 !} x + \frac{\frac{d ^{2}}{d x ^{2}} ( \frac{1}{( 2 + x ) ^{3}} ) ( 0 )}{2 !} x^{2} + \frac{\frac{d ^{3}}{d x ^{3}} ( \frac{1}{( 2 + x ) ^{3}} ) ( 0 )}{3 !} x^{3} + \dots

Evaluate Derivatives

= \frac{1}{8} + \frac{- \frac{3}{16}}{1 !} x + \frac{\frac{3}{8}}{2 !} x^{2} + \frac{- \frac{15}{16}}{3 !} x^{3} + \frac{\frac{45}{16}}{4 !} x^{4} + \dots

Fasse zusammen

= \frac{1}{8} - \frac{3}{16} x + \frac{3}{16} x^{2} - \frac{5}{32} x^{3} + \frac{15}{128} x^{4} + \dots

x \to - \infty lim (3^{- x})

s l o p e (\frac{11}{6}, - \frac{1}{2}), (\frac{1}{3}, \frac{1}{3})

s l o p e (1) (3.5)

\int 19 arctan (x) d x

f (x) = (4 x^{2} - 6 x)^{3}