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

Lösung

taylor

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

1 - 6 x + 24 x^{2} - 80 x^{3} + 240 x^{4} + \dots

Schritte zur Lösung

Apply the Taylor Formula

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

Evaluate Derivatives

= 1 + \frac{- 6}{1 !} x + \frac{48}{2 !} x^{2} + \frac{- 480}{3 !} x^{3} + \frac{5760}{4 !} x^{4} + \dots

Fasse zusammen

= 1 - 6 x + 24 x^{2} - 80 x^{3} + 240 x^{4} + \dots

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