What is taylor 1/(1-x^3),0 ?

The answer to taylor 1/(1-x^3),0 is 1+x^3+x^6+x^9+\ldots

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taylor 1/(1-x^3),0

taylor

\frac{1}{1 - x ^{3}}, 0

1 + x^{3} + x^{6} + x^{9} + \dots

Solution steps

Apply the Taylor Formula

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

Evaluate Derivatives

= 1 + \frac{0}{1 !} x + \frac{0}{2 !} x^{2} + \frac{6}{3 !} x^{3} + \frac{0}{4 !} x^{4} + \frac{0}{5 !} x^{5} + \frac{720}{6 !} x^{6} + \frac{0}{7 !} x^{7} + \frac{0}{8 !} x^{8} + \frac{362880}{9 !} x^{9} + \frac{0}{10 !} x^{10} + \dots

Refine

= 1 + x^{3} + x^{6} + x^{9} + \dots

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

\frac{d}{d x} (\frac{5}{3} x^{3} + x^{2} - 1)

\int \frac{sin ^{7} ( ln ( x ) )}{x} d x

\frac{1}{x ^{8}} y^{^{'}} = 12 + 15 y

\frac{\partial}{\partial x} (x^{2} y + x y^{2})

What is taylor 1/(1-x^3),0 ?
The answer to taylor 1/(1-x^3),0 is 1+x^3+x^6+x^9+\ldots