What is taylor (1+x)^{1/3} ?

The answer to taylor (1+x)^{1/3} is 1+1/3 x-1/9 x^2+5/81 x^3-10/243 x^4+\ldots

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taylor (1+x)^{1/3}

taylor

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

1 + \frac{1}{3} x - \frac{1}{9} x^{2} + \frac{5}{81} x^{3} - \frac{10}{243} x^{4} + \dots

Solution steps

Apply the Taylor Formula

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

Evaluate Derivatives

= 1 + \frac{\frac{1}{3}}{1 !} x + \frac{- \frac{2}{9}}{2 !} x^{2} + \frac{\frac{10}{27}}{3 !} x^{3} + \frac{- \frac{80}{81}}{4 !} x^{4} + \dots

Refine

= 1 + \frac{1}{3} x - \frac{1}{9} x^{2} + \frac{5}{81} x^{3} - \frac{10}{243} x^{4} + \dots

\int 5 e^{2 x} d x

y^{^{''}} + 9 y = t e^{3 t}

y^{^{'}} - y = 8 e^{x} + 18 e^{4 x}

\int 4 sec^{2} (x) tan (x) d x

What is taylor (1+x)^{1/3} ?
The answer to taylor (1+x)^{1/3} is 1+1/3 x-1/9 x^2+5/81 x^3-10/243 x^4+\ldots