What is taylor (1+x)^{(-4)},0 ?

The answer to taylor (1+x)^{(-4)},0 is 1-4x+10x^2-20x^3+35x^4+\ldots

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taylor (1+x)^{(-4)},0

taylor

(1 + x)^{(- 4)}, 0

1 - 4 x + 10 x^{2} - 20 x^{3} + 35 x^{4} + \dots

Solution steps

Apply the Taylor Formula

= 1 + \frac{\frac{d}{d x} ( ( 1 + x ) ^{- 4} ) ( 0 )}{1 !} x + \frac{\frac{d ^{2}}{d x ^{2}} ( ( 1 + x ) ^{- 4} ) ( 0 )}{2 !} x^{2} + \frac{\frac{d ^{3}}{d x ^{3}} ( ( 1 + x ) ^{- 4} ) ( 0 )}{3 !} x^{3} + \dots

Evaluate Derivatives

= 1 + \frac{- 4}{1 !} x + \frac{20}{2 !} x^{2} + \frac{- 120}{3 !} x^{3} + \frac{840}{4 !} x^{4} + \dots

Refine

= 1 - 4 x + 10 x^{2} - 20 x^{3} + 35 x^{4} + \dots

\int \frac{1}{1 - x - 2 x ^{2}} d x

\int_{- 125}^{64} \frac{1}{x ^{\frac{1}{3}}} d x

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

x \to \infty lim (- 0. 5^{x})

\int \frac{4 x ^{4} + 3}{4 x ^{5} + 15 x + 2} d x

What is taylor (1+x)^{(-4)},0 ?
The answer to taylor (1+x)^{(-4)},0 is 1-4x+10x^2-20x^3+35x^4+\ldots