What is taylor 1/x ,-1 ?

The answer to taylor 1/x ,-1 is -1-(x+1)-(x+1)^2-(x+1)^3-(x+1)^4+\ldots

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taylor 1/x ,-1

taylor

\frac{1}{x}, - 1

- 1 - (x + 1) - (x + 1)^{2} - (x + 1)^{3} - (x + 1)^{4} + \dots

Solution steps

Apply the Taylor Formula

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

Evaluate Derivatives

= - 1 + \frac{- 1}{1 !} (x + 1) + \frac{- 2}{2 !} (x + 1)^{2} + \frac{- 6}{3 !} (x + 1)^{3} + \frac{- 24}{4 !} (x + 1)^{4} + \dots

Refine

= - 1 - (x + 1) - (x + 1)^{2} - (x + 1)^{3} - (x + 1)^{4} + \dots

\int \frac{3 x ^{5}}{5 - 12 x ^{6}} d x

\int_{1}^{e} \frac{2}{x} d x

\int (- 1)^{n}

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

f (x) = - 8 sin (4 x) - 0.4 e^{- 0.2 x}

What is taylor 1/x ,-1 ?
The answer to taylor 1/x ,-1 is -1-(x+1)-(x+1)^2-(x+1)^3-(x+1)^4+\ldots