What is taylor f(x)= 1/x ?

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

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taylor f(x)= 1/x

taylor

f (x) = \frac{1}{x}

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_{- 2}^{3} \frac{11}{x ^{4}} d x

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

\int 2 cos (2 t) d t

\frac{d}{d x} (7 + 3 e^{5 x})

(x, y) \to (0, 69) lim (x^{3} y)

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