What is taylor 1-1/(x^2) ?

The answer to taylor 1-1/(x^2) is 2(x-1)-3(x-1)^2+4(x-1)^3-5(x-1)^4+6(x-1)^5+\ldots

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taylor 1-1/(x^2)

taylor

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

2 (x - 1) - 3 (x - 1)^{2} + 4 (x - 1)^{3} - 5 (x - 1)^{4} + 6 (x - 1)^{5} + \dots

Solution steps

Apply the Taylor Formula

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

Evaluate Derivatives

= 0 + \frac{2}{1 !} (x - 1) + \frac{- 6}{2 !} (x - 1)^{2} + \frac{24}{3 !} (x - 1)^{3} + \frac{- 120}{4 !} (x - 1)^{4} + \frac{720}{5 !} (x - 1)^{5} + \dots

Refine

= 2 (x - 1) - 3 (x - 1)^{2} + 4 (x - 1)^{3} - 5 (x - 1)^{4} + 6 (x - 1)^{5} + \dots

x \to 0 lim (\frac{e ^{x}}{1 - e ^{x}})

\frac{\partial}{\partial x} (2 x^{2} - 4 x + 1)

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

f (x) = \frac{1}{4} x^{8}

\frac{d}{d x} (\frac{1 - sin ( x )}{1 - cos ( x )})

What is taylor 1-1/(x^2) ?
The answer to taylor 1-1/(x^2) is 2(x-1)-3(x-1)^2+4(x-1)^3-5(x-1)^4+6(x-1)^5+\ldots