What is taylor (1+x)^{-2} ?

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

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taylor (1+x)^{-2}

taylor

(1 + x)^{- 2}

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

Solution steps

Apply the Taylor Formula

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

Evaluate Derivatives

= 1 + \frac{- 2}{1 !} x + \frac{6}{2 !} x^{2} + \frac{- 24}{3 !} x^{3} + \frac{120}{4 !} x^{4} + \dots

Refine

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

\int_{0}^{\frac{π}{4}} 3 sec^{2} (x) d x

\int \frac{( x ^{2} - 5 x ) ^{2}}{x} d x

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

f (x) = 2 x + 45, at x = 2

49 + 23 sin (\frac{2 π}{52} (t - 10))

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