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Popular Calculus >

taylor (1+x)^{-1}

Solution

taylor

(1 + x)^{- 1}

Solution

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

Solution steps

Apply the Taylor Formula

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

Evaluate Derivatives

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

Refine

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

Popular Examples

tangent of y=-3x^3-2x^2-x+3,(-1,5)

integral of x/((x^2+1))

(\partial)/(\partial x)(6x^{4y})

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derivative of 4-(25/(6x^{1/6)})

Frequently Asked Questions (FAQ)

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