What is maclaurin (1+px)^{-1} ?

The answer to maclaurin (1+px)^{-1} is 1-px+p^2x^2-p^3x^3+p^4x^4+\ldots

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maclaurin (1+px)^{-1}

maclaurin

(1 + p x)^{- 1}

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

Solution steps

Apply the Maclaurin Formula

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

Evaluate Derivatives

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

Refine

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

\frac{8}{ln ( x )}

sec^{2} (x)

\int \frac{cos ( 8 x )}{cos ^{2} ( 4 x )} d x

\int 1 + sin^{2} (x) d x

\int \frac{1}{9 - x} d x

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