What is taylor 1/(1-x),3 ?

The answer to taylor 1/(1-x),3 is -1/2+1/4 (x-3)-1/8 (x-3)^2+1/16 (x-3)^3-1/32 (x-3)^4+\ldots

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taylor 1/(1-x),3

taylor

\frac{1}{1 - x}, 3

- \frac{1}{2} + \frac{1}{4} (x - 3) - \frac{1}{8} (x - 3)^{2} + \frac{1}{16} (x - 3)^{3} - \frac{1}{32} (x - 3)^{4} + \dots

Solution steps

Apply the Taylor Formula

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

Evaluate Derivatives

= - \frac{1}{2} + \frac{\frac{1}{4}}{1 !} (x - 3) + \frac{- \frac{1}{4}}{2 !} (x - 3)^{2} + \frac{\frac{3}{8}}{3 !} (x - 3)^{3} + \frac{- \frac{3}{4}}{4 !} (x - 3)^{4} + \dots

Refine

= - \frac{1}{2} + \frac{1}{4} (x - 3) - \frac{1}{8} (x - 3)^{2} + \frac{1}{16} (x - 3)^{3} - \frac{1}{32} (x - 3)^{4} + \dots

\frac{d}{d x} (\frac{5}{1 - x})

\int \frac{1}{sec ^{2} ( y )} d y

\frac{d}{d x} ((4 - x^{\frac{2}{3}})^{3})

x \to - 8 lim (1 + - 2 x)

\frac{\partial}{\partial x} (x - 2)

What is taylor 1/(1-x),3 ?
The answer to taylor 1/(1-x),3 is -1/2+1/4 (x-3)-1/8 (x-3)^2+1/16 (x-3)^3-1/32 (x-3)^4+\ldots