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

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

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taylor 1/(1-x),-1

taylor

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

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

Solution steps

Apply the Taylor Formula

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

Evaluate Derivatives

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

Refine

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

\int 3 9 x d x

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

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

f (x) = (x^{2} + 1)^{3} - (x^{3} + 1)^{2}

\frac{d}{d x} (- cos^{2} (2 x))

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