What is taylor 1/(2+x) ?

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

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taylor 1/(2+x)

taylor

\frac{1}{2 + x}

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

Solution steps

Apply the Taylor Formula

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

Evaluate Derivatives

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

Refine

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

\int 7 x^{\frac{2}{5}} + 6 x^{- \frac{4}{5}} d x

\frac{d}{d x} (- \frac{1}{x ^{5}} - \frac{1}{x ^{3}} + \frac{1}{x})

\int \frac{2 t - 1}{t ^{2} - 2 t + 1} d t

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

\int \frac{1}{u 1 + u ^{2}} d u

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