What is taylor 1/(1+x),2 ?

The answer to taylor 1/(1+x),2 is 1/3-1/9 (x-2)+1/27 (x-2)^2-1/81 (x-2)^3+1/243 (x-2)^4+\ldots

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taylor 1/(1+x),2

taylor

\frac{1}{1 + x}, 2

\frac{1}{3} - \frac{1}{9} (x - 2) + \frac{1}{27} (x - 2)^{2} - \frac{1}{81} (x - 2)^{3} + \frac{1}{243} (x - 2)^{4} + \dots

Solution steps

Apply the Taylor Formula

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

Evaluate Derivatives

= \frac{1}{3} + \frac{- \frac{1}{9}}{1 !} (x - 2) + \frac{\frac{2}{27}}{2 !} (x - 2)^{2} + \frac{- \frac{2}{27}}{3 !} (x - 2)^{3} + \frac{\frac{8}{81}}{4 !} (x - 2)^{4} + \dots

Refine

= \frac{1}{3} - \frac{1}{9} (x - 2) + \frac{1}{27} (x - 2)^{2} - \frac{1}{81} (x - 2)^{3} + \frac{1}{243} (x - 2)^{4} + \dots

(x e^{- x^{2}})^{^{'}}

\int \frac{( x + 3 )}{( x ^{2} + 6 x ) ^{\frac{1}{3}}} d x

\frac{d y}{d x} = \frac{3 x ^{4} + y ^{3}}{3 x y ^{2}}

\frac{d y}{d t} = \frac{1}{y - 2}

\int_{0}^{1} 4 x y e^{- (2 x + y)} d y

What is taylor 1/(1+x),2 ?
The answer to taylor 1/(1+x),2 is 1/3-1/9 (x-2)+1/27 (x-2)^2-1/81 (x-2)^3+1/243 (x-2)^4+\ldots