What is taylor (1-x)e^{x+(x^2)/2} ?

The answer to taylor (1-x)e^{x+(x^2)/2} is 1-1/3 x^3-1/4 x^4-1/5 x^5-1/9 x^6+\ldots

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taylor (1-x)e^{x+(x^2)/2}

taylor

(1 - x) e^{x + \frac{x ^{2}}{2}}

1 - \frac{1}{3} x^{3} - \frac{1}{4} x^{4} - \frac{1}{5} x^{5} - \frac{1}{9} x^{6} + \dots

Solution steps

Apply the Taylor Formula

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

Evaluate Derivatives

= 1 + \frac{0}{1 !} x + \frac{0}{2 !} x^{2} + \frac{- 2}{3 !} x^{3} + \frac{- 6}{4 !} x^{4} + \frac{- 24}{5 !} x^{5} + \frac{- 80}{6 !} x^{6} + \dots

Refine

= 1 - \frac{1}{3} x^{3} - \frac{1}{4} x^{4} - \frac{1}{5} x^{5} - \frac{1}{9} x^{6} + \dots

x \to 0 + lim (- x^{2} ln (6 x^{3}))

\int \frac{6}{5 + 6 x} d x

\int e^{x} 2 d x

\frac{\partial}{\partial x} (6 e^{x y + 5})

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

What is taylor (1-x)e^{x+(x^2)/2} ?
The answer to taylor (1-x)e^{x+(x^2)/2} is 1-1/3 x^3-1/4 x^4-1/5 x^5-1/9 x^6+\ldots