What is taylor f(x)=x*e^{-x+1} ?

The answer to taylor f(x)=x*e^{-x+1} is ex-ex^2+e/2 x^3-e/6 x^4+e/(24)x^5+\ldots

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taylor f(x)=x*e^{-x+1}

taylor

f (x) = x \cdot e^{- x + 1}

e x - e x^{2} + \frac{e}{2} x^{3} - \frac{e}{6} x^{4} + \frac{e}{24} x^{5} + \dots

Solution steps

Apply the Taylor Formula

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

Evaluate Derivatives

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

Refine

= e x - e x^{2} + \frac{e}{2} x^{3} - \frac{e}{6} x^{4} + \frac{e}{24} x^{5} + \dots

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

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

y = 2 csc (4 x^{4} - 2 x + 5)

x^{^{''}} + 9 x = 0

n = 1 \sum \infty \frac{1}{1 + n ^{2}}

What is taylor f(x)=x*e^{-x+1} ?
The answer to taylor f(x)=x*e^{-x+1} is ex-ex^2+e/2 x^3-e/6 x^4+e/(24)x^5+\ldots