What is taylor e^{-x},x=0 ?

The answer to taylor e^{-x},x=0 is 1-x+1/2 x^2-1/6 x^3+1/24 x^4+\ldots

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taylor e^{-x},x=0

taylor

e^{- x}, x = 0

1 - x + \frac{1}{2} x^{2} - \frac{1}{6} x^{3} + \frac{1}{24} x^{4} + \dots

Solution steps

Apply the Taylor Formula

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

Evaluate Derivatives

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

Refine

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

\int \frac{1}{x x - 4} d x

\int sec^{4} (\frac{x}{3}) d x

y = x^{2} e^{x} - 2 x e^{x} + 5 e^{x}

\int_{- π}^{π} cos^{2} (x) d x

\frac{d}{d x} (\frac{4}{x ^{2} - 1})

What is taylor e^{-x},x=0 ?
The answer to taylor e^{-x},x=0 is 1-x+1/2 x^2-1/6 x^3+1/24 x^4+\ldots