Find the terms in the Maclaurin series for the function $f ( x ) = e ^ { - x }$ , as far as the term in $x ^ { 3 }$ .
A) $1 - x + \frac { 1 } { 2 } x ^ { 2 } - \frac { 1 } { 6 } x ^ { 3 }$
B) $1 + x + \frac { 1 } { 2 } x ^ { 2 } - \frac { 1 } { 6 } x ^ { 3 }$
C) $1 - x + x ^ { 2 } - x ^ { 3 }$
D) $1 + x + x ^ { 2 } + x ^ { 3 }$
E) $1 - x + \frac { 1 } { 2 } x ^ { 2 } - \frac { 1 } { 3 } x ^ { 3 }$
F) $1 + x + \frac { 1 } { 2 } x ^ { 2 } + \frac { 1 } { 3 } x ^ { 3 }$
G) $- x + x ^ { 3 }$
H) $x - x ^ { 3 }$

Question

Find the terms in the Maclaurin series for the function $f ( x ) = e ^ { - x }$ , as far as the term in $x ^ { 3 }$ .
A) $1 - x + \frac { 1 } { 2 } x ^ { 2 } - \frac { 1 } { 6 } x ^ { 3 }$ 
B) $1 + x + \frac { 1 } { 2 } x ^ { 2 } - \frac { 1 } { 6 } x ^ { 3 }$ 
C) $1 - x + x ^ { 2 } - x ^ { 3 }$ 
D) $1 + x + x ^ { 2 } + x ^ { 3 }$ 
E) $1 - x + \frac { 1 } { 2 } x ^ { 2 } - \frac { 1 } { 3 } x ^ { 3 }$ 
F) $1 + x + \frac { 1 } { 2 } x ^ { 2 } + \frac { 1 } { 3 } x ^ { 3 }$ 
G) $- x + x ^ { 3 }$ 
H) $x - x ^ { 3 }$

Quizplus · Accepted Answer

The Maclaurin series for $e^{-x}$ is given by $1 - x + \frac{1}{2}x^2 - \frac{1}{6}x^3 + \cdots$, which matches option A. This series is derived from the general formula for the Maclaurin series of $e^x$, with $x$ replaced by $-x$.

Find the Terms in the Maclaurin Series for the Function f(x)=e−xf ( x ) = e ^ { - x }f(x)=e−x

Find the Terms in the Maclaurin Series for the Function $f ( x ) = e ^ { - x }$