The Maclaurin series of f_{(x)}=e^{-2x} is
f_{(x)}=1+(-2x)+(-2x)^2/{2!}+(-2x)^3/{3!}+ . . .
First Solution Method: The Maclaurin Series of y=e^z is
y=1+z+z^2/{2!}+z^3/{3!}+z^4/{4!}+ . . .
Let z=-2x.
Then \quad f_{(x)}=e^{-2x}=e^z\quad and f_{(x)} has the same Maclaurin series as the one above except we set z=-2x and get
f_{(x)}=1+(-2x)+(-2x)^2/{2!}+(-2x)^3/{3!}+ . . .
I used the well known Maclaurin series for y=e^z to get the answer. If this series has not been discussed in class, you should use the general definition of a Maclaurin series to get the answer.
The Maclaurin series of f_{(x)} is
f_{(x)}= f_((x=0)) \quad +{f'_((x=0))}/{1!}x
\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad +{f''_((x=0))}/{2!]x^2
\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad+ {f'''_((x=0))}/{3!}x^3+. . .
^Couldn't put all terms on same line, sorry for poor formatting.
Anyways, the first term is f_{(x=0)}. Here, \quad f_{(x=0)}=e^{-2(0)}=1.
The second term is {f'_{(x=0)}}/{1!}x={-2e^{-2(0)}}/1x=-2x
The third term is {f''_{(x=0)}}/{2!}x^2={(-2)^2e^{-2(0)}}/{2!}x^2={(-2x)^2}/{2!}
These are the same terms as in the Maclaurin series I wrote above.
By observing a pattern, the n^{th} term of the series is (-2x)^n/{n!}
Using a summation sign, the Maclaurin series of f_{(x)} can be written instead as
f_{(x)}=\Sigma_{n=0}^{n=\infty} [(-2x)^n/{n!} ]
