How do you find the Maclaurin series of #f(x)=e^(-2x)#
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Question

Isabella Ackerman · Accepted Answer

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+. . . #

I apologize for the poor formatting; I was unable to fit all the terms on one line.

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 terms are identical to those in the previously written Maclaurin series.

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!} ]#

William Abdalla · Answer

undefined To find the Maclaurin series of $ f(x) = e^{-2x} $, we use the general formula for the Maclaurin series expansion of $ e^x $:

$$ e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!} $$

Substituting $-2x$ for $x$ in the formula:

$$ e^{-2x} = \sum_{n=0}^{\infty} \frac{(-2x)^n}{n!} $$

Simplifying:

$$ e^{-2x} = \sum_{n=0}^{\infty} \frac{(-1)^n 2^n x^n}{n!} $$

How do you find the Maclaurin series of #f(x)=e^(-2x)# ?