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

May 28, 2015

The Maclaurin series of ${f}_{\left(x\right)} = {e}^{- 2 x}$ 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 $z = - 2 x$.

Then $\setminus \quad {f}_{\left(x\right)} = {e}^{- 2 x} = {e}^{z} \setminus \quad$ and ${f}_{\left(x\right)}$ has the same Maclaurin series as the one above except we set $z = - 2 x$ 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 ${f}_{\left(x\right)}$ is

${f}_{\left(x\right)} = {f}_{\left(x = 0\right)}$ \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 ${f}_{\left(x = 0\right)}$. Here, $\setminus \quad {f}_{\left(x = 0\right)} = {e}^{- 2 \left(0\right)} = 1$.

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 ${n}^{t h}$ term of the series is (-2x)^n/{n!}

Using a summation sign, the Maclaurin series of ${f}_{\left(x\right)}$ can be written instead as

f_{(x)}=\Sigma_{n=0}^{n=\infty} [(-2x)^n/{n!} ]