MHB Finding a maclaurin series for a function with 'e'

[tmt1]

I need to find the Maclaurin series for

$$f(x) = e^{x - 2}$$

I know that the maclaurin series for $f(x) = e^x$ is

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

If I substitute in $x - 2$ for x, I would get

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

However, this is wrong, according to the text. How can I fix this?

 

[Prove It]

I need to find the Maclaurin series for

$$f(x) = e^{x - 2}$$

I know that the maclaurin series for $f(x) = e^x$ is

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

If I substitute in $x - 2$ for x, I would get

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

However, this is wrong, according to the text. How can I fix this?

The reason it's considered wrong in the text is because you have a Taylor series centred around x = 2, while a MacLaurin series is strictly only centred around x = 0.

For this one $\displaystyle \begin{align*} \mathrm{e}^{x - 2} = \mathrm{e}^{-2}\,\mathrm{e}^x = \mathrm{e}^{-2}\,\sum_{n = 0}^{\infty}{ \frac{x^n}{n!} } = \sum_{n = 0}^{\infty}{\frac{\mathrm{e}^{-2}\,x^n}{n!}} \end{align*}$.