Taylor and Maclaurin Series for $f(x)=e^x$ I just came from a final exam where in one question I was asked to derive the Taylor Series for $f(x)=e^{2x}$ centered at $x=1$. I came up with the following:
$$e^{2x}=\sum_{n=0}^{\infty}\frac{2^ne^2}{n!}(x-1)^n$$
I was really confused though, because I was certain that the Maclaurin Series for $e^x$ was $$\sum_{n=0}^{\infty}\frac{x^n}{n!}$$
The problem I found with this, is I wondered where the $e$ went in the Maclaurin Series... Isn't a Maclaurin Series just a special case of the Taylor Series with the center at a=0? If that were the case, wouldn't a Taylor Series expansion of $e^x$ about $a=0$ be $$\sum_{n=0}^{\infty}\frac{e}{n!}x^n$$
Why is there no $e$ in the Maclaurin Series?

Addition: I was asked to show my derivation for $e^{2x}$...
\begin{align}
f(x)&=e^{2x}\\
f'(x)&=2e^{2x}\\
f''(x)&=2^2e^{2x}\\
f^{(n)}(x)&=2^ne^{2x}\\
f^{(n)}(a)&=2^ne^2&&x=a=1\\
f(x)&=\sum_{n=0}^{\infty}\frac{f^{(n)}(a)}{n!}(x-a)^n\\
&=\sum_{n=0}^{\infty}\frac{2^ne^2}{n!}(x-1)^n\\
\end{align}
 A: The coefficient for $(x-a)^n$ in the Taylor series of $f(x)$ is:
$$\frac{f^{(n)}(a)}{n!}$$
Compute a few terms of this for $a=0$ and $a=1$ to see why an $e$ shows up in the latter and not the former.
You are right to say that a Maclaurin series is a special type of Taylor series, namely centered at $0$, but this doesn't say much about how similar our coefficients should look.  Evaluating derivatives at differenct values may give us very different looking coefficients.  Centering $\sin x$ at $0$ makes the even valued coefficients vanish, while centering at $\pi/2$ makes the odd valued coefficients vanish.
A: Your reasoning would be correct if you remembered that $e^0 = 1$... 
A: To calculate the Taylor series of $f(x)=e^{2x}$ at $x=1$ we can apply Taylor's Theorem directly. However, an easier way is to use what you already know about the exponential and use some algebra.
$$ e^{2x} = e^{2(x-1+1)} = e^2e^{2(x-1)} = e^2 \sum_{n=0}^{\infty} \frac{1}{n!}(2(x-1))^n = e^2 \sum_{n=0}^{\infty} \frac{2^n}{n!}(x-1)^n$$
done.
