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}
Best Answer
Your reasoning would be correct if you remembered that $e^0 = 1$...