We are asked to find the coefficients $c_n$ for the Taylor series of $e^x$ centered at $x = 2$. The Taylor series expansion of a function $f(x)$ around $a = 2$ is given by:

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

Here, $f(x) = e^x$, so all the derivatives of $e^x$ are still $e^x$. Therefore, at $x = 2$, $f^{(n)}(2) = e^2$ for all $n$.

Thus, the general form for the Taylor series is:

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

In this case, the coefficients $c_n$ correspond to $\frac{e^2}{n!}$. Let's compute the values for $c_0$ through $c_4$.

1. $c_0 = \frac{e^2}{0!} = e^2$
2. $c_1 = \frac{e^2}{1!} = e^2$
3. $c_2 = \frac{e^2}{2!} = \frac{e^2}{2}$
4. $c_3 = \frac{e^2}{3!} = \frac{e^2}{6}$
5. $c_4 = \frac{e^2}{4!} = \frac{e^2}{24}$

So, the coefficients are:

$c_0 = e^2, \quad c_1 = e^2, \quad c_2 = \frac{e^2}{2}, \quad c_3 = \frac{e^2}{6}, \quad c_4 = \frac{e^2}{24}$

Would you like a detailed explanation on how to find the general formula for Taylor series or any specific part? Here are 5 relative questions to expand:

1. How do you compute higher-order Taylor coefficients?
2. Why does the Taylor series of $e^x$ have all derivatives equal to $e^x$?
3. What does the radius of convergence mean for a Taylor series?
4. How can you approximate functions using a truncated Taylor series?
5. What is the error bound when truncating the Taylor series at a certain degree?

Tip: When working with Taylor series, keep in mind that the closer $x$ is to the center $a$, the faster the series converges to the actual value of the function.