The Maclaurin series is a special case of the Taylor series expansion centered at \( x = 0 \). For the function \( f(x) = e^x \), the Maclaurin series is derived from the function's derivatives evaluated at \( x = 0 \).

The general form of the Maclaurin series for a function \( f(x) \) is:

\[
f(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \frac{f^{(3)}(0)}{3!}x^3 + \cdots + \frac{f^{(n)}(0)}{n!}x^n + \cdots
\]

For \( f(x) = e^x \), all derivatives of \( e^x \) are \( e^x \), and since \( e^0 = 1 \), we have:

\[
f^{(n)}(0) = 1 \quad \text{for all} \, n.
\]

Thus, the Maclaurin series for \( e^x \) is:

\[
e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + \cdots + \frac{x^n}{n!} + \cdots
\]

This is an infinite series that represents \( e^x \) as a sum of powers of \( x \).

Do you have any questions, or would you like more details?

Here are 5 related questions to deepen your understanding:
1. How would the Maclaurin series change if the function were \( e^{2x} \)?
2. How can you use the Maclaurin series to approximate \( e^x \) for small values of \( x \)?
3. What is the general formula for the remainder (error term) in a Taylor or Maclaurin series?
4. How does the convergence of the Maclaurin series for \( e^x \) compare to other functions?
5. Can you derive the Maclaurin series for \( \sin(x) \) or \( \cos(x) \)?

**Tip:** The Maclaurin series is most useful for approximating functions near \( x = 0 \), so it's especially valuable in fields like numerical analysis and physics.

Discover how to derive the Maclaurin series for e^x, using calculus concepts like series expansion and exponential functions. This detailed explanation is suited for high school and undergraduate students, focusing on the infinite series representation of e^x.

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