The Maclaurin polynomial is a type of Taylor polynomial centered at $x = 0$. For the function $f(x) = e^{-2x}$, we need to compute the Maclaurin series up to the 4th degree.

The general form of a Maclaurin series is:

$f(x) = f(0) + f'(0) \frac{x}{1!} + f''(0) \frac{x^2}{2!} + f^{(3)}(0) \frac{x^3}{3!} + f^{(4)}(0) \frac{x^4}{4!} + \cdots$

Step 1: Compute the derivatives of $e^{-2x}$

1. $f(x) = e^{-2x}$
2. $f'(x) = -2e^{-2x}$
3. $f''(x) = 4e^{-2x}$
4. $f^{(3)}(x) = -8e^{-2x}$
5. $f^{(4)}(x) = 16e^{-2x}$

Step 2: Evaluate these derivatives at $x = 0$

1. $f(0) = e^{0} = 1$
2. $f'(0) = -2$
3. $f''(0) = 4$
4. $f^{(3)}(0) = -8$
5. $f^{(4)}(0) = 16$

Step 3: Construct the Maclaurin polynomial

Using the formula for the Maclaurin series, we plug in the values for the function and its derivatives at $x = 0$:

$P_4(x) = 1 - 2 \frac{x}{1!} + 4 \frac{x^2}{2!} - 8 \frac{x^3}{3!} + 16 \frac{x^4}{4!}$

Now simplify each term:

$P_4(x) = 1 - 2x + 2x^2 - \frac{4x^3}{3} + \frac{x^4}{3}$

Thus, the 4th-degree Maclaurin polynomial for $e^{-2x}$ is:

$P_4(x) = 1 - 2x + 2x^2 - \frac{4x^3}{3} + \frac{x^4}{3}$

Would you like a detailed explanation of how each derivative was calculated or clarification on any of the steps?

Related Questions:

1. How is the general Taylor series derived from a function?
2. What are the convergence criteria for a Maclaurin series?
3. Can we derive the Maclaurin series for $e^{2x}$ and compare?
4. What is the relationship between the Taylor series and the function's radius of convergence?
5. How does the error bound in a truncated Taylor series work?

Tip:

When working with exponential functions in a Maclaurin or Taylor series, it’s useful to recognize patterns in their derivatives to speed up computations.