make a Maclaurin series for this x*sin(x)

The Maclaurin series is a Taylor series expansion of a function about \(x = 0\). To find the Maclaurin series for the function \(f(x) = x \sin(x)\), we can follow these steps:

1. **Define the function and its derivatives**.
2. Evaluate each derivative at \(x = 0\).
3. Substitute into the Maclaurin series formula:

\[
f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!} x^n
\]

Let's go through these steps:

### 1. Define the Function and Compute Derivatives

The function is:
\[
f(x) = x \sin(x)
\]

Let's calculate the first few derivatives:

- \( f'(x) = \sin(x) + x \cos(x) \)
- \( f''(x) = 2\cos(x) - x \sin(x) \)
- \( f'''(x) = -3 \sin(x) - x \cos(x) \)
- \( f^{(4)}(x) = -4 \cos(x) + x \sin(x) \)

### 2. Evaluate the Derivatives at \(x = 0\)

Now, substitute \(x = 0\) into each derivative:

- \( f(0) = 0 \)
- \( f'(0) = \sin(0) + 0 \cdot \cos(0) = 0 \)
- \( f''(0) = 2\cos(0) - 0 \cdot \sin(0) = 2 \)
- \( f'''(0) = -3 \sin(0) - 0 \cdot \cos(0) = 0 \)
- \( f^{(4)}(0) = -4 \cos(0) + 0 \cdot \sin(0) = -4 \)

### 3. Write the Maclaurin Series

Substitute these values into the Maclaurin series formula:
\[
f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!} x^n
\]

Using the values we found:
\[
f(x) = 0 + 0 \cdot x + \frac{2}{2!}x^2 + 0 \cdot x^3 + \frac{-4}{4!}x^4 + \cdots
\]

Simplify the terms:
\[
f(x) = x^2 - \frac{x^4}{6} + O(x^6)
\]

### Final Maclaurin Series for \(x \sin(x)\):
\[
x \sin(x) = x^2 - \frac{x^4}{6} + O(x^6)
\]

This series continues with higher-order terms, but the expression above provides the first few non-zero terms of the series.

Would you like more details or have any questions about this process?

Here are 5 related questions to expand your understanding:
1. How do you determine the convergence radius of a Maclaurin series?
2. What is the general formula for the Maclaurin series of a sine function?
3. How does the Maclaurin series of \(x \sin(x)\) compare to that of \(\sin(x)\) alone?
4. How do you derive higher-order derivatives efficiently for series expansion?
5. What are the applications of Maclaurin series in solving differential equations?

**Tip:** For complex functions, calculating just a few terms of the Maclaurin series can often provide a good approximation near \(x = 0\).

Learn how to derive the Maclaurin series for the function x*sin(x) with detailed steps, including the function's derivatives, their evaluation at x=0, and the resulting series. This problem covers calculus concepts, such as series expansions and trigonometric derivatives. Suitable for advanced high school or early university students.

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