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The question asks to find the Maclaurin series for the function:

\[
f(x) = x \cos\left(\frac{1}{2}x^2\right)
\]

### Steps to find the Maclaurin series:

1. **Maclaurin Series for \(\cos(x)\):**
   The Maclaurin series for \(\cos(x)\) is:
   \[
   \cos(x) = \sum_{n=0}^{\infty} (-1)^n \frac{x^{2n}}{(2n)!}
   \]
   Substituting \( x^2 / 2 \) in place of \( x \) in the series:
   \[
   \cos\left(\frac{1}{2}x^2\right) = \sum_{n=0}^{\infty} (-1)^n \frac{\left(\frac{1}{2}x^2\right)^{2n}}{(2n)!}
   = \sum_{n=0}^{\infty} (-1)^n \frac{x^{4n}}{2^{2n}(2n)!}
   \]

2. **Multiply by \( x \):**
   Now multiply the series by \( x \) to obtain the Maclaurin series for \( f(x) = x \cos\left(\frac{1}{2}x^2\right) \):
   \[
   f(x) = x \sum_{n=0}^{\infty} (-1)^n \frac{x^{4n}}{2^{2n}(2n)!} = \sum_{n=0}^{\infty} (-1)^n \frac{x^{4n+1}}{2^{2n}(2n)!}
   \]

### Maclaurin series for \( f(x) = x \cos\left(\frac{1}{2}x^2\right) \):
\[
f(x) = x - \frac{x^5}{2^2 \cdot 2!} + \frac{x^9}{2^4 \cdot 4!} - \cdots
\]
The first few terms of the series are:
\[
f(x) = x - \frac{x^5}{8} + \frac{x^9}{384} + O(x^{13})
\]

Would you like more details or have any further questions?

Here are five related questions for deeper understanding:
1. How do you derive the general Maclaurin series for any function?
2. Can you use the series to approximate values for specific \( x \)?
3. What are the conditions for a function to have a Maclaurin series?
4. How does the convergence of the series depend on \( x \)?
5. How would you extend this method to other trigonometric functions?

**Tip:** Always check the radius of convergence for a power series when applying it to approximate functions.

Find a Maclaurin series for the function f(x) = x * cos(1/2 * x^2).

Find the Maclaurin series for the function f(x) = x * cos(1/2 * x^2). This solution involves using the Maclaurin series for cos(x) and adapting it to handle the composite function. Suitable for advanced calculus students.

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