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: = \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.