To find the Maclaurin series for $f(x) = 5x \cos\left(\frac{1}{4}x^2\right)$, we can start by expanding the $\cos\left(\frac{1}{4}x^2\right)$ function in its Maclaurin series form and then multiply it by $5x$.

Step-by-Step Solution

1. Maclaurin Series for $\cos(x)$: The Maclaurin series for $\cos(x)$ is: $\cos(x) = \sum_{k=0}^{\infty} \frac{(-1)^k}{(2k)!} x^{2k} = 1 - \frac{x^2}{2} + \frac{x^4}{24} - \frac{x^6}{720} + \dots$

2. Substitute $x = \frac{1}{4}x^2$ in $\cos(x)$: Substitute $x$ with $\frac{1}{4}x^2$ in the series for $\cos(x)$: $\cos\left(\frac{1}{4}x^2\right) = 1 - \frac{\left(\frac{1}{4}x^2\right)^2}{2} + \frac{\left(\frac{1}{4}x^2\right)^4}{24} - \dots$

   Expanding each term: $= 1 - \frac{x^4}{32} + \frac{x^8}{1536} - \dots$

3. Multiply by $5x$: Now, we multiply each term in the series by $5x$: $f(x) = 5x \cdot \left(1 - \frac{x^4}{32} + \frac{x^8}{1536} - \dots \right)$

   Distribute $5x$ through the terms: $f(x) = 5x - \frac{5x^5}{32} + \frac{5x^9}{1536} - \dots$

Final Answer

The Maclaurin series for $f(x) = 5x \cos\left(\frac{1}{4}x^2\right)$ is: $f(x) = 5x - \frac{5x^5}{32} + \frac{5x^9}{1536} - \dots$

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

Related Questions

1. How do you derive the Maclaurin series for basic trigonometric functions?
2. Can you expand a Maclaurin series for $\sin\left(\frac{1}{4}x^2\right)$ similarly?
3. What is the general approach to obtain a Maclaurin series for a product of functions?
4. How does the convergence of the series behave near $x = 0$?
5. How can you find the radius of convergence for this series?

Tip: For products of functions, first expand each part separately and then multiply their series expansions term by term.