Question

find the maclaurin series for (f(x)) using the definition of a maclaurin series. assume that (f) has a power series expansion, but you do not need to show that (r_n(x) \to 0).

f(x) = \frac{\cos(5x)}{4}

f(x) = \sum_{n=0}^{\infty} \left( \frac{(-1)^n (5x)^{2n}}{4(2n)!} \
ight)

find the associated radius of convergence (r).

r = \infty

Explanation:

⚡ Using what you learned: Taylor and Maclaurin Series · Interval and Radius of Convergence

Step 1: Recall the standard Maclaurin series for cosine

$$ \cos(u) = \sum_{n=0}^{\infty} \frac{(-1)^n u^{2n}}{(2n)!} $$

Step 2: Substitute and scale the function

Substitute \( u = 5x \) and divide by \( 4 \):

$$ f(x) = \frac{\cos(5x)}{4} = \sum_{n=0}^{\infty} \frac{(-1)^n (5x)^{2n}}{4(2n)!} $$

Step 3: Determine the radius of convergence

The Maclaurin series for \(\cos(u)\) converges for all real numbers:

$$ R_u = \infty $$

Since \( u = 5x \), the interval of convergence remains all real numbers:

$$ -\infty < 5x < \infty \implies -\infty < x < \infty $$

Thus, the radius of convergence is:

$$ R = \infty $$

Answer:

$$ f(x) = \sum_{n=0}^{\infty} \frac{(-1)^n (5x)^{2n}}{4(2n)!} $$

$$ R = \infty $$