Question

Finding Taylor Series at x = 0 (Maclaurin Series)Find the Maclaurin series for the functions in Exercises 11–24.5 cos πx

Accepted Answer

Recall that the Maclaurin series is the Taylor series expansion of a function at $$x = 0$$, given by the formula:  
$$f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!} x^n,$$  
where $$f^{(n)}(0$$)$$ is the n$-th derivative of f$ evaluated at 0. Identify the base function inside the given function. Here, the function is f(x) = 5$$ \cos(\pi x)$$. We can factor out the constant 5 and focus on the Maclaurin series for $$\cos(\pi x)$$. Recall the Maclaurin series for $$\cos u$$ is  
$$\cos u = \sum_{n=0}^{\infty} (-1)^n \frac{u^{2n}}{(2n)!}.$$  
In this problem, substitute u$$ = \pi x$$ to get  
$$\cos(\pi x) = \sum_{n=0}^{\infty} (-1)^n \frac{(\pi x)^{2n}}{(2n)!}.$$ Multiply the entire series by 5 to account for the coefficient in the original function:  
5$$ \cos(\pi x) = 5 \sum_{n=0}^{\infty} (-1)^n \frac{(\pi x)^{2n}}{(2n)!} = \sum_{n=0}^{\infty} 5 (-1)^n \frac{\pi^{2n} x^{2n}}{(2n)!}.$$ Write out the first few terms explicitly by plugging in n=0$$,1,2,\ldots$$ to see the pattern of the series, which helps in understanding the expansion and verifying correctness.

Finding Taylor Series at x = 0 (Maclaurin Series)
Find the Maclaurin series for the functions in Exercises 11–24.
5 cos πx

Verified video answer for a similar problem:

Key Concepts

Maclaurin Series

Derivatives of Trigonometric Functions

Chain Rule in Differentiation