Explain why or why not Determine whether the following statements are true and give an explanation or counterexample.
b. The function f(x) = csc x has a Taylor series centered at π/2.

Verified step by step guidance

1. Recall that a Taylor series of a function \(f(x)\) centered at \(a\) is given by the infinite sum \(\sum_{n=0}^{\infty} \frac{f^{(n)}(a)}{n!} (x - a)^n\), where \(f^{(n)}(a)\) denotes the \(n\)-th derivative of \(f\) evaluated at \(x = a\).
2. To determine if \(f(x) = \csc x\) has a Taylor series centered at \(x = \frac{\pi}{2}\), first check if \(f\) is differentiable at \(x = \frac{\pi}{2}\) and in some open interval around it.
3. Evaluate \(f\left(\frac{\pi}{2}\right) = \csc\left(\frac{\pi}{2}\right) = 1\), which is finite, so the function is defined at \(x = \frac{\pi}{2}\).
4. Next, consider the behavior of \(\csc x\) near \(x = \frac{\pi}{2}\). Since \(\sin x\) is nonzero and smooth around \(\frac{\pi}{2}\), \(\csc x = \frac{1}{\sin x}\) is also smooth and infinitely differentiable there.
5. Therefore, \(f(x) = \csc x\) is infinitely differentiable at \(x = \frac{\pi}{2}\), so it does have a Taylor series centered at \(\frac{\pi}{2}\). The Taylor series converges to \(f(x)\) in some interval around \(\frac{\pi}{2}\) where \(\sin x \neq 0\).

Key Concepts

Taylor Series and Center of Expansion

A Taylor series represents a function as an infinite sum of derivatives at a specific point called the center. The series converges to the function near this point if the function is sufficiently smooth and differentiable there. Understanding the center is crucial to determine if the series exists and converges.

Domain and Differentiability of csc(x)

The function csc(x) = 1/sin(x) is undefined where sin(x) = 0, causing vertical asymptotes. For a Taylor series to exist at a point, the function must be defined and infinitely differentiable there. Checking if csc(x) is defined and smooth at π/2 is essential.

Radius of Convergence and Singularities

The radius of convergence of a Taylor series is limited by the nearest singularity to the center point. Since csc(x) has singularities where sin(x) = 0, the distance from π/2 to these points affects the convergence. Identifying these singularities helps determine if the Taylor series is valid.