Hey there, let's tackle these questions step by step—they get at some key nuances between Taylor series and power series that trip up a lot of folks!

Short answer: Yes, absolutely. Here's why:

• A Taylor series is a specific type of power series—it's centered at some point (a), with coefficients defined by the function's derivatives at (a) (i.e., (\sum_{n=0}^\infty \frac{f^{{(n)}(a)}{n!}(x-a)}n)).
• Even if the Taylor series converges to a function that's different from the original function at some points (a rare but well-known edge case), the series itself is still a power series. The only catch is whether the series sums to the original function (we call functions where this holds analytic at (a)), but the series structure remains a power series regardless.

First, let's nail down the core relationship: If a function can be represented by a power series (\sum_{n=0}^\infty c_n(x-a)^n) on some interval around (a), then this power series is exactly the Taylor series of the function centered at (a). This comes from the uniqueness of power series representations—there's only one way to write a function as a power series around a point, and it's determined by the function's derivatives at that point.

Now let's break down your two follow-ups:

子问题2.1：已知幂级数的收敛半径，是否可将其直接应用于中心位于该半径范围内的泰勒级数？ #

No, you can't assume that. The convergence radius of a Taylor series depends on the function's analyticity (where it's smooth and has no "singularities"), not just the convergence radius of a different power series for the same function.

For example:

• Take (f(x) = \frac{1}{1+x^2}). Its Maclaurin series (Taylor series centered at 0) is (\sum_{n=0}^\infty (-1)^n x^{2n}), with convergence radius (R=1).
• If we center a Taylor series at (a=0.5) (which is inside the original radius of 1), the new Taylor series has a convergence radius of ~1.118—this is the distance from (0.5) to the nearest singularity of (f(x)) in the complex plane ((x=\pm i)).

The key takeaway: Each Taylor series's convergence radius is independent of other power series for the same function. It's determined by how far you can get from the center before hitting a point where the function can't be smoothly extended.

子问题2.2：若由某函数生成泰勒级数，该级数可能无法…… #

I’m guessing you’re referring to whether the series might fail to converge to the original function—if so, yes, this absolutely can happen.

The classic example is this piecewise function:
[
f(x) =
\begin{cases}
e^{-1/x2} & \text{if } x \neq 0, \
0 & \text{if } x = 0.
\end{cases}
]
This function has all derivatives equal to 0 at (x=0), so its Taylor series is (\sum_{n=0}^\infty 0 \cdot x^n = 0). The series converges everywhere on (\mathbb{R}), but only equals (f(x)) at (x=0); for all other (x), the series sum (0) is nothing like (f(x)) (which is positive and approaches 0 as (x\to\pm\infty)).

We call functions like this non-analytic at the center point—they have a valid Taylor series, but the series doesn't capture the function's behavior away from the center.

内容的提问来源于stack exchange，提问作者Sedumjoy