泰勒级数何时可实现完美近似？其适用与偏差场景探讨

Great question—this cuts right to the heart of why Taylor series are such powerful tools, but also where their limits show up. Let’s break this down clearly:

When the Approximation is "Perfect" (Error Can Be Made Arbitrarily Small) #

We can call the approximation "perfect" for a given interval if, for any tiny error threshold $\epsilon > 0$ you pick, there’s always a number of terms $N$ such that using the first $N$ terms of the Taylor series gives a value within $\epsilon$ of the actual function value everywhere in that interval. This happens when:

• The function is analytic at the point $x=a$ (the center of the series). Analytic means two things: the function has infinitely many derivatives at $x=a$, and the remainder term of the Taylor polynomial goes to 0 as we take more and more terms.
  • For example, functions like $e^x$, $\sin x$, $\cos x$ are analytic everywhere on the real line. Their Taylor series (centered at 0, called Maclaurin series) will converge to the original function no matter what $x$ you plug in—keep adding terms, and the error shrinks to as close to 0 as you want.
• Even if the function isn’t analytic everywhere, it might be analytic in a local neighborhood around $x=a$. Take $\ln(1+x)$: its Maclaurin series converges to the original function on the interval $(-1, 1]$. Inside this interval, you can get arbitrarily close to the true $\ln(1+x)$ value with enough terms, but step outside (like $x=2$ or $x=-2$) and the approximation falls apart.

To put it technically, if we use the Lagrange form of the remainder $R_n(x) = \frac{f^{{(n+1)}(c)}{(n+1)!}(x-a)}{n+1}$ (where $c$ is between $a$ and $x$), then the series converges to $f(x)$ exactly when $\lim_{n\to\infty} R_n(x) = 0$ for all $x$ in the interval.

When the Infinite Series Still Has a Deviation #

There are two main scenarios where even an infinite Taylor series won’t match the original function:

• Non-analytic smooth functions: These are functions that have infinitely many derivatives at $x=a$, but their Taylor series doesn’t converge to the function anywhere except at $x=a$ itself. The classic example is:

  f(x) = 
  { 
    e^(-1/x²)  if x ≠ 0,
    0          if x = 0
  }
  

  Every derivative of this function at $x=0$ is 0, so its Taylor series is just the constant 0. But for any $x ≠ 0$, $f(x)$ is positive and non-zero—so the series never gets close to the actual function value, no matter how many terms you add. This is a "flat" function that’s infinitely smooth but not analytic.
• Convergence radius limits: Some Taylor series only converge within a finite interval (called the convergence radius). Outside this interval, the series itself diverges (the partial sums blow up or oscillate wildly), so it can’t possibly approximate the original function. For example, the Maclaurin series for $\frac{1}{1-x}$ is $\sum_{n=0}^\infty x^n$, which only converges when $|x| < 1$. For $|x| ≥ 1$, the series doesn’t converge at all, so it’s useless for approximating the function there.

内容的提问来源于stack exchange，提问作者user525966