Notes on Taylor and Maclaurin Series

As you say, there's no guarantee that even a convergent Taylor series[0] converges to the correct value in any interval around the point of expansion. Though the series is of course trivially[1] convergent at the point of expansion itself, since only the constant term doesn't vanish.

The typical example is f(x) = exp(-1/x²) for x ≠ 0; f(0) = 0. The derivatives are mildly annoying to compute, but they must look like f⁽ⁿ⁾(x) = exp(-1/x²)pₙ(1/x) for some polynomials pₙ. Since exponential growth dominates all polynomial growth, it must be the case that f(0) = f'(0) = f"(0) = ··· = 0. In other words, the Taylor series is 0 everywhere, but clearly f(x) ≠ 0 for x ≠ 0. So the series converges only at x = 0. At all other points it predicts the wrong value for f.

The straight-forward real-analytic approach to resolve this issue of goes through the full formulation of Taylor's theorem with an explicit remainder term[2]:

f(x) = Σⁿf⁽ᵏ⁾(a)(x-a)ᵏ/k! + Rₙ(x),

where Rₙ is the remainder term. To clarify, this is a _truncated_ Taylor expansion containing terms k=0,...,n.

There are several explicit expressions for the remainder term, but one that's useful is

Rₙ(x) = f⁽ⁿ⁺¹⁾(ξ)(x-a)ⁿ⁺¹/(n+1)!,

where ξ is not (a priori) fully known but guaranteed to exist in [min(a,x), max(a,x)]. (I.e the closed interval between a and x.)

Let's consider f(x) = cos(x) as an easy example. All derivatives look like ±sin(x) or ±cos(x). This lets us conclude that |f⁽ⁿ⁺¹⁾(ξ)| ≤ 1 for all ξ∈(-∞, ∞). So |Rₙ(x)| ≤ (x-a)ⁿ⁺¹/(n+1)! for all n. Since factorial growth dominates exponential growth, it follows that |Rₙ(x)| → 0 as n → ∞ regardless of which value of a we choose. In other words, we've proved that f(x) - Σⁿf⁽ᵏ⁾(a)(x-a)ᵏ/k! = Rₙ(x) → 0 as n → ∞ for all choices of a. So this is a proof that the value of the Taylor series around any point is in fact cos(x).

Similar proofs for sin(x), exp(x), etc are not much more difficult, and it's not hard to turn this into more general arguments for "good" cases. Trying to use the same machinery on the known counterexample exp(-1/x²) is obviously hopeless as we already know the Taylor series converges to the wrong value here, but it can be illustrative to try (it is an exercise in frustration).

A nicer, more intuitive setting for analysis of power series is complex analysis, which provides an easier and more general theory for when a function equals its Taylor series. This nicer setting is probably the reason the topic is mostly glossed over in introductory calculus/real analysis courses. However, it doesn't necessarily give detailed insight into real-analytic oddities like exp(-1/x²) [3].

[0]: For reference, the Taylor series of a function f around a is: Σf⁽ᵏ⁾(a)(x-a)ᵏ/k!. (I use lack of upper index to indicate an infinite series as opposed to a sum with finitely many terms.)

[1]: At x = a, the Taylor series expansion is f(a) = Σⁿf⁽ᵏ⁾(a)(a-a)ᵏ/k! = f(a) + f'(a)·0 + f"(a)·0² + ··· = f(a). All the terms containing (x-a) vanish.

[2]: https://en.wikipedia.org/wiki/Taylor%27s_theorem#Taylor's_th...

[3]: Something very funky goes on with this function as x → 0 in the complex plane, but this is "masked" in the real case. In the complex case, this function is said to have an essential singularity at x = 0.

I think it is because being holomorphic is so much stronger even than C^\infty (continuous with all derivatives continuous) real much less just continuous or merely integrable.

To build competence at these problems, you need to exercise the basic skills, of which computational skills are a big part. If you don't exercise your computational skills, you can't become good at mathematics in the sense of (1), (2), or (3). The same way you can't get to be good at boxing without doing a lot of pushups.

Now, not all teachers are good. Many certainly try to force students to do too much computational practice, compared to their level. But both as a student and later as a prof, I discovered that there is value is doing really dumb and complicated calculations by hand sometimes during your education. The reason is that many times when people your computational algorithm fails, you have to do the dirty thing by hand to figure out what went wrong. And depending on what you do, computing the 3rd derivative of a random trig function manually correctly with nuances is exactly what you need to do [1].

Well, not everyone. Many/most people do jobs that only require low levels of maths, and they might never need to do any of that. But a math course taught to everyone has to take a cookie cutter approach and that usually requires setting a high enough bar, so people who have to do the hard math at their job, have those skills. This is an unfortunate fact of resource optimization.

[1] I am an industry researcher now, and have to do all kinds of these crappy calculations.