Given the following Taylor series:

$1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+\frac{x^8}{8!}- \dots$

We know that:

  1. It converges for all of $x$
  2. It converges to the function $\cos x$

The Taylor series converge for all of $x$ if for a fixed value of $x$, the partial sums converge to a limit, $L$.

The Taylor series converge to $\cos x$ if its error term is $0$ as $n$ (the number of terms in the Taylor series) goes to infinity.

My question is:

Are these two concepts related?

I thought point 1 would be useful when proving point 2. But when doing the proof of 2, I don't see any connection to point 1. If the two concepts are not related, then why is it useful to know the interval of convergence of a Taylor series (or any series for that matter)?


Clearly they are related: 2. implies 1. But you are correct that using Taylor's Theorem with Remainder, it is sometimes possible to show 2. directly without first establishing the interval of convergence of the power series.

As for 1. Well, for instance what if the function is defined as a power series? E.g. this is perhaps the cleanest definition for the transcendental elementary functions like $e^x$, $\sin x$, etc.

Also, in practice it is usually much easier to determine the radius of convergence of the Taylor series than to apply Taylor's Theorem with Remainder. If one knows for some other reason that a function is analytic -- e.g. if it satisfies a suitably nice differential equation -- then one gets 2. from 1. more or less for free.

Added: Although I feel that the other answers may be missing a bit of the richness of the question (it's a very good question), they do make one good point: the interval of convergence of the Taylor series is certainly an upper bound for the interval on which the Taylor series could converge to $f$. When this interval is not simply all of $\mathbb{R}$ it is useful to know what it is. Consider for instance the case of the Binomial Series: one has to work a bit to get the full interval of convergence out of Taylor's Theorem with Remainder (although it can be done). If we didn't know that the radius of convergence is $1$, how would we know what we are trying to show? You can see the details worked out in $\S$ 12.4.3 of these notes.

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  • $\begingroup$ Could you please clarify what do you mean to say in paragraph 2? Not sure why you wanted to mention if the function is defined as a power series. Also, do you know of any notes that attempt to proof 2 by first assuming 1? I'm now convinced that 2 implies 1 but I don't see how if I assume 1, I can get to 2. $\endgroup$ – mauna Nov 4 '13 at 19:59
  • $\begingroup$ @mauna: If a function is defined as a power series, it is automatically (or, perhaps, easily shown once and for all) to be equal to its Taylor series. As for 1. implies 2.: it is not always true, of course; there are standard counterexamples. As I said, you can show that 1. implies 2. e.g. for solutions of differential equations. This applies to $\cos x$, for instance. $\endgroup$ – Pete L. Clark Nov 4 '13 at 20:59
  • $\begingroup$ "the interval of convergence of the Taylor series is certainly an upper bound for the interval on which the Taylor series could converge to f" Wow!Thank you very much $\endgroup$ – viru Sep 21 '18 at 16:07

When you say that a series converges you are guaranteeing that you can keep it within some range you define. When you say that it converges to a function, then you are saying that within any error bounds you can choose enough terms of the sequence to get within the desired error bounds. When you are proving that the sequence converges to a specific function you can assume that it is that function and then find the desired number of terms needed to converge to that function. Basically, #1 is useful for proving #2 because you know it does converge so you can try to find what it converges to. The interval of convergence of a Taylor series is important because you can only use the Taylor series to approximate the function in the interval of convergence, which is why you use Taylor series. If you try to use the Taylor series outside of this interval your series won't converge.

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Taylor series for functions with poles in the complex plane have a finite radius of convergence. For example, you can show that the Taylor series \begin{equation} \frac{1}{1 + x^2} = \sum_n (-1)^n x^{2n} \end{equation} converges to $(1 + x^2)^{-1}$, but only for $|x| < 1$. Thus, it has (2) without having (1). (1) guarantees that your Taylor series is good anywhere, while (2) tells you what the Taylor series is representing somewhere. So if you are trying to prove (2), you have to be sure that unless (1) holds you are specifying where the series converges to the function (e.g. on the interval $x \in (-1, 1)$).

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  • $\begingroup$ It seems to me that a more useful interpretation of 1. is determining the interval of convergence of the Taylor series. In the given example it is all of $\mathbb{R}$ (or $\mathbb{C}$, if you like), but I didn't see anything the question to suggest that the OP thinks this is always the case. $\endgroup$ – Pete L. Clark Nov 4 '13 at 18:44

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