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When demonstrating the convergence of the Taylor series of a function like $e^x$, to the function (i.e. $R_n(x) \to 0$) is it sufficient to prove convergence of the Maclaurin series?

If the radius of convergence is non-infinite, then I can imagine it would matter; however, for functions like $e^x$, $\sin(x)$ and $\cos(x)$, where their radii are infinite, it somehow seems less relevant.

Practically speaking, I know the difference in the proof is minor, so my question is really aimed at understanding how one determines whether a given simplification in a proof affects the generality of the result, in this particular case.

Thank you in advance for your time and help!


Edit for further clarification

e.g. to show that the Maclaurin series for $e^x$ converges to $e^x$, we must show that $f(x) = \lim_{n\to\infty} T_n(x)$, which is equivalent to showing that the remainder $R_n(x) = f(x)-T_n(x) \to 0$.

As such, we can use Taylor's inequality:

$$ |R_n(x)|= f^{(n+1)}(z) \; \frac{|x|^{n+1}}{(n+1)!} \leq M \; \frac{|x|^{n+1}}{(n+1)!} $$

Noting that, for $z \in \mathrm{I\!R}$, $f^{(n+1)}(z) = e^z$. So, for any $I = [-d, d]$, we can take $M = e^d$. Plugging this into the above inequality, and taking the limit, we have:

$$ \begin{aligned} \lim_{n\to\infty} |R_n(x)| & \leq \lim_{n\to\infty} e^d \; \frac{|x|^{n+1}}{(n+1)!}, \qquad \textrm{for $x \in I$} \\[10pt] % % & \leq e^d \lim_{n\to\infty} \frac{|x|^{n+1}}{(n+1)!} \\[10pt] % % & = 0 \end{aligned} $$

The proof that $T_n(x) \to f(x)$ in the general case of expansion about $x = c$ is entirely analogous.

This lead me to wonder if:

  • a) proving the case for the Maclaurin series is sufficient to imply the result for the general Taylor series (without further work) and
  • b) (if so) does this hold for other Taylor series whose radii of convergence are infinite.
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  • $\begingroup$ Are you asking, "If the radius of convergence of the Maclaurin series is infinite, does it follow that the Taylor series at every point has infinite radius of convergence?" If so, the answer is yes. $\endgroup$
    – saulspatz
    Jun 20, 2019 at 14:01
  • $\begingroup$ No, I'm asking "if the Maclaurin series converges to the function, does that imply that the Taylor series will also converge to the function," ie is convergence affected by the point about which we expand. The detail about the radii is just a guess that the answer will depend on whether or not the radius is infinite. $\endgroup$
    – Rax Adaam
    Jun 20, 2019 at 14:23
  • $\begingroup$ The Taylor series will converge if and only if you expand about a point in the interval of convergence of the Maclaurin series. $\endgroup$
    – saulspatz
    Jun 20, 2019 at 14:26
  • $\begingroup$ Sorry, as I mentioned in my last comment, I mean "converges to the function," I'm not talking about convergence of the series, as a series; rather, convergence of the series to the function (i.e. proving that the function is equal to its Tyalor series). I have edited the question to emphasize this; apologies for the lack of clarity. $\endgroup$
    – Rax Adaam
    Jun 20, 2019 at 14:31
  • $\begingroup$ Ok, I see; I was only thinking of the case where the Maclaurin series represents the function. I'm not sure what, if anything, one can say if it doesn't. $\endgroup$
    – saulspatz
    Jun 20, 2019 at 14:40

2 Answers 2

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If the Maclaurin series converges, but does not represent the function, then nothing can be deduced about the Taylor series, even if the radius of convergence of the Maclaurin series is infinite.

Consider the standard example $$f(x)=\cases{e^{-1/x^2}, &$x\neq0$\\ 0, &$x=0$}$$

It's not hard to show that all the coefficients of the Maclaurin series are $0$. (The derivatives of $f$ are all of the form $e^{-1/x^2}P_n(1/x)$ for polynomials $P_n$.) Therefore, the radius of convergence is infinite. In this case, the Taylor series about $x=a$ represents the function in an interval of positive length, for every real number $a\neq0$.

However, the Maclaurin series only depends on the values of the function in an arbitrarily small interval about the origin. Given $a\neq0$ we can modify $f$ to get a function $f_a$ that agrees with $f$ in a small interval about $0$, but exhibits the same behavior near $a$ that $f$ does near $0$. The Maclaurin series of $f_a$ is still identically $0$, and the Taylor series about $x=1$, which does not represent the function.

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  • $\begingroup$ Thank you for answering that related question: I wouldn't have expected to be able to deduce anything about the convergence of the series to the function if a special case didn't converge, but it's a useful question to consider in parallel. Do you know of any more examples like the above? I've always found that example rather artificial, as it seems constructed-to-make-the-point. I'm familiar with the "bump" function, as well, but I'd be curious to see an example of a non-piecewise function, if you know one. Thanks again for your time and help! $\endgroup$
    – Rax Adaam
    Jun 20, 2019 at 15:03
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    $\begingroup$ No, I don't know of any other examples. The bump functions are used to construct $C^{\infty}$ partitions of unity, which are important tools in advanced analysis. I don't know if they were constructed first as counterexamples, or as tools. $\endgroup$
    – saulspatz
    Jun 20, 2019 at 15:09
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Considering $f(x) = exp(x)$ it does not matter because of the following argument.

So let's say you have proven that the convergence radius of the Maclaurin series of the function $f(x)$ is infinite. Therefore $f(x) = \Sigma_{i = 0}^{\infty}a_ix^i$ for all $x \in R$. Now let $x_0 \in R$ be arbitrary. Then $f(x) \overset{f = exp}=f(x_0)f(x - x_0) = f(x_0)\Sigma_{i = 0}^{\infty}a_i(x- x_0)^i = \Sigma_{i = 0}^{\infty}b_i(x - x_0)^i$ where the last expression corresponds to the taylor series around $x_0$ of $f$. Thus the taylor series has infinite radius of convergence everywhere.

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    $\begingroup$ I'm not sure about that last paragraph - the coefficients for a general function still depend on the point about which one expands. Also, to be clear, I'm talking about proving that the series converges to $f(x)$, so it is the handling of the above in terms of the remainder that I think is important (however, I think it is a direct extension of what you've presented). Anyhow, very much appreciate the presentation for $e^x$ - very clever. Thank you for your help! $\endgroup$
    – Rax Adaam
    Jun 20, 2019 at 14:28
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    $\begingroup$ You are right I made a bad mistake there. Thanks for pointing it out. I took it away. $\endgroup$
    – sebastian
    Jun 20, 2019 at 14:32

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