Is Convergence of the Maclaurin Series Sufficient to Prove the Convergence of the Taylor Series?

Question

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!

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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.

Answer 1

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.

Answer 2

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.