What is the connection between Taylor's Theorem and Taylor Series

Question

In my calculus classes, we have covered Taylor's Theorem and Taylor Series. But the way my lecturer covered them, they were disconnected from each other. However, just by looking at it, I know it must be the case that the Taylor Series is some special case of Taylor's theorem, I just can't quite get how.

From my understanding, for some $f: \mathbb{R} \to \mathbb{R}$ which is $n$-times differentiable at $a \in \mathbb{R}$, there exists some function $h_k: \mathbb{R} \to \mathbb{R}$ such that: $$ \begin{gathered} f(x) = \sum^{n}_{i=0} \frac{f^{(i)}(a)}{i!}(x-a)^i + h_n(x)(x-a)^{n} \\ \\ \text{and also} \\ \\ \lim_{x \to a} h_n(x) = 0 \end{gathered} $$ And then the last term, $R_n(x) = h_n(x)(x-a)^{n}$ is called the remainder, and under stronger assumptions, the remainder even has more precise forms, like Lagrange form etc.

Now, when we are talking about Tailor Series, we are talking about infinitely differentiable functions, which is what confuses me.

So, for some $f: \mathbb{R} \to \mathbb{R}$ which is $\infty$-times differentiable at $a \in \mathbb{R}$, does there exists some function $h_{\infty}: \mathbb{R} \to \mathbb{R}$? What would that even mean? And I understand that for a function to be equal to its Tailor Series expansion, the remainder ($R_{\infty}(x)$??) will have to be equal to zero, so that the whole function is described entirely by the polynomial terms. But how would one write this condition/special case? Is it a limit of some kind? I've seen in other answers someone writing this condition as:

$$ \lim_{n \to \infty} R_n(x) = 0 $$

But that doesn't make sense to me, because, what is $n$? No idea. If $n$ is the differentiability of $f$ then $n$ IS $\infty$, so would it be $\infty \to \infty$?

What is the clear and correct way to extend the $R_n$ and $h_n$ to infinitely differentiable functions, and what is the correct way to express the condition that $R_{\infty}$?? should be zero?

Answer 1

Formal Taylor series around a point $c\in\mathbb{R}$ exists only for functions $f$ for which there is an interval $U$ that contains $c$ on which $f$ has derivatives of all orders, that is $f^{(n)}(x)$ exists for any $x\in U$ and $n\in\mathbb{Z}_+$. If $f$ satisfies such a condition, then there is a formal power (or Taylor) series of the given by $$\sum^\infty_{n=0}\frac{f^{(n)}(c)}{n!}(x-c)^n=f(c)+\frac{f'(0)}{1!}(x-c)+\ldots +\frac{f^{(n)}(c)}{n!}(x-c)^n+\ldots$$

The issue are whether

• (1) The formal series converges at any point in $U$ (other that $c$) and if so,
• (2) is $$f(x)=\sum^\infty_{n=0}\frac{f^{(n)}(c)}{n!}(x-c)^n$$ for all $x\in U$?

Functions for which (1) and (2) hold are very special. They are the real analytic functions (the concept of derivative, convergence and power series is carried out to functions defined complex domain with complex values, which is the subject of Complex Analysis).

Sticking to the real line, there are functions that have derivatives of any order in an interval and for which (2) do a not hold. The classic example is $f(x)=e^{-1/x}$ for $x>0$ and $f(x)=0$ for $x\leq0$. IT is a good exercise to show that $f^{(n)}(x)$ exists for any $x\in\mathbb{R}$ and that $f^{(n)}(0)=0$ for all $n\in\mathbb{Z}_+$. It follows $$f(x)\neq \sum^\infty_{n=0}\frac{f^{(n)}(0)}{n!}x^n \quad\text{if}\quad x>0$$ Thus, $f$ is not analytic! Of course, in this case $f(x)=R_n(x;0)$ for all $x\in\mathbb{R}$ and $n\in\mathbb{N}$.

On the other hand, there is the well known result in Calculus that states that if $f$ has derivatives of order up to $N+1$ in some interval $(a,b)$, then for any $c\in (a,b)$ $$f(x)=f(c)+\frac{f'(c)}{1!}(x-c)+\ldots+\frac{f^{(n)}(c)}{n!}(x-c)^n + R_N(x;c)$$ where $$\frac{|R_n(x;c)|}{|x-c|^n}\xrightarrow{x\rightarrow c}0$$ There are different expressions for the residue $R_c(x;c)$ (Lagrange, Cauchy, Schömich-Roché, etc) which can be ontained by different applications of the mean value theorem.

If $f$ is (real) analytic at $c$ then by convergence of series considerations it follows that $$R_n(x;n)=\sum^\infty_{k={n+1}}\frac{f^{(k)}(c)}{k!}(x-c)^k$$ and $\frac{|R_n(x;c)|}{|x-c|^n}\xrightarrow{x\rightarrow c}0$

Answer 2

Thanks to Paul, mihaild and Paul Sinclair for making it click. I will quickly answer my own question with their insight.

For some $f: \mathbb{R} \to \mathbb{R}$ to be $\infty$-times differentiable at $a \in \mathbb{R}$ means that for all $n \in \mathbb{N}$, $f$ is $n$-times differentiable, and Tailor Theorem applies for every $n$. Hence you have: $$ \begin{gathered} f(x) & = & f(a) + h_0(x) \\ & = & f(a) + f^{'}(a)(x-a) + h_1(x)(x-a) \\ & = & f(a) + f^{'}(a)(x-a) + \frac{f^{''}(a)}{2}(x-a)^2 + h_2(x)(x-a)^{2} \\ &=&\dots \\ & = & \sum^{n}_{i=0} \frac{f^{(i)}(a)}{i!}(x-a)^i + h_n(x)(x-a)^{n} \\ &=&\dots \end{gathered} $$ And thus you can define a sequence $( \ R_{n}(x) \ )_{n\geq 0} = h_n(x)(x-a)^{n}$ that captures all these different remainders for each application of the Tailor Series, and using this sequence, we have it that IF for all $x$ $$ \lim_{n \to \infty} R_n(x) = 0 $$ Then the Tailor Series expansion of the function $f$ equals exactly to the function $f$, i.e. $f$ is analytic.

This is my understanding I get from these comments, If this is incorrect, please correct it.