0
$\begingroup$

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?

$\endgroup$
4
  • $\begingroup$ If it is infinitely differentiable then it is n times differentiable so that Taylor's Theorem applies. An infinite series is by definition a limit of partial sums so you create a sequence of remainder terms which converge to 0. $\endgroup$
    – Paul
    Jan 9 at 14:54
  • $\begingroup$ "what is $n$?" - we write Talor formula for different number of terms, for each $n$ we get a reminder (it's always well-defined), and not we have just sequence of functions, which we require to converge (pointwise) to $0$. Note that the reminder for the whole series isn't necessary well-defined, because it's possible that the series doesn't converge anywhere but in $x = a$. $\endgroup$
    – mihaild
    Jan 9 at 14:55
  • $\begingroup$ To say that a function is "$n$-times differentiable" means that you can take $n$ derivatives. No more than that. You may or may not be able to take additional derivatives beyond $n$. An infinitely differentiable function is also an $n$-differentiable function for each (finite) natural number $n$. The first $n+1$ terms of the function's Taylor series is the $n$-th Taylor polynomial for the function, and (assuming the Taylor series actually converges to the function), the remainder $R_n$ is the sum of the remaining terms of the Taylor series. $\endgroup$ Jan 10 at 17:43
  • $\begingroup$ Thankyou. I understand it now $\endgroup$ Jan 10 at 23:28

2 Answers 2

2
$\begingroup$

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 $\frac{|R_n(x;c)|}{|x-c|^n}\xrightarrow{x\rightarrow c}0$

$\endgroup$
3
  • 1
    $\begingroup$ FYI, there are also functions having derivatives of all orders at $x=c$ whose Taylor series converges ONLY for $x=c.$ Thus, there are two main ways for a function that is infinitely differentiable at $x=c$ to fail to be real analytic at $x=c$ -- (1) the Taylor series converges only for $x=c$; (2) the Taylor series converges in a neighborhood of $x=c$ but differs from the function used to form the Taylor series in every neighborhood of $x=c.$ For more details about this (probably too advanced for the OP, however), see this MSE answer. $\endgroup$ Jan 10 at 23:42
  • $\begingroup$ @DaveL.Renfro: Thanks very much for the link. I am aware of two interesting facts (exercises in 12 and 13, Chapter 19 in Rudin's RCA) which go along the lines of your comment. I will take a look at your two part essays for my own mathematical amusement. I did not add anything about the result by Borel since I thought it was already too much for the OP. Perhaps I will add it to the Community wiki posting. If you feel like adding more to it, I'll be honored too. $\endgroup$
    – Mittens
    Jan 11 at 1:43
  • $\begingroup$ Probably more relevant to the OP is how Taylor's theorem is often used, or at least how to justify the uses that are often made, such as in finding limits of quotients by keeping the most significant/dominant terms -- see "Evaluate the following limit by the Taylor series method" here (Taylor's theorem justifies jettisoning tails of Taylor expansions) -- and by obtaining (besides linear) best quadratic, cubic, etc. approximations to a function at a point (examples). $\endgroup$ Jan 11 at 10:19
1
$\begingroup$

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.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .