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I'm quite new to Taylor series and I have some questions.

  1. Given a function $f$, can it be approximated with Taylor polynomials ($n$-degree) iff $f$ is $n$ times differentiable. Right? Or should I think about the interval convergence of Taylor series?

  2. As the $n$ gets bigger, I can approximate further than the expansion point $x_0$, right?

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  • $\begingroup$ Welcome to stackexchange - you are new both here and in your study of Taylor series. The answer to (1) is "no". For (2) it's a qualified "yes". Those are both good questions - you will understand the answers better as your studies progress and you learn about estimating the remainder in a Taylor series expansion. You may get an answer here in the meanwhile spelling out my brief one above. $\endgroup$ – Ethan Bolker Nov 4 '17 at 13:45
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Given a function $f$, can it be approximated with Taylor polynomials ($n$-degree) iff $f$ is $n$ times differentiable.

No, a function can be infinitely differentiable, but the Taylor series can still fail to converge to it, meaning that the $n$th Taylor series can fail to be a good approximation.

The textbook example of this phenomenon is $$f(x)=\begin{cases}e^{-\frac{1}{x^2}} & x>0\\0 & x \leq 0\end{cases}.$$ At $x=0$ all its derivatives are $0$, but it is not the zero function. Notice that the part of the function defined on the positive real line would have a singularity (division by $0$ error) at $x=0$. That is important.

As the $n$ gets bigger, I can approximate further than the expansion point $x_0$, right?

Higher order Taylor approximations are indeed better approximations, as long as you are within the radius of convergence for a well-behaved function. For example the Taylor series for $f(x) = \frac{1}{1-x}$ is $1+x+x^2+\dotsb$. Its radius of convergence is $|x|<1.$ As long as you are considering $x$ values in this range, then higher order Taylor polynomials give better approximations, farther away from $0$.

But once you go past $x=1$, the Taylor series fails to converge, and the Taylor polynomials fail to approximate the function. Notice that the function has a pole at $x=1.$

If a function has a pole, this will typically spoil the nice convergence properties of the Taylor series of the function.

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The answer to both questions is negative. A classical example is this one: considere the function $f\colon\mathbb{R}\longrightarrow\mathbb R$ defined by$$f(x)=\begin{cases}e^{-1/x^2}&\text{ if }x\neq0\\0&\text{ otherwise.}\end{cases}$$It can be proved that $f$ is $n$ times differentiable, for any natural $n$. And it can also be proved that $(\forall n\in\mathbb{N}):f^{(n)}(0)=0$. Therefore, the $n$th Taylor polyonomial of $f$ at $0$ is the null polynomial. So, it is clear that the Taylor polynomials don't get more and more close to $f(x)$ (for any $x\neq0$) as $n\to\infty$.

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  • $\begingroup$ So, what should I check in order to be like I said? $f^{(n)}(x_0) \neq 0$ for all $n$? $\endgroup$ – user_anon Nov 4 '17 at 13:52
  • $\begingroup$ No. What you should check is if the Taylor remainder converges to $0$ or not. $\endgroup$ – José Carlos Santos Nov 4 '17 at 13:55
  • $\begingroup$ This is the classic counterexample for (1). It's also a counterexample for (2) but I don't think that matches the spirit of the OP's question. If the series has a positive radius of convergence then to get some prescribed accuracy uniformly (i.e. 'further away") you do take more terms. $\endgroup$ – Ethan Bolker Nov 4 '17 at 13:55
  • $\begingroup$ @EthanBolker I understand what you mean, but you didn't express yourself correctly. In the case of my example, the radius of convergence of the Taylor series is $+\infty$. Nevertheless, the approximation doesn't get better as $n$ growths. $\endgroup$ – José Carlos Santos Nov 4 '17 at 13:57
  • $\begingroup$ Right. I should have said "a circle of convergence in which the series represents the function". $\endgroup$ – Ethan Bolker Nov 4 '17 at 14:00
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In general Taylor’s theorem comes in a few different phrasings but in general it looks like this:

Given some function $f$ which is $n$ times differentiable at [or on a open set containing] $x$, we have: $$f(x+t) = f(x) + \cdots + \frac{f^{(n-1)}(x)t^{n-1}}{(n-1)!} + R_n(x,t)$$ where the error $R_n(x,t)$ is ...

Different phrasings may give different bounds on the error but as you have seen there are pathological cases where the error does not go to zero as $n\to\infty$. There is a theorem which says that any power series [potentially including negative powers] has a radius [resp. annulus (ring)] of convergence in the complex plane but just because the Taylor series converges to some value, it does not mean that the series converges to the correct value.

One phrasing of the theorem gives $R_n(x,t)=\frac{t^n}{n!}f^{(n)}(\xi)$ for some $\xi\in(x,x+t)$ which depends on $n,x,t.$ This implies that if one can bound all the derivatives of $f$ within some region as one may do with most functions, then the Taylor series must converge to the correct values.

A corollary to the fundamental theorem of applied maths (that “if it looks right then it is”) is that all Taylor series in the real world converge to the right values.

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A standard counter-example is $f(x)= e^{-1/x^2}$, f(0)= 0. It is easy to see that this function is infinitely differentiable, the nth derivative of f is $e^{-1/x^2}$ times a polynomials and that all derivatives are 0 at x= 0. That is, the MacLaurin series, the Taylor series at x= 0, is identically 0 although f is 0 only at x= 0. A function that has a Taylor's series at a point, and that is equal to its Taylor's series in some neighborhood of a point is called "analytic" at that point. Obviously, if a function is "analytic" at a point, it is infinitely differentiable there but infinitely differentiable does NOT imply analytic.

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