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If I want to check whether a Taylor series converges to its function f(x). Is it enough to check if the radius of convergence of the Taylor series is infinite?

Or do I have to use the remainder theorem and show the remainder converges to zero?


Edit: I'm still confused about convergence of taylor series. If I have a function f(x) which is defined on $x\in ]-R,R[$, and I found it's taylor series T(x) with a radius of convergence R. Doesn't the taylor series converge to f(x)?

Why do I necessarily need to check that the remainder $R_n(x)$ converges to zero for it to be true?

In other words, what does the taylor series converge to, if it doesn't converge to f(x)? (Because I thought if a taylor series was convergent it would always be towards f(x)? But that's probably where the whole issue is?).

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    $\begingroup$ $f(X)=e^{-1/x}$ for $x >0$ $f(x)=0$ for $x \leq 0$. All the terms in the Taylor series are $0$ for this function. $\endgroup$ Nov 26 '20 at 23:43
  • $\begingroup$ @KaviRamaMurthy, but what if I work with regular functions like e^x, cosx, ln(x), sinx*cosx, etc. If I found the radius of convergence to be the same as the interval that the original function is defined. Would the series converge? $\endgroup$
    – sjm23
    Nov 26 '20 at 23:49
  • $\begingroup$ For $\mathrm e^x, \sin x,\cos x, \sinh x,\cosh x$, it does converge. $\endgroup$
    – Bernard
    Nov 26 '20 at 23:53
  • $\begingroup$ @Bernard but can I conclude that because the radius of convergence is infinite? Or that isn't enough? $\endgroup$
    – sjm23
    Nov 26 '20 at 23:58
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    $\begingroup$ @sjm23 If by "normal" you mean "analytic", then yes by definition. $\endgroup$
    – user239203
    Nov 26 '20 at 23:58
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An infinitely differentiable function of a real variable is not necessarily the sum of its Taylor series. I'll give an example below. It it is the sum of its Taylor series, it is called an analytic function. Contrary to the real case, a function of a complex variable which is differentiable is infinitely differentiable and analytic.

Example of a non-analytic infinitely differentiable function of a real variable: $$f(x)=\begin{cases} \mathrm e^{-1/x^2}&\;\text{ if }x\ne 0,\\ 0&\;\text{ if }x=0. \end{cases}$$ It is easy to show that $f^{(n)}(0)=0$ for all $n$, hence its Taylor series at $0$ is $0,\:\ne f(x)$ if $x\ne 0$.

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  • $\begingroup$ Arrh, that makes sense now. I study physics, and have always been told that you can just approximate a function by a taylor polynomial. But I guess that's only true, if the taylor series converges to the function on the interval you want to approximate the function? So I am right, if I say that in general a taylor polynomial will be a bad approximation of a non-analytic function. And in theory, whenever I use taylor polynomials of a function, I should check if the taylor series converges to that function (In the interval I want to approximate)? $\endgroup$
    – sjm23
    Nov 27 '20 at 19:24
  • $\begingroup$ And also, so the radius of convergence only tells me that the taylor series converges to some function. But it doesn't tell me which one. So if I want to make sure it converges to f(x), I should check if the remainder converges to zero? $\endgroup$
    – sjm23
    Nov 27 '20 at 19:28
  • $\begingroup$ Yes, absolutely (unless you have some good reason to know the function is analytic). $\endgroup$
    – Bernard
    Nov 27 '20 at 19:52

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