1. Appplying taylor to any functions its not necessary that that polynomial which gets formed from taylor will work for all x , taylor doesnt gives a condition for when the series is always equal to the function right ?

  2. We would need some other methods to check whether the series always matches with the function right ? Like in case of ln(1+x) where expanding using taylor around x=0 we will get a series which is only valid for |x|<=1 , or is it that we can have a series around some number like x= 5 for ln(1+x) then the series will always converge ? If yes when and when not to check for series convergence ?

  3. Is there a deep reason as to why taylor fails to give the whole function in terms of polynomial when expanded around x= 0 but ut might work when expanded around some other point ?

  • $\begingroup$ Taylor series fail for some functions when there is a pole of that function in the complex plane. That is, Taylor series work everywhere if and only if the function is entire (no poles). See en.wikipedia.org/wiki/Laurent_series for more information $\endgroup$
    – QC_QAOA
    Commented May 6, 2022 at 18:15
  • 1
    $\begingroup$ @QC_QAOA Poles are not the only type of complex singularity. $\endgroup$
    – Ian
    Commented May 6, 2022 at 18:22
  • $\begingroup$ @QC_QAOA can this working and not working can be showed from the derivation of taylor series itself? I mean from the assumption taken while deriving taylor series we may get why pole might cause problem etc.? $\endgroup$ Commented May 6, 2022 at 18:22
  • 1
    $\begingroup$ @ProblemDestroyer Not really. The fact that $i$ has anything to do with the failure of the Maclaurin series of $\frac{1}{x^2+1}$ to converge at $x=2$ is not obvious. $\endgroup$
    – Ian
    Commented May 6, 2022 at 18:24
  • $\begingroup$ I see understood @Ian thanks $\endgroup$ Commented May 6, 2022 at 18:25

1 Answer 1


Surprisingly, the answers to your questions require some complex analysis, even though your questions are ostensibly about real-valued functions of a real variable.

That said:


Taylor's theorem with the Lagrange remainder says that if $\lim_{n \to \infty} \frac{|f^{(n)}(x_1)|}{n!} (x-x_0)^n = 0$ whenever $x \in (a,b)$ and $x_1$ is between $x$ and $x_0$ then the Taylor series centered at $x_0 \in (a,b)$ converges to $f$ at all points of $(a,b)$. However this result can be extremely hard to check.


Complex analysis tells us that the radius of convergence of the Taylor series at a given $x_0$ is the distance from $x_0$ to the nearest complex singularity. For $\operatorname{Log}(z)$ (the principal complex logarithm) for example this is the distance from $x_0$ to $0$, which is a singularity of $\operatorname{Log}$ because it is a branch point. Thus for instance $\ln(1+x)$ centered at $x_0=5$ has radius of convergence $6$.

Complex analysis also tells us that if $f$ is complex differentiable on a disk of radius $r$ centered at $z_0$, then the Taylor series of $f$ at $z_0$ converges to $f$ on a disk of radius $R=\min \left \{ r,\left ( \limsup_{n \to \infty} \left ( \frac{|f^{(n)}(z_0)|}{n!} \right )^{1/n} \right )^{-1} \right \}$, where $0^{-1}$ is understood as $+\infty$. We can't freely drop this "if $f$ is complex differentiable..." assumption. If we do drop it, we can still say the series converges on a disk of radius $\left ( \limsup_{n \to \infty} \left ( \frac{|f^{(n)}(z_0)|}{n!} \right )^{1/n} \right )^{-1}$ but we cannot be sure that the limit is actually $f$.


I think this is answered by my answer to #2.

This gets much more counterintuitive with functions that look completely innocuous on the real line, such as $\frac{1}{x^2+1}$. Here the series at $x_0=0$ has radius of convergence $1$ even though the restriction to the real line has no singularities anywhere. In some sense the restriction "knows" about the singularity at $\pm i$.

  • $\begingroup$ Sir actually i dont know much about poles and basically the complex analysis , i was into real numbers only and also till high school level , but yeah a bit real analysis i know . Can this be a bit easily explained using the knowledge without complex analysis ? $\endgroup$ Commented May 6, 2022 at 18:24
  • $\begingroup$ @ProblemDestroyer Only at the level of my point #1. The "deep explanation" needs complex analysis, period. $\endgroup$
    – Ian
    Commented May 6, 2022 at 18:25
  • $\begingroup$ I think i got a bit of understanding from #1 thanks Sir . Although its possible right that the series might converge at some other values of a where its being evaluated (expanded) at? For all values of x ? $\endgroup$ Commented May 6, 2022 at 18:27
  • 1
    $\begingroup$ @ProblemDestroyer What I said in #1 will guarantee convergence when it works. I don't think it guarantees divergence when it doesn't work (except in the trivial setting where this limit isn't zero at $x=x_0$, in which case the Taylor series doesn't converge to anything, much less the original function). $\endgroup$
    – Ian
    Commented May 6, 2022 at 18:31
  • 1
    $\begingroup$ @ProblemDestroyer I looked back and spotted another error, fixed now. $\endgroup$
    – Ian
    Commented May 11, 2022 at 18:25

You must log in to answer this question.

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