# Taylor series expansion limitation

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 ?

• 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 Commented May 6, 2022 at 18:15
• @QC_QAOA Poles are not the only type of complex singularity.
– Ian
Commented May 6, 2022 at 18:22
• @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.? Commented May 6, 2022 at 18:22
• @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.
– Ian
Commented May 6, 2022 at 18:24
• I see understood @Ian thanks Commented May 6, 2022 at 18:25

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:

1:

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.

2:

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 $$f$$.

3:

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 $$\pm i$$.

• 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 ? Commented May 6, 2022 at 18:24
• @ProblemDestroyer Only at the level of my point #1. The "deep explanation" needs complex analysis, period.
– Ian
Commented May 6, 2022 at 18:25
• 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 ? Commented May 6, 2022 at 18:27
• @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).
– Ian
Commented May 6, 2022 at 18:31
• @ProblemDestroyer I looked back and spotted another error, fixed now.
– Ian
Commented May 11, 2022 at 18:25