B Complex Numbers and Real Taylor Series

[FranzS]

TL;DR

How complex singularities affect Taylor series

Hello,

I've always been fascinated by the fact that even strictly-complex singularities determine the radius of convergence of the (real) Taylor series of a (real) function that doesn't have any singularities in $\mathbb{R}$.
Can anyone provide an intuitive explanation besides the rigorous math behind this (assuming the rigorous math is not intuitive, but I may be wrong)?
Thanks!

 

[fresh_42]

I am not sure I understand what you mean. An example might be helpful here.

In general, there is "the trick with the geometric series"
\begin{align*}\dfrac{1}{\zeta – z}&=\dfrac{1}{(\zeta -z_0)-(z-z_0)}=\dfrac{1}{\zeta -z_0}\cdot \dfrac{1}{1-\dfrac{z-z_0}{\zeta-z_0}}=\dfrac{1}{\zeta -z_0}\cdot \sum_{n=0}^\infty \left(\dfrac{z-z_0}{\zeta-z_0}\right)^n\end{align*}
that connects all these terms. But as I said, I'm not sure whether you meant this.

Source: https://www.physicsforums.com/insights/an-overview-of-complex-differentiation-and-integration/

 

[martinbn]

If a series $a_0+a_1x+a_2x^2+\cdots$ with real coefficients converges for some $x=r$, then it will converge for all $|z|<r$. Here $z$ can be complex. So if you consider it for complex values and if that function has a singularity at $z_0$, then clearly the radius of convergence of the real series cannot be more than $|z_0|$.

 

[FranzS]

Thanks for your replies. Actually, using a classic specific example, what I meant is: how does a real function like...
$$
f(x)=\frac{1}{1+x^2}
$$
... have a MacLaurin series $m(x), \ x \in \mathbb{R}$ with finite radius of convergence when it has no real singularities? Why do complex singularities (in this specific case, strictly imaginary singularities $i$ and $-i$) affect the convergence of the real MacLaurin power series? Again, I'm interested in possible intuitive explanations.

 

[FranzS]

By the way, I have clear in mind the "circle of convergence" in the complex plane, I was just wondering if there were some other sort of intuitive explanation.

 

[fresh_42]

I don't think you need the complex singularities for an explanation. We have a series $f(x)=1-x^2+x^4+O(x^6)$ at the origin. It is obvious that $|x|<1$ has to hold for convergence.

If you insist on complex numbers, then @martinbn 's post #3 explains it.

The link has an example of how to integrate this with using residues.

 

[FranzS]

I don't think you need the complex singularities for an explanation. We have a series $f(x)=1-x^2+x^4+O(x^6)$ at the origin. It is obvious that $|x|<1$ has to hold for convergence.

If you insist on complex numbers, then @martinbn 's post #3 explains it.

The link has an example of how to integrate this with using residues.

This is already a very good explanation to me, thank you

 

[martinbn]

Why is what I wrote not what you were looking for!

 

[FranzS]

Why is what I wrote not what you were looking for!

It was a backward explanation to me, so to speak. In @fresh_42 specific example it was clear that the power series diverges for $x \geq 1$. Sorry

 

[Office_Shredder]

I'm just summarizing what has been said but maybe rewording it helps. I'm going to just use power series around 0 for notational simplicity.

Suppose $\sum a_n R^n$ converges for some radius $R$. Then $|a_n R^n|$ must converge to 0 and hence it's eventually bounded by 1. Then if n is large and $r<R$, $|a_n r^n| = |a_nR^n| |\frac{r}{R}|^n < (r/R)^n$. So by the comparison test with a geometric series if the series converges for some radius R it converges absolutely on the entire complex plane inside of that radius.

Now consider the maximum value of $R$ for which it converges absolutely for all $r<R$. There are two possibilities: one is it converges absolutely for $R$ also. $x^n/n^2$ is an example of this, it has a radius of convergence of 1 but converges absolutely when $|x|$=1.


The other possibility is it doesn't converge absolutely. Then it might converge conditionally for some choices of $x$. The interesting question is can it converge conditionally for every value of $x$ even though it doesn't converge absolutely?

And the answer is at least usually no, that would be a crazy coincidence. Most simple examples you can just pick a value of x in $\mathbb{C}$ which forces all the terms to be the same sign and hence it obviously must fail to converge. I don't remember if there's a theorem that guarantees it must fail to converge at some point on the circle of radius $R$, or if it's just very likely in examples that people write down.