B Complex Numbers and Real Taylor Series

  • B
  • Thread starter Thread starter FranzS
  • Start date Start date
FranzS
Messages
86
Reaction score
26
TL;DR Summary
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!
 
Physics news on Phys.org
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/
 
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|##.
 
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.
 
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.
 
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.
 
fresh_42 said:
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
 
Why is what I wrote not what you were looking for!
 
martinbn said:
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
 
  • #10
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.
 
Last edited:
Back
Top