Is there a deeper pattern in the radius of convergence?

I'm curious why the Taylor expansion of $\sin(x)$ and $\cos(x)$, for example, converge everywhere on the domain. But the expansion of $\frac{1}{1-x}$ converges for $|x| < 1$. I can compute it trivially using the ratio test and understand that the terms must decrease geometrically with the expansion for $\frac{1}{1-x}$ to convergence. But my question is more geared towards why this is the case: why some functions have power series' that converge everywhere and some that converge only on a certain interval? In other words, besides doing the ratio test, I don't see any sort of pattern or reasoning for why this is the case. Is there some higher level explanation someone can give to a person like myself who hasn't taken real or complex analysis, if the answer lies in those fields? I'm already a bit familiar with the fact that it is called the radius of convergence because it converges on a disc in the complex plane, but nothing further. Thank you.

• The pattern is that the radius of convergence for an "analytic function" (i.e. one with a convergent power series) is the distance from the point of expansion (say $x=0$ in your examples) to the nearest singularity (point in the complex plane where the derivative fails). – hardmath Jul 23 '17 at 3:42
• @hardmath thank you - now how does this align with the idea that $ln(1+x)$ converges for $|x|<1$ also? Does it have to do with the fact that because convergence is symmetric, and $ln(0)$ doesn't exist, that the power series also doesn't converge for $f(x)=ln(1+x)$ for all $x ≥ 1$ – rb612 Jul 23 '17 at 3:48
• Yes, that is the right idea. In this case $\ln(1+x)$ is singular at $x=-1$, and this is the nearest singularity to the point of expansion $x=0$. The distance is $1$ and that is the radius of convergence. – hardmath Jul 23 '17 at 3:52
• For a nice explanation read chapter 2 of Visual Complex Analysis by Tristan Needham – Ovi Jul 23 '17 at 12:53

The answer is yes, there is some simple explanation of why this happens. To understand this better, you need to take a class in Complex Analysis.

Fact 1 A real function is equal to a power series $\sum_{n=0}^\infty a_n(x-z)^n$ on $(a-R, a+R)$ if and only if, there exists a complex function $g$, which is differentiable on $|z-a|<R$ such that $f=g$ on $(a-R,a+R)$.

Fact 2 A complex function is equal with its Taylor series on $|z-a|<R$ if and only if it is differentiable on $|z-a|<R$.

It follows from Fact 2, that the radius of convergence for an analytic function (real or complex) is the distance to the closest point in the complex plain where the function is not differentiable.

Since $\sin(x), \cos(x)$ can be extended to complex functions which are analytic everywhere, the radius of convergence is $\infty$.

For $\frac{1}{1-x}$, or even more interesting $\frac{1}{1+x^2}$, the radius of convergence at $a=0$ is simply the distance to $1$ respectively $\pm i$, which is $1$.

• @N.S. This is great! Thank you! So like in your example of $\frac{1}{1+x^2}$, the function is not analytic for $x=i, x=-i$, I'm still a bit confused on how you then got $1$ as your radius of convergence. Wouldn't the radius of convergence then be $i$ instead? – rb612 Jul 23 '17 at 3:53
• @rb612 It is the distance: $d(0,i)=|0-i|=|i|=1$. – N. S. Jul 23 '17 at 3:54
• @N.S., ah thank you - but isn't there supposed to be a squared term in the distance formula? Because it would be Pythagorean distance, no? – rb612 Jul 23 '17 at 3:56
• @rb612 The distance/modulus are in complex plane. The complex number $i$ has "coordinates" $(0,1)$ if you want to use the "Pytagorean" distance – N. S. Jul 23 '17 at 4:09
• I think the first fact is false or not what you meant. For instance $f(x) = \frac{1}{1+x^2}$ is analytic on $(-5, 5)$ but clearly there is no analytic function $g(z)$ on $|z|<5$ which agrees with $f$ on $(-5, 5)$. – Adayah Jul 23 '17 at 9:02