# Radius of convergence and intuition

I'm a math student on a quest of relearning Calculus with an emphasis of developing my intuition. My question is: why do some functions converge on the whole $\mathbb{R}$, and some don't?

Let me expand. So far, I found good intuitive approach about main characteristics of functions. Example: what does it mean if a function isn't uniformly continuous on some open interval $(a,b)$? Well, there could be two ''problems'' with that function: it's oscillating too wildly or it's growing too fast. What does it mean if a function isn't differentiable at some point $a$? Well, the function isn't really ''smooth'' at that point, it has a ''horn''. I hope you see what kind of intuition I hope to get.

Let's get to the point: Why doesn't $\ln(1+x)$ converge on whole $\mathbb{R}$? I could try to explain it like this. Every form of Taylor remainder (Cauchy, integral, Lagrange) looks something like this n-th derivative at some point $\cdot$ distance from the main point to the n-th power $\cdot \frac{1}{n!}$

Every time you derivate $\ln(1+x)$, a constant factor pops out, so it creates counterbalance to that $\frac{1}{n!}$, and the deciding factor is distance from the main point.

The problem is I don't understand what quality of a function ''there is a constant factor that grows when you take repeated derivatives'' represents? There is a similar story about $\arctan$ (also a function that has a finite radius of convergence). The similarity between $\arctan$ and $\ln$ is that both of functions grow very slowly after initial ''boom'', but I can't connect that with high-order derivatives?

Dual to that, $e^x$, $\cos$ and $\sin$ have a good characteristic: their constant factors are bounded, but I also can't connect that with some visual quality of those functions.

If you type "a sin b" instead of "a\sin b", then you see $a sin b$ instead of $a\sin b$, and similar comments apply to \ln, \arctan, and \cos. The backslash not only prevents italicization but also results in proper spacing. That is standard usage. I fixed these in the question. –  Michael Hardy Dec 10 '12 at 14:02
If you want a visual grasp of what is behind the radius of convergence, then you really need to look at the functions as functions of a complex variable. In that case, the radius of convergence of the series for a function (centred at $0$) is normally the largest circle you can take centred at $0$ which does not contain any singularities of the function.
For example, in the real line, it is easy to see that the series for $\frac{1}{x^2-1}$ has radius of convergence equal to 1, since there is clearly a singularity at $x=1$, but when you consider the function (of a real variable) $\frac{1}{x^2+1}$, then there is no singularity in the (finite) real line, and hence no obvious reason for the radius of convergence to be $1$ also. However, when you consider the function $\frac{1}{z^2+1}$ in the complex plane, we see that there are poles at $\pm i$, and it is clear that the circle of convergence cannot include those points (at distance 1 from the origin).