# Power series representation of arctangent: fails to converge everywhere

My understanding of power series turns out to be less-well-formed than I thought. To confess, I took my two courses in analysis in grad school (one real, one complex) and got out.

Since this is my Calc II class, let's keep everything in real variables, please. It's not hard to derive the power series for $\arctan(x)$ as $$\arctan(x) = \sum_{n=0}^\infty \frac{(-1)^n}{2n+1} x^{2n+1}, \ -1 \leq x \leq 1.$$ Also not hard to work out the interval of convergence for the right-hand side. So far, so good.

Here's my question and why I suddenly see how naive I am. I tend to think of $\arctan$ as an incredibly nice function, so I expect its power/Taylor series to converge everywhere. In short, I view $\arctan$ as being just as nice as $f(x) = e^x$, whose power series representation converges everywhere (domain of the power series matches the domain of the function). Same story for $\sin(x)$ and $\cos(x)$. They're "nice" so their power series converge on their entire domain.

When the power series for something like $\ln (x)$ or $\frac{1}{x}$ has finite radius, I'm completely fine with that as there is an obvious discontinuity that you bump into as you work your way out from the center. But why does the power series for $\arctan(x)$ have a finite radius? I know that something goes wrong with Taylor's remainder and this is what prevents the series from representing $\arctan(x)$ everywhere, but I would appreciate an explanation from the point of view of properties of $\arctan(x)$ and not its power series: what is it about $\arctan(x)$ that prevents its power series from being optimally "nice"?

• You can't explain that if you stay on the real line. There, $\arctan(x)$ is "nice". But it has singularities at $\pm i$, having a distance of $1$ from the origin, that's why the radius of convergence is $1$. – user436658 Nov 27 '17 at 13:06
• Oh, this is interesting. Thanks. – Randall Nov 27 '17 at 13:56
• @ProfessorVector "You can't explain that if you stay on the real line" -- that can't be true, though it is doubtless true that the most satisfactory explanation involves complex variables. It isn't like this is a mystery in the theory of real analysis. – John Coleman Nov 27 '17 at 14:54
• @John Coleman Is another "nice" function for real arguments, $\arctan(e^x)$, a mystery in real analysis? Can you derive the radius of convergence of its Taylor series around $0$ with the tools of real analysis (it doesn't have to be satisfactory, just give the correct result)? – user436658 Nov 27 '17 at 15:09
• @ProfessorVector: It seems the last paragraph of the top answer gives you such a method? – user541686 Nov 28 '17 at 2:46

## 2 Answers

Your insistence "let's keep everything in real variables, please" is precisely the problem: the cause of the finite radius of convergence is due to the function's behavior in $\mathbf C$, not $\mathbf R$.

A much simpler example than $\arctan x$ is $1/(1+x^2)$, which is defined and infinitely differentiable on the whole real line but its power series at $0$ (a geometric series with $-x^2$ in place of $x$) has radius of convergence $1$, not $\infty$. To use your language, "there is an obvious discontinuity that you bump into as you work your way out from the center," namely at $x = \pm i$ where the function blows up. In fact, if you expand $1/(1+x^2)$ into a power series at a real number $a$, not necessarily at $0$, the radius of convergence will be $\sqrt{a^2+1} = |a-i|$ -- the distance from the center out to $i$. This phenomenon is bewildering if you refuse to use complex numbers and extremely clear if you use them. Choose wisely.

If $f(x)$ is a rational function in reduced form with a nonconstant denominator and its denominator does not vanish at $a$, its power series at $a$ has radius of convergence $|a-\rho|$ where $\rho$ is the root of the denominator in $\mathbf C$ that is closest to $a$. This simple geometric result can not be explained in terms of real variables if the roots of the denominator are not all real.

To reinforce how poorly the real numbers are compared to the complex numbers as a predictive tool for the radius of convergence, there are functions $\mathbf R \rightarrow \mathbf R$ that are infinitely differentiable on the whole real line but their power series at each real number $a$ has radius of convergence zero for all $a$ in $\mathbf R$.

Strictly speaking, the real numbers have enough information in principle to compute the radius of convergence $R$ of a power series $\sum c_n(x-a)^n$ with all real coefficients using Hadamard's formula $1/R = \varlimsup\limits_{n\to\infty} \sqrt[n]{|c_n|}$, but this formula is often not feasible to compute in practice.

• Can you please give the example of functions that are $C^\infty$ on all of $\mathbb{R}$ but who fail to be analytic anywhere. – Clclstdnt Nov 27 '17 at 13:49
• Excellent explanation. Thank you. I get it now. – Randall Nov 27 '17 at 13:57
• @Clclstdnt see the Wikipedia page for "non-analytic smooth functions". – KCd Nov 27 '17 at 14:12
• That last paragraph is pretty cool. To be clear, is that the superior limit as $n \to \infty$? – user541686 Nov 28 '17 at 2:47
• +1 for touching the heart of the issue. The more interesting idea which can be built up using this same concept is analytic continuation. – Paramanand Singh Nov 28 '17 at 5:53

Let $z$ be an arbitrary complex number. Then $\arctan(z) = \frac12 i \log(1 - i z) - \frac12 i \log(1 + i z).$

Notice that $1-iz=0$ if $z=-i,$ and $1+iz=0$ if $z=i.$ So $\arctan(z)$ has singularities at $z=-i$ and $z=i.$

A thing about the radius of convergence in complex numbers is that if you plot it on an Argand diagram, it really is a radius of a circle. Inside the circle the series always converges, outside it diverges. The series of the real-number function $\arctan(x)$ is just $\arctan(z)$ restricted to the real number line, so it has the same behavior.