Expanding Arcus Tangens as Taylor Series around 1: Symmetries

Question

Expanding $\arctan$ around 0 as a Taylor series yields $$ \arctan x = \sum_{n=0}^\infty (-1)^n\frac{x^{2n+1}}{2n+1} $$ for $x$'s that are in the region of convergence. All even terms vanish, which is due to the symmetry $\arctan (-x)=-\arctan x$, i.e. $\arctan$ is an odd function.

Now expanding around 1 (with radius of convergence $\sqrt2$) yields something like:

$$\begin{align} \arctan(x+1) - \arctan(1) =& \frac12x - \frac1{4}x^2 + \frac1{12}x^3 - \frac1{40}x^5 + \frac1{48}x^6 - \frac1{112}x^7 + \frac1{288}x^9 - \frac1{320}x^{10}\\ & {} + \frac1{704}x^{11} - \frac1{1664}x^{13} + \frac1{1792}x^{14} - \frac1{3840}x^{15} + \frac1{8704}x^{17} - \frac1{9216}x^{18} + \frac1{19456}x^{19} \cdots \end{align}$$ As you can see, every 4-th term vanishes, i.e. powers of $x$ with exponents of $0\bmod4$ do vanish.

Questions:

• What does this tell us about symmetries of $\arctan$ or any other function that shows such behaviour?
• Is there a straight forward way to see that every 4th term does actually vanish without carriying out fancy computations?

$\arctan$ satisfies $\arctan x+ \arctan(1/x)=\pi/2$, but I do not see how (or whether) this symmetry has something to do with the question.

Answer 1

We have $$ \frac{1}{(x + 1)^2 + 1} = \frac{1}{2i}\left(\frac{1}{(x + 1) - i} - \frac{1}{(x + 1) + i}\right), \\ \arctan(x + 1)^{(n)} = c_n\left(\frac{1}{((x + 1) - i)^n} - \frac{1}{((x + 1) + i)^n}\right),\quad n\ge 0, $$ where $c_n$ is a constant. For $x = 0$ the latter becomes $\displaystyle \frac{1}{(1 - i)^n} - \frac{1}{(1 + i)^n}$. This difference equals zero if $n$ is divisible by $4$ hence every 4th term in the Taylor series of $\arctan$ vanishes.

Answer 2

Pavel Gubkin has answered your second question. To more explicitly address your first question: The oddness of a function, that is, the vanishing of coefficients of $x^n$ with $n$ a multiple of $2$, can be expressed as $f(x)+f(-x)=0$. This can be generalized to the vanishing of coefficients with $n$ a multiple of any integer $k$ as

$$ \sum_{j=0}^{k-1}f\left(\omega^j x\right)=0\;, $$

where $\omega$ is any primitive $k$-th root of unity. Thus, with $\omega=\mathrm i$ a primitive fourth root of unity, the vanishing of the coefficients with $n$ a multiple of $4$ corresponds to the functional equation

$$ f(x)+f(\mathrm ix)+f(-x)+f(-\mathrm ix)=0 $$

(or a constant if the coefficient of $x^0$ doesn’t vanish). In this case,

$$ \arctan(1+x)+\arctan(1+\mathrm ix)+\arctan(1-x)+\arctan(1-\mathrm ix)=\pi\;. $$

I was at first hoping to relate this to the other equation you mention, $\arctan x+\arctan\frac1x=\frac\pi2$, by writing that as

$$ \arctan\mathrm e^y+\arctan\mathrm e^{-y}=C\;, $$

expressing the fact that $\arctan\mathrm e^y$ is an odd function up to a constant; but this doesn’t imply the vanishing of the coefficients with $n$ a multiple of $4$. For instance $x-\frac1x$ also has this property, but none of the coefficients in its series expansion at $x=1$ vanish.

By the way, $2(\arctan(1+x)-\arctan1)$ is the exponential generating function of OEIS sequence A217260, which is the series reversion of OEIS sequence A000111, the Euler numbers.