$$
\begin{align}
\arctan(x) & = \arctan(1) + \arctan\left(\frac{x-1}{2}\right) \\
& {} - \arctan\left(\frac{(x-1)^2}{4} \right) + \arctan\left(\frac{(x-1)^3}{8}\right) - \cdots
\end{align}
$$
Is this known?
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I have no references for this particular series, but here's some hints to get a closed formula for the coefficients listed above by Michael Renardy. If we let $u:=\frac {1-x} 2$, an expansion $$\arctan(1-2u)=\arctan(1) + \sum_{k=1}^\infty \frac {c_k} k \arctan(u^k)$$ can be obtained by term-wise integration over on $[0,u]$ of a (somehow more common) expansion into rational fractions $$\frac 2 {1 + (1-2u)^2}= \sum_{k=1}^\infty \ c_k \frac {u^{k-1}} {1+u^{2k} }\ ,$$ (such expansions have a role in number theory, and are related to Dirichlet series). Here the coefficients may be identified expanding formally the geometric series $ (1+ u^{2k} )^{-1}$ and rearranging into a series of powers of $u$, to be compared with the power series of the LHS. One finds an equality with an arithmetic convolution, that inverted gives the $c_k$'s. The exponential growth of the $c_k$ give a positive radius of convergence (I guess $1/\sqrt 2$), that in particular allows the term-wise integration. Note that $\frac 2 {1 + (1-2u)^2}= \mathrm{Im} { \frac 2 {1-2u+i} }$, that simplifies things a bit. |
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I've voted up Pietro Majer's incomplete answer and Michael Renardy's incomplete answer in the "comments" section. Here's my own incomplete answer. Here's how I got this series: start with the identity $$ \arctan a - \arctan b = \arctan \frac{a-b}{1+ab}. $$ From this we get $$ \arctan x = \arctan 1 + \arctan\frac{x-1}{1+x}. $$ Substituting 1 for $x$ everywhere in the last expression except the power of $x-1$, we get the 1st-degree term. So we need to replace the last term above by the 1st-degree term plus another arctangent by using the basic identity above, and we get $$ \arctan\frac{x-1}{1+x} = \arctan\frac{x-1}{2} + \arctan\frac{-(x-1)^2}{2(1+x) +(x-1)^2}. $$ Then again substitute 1 for $x$ everwhere in the last term except in the power of $(x-1)$ in the numerator, to get the 2nd-degree term, and then write the last term above as the sum of the 2nd-degree term and another arctangent of a yet more complicated rational function. And so on. Does the sequence of arctangents of rational functions go to 0? In some sense? I don't know, nor do I know the general pattern. I actually tried this first with $x-2$ instead of $x-1$; then I decided that $x-1$ already has enough initial unclarity. I don't even know whether in some reasonable sense the process goes on forever. |
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