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$$ \begin{align} \arctan(x) & = \arctan(1) + \arctan\left(\frac{x-1}{2}\right) \\ & {} - \arctan\left(\frac{(x-1)^2}{4} \right) + \arctan\left(\frac{(x-1)^3}{8}\right) - \cdots \end{align} $$

Is this known?

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What is the rule behind the -+ signs? –  darij grinberg Dec 27 '10 at 10:35
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Michael, is this from Ramanujan's letter to Hardy (i.e., you)? At least the style is borrowed... There is no obvious way to reconstruct a correct formula from yours: it's false when the RHS is understood as $\arctan(1)+\sum_{n=1}^\infty(-1)^{n-1}\arctan((x-1)/2)^n$. –  Wadim Zudilin Dec 27 '10 at 11:28
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You can reexpand the Taylor series of the arctan function. I am not sure what the pattern is. Up to tenth order, I get $$\arctan(x)=\arctan(1)+\arctan((x-1)/2)-\arctan((x-1)^2/4)+\arctan((x-1)^3/8)$$ $$-\arctan((x-1)^5/32+\arctan((x-1)^6/64)-\arctan((x-1)^7/128$$ $$+\arctan((x-1)^9/256)-\arctan(3(x-1)^{10}/1024).$$ So up to this point, we get the coefficient sequence $$1, -1, 1, 0, -1, 1, -1, 0, 2, -3.$$ –  Michael Renardy Dec 27 '10 at 11:51
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oeis.org/A123221 This sequence begins with the absolute values of the sequence of coefficients given above. –  Michael Hardy Dec 27 '10 at 19:25
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No, the OEIS stuff does not pan out. The modulus of the next coefficient is 3, not 5. –  Michael Renardy Dec 27 '10 at 21:21

2 Answers 2

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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Modern mathematicians seem very much accustomed to power series but not to trigonometric identities. In the 18th century, Euler began with the identity $$ \sin\left(\sum_k \theta_k\right) = \sum_{\text{odd }n \ge 1} (-1)^n \sum_{|A|=n} \prod_{i\in A} \sin\theta_i \prod_{i\not\in A} \cos\theta_i $$ and derived the power series for sine from it by saying $$ \theta = \frac{\theta}{n} + \cdots + \frac{\theta}{n} $$ where $n$ is an infinitely large integer, then applying the identity above, then saying that since $n$ is infinitely large, $\sin(\theta/n) = \theta/n$ and $\cos(\theta/n) = 1$. –  Michael Hardy Dec 27 '10 at 19:47
    
I suppose I should add a qualification. The identity would be $$ \arctan a - \arctan b = \text{one of the values of }\arctan\frac{a-b}{1+ab}. $$ –  Michael Hardy Dec 27 '10 at 21:38
    
Michael: nice identity, but the sums are over $n\ge0$ (odd or even) and $|A|=2n+1$. –  Did Dec 28 '10 at 11:12
    
Typo. Let's try again: $$ \sum_{\text{odd }n \ge 1} (-1)^{(n-1)/2} \cdots\cdots. $$ –  Michael Hardy Dec 28 '10 at 17:59
    
So the full identity is $$ \sin\left( \sum_k \theta_k \right) = \sum_{\text{odd }n\ge 1} (-1)^{(n-1)/2} \sum_{|A|=n} \prod_{i\in A} \sin\theta_i \prod_{i\not\in A} \cos\theta_i. $$ –  Michael Hardy Dec 28 '10 at 18:01

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