# new $\arctan$ series working for any $x$?

Using Log series, we can write:

$$\frac{1}{2 i} \left( \text{Log} \left( 1 + e^{\text{i2t}} \right) - \text{Log} \left( 1 + e^{\text{-i2t}} \right) \right) = \frac {1} {2 i}\left ( \sum _ {k = 1}^{\infty}\frac {(-1)^{k + 1}} {k} e^{\text{i2t}^k} - \sum _ {k = 1}^{\infty}\frac {(-1)^{k + 1}} {k} e^{-\text{i2t}^k} \right)$$ $$\frac {1}{2 i} \text {Log} \left( \frac{1 + e^{\text {i2t}}}{1 + e^{-\text {i2t}}} \right) = \frac{1}{2 i} \sum_{k = 1}^{\infty}\frac{(-1)^{k + 1}}{k} \left( e^{\text{i2kt}} - e^{-\text{i2kt}} \right)$$ $$\frac {1}{2 i} \text {Log} \left( \frac{e^{\text{i2t}} \left( e^{-\text{i2t}}+1 \right)}{1+e^{-\text{i2t}}} \right) = \sum_{k = 1}^{\infty} \frac{(-1)^{k + 1}}{k} \left( \frac{e^{\text {i2kt}} - e^{-\text {i2kt}}}{2 i} \right)$$ $$\frac{1}{2i} \text{Log} \left( e^{\text{i2t}} \right) = \sum_{k = 1}^{\infty}\frac {(-1)^{k + 1}} {k}\text {Sin} (2kt)$$ $$t = \sum_{k = 1}^{\infty}\frac {(-1)^{k + 1}} {k}\text {Sin} (2kt)$$

Here is the plot of the sum in red with sucessive aproximations and y=t in blue:

Choosing $t = \text{ArcTan} (x)$, we have: $$\text{ArcTan} (x) = \sum_{k = 1}^{\infty} \frac {(-1)^{k + 1}} {k} \text{Sin} ( 2k \text{ ArcTan} x )$$

Using $z=1 + i x=|z| e^{i \text{arcTan}x }$ and its complex conjuguate $z^*=1 - i x=|z| e^{-i \text{arcTan}x }$, with $|z|=\sqrt{1 + x^2}$, we can write: $$\text{Sin} ( 2k \text{ Arctan} x) = \frac {z^{2 k} - z^{*2 k}} {2i |z| ^{2 k}} = \frac {(1 + ix)^{2k} - (1 - ix)^{2k}} {2i (1 + x^2)^k} = \frac{\sum_{q=0}^k (-1)^q \left( \begin{array}{c} 2k \\ 2q+1 \end{array} \right) x^{2 q+1}}{\left(1+x^2\right)^k}$$

Thus: $$\text{ArcTan} (x) = \sum_{k = 1}^{\infty} \frac {(-1)^{k + 1}} {k} \frac{\sum_{q=0}^k (-1)^q \left( \begin{array}{c} 2k \\ 2q+1 \end{array} \right) x^{2 q+1}}{\left(1+x^2\right)^k}$$

Here are the plots of $\text{ArcTan}(x)$ in blue, and successive aproximations to the classic ArcTan series $\text{ArcTan}(x)=\sum_{k = 0}^{\infty} \frac {(-1)^k} {2k + 1} x^{2k + 1}$ in green, working for $x<1$, and the one I gave in red, working for any $x$ as the sum goes to infinity:

is that result known?

• Did you at Fourier series for the first ? I do not know if these results are known but they are beautiful and very interesting. Congratulations. Go ahead. – Claude Leibovici Nov 24 '13 at 9:13
• By the way, I am in the same country. Any way to be in touch ? – Claude Leibovici Nov 24 '13 at 9:15
• Sure, @claude! my email: eddykem@gmail.com. are you french, cause I'am! – Eddy Khemiri Nov 24 '13 at 12:10

I do not know whether your series for $\arctan$ is new (it may be), but when evaluating its advantages, you should consider alternatives other than the classical McLaurin series. For example, the identity $\tan 2\theta=\frac{2\tan \theta}{1-\tan^2\theta}$ can be turned into $$\arctan x = 2\arctan \frac{x}{1+\sqrt{1+x^2}}$$ Note that the argument of $\arctan$ on the right is always less than $1$ in absolute value. Hence,
$$\arctan x = 2 \, \sum_{k=0}^\infty \frac{(-1)^k}{2k+1}\left(\frac{x}{1+\sqrt{1+x^2}}\right)^{2k+1},\quad x\in\mathbb R$$ The content of parentheses is an irrational function of $x$ but it only needs to be computed once. Partial sums up to $k=0,1,2,3$ are shown below; the purple curve is the arctangent itself.
(The horizontal window is $[0,4]$, as in your post.)