While messing around with the formula $$\arctan x+\arctan y=\arctan\frac{x+y}{1-xy},\tag1$$ I decided to iteratively find expressions for $n\arctan x$ by applying the formula $(1)$ $n$ times. I was able to reach $n=9$ before it got too annoying. Here they are: $$\begin{align} 2\arctan x&=\arctan\left(2x\cdot\frac{1}{1-x^2}\right), &|x|&<1\\ 3\arctan x&=\arctan\left(x\frac{3-x^2}{1-3x^2}\right),&|x|&<1/\sqrt3\\ 4\arctan x&=\arctan\left(4x\frac{1-x^2}{1-6x^2+x^4}\right),&|x|&<\sqrt2-1\\ 5\arctan x&=\arctan\left(x\frac{5-10x^2+x^4}{1-10x^2+5x^4}\right),&|x|&<\sqrt{1-2/\sqrt{5}}\\ 6\arctan x&=\arctan\left(2x\frac{3-10x^2+3x^4}{1-7x^2+7x^4-x^6}\right),&|x|&<2-\sqrt{3}\\ 7\arctan x&=\arctan\left(x\frac{7-35x^2+21x^4-x^6}{1-21x^2+35x^4-7x^6}\right),&|x|&<\alpha_7\\ 8\arctan x&=\arctan\left(8x\frac{1-7x^2+7x^4-x^6}{1-28x^2+70x^4-28x^6+x^8}\right),&|x|&<-1+\sqrt{2}\left(-1+\sqrt{2+\sqrt2}\right)\\ 9\arctan x&=\arctan\left(x\frac{9-84x^2+126x^4-36x^6+x^8}{1-36x^2+126x^4-84x^6+9x^8}\right),&|x|&<\alpha_9 \end{align}$$ Here, $\alpha_7$ and $\alpha_9$ are the smallest positive roots of $1-21x^2+35x^4-7x^6$ and $1-36x^2+126x^4-84x^6+9x^8$ respectively.

There are clearly some patterns going on here. Taking $R_n(x)=\tan(n\arctan x)=p_n(x)/q_n(x),$ we can explore some of them.

Firstly, it is easy to see that $$\arctan R_{n+1}(x)=(n+1)\arctan x=n\arctan x +\arctan x\\ =\arctan R_n(x)+\arctan x=\arctan\frac{x+R_n(x)}{1-xR_n(x)},$$ so that $$R_{n+1}(x)=\frac{x+R_n(x)}{1-xR_n(x)}=\frac{xq_n(x)+p_n(x)}{q_n(x)-xp_n(x)}.$$ From this, it is clear to see that $x|p_n(x)$ for all $n\ge1$, as $$x|p_n(x)\Rightarrow x|(xq_n(x)-p_n(x))\Rightarrow x|p_{n+1}(x).$$

Furthermore, the relation $p_n(1/x)=(-1)^{\tfrac{n-1}{2}}x^{-n}q_n(x)$ is satisfied by the cases $n=3,5,7,9$. Similarly, the relations $p_n(1/x)=(-1)^{\tfrac{n}2-1}x^{-n}p_n(x)$ and $q_n(1/x)=(-1)^{n/2}q_n(x)$ are satisfied by $n=2,4,6,8$. Put more simply, $$\begin{align} R_2(1/x)&=-R_2(x), &R_3(1/x)&=1/R_3(x),\\ R_4(1/x)&=-R_4(x), &R_5(1/x)&=1/R_5(x),\\ R_6(1/x)&=-R_6(x), &R_7(1/x)&=1/R_7(x),\\ R_8(1/x)&=-R_8(x), &R_9(1/x)&=1/R_9(x).\\ \end{align}$$ Indeed, this pattern can be shown to continue. Specifically, if we suppose that $R_{2k}(1/x)=-R_{2k}(x)$ for some $k\ge1$, then $$R_{2k+1}(\tfrac1x)=\frac{\tfrac1x+R_{2k}(\tfrac1x)}{1-\tfrac1x R_{2k}(\tfrac1x)}=\frac{\tfrac1x-R_{2k}(x)}{1+\tfrac1x R_{2k}(x)}=\frac{1-xR_{2k}(x)}{x+R_{2k}(x)}=\frac{1}{R_{2k+1}(x)}.$$ Similarly, if $R_{2k-1}(1/x)=1/R_{2k-1}(x)$ for some $k\ge1$, we get $$R_{2k}(\tfrac1x)=\frac{\tfrac1x+R_{2k-1}(\tfrac1x)}{1-\tfrac1xR_{2k-1}(\tfrac1x)}=\frac{\tfrac1x+\tfrac1{R_{2k-1}(x)}}{1-\tfrac1{xR_{2k-1}(x)}}=\frac{R_{2k-1}(x)+x}{xR_{2k-1}(x)-1}=-R_{2k}(x).$$

Another thing we may notice is the following. Here I will abbreviate $R_s(x)$ as $R_s$ for simplicity. $$\arctan R_{n+m}=(n+m)\arctan x=n\arctan x+m\arctan x=\arctan R_n+\arctan R_m=\arctan\frac{R_n+R_m}{1-R_nR_m}.$$ That is $$R_{n+m}(x)=\frac{R_n(x)+R_m(x)}{1-R_n(x)R_m(x)},$$ and consequentially $$\frac{p_{n+m}(x)}{q_{n+m}(x)}=\frac{p_n(x)q_m(x)+p_m(x)q_n(x)}{q_n(x)q_m(x)-p_n(x)p_m(x)}.$$

These are some pretty remarkable identities, so I can't possibly be the only person to consider these rational functions. Do they have a name? A generating function? A general closed form (other than $\tan(n\arctan x)$ and similar)? Where can I learn more about them?

  • $\begingroup$ They seem similar to Chebyshev polynomials. Other than that, I think you are even more qualified than me to answer the question. $\endgroup$ Oct 20, 2023 at 1:26
  • $\begingroup$ A closed form is $\frac{2i}{(\frac{i-x}{i+x})^n+1}-i= \frac{2i}{(\frac{2i}{x+i}-1)^n+1}-i$ $\endgroup$ Oct 20, 2023 at 1:35

2 Answers 2


Consider the complex number $z = 1 + ix$

$$z = |z|e^{i\theta} = |z|(\cos \theta + i\sin \theta)$$ $$\frac {z}{|z|\cos\theta} = (1+i\tan\theta) = z$$ $$\tan \theta = x$$ $$\theta = \arctan x$$

$z^n = |z|^ne^{in\theta}$

$$\tan n\theta = \tan (n\arctan x) = \frac {\text{Im}(z^n)}{\text{Re}(z^n)}$$

And now we just need the binomial theorem

$$\tan (n\arctan x)= \cfrac {\displaystyle\sum_{k=0}^{\lfloor{\frac{n-1}{2}\rfloor}} (-1)^k{n\choose 2k+1}x^{2k+1}}{\displaystyle\sum_{k=0}^{\lfloor{\frac{n}{2}\rfloor}} (-1)^k{n\choose 2k}x^{2k}}$$

  • $\begingroup$ oh wow okay this is quite simple. Thanks! $\endgroup$
    – clathratus
    Oct 20, 2023 at 1:53
  • 5
    $\begingroup$ @Gary Nice edit. I usually don't like it when people edit my posts, but this really did improve the readability. Thanks. $\endgroup$
    – user317176
    Oct 20, 2023 at 2:50

By Binomial Theorem, we have $$ (\cos \theta+i \sin \theta)^n=\sum_{k=0}^n\left(\begin{array}{c} n \\ k \end{array}\right) \cos ^{n-k} \theta(i \sin \theta)^{k} $$ Using De Moivres’ Theorem, $$ \begin{aligned} \displaystyle \cos n \theta +i \sin n \theta= & \sum_{k=0}^{\left[\frac{n}{2}\right]}\left(\begin{array}{c} n \\ 2 k \end{array}\right) \cos ^{n-2 k} \theta(-1)^k \sin ^{2 k} \theta +i \sum_{k=0}^{[\frac{n-1}{2} ]}\left(\begin{array}{l} \quad n \\ 2 k+1 \end{array}\right) \cos ^{n-2 k-1} \theta(-1)^k \sin^{2k+1} \theta \end{aligned} $$ Comparing their real and imaginary parts, we have $$ \displaystyle \begin{aligned} \tan n \theta&=\frac{\sin n \theta}{\cos n \theta} \\&=\frac{\displaystyle \sum_{k=0}^{[\frac{n-1}{2} ]}\left(\begin{array}{l} \quad n \\ 2 k+1 \end{array}\right) \cos ^{n-2 k-1} \theta(-1)^k \sin^{2k+1} \theta}{\displaystyle \sum_{k=0}^{\left[\frac{n}{2}\right]}\left(\begin{array}{c} n \\ 2 k \end{array}\right) \cos ^{n-2 k} \theta(-1)^k \sin ^{2 k} \theta} \\&= \frac{\displaystyle \sum_{k=0}^{\left[\frac{n-1}{2}\right]}(-1)^k\left(\begin{array}{c} n \\ 2 k+1 \end{array}\right) \tan ^{2 k+1} \theta}{\displaystyle \sum_{k=0}^{\left[\frac{n}{2}\right]}(-1)^k\left(\begin{array}{c} n \\ 2 k \end{array}\right) \tan ^{2 k} \theta}\end{aligned} $$ Putting $x=\tan\theta$ yields $$ \tan \left(n \tan ^{-1} x\right)= \frac{\displaystyle \sum_{k=0}^{\left[\frac{n-1}{2}\right]}(-1)^k\left(\begin{array}{c} n \\ 2 k+1 \end{array}\right) x^{2 k+1}}{\displaystyle \sum_{k=0}^{\left[\frac{n}{2}\right]}(-1)^k\left(\begin{array}{c} n \\ 2 k \end{array}\right) x^{2 k}} $$


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .