While trying to answer this MSE question, I found that arctangents of many odd powers of the golden ratio $\varphi=\frac{1+\sqrt5}2$ are expressible as rational linear combinations of arctangents of positive integers: $$\begin{align} \arctan\varphi&=2\arctan1-\frac12\,\arctan2\\ \arctan\varphi^3&=\arctan1+\frac12\,\arctan2\\ \arctan\varphi^5&=4\arctan1-\frac32\,\arctan2\\ \arctan\varphi^7&=3\arctan1+\frac12\,\arctan2-\arctan5\\ \arctan\varphi^9&=\arctan1-\frac12\,\arctan2+\arctan4\\ \arctan\varphi^{11}&=5\arctan1+\frac12\,\arctan2-\arctan5-\arctan34\\ \arctan\varphi^{13}&=3\arctan1-\frac12\,\arctan2+\arctan4-\arctan89\\ \arctan\varphi^{15}&=-2\arctan1+\frac32\,\arctan2+\arctan11 \end{align}$$ I was not able to find such a representation for $\arctan\varphi^{17}$ though.

Question 1. Can we prove that it does not exist?

I also could not find such a representation for any positive even power.

Question 2. Can we prove that it does not exist for any positive even power?

Question 3. Is there a simple way to determine if such a representation exists for a given power?


2 Answers 2


Q1, Q2: Such a representation exists for all odd powers of $\varphi$ as we can show by induction. Using the arctangent identity, let us first write:

$\arctan \varphi^{2n+1} = \arctan\frac{\varphi^{2n+1}+v}{1-\varphi^{2n+1}v} - \arctan v$

Next, we set the argument $\frac{\varphi^{2n+1}+v}{1-\varphi^{2n+1}v}$ equal to $\varphi^{2n-1}$ in order to set up an induction and solve for $v$. After moving terms around and simplifying via identities involving $\varphi$, we end up with $v = \frac{\varphi^{2n-1}-\varphi^{2n+1}}{1+\varphi^{4n}} = -\frac{1}{L_{2n}}$ where $L_n$ is the $n^{th}$ Lucas number. As $\arctan\left(-\frac{1}{x}\right) = -(2\arctan 1-\arctan x)$, this proves the claim for odd powers of $\varphi$.

Q3: For even powers if we set up the analogous equation:

$\arctan \varphi^{2n} = \arctan\frac{\varphi^{2n}+v}{1-\varphi^{2n}v} - \arctan v$

However in this case, setting $\frac{\varphi^{2n}+v}{1-\varphi^{2n}v} = \varphi^{2n-2}$ and solving for $v$ gives $v=\frac{-1}{F_{2n-1}\sqrt{5}}$. This implies there cannot be a rational expression for $\arctan\varphi^{2n}$ in terms of $\arctan\varphi^{2n-2}$ and arctangents of integers for, if there were such an expression, the arctangent identity could then be applied to these terms to express $\frac{-1}{F_{2n-1}\sqrt{5}}$ as a rational number. Similar problems with $\sqrt{5}$ appearing in the expression for $v$ occur if we try setting the argument $\frac{\varphi^{2n}+v}{1-\varphi^{2n}v} = \varphi^k$ for other powers $k$ such as $k=2n-1$, so if any $\arctan \varphi^{2n}$ happen to be rational combinations of arctangents of integers, such expressions ought to be unrelated to one another unlike the situation for odd powers.

As a side note, this difference between the even and odd powers reminds me of the situation of values of the $\zeta$ function at integers where the $\zeta(2n)$ have simple closed form expressions while it is unknown if any $\zeta(2n+1)$ have a closed form expression. Perhaps someone with more familiarity of $\zeta$ values can comment on whether we might expect such similar behavior here.

  • 5
    $\begingroup$ As an applied mathematician, I used this method to calculate $\arctan(\phi^{17})$, and it turns out to be $$18\arctan(1)-\frac{1}{2}\arctan(2)-\arctan(3)-\arctan(7)-\arctan(18)$$ $$-\arctan(47)-\arctan(123)-\arctan(322)-\arctan(843)-\arctan(2207)$$ Of course it can be written differently to make the arguments of the arctangens smaller (like in the question), but I like to think this shows the structure more clearly. $\endgroup$
    – user75414
    Jun 26, 2015 at 7:52
  • 15
    $\begingroup$ Or, explicitly: $$\arctan(\phi^{2n+1})=(2n+2)\arctan(1)-\frac{1}{2}\arctan(2)-\sum_{i=1}^{n} \arctan(L_{2n})$$ If I am not mistaken. $\endgroup$
    – user75414
    Jun 26, 2015 at 8:01
  • $\begingroup$ @Kugelblitz: it appears that Pakk and I both are using the convention that $L_n = \phi^n+\phi^{-n}$, whereas you seem to be shifting your indexing for some reason. $\endgroup$
    – ARupinski
    Jun 28, 2015 at 0:53
  • $\begingroup$ My bad. My thoughts were muddled; you're right about the unnecessary index shift. I will delete my comments so as to prevent unnecessary confusions it may serve to procreate to other readers. I had a doubt originally, as to why Pakk mentioned i=1, but never mentioned a term 'i' in argument of the summation, and instead stated $L_{2n}$. Shouldn't it be $L_{2i}$ ? $\endgroup$ Jun 28, 2015 at 3:34
  • $\begingroup$ Yes, probably it should have been $L_{2i}$, my bad. $\endgroup$
    – user75414
    Nov 20, 2015 at 14:48

It may be helpful to transform this from a problem involving tangents to one involving complex exponentials. Note that $\exp(2 i \arctan(x)) = \dfrac{i-x}{i+x}$. Thus an identity $2 \arctan(x) = \sum k_j \arctan(x_j)$ for integers $k_j$ implies $$ \left(\dfrac{i-x}{i+x}\right)^2 = \prod_j \left( \dfrac{i-x_j}{i+x_j} \right)^{k_j} \tag{1}$$ Conversely, (1) implies $2 \arctan(x) = \sum_{j} k_j \arctan(x_j) + n \pi$ for some integer $n$.

Thus to have an identity for $2 \arctan(x)$ as an integer linear combination of $\arctan(x_j)$ (with $x_1 = 1$ to take care of the $n\pi$), we need $((i-x)/(i+x))^2$ to be in the group generated by $((i-x_j)/(i+x_j))$.

In the case at hand, for odd $n$ I get

$$ \left( \dfrac{i - \phi^n}{i+\phi^n}\right)^2 = \dfrac{L_{2n} - 6 - 4 i L_n}{L_{2n}+2} = \dfrac{i+2}{i-2} \prod_{k=1}^{(n-1)/2} \left(\dfrac{i+L_{2k}}{i-L_{2k}}\right)^2$$

(I admit I might not have found these formulas without seeing Pakk's comment to ARupinski's answer). For even $n$, on the other hand, I get

$$\left(\dfrac{i-\phi^n}{i+\phi^n}\right)^2 = \dfrac{L_{2n} - 6 - 4 i \sqrt{5} F_n}{L_{2n}+2}$$

and clearly you need some factors involving $\sqrt{5}$ to have any hope.


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