This was inspired by similar posts like this one. Define the function,

$$F(p) = \lim_{n\to\infty}2^n\sqrt{2-\underbrace{\sqrt{2+\sqrt{2+\dots+\sqrt{p}}}}_{n \textrm{ square roots}}}$$

We know that,

$$F(2) = \frac{\pi}{2},\quad F(3) = \frac{\pi}{3}$$

I was wondering what it evaluates to if we use other integers. Some numerical computation and the Inverse Symbolic Calculator suggests that,

$$\begin{aligned} F(5) &= 2\ln\big(\tfrac{1+\sqrt{5}}{2}\big)\,i\\ F(6) &= \ln\big(2+\sqrt{3}\big)\,i\\ F(7) &= \ln\big(\tfrac{5+\sqrt{3\times7}}{2}\big)\,i\\ \vdots\\ F(11) &= \ln\big(\tfrac{9+\sqrt{7\times11}}{2}\big)\,i\\ \vdots\\ F(17) &= \ln\big(\tfrac{15+\sqrt{13\times17}}{2}\big)\,i \end{aligned}$$

Note that the radical arguments are fundamental units. If we use negative $p$,

$$\begin{aligned} F(-1) &= \pi-2\ln\big(\tfrac{1+\sqrt{5}}{2}\big)\,i\\ F(-2) &= \pi-\ln\big(2+\sqrt{3}\big)\,i\\ F(-3) &= \pi-\ln\big(\tfrac{5+\sqrt{3\times7}}{2}\big)\,i\\ \vdots\\ F(-7) &= \pi-\ln\big(\tfrac{9+\sqrt{7\times11}}{2}\big)\,i\\ \vdots\\ F(-13) &= \pi-\ln\big(\tfrac{15+\sqrt{13\times17}}{2}\big)\,i \end{aligned}$$

and so on. It seems $F(2+m)+F(2-m) = \pi$. I also observed that if $m\pm2$ are primes, then,

$$F(2+m) = \pi-F(2-m) = \ln\Big(\tfrac{m+\sqrt{(m-2)(m+2)}}{2}\Big)\,i\tag1$$

though the form of $(1)$ is only conjectural.

Question: What is then the formula for $F(p)$ using general $p$?

  • $\begingroup$ May be $F(5)=2\ln(\eta(5))i$ and $F(4+k)=\ln(\eta(5k+k(k-1)))i$? I checked this for the $k=2,3,4,7$. Here $\eta(D)$ is fundamental unit. $\endgroup$ – grizzly Sep 2 '15 at 17:48

The (hyperbolic) cosine bisection formula gives: $$2\cos\frac{x}{2}=\sqrt{2+2\cos x},\qquad 2\cosh\frac{x}{2}=\sqrt{2+2\cos x}$$ hence assuming $a_0=2\cosh(u_0)=\sqrt{p}$ and $a_{n+1}=\sqrt{2+a_n}$ we have: $$ a_n = 2 \cosh\left(\frac{u_0}{2^n}\right),\quad \sqrt{2-a_n}= 2\sinh\left(\frac{u_0}{2^{n+1}}\right)$$ so: $$ \lim_{n\to +\infty} 2^n\sqrt{2-a_n} = u_0 = \text{arccosh}\left(\frac{\sqrt{p}}{2}\right). $$ If $p$ is a positive real number less than $4$ it is enough to replace $\cosh$ with $\cos$ and $\text{arccosh}$ with $\arccos$. If $p$ is negative, we have to be careful in defining what the square root of a complex number is, but the trick is just the same.

| cite | improve this answer | |
  • $\begingroup$ Thanks. Can you give an example, say, how to arrive at the value of $F(17)$ given above? $\endgroup$ – Tito Piezas III Sep 2 '15 at 16:43
  • $\begingroup$ @TitoPiezasIII: to compute $\text{arccosh}\left(\frac{\sqrt{17}}{2}\right)$ is equivalent to solving $\cosh(x)=\frac{\sqrt{17}}{2}$ or $e^{x}+e^{-x}=\sqrt{17}$ that is a quadratic equation in $e^x$. Logarithms arise from there. $\endgroup$ – Jack D'Aurizio Sep 2 '15 at 16:46
  • $\begingroup$ Obviously you have to adjust something in the above lines if $a_n>2$, since in such a case $\sqrt{2-a_n}$ is not a real number. Anyway, the point is that the structure of the nested radical is related with the cosine bisection formula, hence the limit depends on the inverse function of $\cos$/$\cosh$. $\endgroup$ – Jack D'Aurizio Sep 2 '15 at 16:50
  • $\begingroup$ Hm. By the way, can you confirm if the form of $(1)$ (the factorization of the discriminant) holds true when $m\pm2$ are primes? $\endgroup$ – Tito Piezas III Sep 2 '15 at 16:50
  • $\begingroup$ @TitoPiezasIII: it looks reasonable as a consequence of the (hyperbolic) cosine addition formulas, but I see no reason for that identity to depend on the primality of $m$. Maybe just on the parity. $\endgroup$ – Jack D'Aurizio Sep 2 '15 at 16:52

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.