Formula for Nested Radicals

Question

I know that: $$\sqrt{2+\sqrt{2+\sqrt{2+...\sqrt{2}\; (upto\; n\; times)}}}=2\cos(2^{-n-1}\:\pi)$$

I was wondering whether such a formula exists for $$\sqrt{3+\sqrt{3+\sqrt{3+...\sqrt{3}\; (upto\; n\; times)}}}$$

or in general for, $$\sqrt{k+\sqrt{k+\sqrt{k+...\sqrt{k}\; (upto\; n\; times)}}}$$

I've tried scaling the formula for 2 to get an approximate result.
For example for the 6 case: $$\sqrt{6+\sqrt{6+\sqrt{6+...\sqrt{6}\; (upto\; n\; times)}}}\approx\left(\frac{2\cos(2^{-n-1})-\sqrt2}{2-\sqrt2}\right)(3-\sqrt6)+\sqrt6$$

For the 6 case, if n is infinite, then the value is 3. That's the upper limit. the lower limit is $\sqrt6$. So, the results for the general case must lie between the two. After that, I simply scaled it up. — Arya Das, Oct 26 '15 at 10:44
I highly doubt that that formula can be generalised for $k$ as the choice of $k=2$ is a special case that relates to trigonometric values. — Ali Caglayan, Oct 26 '15 at 10:51

score 11 · Answer 1 · answered Oct 26 '15 at 10:50

11

Let $$f(k,n)=\underbrace{\sqrt{k+\sqrt{k+\sqrt{\ldots+\sqrt k}}}}_n $$ We know that $\cos\frac x2=\pm\frac12\sqrt{1+\cos x}$. Therefore if $f(n,k)=a\cos b$ then $$f(n+1,k)=\sqrt{k+f(n,k)}=2\sqrt k\cdot\frac12\sqrt{1+\frac{f(n,k)}{k}}=2\sqrt k\cdot\frac12\sqrt{1+\frac ak\cos b}. $$ This works out nicely with our haf-angle formula only if $a=k$ and then becomes $$ f(n+1,k)=2\sqrt k\cdot\frac12\sqrt{1+\cos b}=2\sqrt k\cos\frac b2.$$ We'd like to have $2\sqrt k=a=k$ again, but that works only if $k=2$. In other words, there won't be formulas of the form $f(n,k)=a_k\cdot \cos 2^{-n}\alpha_k$ except for $k=2$.

answered Oct 26 '15 at 10:50

Hagen von Eitzen

374,180

2

Perfect answer. – Ali Caglayan Oct 26 '15 at 10:52
But what about a formula that doesn't use trigonometry. – Arya Das Oct 26 '15 at 10:53
3

Well, there is a formula using nested radicals. – Klaus Draeger Oct 26 '15 at 12:28

score 1 · Accepted Answer · 2015-10-26T11:49:39.923

Hint:

The sequence tends to

$$t=\frac{1+\sqrt{1+4k}}2.$$

The iterations converge quickly and after a few of them, $t_n=t-\delta_n$, where $\delta_n\ll t$.

Then

$$t-\delta_{n+1}=\sqrt{k+t-\delta_n},\\ t^2-2t\delta_{n+1}+\delta_{n+1}^2=k+t-\delta_n,\\$$ simplifying and neglecting the square term, $$\delta_{n+1}\approx\frac{\delta_n}{2t}.$$

The error decays exponentially,

$$t_n\approx t-\frac c{(2t)^n}.$$

This is in agreement with the case $k=2$, such that $t=2$, and by Taylor

$$2\cos(2^{-n-1}\pi)\approx 2-\frac{\pi^2}{4\cdot(2\cdot2)^n}.$$

Formula for Nested Radicals

2 Answers2