Sign up ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

In 1987, R. Paris proved that the nested radical expression for $\phi$,


approaches $\phi$ at a constant rate. For example, defining $\phi_n$ as using $n = 5, 6, 7$ "ones" respectively, then,

$$(1/2)(\phi-\phi_5)(2\phi)^5 = 1.0977\dots$$

$$(1/2)(\phi-\phi_6)(2\phi)^6 = 1.0983\dots$$

$$(1/2)(\phi-\phi_7)(2\phi)^7 = 1.0985\dots$$

which is approaching the Paris constant $R = 1.09864196\dots$. It seems it can be generalized. Define,


for integer $k>1$ and the equations,

$$x^k = x+1\tag{2}$$

$$y = \frac{1}{x}+1\tag{3}$$

where $x$ is the root of $(2)$ such that $x = x_n$ as $n \to \infty$. Then one can conjecture that,

$$\lim_{n\to\infty}(1/2)(x-x_n)(ky)^n = C_k\tag{4}$$

for some constant $C_k$. The Paris constant is simply the case $C_2$.

I tested it for increasing large $k$. The sequence of $C_k$ seem to be themselves approaching a constant. The rate is very slow, so for much higher $k = 10^{14},10^{15},10^{16}$,

$$C_{10^{14}} = 0.6931471805599500\dots$$

$$C_{10^{15}} = 0.6931471805599457\dots$$

$$C_{10^{16}} = 0.6931471805599454\dots$$


  1. Does $C_k \to \ln 2 = 0.6931471805599453\dots$ as $k \to \infty$?

$\color{blue}{Edit,\; Nov.\;25}$

More generally, define,


for integers $a\ge 1,\;k>1$ and,

$$x^k = x+a\tag{6}$$

$$y = \frac{a}{x}+1\tag{7}$$

Then it seems,

$$\lim_{n\to\infty}(1/2)(x-x_n)(ky)^n = C_{a,k}\tag{8}$$

The Paris constant is the case $C_{1,2}$. Is it true that as $k \to \infty$, then,

$$\lim_{k\to \infty} C_{1,k} = \ln 2$$

$$\lim_{k\to \infty} C_{2,k} = \tfrac{3}{2} \ln \tfrac{3}{2}$$

$$\lim_{k\to \infty} C_{3,k} = \tfrac{4}{2} \ln \tfrac{4}{3}$$

$$\lim_{k\to \infty} C_{4,k} = \tfrac{5}{2} \ln \tfrac{5}{4}$$

and so on?

P.S. The only known closed-form in terms of transcendental constants is $C_{2,2} = \pi^2/8$.

share|cite|improve this question
It is an iteration of a holomorphic function near an attracting fixed point, and the asymptotics should be determinable from the leading term or two of the power series near the fixed point. I would not expect that there is anything special about iterating radicals other than the power series behavior near $x$. Thus, it is very believable that there is a formula as you wrote, including for non-integer $k$. – zyx Nov 24 '13 at 20:12
From the definition of $C_k$, as $k\to\infty$, is it expected that $C_k$ is finitely bounded (instead of getting smaller and smaller)? Furthermore, why is the bound apparently $\ln2$? – Tito Piezas III Nov 24 '13 at 20:24
I suppose it might be worth knowing that $$x = 1 + \frac{\log 2}{k} + \frac{\log 2 + (\log 2)^2}{2k^2} + O\left(\frac{1}{k^3}\right).$$ – Antonio Vargas Nov 24 '13 at 20:36
Hm. Interesting that the appropriate root of $x^k=x+1$ can be approximated by that formula... – Tito Piezas III Nov 24 '13 at 20:43

3 Answers 3

up vote 5 down vote accepted

Antonio Vargas's observation means that $1$ starts closer and closer to the fixpoint, so that maybe there is less and less difference between $C_k$ and the first term in the sequence defining it ; and maybe that first term converges to $\log 2$.

Let $f_k(x) = \sqrt[k]{1+x}$ for $x \ge 0$ and $k > 1$. Let $\alpha_k$ the unique positive fixpoint of $f_k$ (it is the positive root of $\alpha_k^k = \alpha_k+1$). Define $c_{k,n} = (1/2)(\alpha_k - f_k^{n-1}(1))(f_k'(\alpha_k))^{-n}$ for $n \ge 1$.

Now your constant $C_k$ is defined by $C_k = \lim_{n \to \infty} c_{k,n}$, and we want to give an estimation of $C_k/c_{k,1}$.

We have $c_{k,{n+1}}/c_{k,n} = f_k'(\alpha_k)^{-1}(f_k(\alpha_k) - f_k(f_k^{n-1}(1)))/(\alpha_k - f_k^{n-1}(1)) = f_k'(\alpha_k)^{-1}f_k'(z_{k,n})$ for some $f_k^{n-1}(1) \le z_{k,n} \le \alpha_k$.

Since $f_k'$ is decreasing, we obtain $1 \le c_{k,n+1}/c_{k,n} \le f_k'(f_k^{n-1}(1))f_k'(\alpha_k)^{-1}$

Some crude estimates gives us $\alpha_k \ge f_k^n(1) \ge \alpha_k - (\alpha_k - 1)f_k'(1)^n$,
and then ($f''_k$ is increasing), $f'_k(\alpha_k) \le f_k'(f_k^n(1)) \le f_k'(\alpha_k) - (\alpha_k-1)f'_k(1)^nf_k''(1)$,
and finally $1 \le c_{k,n+1}/c_{k,n} \le 1 + (\alpha_k-1)f'_k(1)^{n-1}(-f''_k(1))f_k'(\alpha_k)^{-1} $.

Using $1+x \le \exp(x)$ and taking the product, we obtain $1 \le C_k/c_{k,1} \le \exp((\alpha_k-1)(-f''_k(1))f'_k(\alpha_k)^{-1}/(1-f'_k(1))) $

Since $\alpha_k = 1 + \log 2/k + O(k^{-2})$, we have

$\alpha_k-1 \sim \log2 / k$
$f'_k(\alpha_k) = \alpha_k(1 + \alpha_k)^{-1}/k \sim 1/2k$
$c_{k,1} = (1/2)(\alpha_k-1)f'_k(\alpha_k)^{-1} \sim (1/2)(\log 2/k)(2k) = \log 2$
$f_k'(1) = 2^{1/k-1} \frac 1k \sim 1/2k$
$f''_k(1) = 2^{1/k-2} \frac 1k (\frac 1k -1) \sim -1/4k$
$(\alpha_k-1)(-f''_k(1))f'_k(\alpha_k)^{-1}/(1-f'_k(1)) \sim \log 2/2k \to 0$

This shows that $C_k \sim c_{k,1} \to \log 2$

For the more general case, we start from $x_1 = a^{1/k} = 1 + \log(a)/k + \ldots$, while $\alpha = 1 + \log(a+1)/k + \ldots$, which are again close to each other.

$\alpha - a^{1/k} \sim \log(\frac{a+1}a)/k$
$f'(\alpha) = \alpha/k(a+\alpha) \sim 1/(a+1)k$
$c_1 = (1/2)(\alpha - a^{1/k})/f'(\alpha) \sim \frac{a+1}2\log(\frac{a+1}a)$

Since $f'(\alpha^{1/k})$ and $f''(\alpha^{1/k})$ are on the order of $1/k$, we have $C = \frac{a+1}2\log(\frac{a+1}a)$

share|cite|improve this answer
Thanks, Mercio. I'll accept your answer soon. However, I was wondering if you know how to tweak it to cover the more general case $x_n=\sqrt[k]{a+\sqrt[k]{a+\sqrt[k]{a+\sqrt[k]{a_n+\dots}}}}$? (Kindly see the Nov.25 edit.) – Tito Piezas III Nov 26 '13 at 0:17

Hint : $\displaystyle x(n)=\underset{k=0}{\overset\infty{\Large\Xi}}\left(a,b\,;\tfrac1n\right)\iff x^n=a+bx\iff n(x)=\frac{\ln(a+bx)}{\ln x}\iff n(1)=\infty$ , $n(\infty)=1$ . Now show, using l'Hopital, that $\displaystyle\lim_{x\to1}\Big[n(x)\cdot(x-1)\Big]=\ln2$.

share|cite|improve this answer
Could you please explain/define the notation $\Xi_{k=0}^\infty\left(a,\,b;\,1/n\right)$? – Ian Mateus Nov 24 '13 at 22:40
Sorry, I thought it was self-evident: $$\underset{_{\text{k}\,=\,0}}{\overset{n}{\mathbf\Xi}}\ \Bigg(a_{_\text{k}}\ ,\ b_{_\text{k}},\ \frac1{N_{_\text{k}}}\Bigg)\ =\ \sqrt[^{N_{_\text{0}}}]{a_{_\text{0}}\ +\ b_{_\text{0}}\sqrt[^{N_{_\text{1}}}]{a_{_\text{1}}\ +\ b_{_\text{1}}\sqrt[^{N_{_\text{2}}}]{\ldots\ \sqrt[^{N_{_{n}}}]{a_{_{n}}}}}}$$ which, for $N_{_\text{k}}=-1$, becomes the continued fraction $$\underset{_{\text{k}\,=\,0}}{\overset{n}{\mathbf\Xi}}\ \Big(a_{_\text{k}}\ ,\ b_{_\text{k}},\ -1\Big)\ =\ \cfrac1{a_{_0}\ +\ \cfrac{b_{_0}}{a_{_1}\ +\ \cfrac{b_{_1}}{\ddots\ {a_{_n}}}}}$$ – Lucian Nov 30 '13 at 19:58

You might notice that if instead of:

$\lim_{n\to\infty}(1/2)(x-x_n)(ky)^n = C_{a,k}^+\tag{8}$

You don't divide by 2:

$\lim_{n\to\infty}(x-x_n)(ky)^n = C_{a,k}^+\tag { }$

The known closed form constant is then $C_{2,2}^+=\frac{pi^2}{4}$, while the other version (below) with $a=x^k+x$ has a closed form constant of $C_{2,2}^{-}=\frac{pi}{2\sqrt{3}}$.

other version: $x_n=\sqrt[k]{a-\sqrt[k]{a-\sqrt[k]{a-\sqrt[k]{a_n-\dots}}}}\tag { }$

C+ approaches 1 from above: $\lim_{k\to \infty} C_{1,k}^+ = 2 \ln 2\tag{}$

$\lim_{k\to \infty} C_{2,k}^+ = 3 \ln \tfrac{3}{2}\tag{}$

$\lim_{k\to \infty} C_{3,k}^+ = 4 \ln \tfrac{4}{3}\tag{}$

C- approaches 1 from below:

$\lim_{k\to \infty} C_{2,k}^- = \ln 2 \tag{}$

$\lim_{k\to \infty} C_{3,k}^- = 2 \ln \frac{3}{2}\tag{}$

$\lim_{k\to \infty} C_{4,k}^- = 3 \ln \frac{4}{3} \tag{}$

You'll notice that the second form (with $a=x^k+x$) has a C that approaches 1 from below, and the first one (you posted about) approaches 1 from above.

$\lim_{k\to \infty} C_{a,k}^+ = (a+1) \ln \frac{a+1}{a} \tag{}$

$\lim_{k\to \infty} C_{a,k}^- = (a-1) \ln \frac{a}{a-1} \tag{}$

Coincidentally :D $\lim_{a\to\infty} (\frac{a}{a-1})^a \to e^+$ from above, and $\lim_{a\to\infty} (\frac{a}{a-1})^{a-1} \to e^-$ from below.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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