# Limit of $\lim_{x\rightarrow 1}\sqrt{x-\sqrt[3]{x-\sqrt[4]{x-\sqrt[5]{x-.....}}}}$

I known how to find the limit of my last question which was Finding $\lim_{x\rightarrow 1}\sqrt{x-\sqrt{x-\sqrt{x-\sqrt{x....}}}}$, but I couldn't how to start to find the limit of $$\lim_{x\rightarrow 1}\sqrt{x-\sqrt[3]{x-\sqrt[4]{x-\sqrt[5]{x-.....}}}}$$ so I used a program written in visual basic 6 to calculate it.I found the limit about equal to $29/30$ but I am not sure if this value is right because the accuracy of this language not enough to give me a trust value. Anyhow,I want to know how to find the limit analytically.

• If you try to calculate this with a program, the result depends on what initial value you have for the roots, if the initial value is $0$, the final result is $1$, if the initial value is $1$ the final result is $0$, if it's between $0$ and $1$ it converges to $0.581881$ and it fails otherwise. I'm guessing the actual answer is $0.581881$ but who knows. Nov 30, 2014 at 19:07
• I tried just evaluating it with $x=1$ and $100000$ nested roots. The initial value I was talking about is the value I assigned to the $1000001$ root since I cant evaluate all infinite roots. Nov 30, 2014 at 19:29
• However I get the same result with $x=1.0001$, with $x<1$ I get undefined and with $x=2$ I get $1$ Nov 30, 2014 at 19:37
• Can you post the program you used to evaluate the limit? Nov 30, 2014 at 19:39
• I get $L=0.5818805230597856271212893881\ldots$ Dec 1, 2014 at 7:06

It is not proved that a limit exists.

The figure below will give a new view on this question of limit or no limit. We consider an infinite set of functions $y_1(x)$ , $y_2(x)$ , ... , $y_n(x)$ , ... where the index $n$ is the number of radicals :

In case of the functions with odd index, the limit for $x$ tending to $1$ is $y(1)=1$

In case of the functions with even index, the limit for $x$ tending to $1$ is $y(1)=0$

When the number of radicals tends to infinity without specified parity, the limit is not defined. So we could say that there is no limit for the expression given in the wording of the question.

But, what is interresting is not the set of functions and corresponding curves, but what is the boundary curve of the infinite set of curves. This curve is not included into the set itself. The corresponding function might be on the form $y_{boundary}(x)=f(x)H(x-1)$ where H is the symbol of the Heaviside step function. Looking for $f(x)$ is something else: This is a challenge.

We can express the firsts terms of the series expansion around $x=1$.

Let $x=1+\epsilon$ , with $\epsilon$ close to $0$.

$y_1=1+\frac{\epsilon}{2}+O(\epsilon^2)$

$y_2=(\frac{2\epsilon}{3})^{1/2}+O(\epsilon^{2/2})$

$y_3=1-(\frac{3\epsilon}{32})^{1/3}+O(\epsilon^{2 /3})$

$y_4=(\frac{4\epsilon}{405})^{1/8}+O(\epsilon^{2/8})$

$y_5=1-(\frac{5\epsilon}{201326592})^{1/15}+O(\epsilon^{2/15})$

$y_6=(\frac{2\epsilon}{ 10296910184203125})^{1/48}+O(\epsilon^{2/48})$

etc.

For $\epsilon=0$, this confirms that $y_n$ is alternatively equal to $0$ and to $1$ when $n$ tends to infinity.

When $n$ tends to infinity, the curves corresponding to the functions $y_n(x)$ tend asymptotically to a curve $f(x)$ for $x>1$. For $x=1$, $y_n(1)$ tends to $1$ if $n$ is odd, or tends to $0$ if $n$ is even.

The bounding curve corresponding to $f(x)$ is drawn in red on the figures below, represented in various enlargements. Of course, the curve in red is the same for all the figures. It looks like a straight line if we consider a range of $x$ close to $x=1$, but it is a misleading impression : $f(x)$ is not a linear function. The computation for drawing $f(x)$ was made with an algorithm similar to that given by Han de Bruijn.

• What are the curves you have plotted there? I don't get the same result for $x<1$, the function is complex for odd index $>1$ there, and neither real nor imaginary part resembles what you've plotted. Looks like you've plotted something like $\sqrt x+f_i(x)$ for $x<1$ and $i>1$, $i$ odd. Dec 23, 2014 at 14:49
• $y_3$ is real if $x>1$ because $x>x^{1/3}>(x-x^{1/4})^{1/3}$ and $x>x^{1/4}$ if x>1. OTOH, if $x<1$ then $x-x^{1/4}<0$ so $x-(x-x^{1/4})^{1/3}>x$ and it has a real square-root, so $y_3$ is real again. Dec 23, 2014 at 15:07
• @Ruslan: The seven curves ploted on the figure correspond to the seven equations written on the same figure : $y_1(x)$ , $y_2(x)$ , ..., $y_7(x)$. Of course, they are ploted in real domains for $x$ and $y$. Dec 23, 2014 at 16:40
• @Michael oh, indeed. Was basing my comment on Mathematica plot. It appears to take the wrong branch of cubic root, funny!.. Dec 23, 2014 at 17:58
• @Ruslan, if you raise a negative number to 0.33333, it has to multiply the angle $\pi$ by 0.33333 Is there an nth-root command in Mathematica? Dec 23, 2014 at 18:54

Disclaimer. Please read the answer by JJacquelin . Without it, the answer you are reading now wouldn't even come into existence - literally : the possibility of no-limit must not be rejected. It must be admitted that the current answer rests upon the assumption that the limit does indeed exist. But the question is: does it exist?

From a computational point of view , this question is related to :

And via this reference it is related to at least three other questions. The common denominator is that backward recursion may be employed here too. For the problem at hand, this backward recursion goes as follows: $$a_{k-1} = \sqrt[k]{x - a_k} = e^{\ln(x - a_k)/k} \qquad ( x-a_k > 0 )$$ Starting "from infinity" with $\,x=1\,$ (where $\,\infty = 100\,$ is large enough for our purpose) , but with arbitrary $\;0 < a_\infty < 1$ , the recursion invariably leads to the outcome in the comment by Lucian and Kristoffer Ryhl : $$L = 0.5818805230597856271212893881$$ Here is the Pascal program snippet that does the job:

program Lucian;
procedure test;
const
n : integer = 100; { = oo }
var
k : integer;
a,x,d : double;
begin
while true do
begin
a := Random;
x := 1; d := 1;
{ Backward recursion }
for k := n downto 2 do
begin
a := exp(ln(x-a)/k);
{ Error analysis }
d := d*a/(x-a)/k;
end;
Writeln(a,' +/-',d);
end;
end;
begin
test;
end.

Output:
 5.81880523059786E-0001 +/- 2.39470132977704E-0043
5.81880523059786E-0001 +/- 6.60618483531661E-0045
5.81880523059786E-0001 +/- 4.15120583976206E-0044

The generality of $\;0 < a_\infty < 1$ sounds more impressive than it is. Because, apart from the first one, the iterations actually start with numbers close to and somewhat smaller than $1$ ($0 < x-a_n < 1$) : $$\lim_{n\to\infty} \sqrt[n]{x - a_n} = 1$$ An estimate for the errors may be obtained by differentiation: $$d a_{k-1} = d \sqrt[k]{x-a_k} = d a_k \frac{1}{k} \left(x-a_k\right)^{1/k-1} = \frac{a_{k-1}}{k(x-a_k)}d a_k$$ starting with $\,da_k = 1\,$ for some sufficiently large value $n$ of $k$ , in our case $\,n=100$ . It is observed that the error in $\;a_1\;$ can be calculated (more or less) by backward recursion as well. It may well be conjectured that some reasonable bound goes like $\,1/n!$ , meaning that convergence is rather fast. Apart from the first iterations, though, where there are large denominators, due to the fact that $a_{\infty-1}$ is close to $x=1$ .

Notes.
If we change x := 1 in the program by a slightly different value below $1$ , say x := 0.99 , then the iterations quickly become unstable for some $\;a_\infty$ , even if reasonable precautions are taken.
So it seems that the limit should better be approached from above : $\;x \downarrow 1$ in the first place.

Furthermore, <quote> it is supposed that $a\ne 0$ and $a\ne 1$ . This is a not proved assumption.
Doing this is something like using the result to be proved, to prove it. If $\,a=1\,$ or $\,a=0\,$ is the starting value for the recursion (case $\,x=1$ ), the result will be $\,0\,$ or $\,1\,$ (depending on the parity of $\,n\,$ in the program).</quote>

Killing the fly in the soup ?   It's the difference between a closed interval $\,0 \le a \le 1\,$ and an open interval $\,0 < a < 1$ . I know it's a distortion of the original problem, but my proposal would be to replace all radicals $\,\sqrt[n]{a}\,$ by exponents $\,\exp(\ln(a)/n)$ , like I've done in the program ( thus simply obviating the special cases $a \in \{0,1\}$ ) . Wonder if that's acceptable for the OP.

BONUS. Making a graph of the following more general function $f(x)$ might be interesting. $$f(x) = \lim_{n\to \infty}\sqrt[2]{x-\sqrt[3]{x-\sqrt[4]{x-\cdots-\sqrt[n]{x}}}}$$ So here goes, for $\;1 < x < 10\;$ and $\;0 < y < 5\;$ :

It is conjectured that $f(2) = 1$ . Proof :
Start with $a_n = 1$ , then $\,a_{n-1} = \sqrt[n]{2 - 1} = 1$ and so on for all $\,a_k\,$ until $\,a_1$ .

• The limit $L=0.581880523..$ obtained with this method of computation or other numerical method is the numerical approximation of $f(1)$ according to the definition of the function $f(x)$ in my answer. The equation of the bounding function is $f(x)H(x-1)$ where H is the Heaviside's step function. Of course, $f(x)$ has a limit for $x$ tending to $1$. But it is not a limit for the infinite nested radicals function which should be called an improper "limit" because it is an asymptotic limit to the step function $f(1)H(x-1)$ for $x$ tending to $1$. Dec 23, 2014 at 12:25
• @JJacquelin: my current interpretation (subject to change) is that you are using forward recursion, which is known to be (numerically) unstable. But I do not deny beforehand that there may be a can of worms here. Dec 23, 2014 at 12:43
• @JJacquelin: I've updated my answer. Which doesn't mean that I don't listen, but I cannot pinpoint exactly where my reasoning could possibly go wrong. Is it perhaps in the denominators of the error estimates? Dec 23, 2014 at 13:05
• I think that your reasoning is correct, depending of what we are looking for. Your computer program is well adapted to the numerical computation of the so called $f(x)$ function : instead of starting with $x=1$ which leads to a good approximate of $f(1)$, we could start with other values of $x$ and draw the curve representing $f(x)$. So, your approach is in agreement with what I consider as the most interesting challenge: is there a closed form for this function ? . On the other hand, ... to be continued... Dec 23, 2014 at 13:53
• On the other hand, if we consider the question of the limit or no-limit of the infinite nested radicals function, by using the backward recursion, there is a fly in the soupe. The program starts with random values $0<a<1$. So it is supposed that $a\neq 0$ and $a\neq 1$. This is a not proved assumption. Doing this is something like using the result to be proved, to prove it. If $a=1$ or $a=0$ is the starting value for the recursion (case $x=1$), the result will be $0$ or $1$ (depending of the parity of $n$ in the program). This proves that the possibility of no-limit must not be rejected. Dec 23, 2014 at 14:15