How to find value of $\sqrt{1-\sqrt{1+\sqrt{1-\sqrt{1+\cdots\sqrt{1-\sqrt{1+1}}}}}}$ ?
I've calculated it by MATLAB for some finite terms and I've got : $0.3001 - 0.4201i$, but I don't know how to find the value analytically! Would you mind helping me find it? Thanks

  • 1
    $\begingroup$ Do you mean to be asking about the limit of this expression as the number of radicals goes to infinity? $\endgroup$
    – 2'5 9'2
    Commented Jul 11, 2013 at 19:06
  • $\begingroup$ @alex.jordan Yes! Does the notation have any other interpretation? $\endgroup$ Commented Jul 11, 2013 at 19:21
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    $\begingroup$ @Mahdi: The usual literal interpretation of what you wrote would be that there is a finite but unspecified number of radicals. Compare: $0.9999999\ldots$ versus $0.9999999\ldots 99$. The first is equal to $1$, the second is equal to $1-10^{-k}$ for some unspecified positive integer $k$. In this case, there may not be an easy way to put the final ellipses in a good place, but you can use limit notation to make this clearer, or perhaps just include the phrase "as the number of radicals goes to infinity". $\endgroup$ Commented Jul 11, 2013 at 19:36
  • $\begingroup$ If you change 1 to 7 the answer is 2 see mks.mff.cuni.cz/kalva/putnam/psoln/psol536.html $\endgroup$
    – igumnov
    Commented Jul 11, 2013 at 20:38
  • $\begingroup$ I don't understand why people are so fond of writing seemingly infinite expressions that simply make no sense. If you want to compute the limit of a recursively defined sequence, write down the recurrence instead of some expression with ellipses that gives no clue where to start (in this case inside or outside). Someday I would like to think up of a nice example where multiple equally valid interpretations all give well defined convergent but distinct limits. $\endgroup$ Commented Aug 13, 2013 at 9:32

4 Answers 4


First of all let's assume the series is convergent. Looking for fixed points we have:


Now we will try to solve this equation. First squaring both sides:

$$1-x^2=\sqrt{1+x} \\ \left(\left( 1-x\right)\left( 1+x\right) \right)^2=1+x$$

Note that $x$ must be nonnegative, therefore:

$$(1-x)^2(1+x)-1=0 \\ \Rightarrow x^3-x^2-x=0$$

So $x=0$ is a solution. The other solutions are:

$$x^2-x-1=0 \Rightarrow x=\frac{1 \pm \sqrt{5}}{2}$$

where only $x=\frac{1 + \sqrt{5}}{2}$ is greater than or equal to zero and may look valid. But as people pointed out, one has to check if the answers actually fit into the initial equation. In this case $\frac{1 + \sqrt{5}}{2}$ doesn't, therefore the only fixed point we have found is $x=0$.

But $x=0$ cannot be the convergence limit(it doesn't converge smoothly). Assume we deflect $x=0$ with the tiny amount of $\epsilon$(or rather starting with a tiny $x_1=\epsilon$). Putting it back into our initial equations and getting the next $x$:

$$x_2=\sqrt{1-\sqrt{1+\epsilon}}\approx \sqrt{1-\left( 1+\frac{\epsilon}{2}\right)} \approx \frac{i\sqrt{\epsilon}}{\sqrt{2}}$$

Now for $\epsilon < \frac{1}{2}$, $|x_2|>|x_1|$; ergo $x=0$ cannot be the convergence limit. We have proved that this infinite radicals doesn't have a single limit.

  • $\begingroup$ How do you know that it is converging to a certain value? $\endgroup$
    – EricAm
    Commented Jul 11, 2013 at 19:31
  • $\begingroup$ @AdamYac Good point, I will add a note about that. $\endgroup$
    – Ali
    Commented Jul 11, 2013 at 19:33
  • $\begingroup$ This does not satisfy$x = \sqrt{1-\sqrt{1+x}}$. See my answer for the reason. $\endgroup$ Commented Jul 11, 2013 at 19:46
  • $\begingroup$ @martycohen yep, I was going to do something about that. $\endgroup$
    – Ali
    Commented Jul 11, 2013 at 19:49
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    $\begingroup$ It might be interesting to think about why the golden number insists on showing up even when not wanted. $\endgroup$ Commented Jul 11, 2013 at 19:50

The iteration of function $f(x) = \sqrt{1-\sqrt{1+x}}$ starting at $x=1$ approaches a 2-cycle of $.2229859448+.4133637969 i$ and $.2229859448-.4133637969 i$.

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    $\begingroup$ Note: These complex numbers $a\pm bi$ satisfy $$a^4-6a^2b^2-2a^2+b^4+2b^2-a=4a^3b-4ab^3-4ab+b=0$$ but Wolfram Alpha doesn't turn this into a radical expression. Might be tricky. $\endgroup$
    – MvG
    Commented Jul 11, 2013 at 21:03
  • $\begingroup$ @GEdgar Thank you for the answer. You mean a fixed point algorithm? $\endgroup$ Commented Jul 12, 2013 at 11:08

This $cannot$ have a positive real root, because, if $x$ is such a root, then $\sqrt{1+x} > 1$ so $1-\sqrt{1+x} < 0$, which means that $\sqrt{1-\sqrt{1+x}}$ is complex, not real.

In other words, finding a fixed point does not work - there is $no$ fixed point.

The best that can be done is to find a two-cycle as GEdgar has done. This involves solving $x = f(f(x))$, which is much more complicated.

Therefore Ali's work, which I duplicated, is wrong. It solves $f(x) = x$, but $f$ does not have a limit, it has a two-cycle.

  • $\begingroup$ I have modified my answer. $\endgroup$
    – Ali
    Commented Jul 12, 2013 at 7:05

This question was already answered and one answer was accepted. This answer is just to add to the completeness for some later reader.

If one looks at the function $f(x) = \sqrt{1-\sqrt{1+\sqrt{1-\sqrt{1+x}}}}$ then this function has true fixpoints; the fixpoints are $\small t_0 = 0$, $ \small t_1 \approx 0.222985944830 + 0.413363796251 i$ and $ \small t_2 \approx 0.222985944830 - 0.413363796251 i$.

Around the fixpoint $t_0$ we get a real power series with a fractional cofaktor but without constant term due to wolframalpha, see my related question.

Developing around the complex fixpoints, say $t_1$ we get a usual (though with complex coefficients) power series, $ \small f_4(x+t_1)-t_1 = g(x) \approx 0.219471696356 x + $ $ \small (-0.102599142965 + 0.177579081091 i) x^2 + $ $ \small(-0.133743852861 - 0.220457785139 i) x^3 $ $ \small+ (0.375580757794 - 0.0217915275445 i) x^4 + O(x^5) $ where the absolute value of the coefficient of the linear term is smaller than 1 - which means, that this fixpoint $t_1$ is also attracting.

Thus infinite iteration of $f(x)$ beginning from any complex value (except 0) shall converge to that fixpoint $t_1$. This solves the first problem: is that OP's notation converging at all.

After that we can simply state, that any finite truncation of the OP's expression approximates one of the cycling fixpoints :

t1=0.222985944830 + 0.413363796251*I
x0=sqrt(1+t1)     :    %2164 = 1.1211469 + 0.18434863*I
x0=sqrt(1-x0)     :    %2165 = 0.22298594 - 0.41336380*I
x0=sqrt(1+x0)     :    %2166 = 1.1211469 - 0.18434863*I
x0=sqrt(1-x0)     :    %2167 = 0.22298594 + 0.41336380*I  \\ cycling occurs, approxi-
x0=sqrt(1+x0)     :    %2168 = 1.1211469 + 0.18434863*I   \\ mating the four-point
x0= ...           :     ...  =  ...                       \\  cycle

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