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Convergence of $sqrt{1-sqrt{{1}-sqrt{{1}-sqrt{{1-sqrt{{1}}}...}}}}$

Mathematics Asked on November 24, 2021

Recently, as is evident from many of my recent questions, I have been very interested in nested radicals. Recently I attempted to investigate the following infinite nested radical and arrived at a strange, counter-intuitive result.
$$sqrt{1-sqrt{{1}-sqrt{{1}-sqrt{{1-sqrt{{1…}}}}}}}$$
At first glance, it would seem that the value of the nested radical must be either $0$ or $1$. However, if we make the assumption that a single value can be assigned to it, ie it converges, we arrive at a different value for it:
Let
$$x=sqrt{1-sqrt{{1}-sqrt{{1}-sqrt{{1-sqrt{{1…}}}}}}}implies x=sqrt{1-x}implies x^2=1-x implies x^2+x-1=0$$
Using the quadratic formula, we get $$x=frac{-1+sqrt5}{2}$$
as $x$ is necessarily positive. Can this result be true? Or is my assumption that allowed me to label the radical as $x$ and assume it converges incorrect?

One Answer

Whenever we find $ldots$ in a formula we must be clear what they are meaning. My interpreation of a formula like $$sqrt{1-sqrt{1-sqrt{1-sqrt{1-ldots}}}}$$ is that this is the limit of the sequence $$x_1=sqrt{1}\ x_2=sqrt{1-sqrt{1}}\ x_3=sqrt{1-sqrt{1-sqrt{1}}}\ ldots\ x_{n+1}=sqrt{1-x_n}$$

If this is the meaning of this expression then we have

$$x_1=1\ x_2=0\ x_3=1\ x_4=0\ x_5=1\ x_6=0\ ldots $$

and this sequence does not converge at all.


Your calculation says:

If $x_n$ converges, then the limit will be $frac{-1+sqrt5}{2}$ or $frac{-1-sqrt5}{2}$

This is true. But $x_n$ does not converge, so this result is useless.

Answered by miracle173 on November 24, 2021

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