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Do we have nested square roots with initial and final term and infinite terms in between? For example $$\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{2+\cdots+\sqrt{3}}}}}$$ which happens in modified Viète nested radical

Or nested radicals like general form as follows $$\sqrt{a-\sqrt{a+\sqrt{a+\sqrt{a+\cdots+\sqrt{b}}}}}$$ If there are, what is the proof?

Thanks in advance

Andrew Chin
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Sivakumar Krishnamoorthi
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  • If there is a meaning $x$ for it, then $(x^2-2)^2+x^2=4,$ leaving zero relevance to the $\sqrt{3}$ part. – Thomas Andrews Sep 19 '19 at 15:18
  • If what should lay where the sign $(\ldots)$ is, has infinitely many opèrations, I can't figure what is the exact meaning of this. It resembles imaginative expressions by my pupils, when they are asked about the least element in $(0,1)$; some of them say that is $0.0\underbrace{00000000000\ldots0}_{\text{infinitely many zeros}}1$ :) – ajotatxe Sep 19 '19 at 15:24

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It has been mention in the comments but the actual value of the infinityth root does not matter, and does not really exist. To that note we may as well say: $$y=\sqrt{a-\sqrt{a+\sqrt{a+...\sqrt{b}}}}=\sqrt{a-\sqrt{a+\sqrt{a+...\sqrt{a}}}}$$ This can be easily solved by saying: $$x=\sqrt{a+\sqrt{a+...}}$$ $$x=\sqrt{a+x}\Rightarrow x^2-x-a=0$$ Which I believe has the solution: $$x=\frac{1+\sqrt{4a+1}}{2}$$ Now we can substitute this in to get: $$y=\sqrt{a-\frac{1+\sqrt{4a+1}}{2}}$$

Henry Lee
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