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Is there some interconnection between these two topics?

A sort of classification of the possibile types of nested radicals and maybe some way (hopefully bijective, in some sense) to pass from a nested radical to a partial fraction and vice versa?

I know this is vague, but I didn't found nothing about it.

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    $\begingroup$ Continued fractions are nested radicals of order $-1.$ $\endgroup$
    – Lucian
    Feb 21, 2016 at 4:48
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    $\begingroup$ Technically, radicals is not a correct term in this case. Nested powers seems more appropriate $\endgroup$
    – Yuriy S
    Feb 21, 2016 at 8:04
  • $\begingroup$ See also this question math.stackexchange.com/q/1454204/269624 $\endgroup$
    – Yuriy S
    Feb 21, 2016 at 18:16

1 Answer 1

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Yes, there is. Consider a continued fraction in the form:

$$x=\cfrac{a}{b+\cfrac{a}{b+\cfrac{a}{b+\cdots}}}$$

Assume the limit exists and find it:

$$x=\cfrac{a}{b+x}$$

$$x^2+bx-a=0$$

$$x=\frac{\sqrt{b^2+4a}-b}{2}$$

Now consider the nested radical:

$$x=\sqrt{c+d\sqrt{c+d\sqrt{c+\cdots}}}$$

Assume the limit exists and find it:

$$x=\sqrt{c+dx}$$

$$x^2-dx-c=0$$


If we set $d=-b$ and $c=a$ we get exactly the same value of the limit. I assumed that $b>0$ so in this case $d<0$ and we get the radical:

$$x=\sqrt{a-b\sqrt{a-b\sqrt{a-\cdots}}}=\cfrac{a}{b+\cfrac{a}{b+\cfrac{a}{b+\cdots}}}$$

With the condition $a>b>0$, of course.


This is the most simple connection we could find, but of course there may be countless others.

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    $\begingroup$ This was beautiful. Did you find it yourself ? $\endgroup$
    – Saikat
    Feb 21, 2016 at 3:44
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    $\begingroup$ No, I've seen it somewhere. But it is fairly easy to guess $\endgroup$
    – Yuriy S
    Feb 21, 2016 at 8:03

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