# A constant bizarrely related to the Fibonacci Numbers

For roughly the past month, I have been studying denesting radicals. For example: the expression $$\sqrt{\sqrt2-1}$$ is a radical expression that contains another radical expression, so this radical is nested. Is there a way to express this with radicals that are not (or not as) nested? Writing it in this way is referred to as denesting, and indeed, there is one such way. Ramanujan mysteriously found that $$\sqrt{\sqrt2-1}=\sqrt{\frac 19}-\sqrt{\frac 29}+\sqrt{\frac 49}.$$ I found that this is also equal to $$\sqrt{\sqrt{\frac 43}-\sqrt{\frac 13}}$$, which does not denest it, but it does write the expression under a radical of a coprime degree, which fascinates me.

I have found an abundance of results, one of which fascinates me the most, and is on the constant, $$1-\sqrt{\frac 12}+\sqrt{\frac 14}.$$ This constant is equal to all of the expressions below. Note that the degree of the radicals are, lo and behold, the Fibonacci numbers! $$\sqrt{\frac 32\bigg(\sqrt2-\frac 1{\sqrt2}\bigg)}$$ $$\sqrt{\frac{3^2}{2^2}\big(\sqrt2-1\big)}$$ $$\sqrt{\frac{3^3}{2^3}\bigg(\frac 3{\sqrt2}-\sqrt2-1\bigg)}$$ $$\sqrt{\frac{3^5}{2^5}\bigg(4-\frac{5}{\sqrt2}\bigg)}$$ $$\sqrt{\frac{3^8}{3^8}\bigg(1+\frac{17}{\sqrt2}-\frac{23}{\sqrt4}\bigg)}$$ and, slightly breaking the pattern, $$\sqrt{\frac{3^{14}}{2^{14}}\big(41-59\sqrt2+21\sqrt4\big)}$$ and presumably, this list goes on forever (but the numbers start becoming pretty big).

So... what on earth is going on here? It appears that for some $$n$$th Fibonacci number $$F_n$$, this is equal to (for at least most of the time), $$\sqrt[F_n]{\frac{3^{F_{n-1}}}{2^{F_{n-1}}}\big(a+b\sqrt2+c\sqrt4\big)}$$ for some $$\{a, b, c\}\subset \mathbb R$$, with a minimal polynomial of $$4x^3-12x^2+18x-9$$.

Can anybody explain these wild affairs?

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• Your element is equal to $(1 - \sqrt{2}+\sqrt{4})/\sqrt{4}$, i.e., to the quotient of elements with norms $3$ and $4$ in the number field ${\mathbb Q}(\sqrt{2})$. If you raise this to the $n$-th power, you get an $n$-th power, which is hardly surprising. Jun 10 '20 at 9:44
• @FranzLemmermeyer should you be referring to galois theory, I have not examined this part to construct nested radicals, and have purely been tackling this from an algebraic approach, given thus far my limited knowledge of matrices, vector spaces and linear transformations; however, does your comment explain the forms possessed by the fibonacci-type nested radicals? Jun 10 '20 at 10:12
• You've only given examples with Fibonacci roots. Are you sure that there are none with 4th, 7th or other roots? Jun 10 '20 at 11:52
• @FranzLemmermeyer there appear to be none of this particular form. Although similar, the exponents raised by $3/2$ does not seem to share a pattern relative to the degree of the radical put over it, as in the case of the fibonacci degrees. In fact, sometimes the radicands would have $3^m/2^n$ for $m\neq n$ unlike in the fibonacci-type examples. Jun 10 '20 at 12:36
• $14$ is not a Fibonacci number. Was the second-last display meant to involve $3^{13}/2^{13}$? Jun 10 '20 at 23:21

First of all we get $$A=1-\sqrt{\frac{1}{2}}+\sqrt{\frac{1}{4}}=\frac{3\sqrt{2}}{2(\sqrt{2}+1)}$$......(1)

And from here you can proof by induction...

$$A^{F_n}=A^{F_{n-1}}.A^{F_{n-2}}$$

If $$A^{F_{n-1}}=(\frac{3}{2})^{F_{n-2}}(a_{n-1}+b_{n-1}\sqrt{2}+c_{n-1}\sqrt{4})$$

and

$$A^{F_{n-2}}=(\frac{3}{2})^{F_{n-3}}(a_{n-2}+b_{n-2}\sqrt{2}+c_{n-2}\sqrt{4})$$

From (1) we get, $$A^{F_2}, A^{F_3}$$, so we shall get the rest by induction.

Where, $$F_2=2, F_3=3$$.

$$\frac{3}{2}(\frac{\sqrt{2}}{\sqrt{2}+1})^2=\sqrt{2}-\sqrt{\frac{1}{2}}$$

And,

$$\frac{3}{2}(\frac{\sqrt{2}}{\sqrt{2}+1})^3=\sqrt{2}-1$$