Ramanujan found many awe-inspiring nested radicals, such as...

$$\sqrt{\frac {1+\sqrt[5]{4}}{\sqrt[5]{5}}}=\frac {\sqrt[5]{16}+\sqrt[5]{8}+\sqrt[5]{2}-1}{\sqrt[5]{125}}\tag{1}$$$$\sqrt[4]{\frac {3+2\sqrt[4]{5}}{3-2\sqrt[4]{5}}}=\frac {\sqrt[4]{5}+1}{\sqrt[4]{5}-1}\tag{2}$$$$\sqrt[3]{\sqrt[5]{\frac {32}{5}}-\sqrt[5]{\frac {27}{5}}}=\frac {1+\sqrt[5]{3}+\sqrt[5]{9}}{\sqrt[5]{25}}\tag{3}$$$$\sqrt[3]{(\sqrt{2}+\sqrt{3})(5-\sqrt{6})+3(2\sqrt{3}+3\sqrt{2})}=\sqrt{10-\frac {13-5\sqrt{6}}{5+\sqrt{6}}}\tag{4}$$$$\sqrt[6]{4\sqrt[3]{\frac {2}{3}}-5\sqrt[3]{\frac {1}{3}}}=\sqrt[3]{\sqrt[3]{2}-1}=\frac {1-\sqrt[3]{2}+\sqrt[3]{4}}{\sqrt[3]{9}}\tag{5}$$

And there's more!

Question: Is there a nice algebraic way to denest each radical such as above?

For me, I've only been able to prove such identities by raising both sides to the appropriate exponents and use Algebra to simplify them. But sometimes, that can be very difficult for identities such as $(1)$.

  • 5
    $\begingroup$ This is like advanced arithmetic, hey, no variable! I need to write these down for those who think that there is no math without letters.... $\endgroup$
    – imranfat
    Jul 29 '16 at 17:50
  • $\begingroup$ I believe the $(4)$th one is not Ramanujan's $\endgroup$
    – Mourad
    Jan 22 '20 at 1:15
  • $\begingroup$ I just noticed that $\bigg(\dfrac {\sqrt[4]{5}+1}{\sqrt[4]{5}-1}\bigg)^4$ is almost an integer, $643.992\ldots$ Coincidence? $\endgroup$
    – Mr Pie
    Mar 3 '20 at 9:41
  • $\begingroup$ @MrPie I noticed you like seeing these patterns, well you may also like to see/prove why the following may also be true $$\text{Eq.}(4)\approx \pi\tag{1}$$ $$\left(\frac {\sqrt[4]{5}+1}{\sqrt[4]{5}-1}\right)^4\approx 2\varphi^{12}\tag{2}$$ Where $\varphi=\frac{1+\sqrt{5}}{2}$, Proving $(2)$, is the answer to your doubt since $\varphi^n \rightarrow \text{Integer}$. The great thing is that you can prove the $(2)$, by using a much more beautiful nested radical found by Ramanujan which is not in this post. $\endgroup$
    – Mourad
    Mar 23 '20 at 14:30

Landau's algorithm: http://ieeexplore.ieee.org/xpl/articleDetails.jsp?arnumber=63496 $\qquad\qquad$




$$(\sqrt[4]{a} \pm \sqrt[4]{b})^{4}= a+b+6\sqrt{ab} \pm 4\sqrt[4]{ab}(\sqrt{a}+\sqrt{b})$$

Making a factor in the form of $\sqrt{m}+\sqrt{n}$, $$\displaystyle \frac{6\sqrt{ab}}{a+b}=\frac{\sqrt{a}}{\sqrt{b}} \implies \frac{a}{b}=5$$

Then $$\left( \frac{\sqrt[4]{5}+1}{\sqrt[4]{5}-1} \right)^{4}= \frac{(\sqrt{5}+1)(6+4\sqrt[4]{5})} {(\sqrt{5}+1)(6-4\sqrt[4]{5})}= \frac{3+2\sqrt[4]{5}} {3-2\sqrt[4]{5}}$$

Note on the symmetric roles of $a$ and $b$: $$\displaystyle \frac{6\sqrt{ab}}{a+b}=\frac{\sqrt{b}}{\sqrt{a}} \implies \frac{b}{a}=5$$ which gives the same result.

Hence, there's no other cases similar to $(2)$.


Once you have the formula, it's routine (if tedious) to check it by expanding out. Finding a nice formula is more challenging. In Maple, you could denest the left side of (1) as follows:

Q:= sqrt((1+4^(1/5))/5^(1/5));

$$ \frac{1}{10}\,{5}^{2/5} \left( {4}^{4/5}+{4}^{3/5}+2\cdot {4}^{2/5}-2 \right) $$


Now let's try some reverse engineering. Say you wanted to find a nice formula for a square root where the numerator of the right side involved some linear combination of fifth roots of powers of $2$ with small integer coefficients. We might look at

$$R = a_0 + 2^{1/5} a_1 + 2^{2/5} a_2 + 2^{3/5} a_3 + 2^{4/5} a_4$$ We can square this and extract the coefficients of $2^{i/5}$, $i=0\ldots 4$: $R^2 = \sum_{i=0}^4 c_i 2^{i/5}$ where $$ \eqalign{c_0 &= {a_{{0}}}^{2}+4\,a_{{1}}a_{{4}}+4\,a_{{2}}a_{{3}}\cr c_1 &= 2\,a_{{0}}a_{{1}}+4\,a_{{2}}a_{{4}}+2\,{a_{{3}}}^{2}\cr c_2 &= 2\,a_{{0}}a_{{2}}+{a_{{1}}}^{2}+4\,a_{{3}}a_{{4}}\cr c_3 &= 2\,a_{{0}}a_{{3}}+2\,a_{{1}}a_{{2}}+2\,{a_{{4}}}^{2}\cr c_4 &= 2\,a_{{0}}a_{{4}}+2\,a_{{1}}a_{{3}}+{a_{{2}}}^{2} }$$ We want most of these (say at least $3$ of the $5$) to be $0$. It's not easy to solve such a system of Diophantine equations, but we can resort to brute force: try all cases where $a_i \in \{-2,-1,0,1,2\}$. In Maple it takes practically no time. Thus one such case is Ramanujan's $$ (a_0, \ldots, a_4) = (-1,1,0,1,1)$$ which makes $$ (c_0, \ldots, c_4) = (5,0,5,0,0)$$ Another is $$ (a_0, \ldots, a_4) = (2, 0, 2, 2, -1)$$ which makes $$ (c_0, \ldots, c_4) = (20,0,0,10,0) $$ i.e. $$ 20 + 10 \cdot 2^{3/5} = (2 + 2 \cdot 2^{2/5} + 2 \cdot 2^{3/5} - 2^{4/5})^2 $$ Divide by $100$ and take square roots: after checking the sign is right, this says $$ \sqrt{\dfrac{1 + 2^{-2/5}}{5}} = \dfrac{1 + 2^{2/5} + 2^{3/5} - 2^{-1/5}}{5}$$ I don't know if it's as nice as Ramanujan's formula, but I like it.

Or maybe you'd prefer

$$ \sqrt{\dfrac{4\cdot 2^{3/5} - 3 \cdot 2^{2/5}}{5}} = \dfrac{4 - 2^{1/5} - 2 \cdot 2^{2/5} + 2\cdot 2^{3/5}}{5} $$


$$ \sqrt{8+5 \cdot 3^{1/6}+3^{1/2}} = \frac{1+2 \cdot 3^{1/6}+3^{1/3} - 3^{1/2} + 3^{5/6}}{\sqrt{2}}$$

  • $\begingroup$ Hm.. But I don't have Maple. I do use Wolfram Alpha. But I don't think it can simplify those expressions like Maple can... $\endgroup$
    – Frank
    Jul 29 '16 at 18:16
  • $\begingroup$ Alternately, once you have the formula is it routine (and still tedious) to verify that both sides satisfy the same polynomial equation. $\endgroup$
    – GEdgar
    Jul 29 '16 at 18:33
  • 1
    $\begingroup$ The first nested radical $\sqrt{\dfrac{1 + 2^{-2/5}}{5}}$ , is just a variation of the radical in the question. The second radical; $\sqrt{\dfrac{4\cdot 2^{3/5} - 3 \cdot 2^{2/5}}{5}}$, was also recorded by Ramanujan. You can view it in this paper $\endgroup$
    – Mourad
    Jan 28 '20 at 6:38

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