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 $(1)$.

  • 4
    $\begingroup$ 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

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

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