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Here I have a function

$$\sqrt{\left( x - 2 -\sqrt{(x-2)^2-4} \right)}$$ but, is there a way the outer-most radical cancels via a $(\sqrt{u + \sqrt{v}})^2 = u + \sqrt{v}$?

Maybe this is an application of one of Ramanujan's nested radical formulas.

Thank you for considering this!

Blue
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StackQuest
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    Have you looked for a possible application of this "denesting"? Ramanujan is linked there with a pdf-file. – Dietrich Burde Jan 19 '22 at 17:42
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    If $a, b \geq 0$ with $a^2 \geq b$ then : $$\sqrt{a + \sqrt{b}} = \sqrt{\dfrac{a + \sqrt{a^2 - b}}{2}} + \sqrt{\dfrac{a - \sqrt{a^2 - b}}{2}}$$ and : $$\sqrt{a - \sqrt{b}} = \sqrt{\dfrac{a + \sqrt{a^2 - b}}{2}} - \sqrt{\dfrac{a - \sqrt{a^2 - b}}{2}}$$ – Essaidi Jan 19 '22 at 18:05
  • You can also apply these formulae with $b<0$, in which case the square root of a complex number is resolved into real and imaginary parts. Higher-order radicals (except powers of 2) do not offer analogous identities, which is connected with the famed "casus irreducibilis" in solving cubic or higher-degree equations. – Oscar Lanzi Jan 19 '22 at 19:06

2 Answers2

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Not a single radical, but a pair of uncoupled radicals.

Suppose $\sqrt{a-\sqrt{b}}=\sqrt u-\sqrt v$. Then

$(\sqrt u-\sqrt v)^2=a-\sqrt{b}=(u+v)-2\sqrt{uv}$

If $u$ and $v$ are both rational then we must have $u+v=a, uv=b/4$ from which $(u-v)^2=(u+v)^2-4uv=a^2-b$. So with $u>v$ for a positive square root, we have

$\sqrt{a-\sqrt b}=\sqrt{\dfrac{a+\sqrt{a^2-b}}{2}}-\sqrt{\dfrac{a-\sqrt{a^2-b}}{2}}$

where to get rational radicands we need $a^2-b$ to be a rational square.

In this case $a=x-2$ and $a^2-b=4=2^2$, so we can render

$\sqrt{(x-2)-\sqrt{(x-2)^2-4}}=\sqrt{\dfrac{x}{2}}-\sqrt{\dfrac{x-4}{2}}.$

Oscar Lanzi
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    Thank you Oscar, I'll see if I can get this to work. Is there a name for this? Is this one of Ramanujan's de-nesting methods? – StackQuest Jan 19 '22 at 18:35
  • I have not seen Ramanujan's name associated with this method, nor have I seen a name for it. – Oscar Lanzi Jan 19 '22 at 18:57
  • Okay, so suppose $u$ and $v$ are rational functions. Why must their sum $=a$ and their product suddenly equal this other $b/4$? And why couldn't this work for non-rational functions? What if we had $a = \ln(x)$ and $b = x$? – StackQuest Jan 19 '22 at 20:40
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Well it helps to note that $(x-2)^2-4= [(x-2)+2][(x-2)-2] = x(x-4)$

So $\sqrt{(x-2)-\sqrt{(x-2)^2 -4}}=\sqrt{x-2 -\sqrt{x(x-4)}}$

Now we kind of want $\sqrt{x-2 -\sqrt{x(x-4)}} =a - b$ so we'd like $x-2 - \sqrt{x(x-4)} = a^2 + b^2 - 2ab$.

We can get $x-2 + \sqrt{x(x-4)} = (x-2) - 2\sqrt{\frac {x(x-4)}4}$ so it'd be really nice if we could find and $ab$ so that $ab =\sqrt{\frac {x(x-4)}4}$ and $a^2 + b^2 = x-2$.

The first thing to try is $a=\sqrt\frac x2$ and $b=\sqrt\frac {x-4}2$ which is really nice because if so $a^2 + b^2 = \frac x2 + \frac {x-4}2 = x-2$ just as we hoped for.

$\sqrt{(x-2)-\sqrt{(x-2)^2 -4}}=\sqrt{x-2 -\sqrt{x(x-4)}}=$

$\sqrt{\frac x2 + \frac {x-4}2 - 2\sqrt{\frac{x(x-4)}4}}=$

$\sqrt{(\sqrt \frac x2 -\sqrt\frac {x-4}2)^2}=$

$\sqrt\frac x2 -\sqrt\frac{x-4}2$

fleablood
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  • Hi fleablood, what is the intuition behind factoring the 2 out from the radical in the 5th line from the top? And why would it be nice to find an $ab$ of the form you specify? Is there some type of completing the square behind the scenes? You suppose the radical expression is equal to $(a+b)^2,$ are you equating polynomial terms? – StackQuest Jan 19 '22 at 20:24
  • Well $(a-b)^2 = a^2 +b^2 - 2\sqrt ab$. So if I want $\sqrt{x(x-4)} = 2\sqrt{ab}$ then $\sqrt{ab} =\frac 12 \sqrt{x(x-4)} = \sqrt{\frac {x(x-4)}4}$. – fleablood Jan 19 '22 at 20:27
  • I'm not sure I follow. $(a-b)^2 = a^2 -2ab + b^2.$ Also, why did you choose $(a - b)^2$ instead of $(a+b)^2$? Why do you want $\sqrt{x(x-4)} = 2\sqrt{ab}$? – StackQuest Jan 19 '22 at 20:28
  • "You suppose the radical expression is equal to (a+b)2, are you equating coefficients somewhere?" I'm hoping and praying. – fleablood Jan 19 '22 at 20:28
  • Well I'm looking to learn how to "derive" $a$ and $b$ to rely on this method for more general cases, but you knew something particular about the solution in advanced and assumed the solution out of nowhere in your reasoning. How do you not just randomly guess that $a = \sqrt{x/2}$ and $b = \sqrt{(x-4)/2}$? – StackQuest Jan 19 '22 at 20:31
  • Why $(a-b)^2$ instead of $(a+b)^2$? Because we have $(x-2) \underbrace{\color{red}{\large{-}}}\sqrt{(x-2)^2-4}$. We can use $a+b$ but then we want $2ab = \color{red}{\large{-}}\sqrt{(x-2)^2-4}$ and I'd figure we'd just skip those distracting negative values. – fleablood Jan 19 '22 at 20:33
  • "How do you not just randomly guess that a=x/2−−−√ and b=(x−4)/2−−−−−−−−√?" Um.... I thought I did just randomly guess that. .... Oh, you mean why did I assign one to the label $a$ and the other to $b$ and not the other way around? It doesn't matter what I call them. $a$ and $b$ are just labels. – fleablood Jan 19 '22 at 20:35
  • No I do mean precisely how do you go about this method formulaically so that it isn't a guess but a derivation. – StackQuest Jan 19 '22 at 20:36
  • Well... I guess as $ab = \sqrt{stuff}$ I figured $a,b =\sqrt{thing}$ so $a^2,b^2 = thing$ so I need $a^2 + b^2 = thing_1 + thing_2= x-2$ and $ab =thing_1\cdot thing_2 = \frac{x(x-4)}4$. As the product involves $x$ to a square and the sum involves $x$ to a single power we basically have $things$ be $x$s to a single power. If I were to formalize I might try to solve for $(ax+b)+(cx+d)=x-2$ so $a+c=1$ while $b+d=-2$ and $(ax+b)(cx+d)\frac {x(x-2)}4$ so $ac=\frac 14;bd=-\frac 12;bc+ad=0$ but as $\frac{x(x-2)}4$ factors cleanly to $\frac x2\times \frac{x-2}2$ I just tried those off the bat. – fleablood Jan 19 '22 at 21:35