Can Higher Degree Nested Radicals Be Simplified?

[renormalize]

You define $R=\dfrac{1}{4}(\sqrt{5}-1)$ and then operate with $R=\sqrt[5]{\dfrac{5\sqrt{5}-11}{64}}$ and conclude $R\overline{R}=4.$ By that, you also assumed that $\overline{R}$ has the same properties. None of which is known at that stage of the calculation.

Sorry, I still don't understand. I define big-$R$ to be $R=\left(\frac{5\sqrt{5}-11}{2^{6}}\right)^{\frac{1}{5}}$ and solve for little-$r$, defined as $r\equiv a\sqrt{5}+b$, to be $\frac{1}{4}\left(\sqrt{5}-1\right)$, and finally verify that $R=r$.

 

[fresh_42]

Sorry, I still don't understand. I define big-$R$ to be $R=\left(\frac{5\sqrt{5}-11}{2^{6}}\right)^{\frac{1}{5}}$ and solve for little-$r$, defined as $r\equiv a\sqrt{5}+b$, to be $\frac{1}{4}\left(\sqrt{5}-1\right)$, and finally verify that $R=r$.

There is no little $r$ on my screen in post #13.

Edit: also $R\overline{R}=r\overline{r}=4$ does not mean $R=r$ and $\overline{R}=\overline{r}$ because $2\cdot 2=4,$ too.

 

[renormalize]

There is no little $r$ on my screen in post #13.

My apologies, I introduced little-$r$ in post #23.
Let me present a revised compilation of my step-by-step reasoning:

1. Define the radical $R$, its conjugate $\overline{R}$, and find their product $R\overline{R}$:$$R\equiv\left(\frac{5\sqrt{5}-11}{2^{6}}\right)^{\frac{1}{5}},\quad\overline{R}\equiv\left(\frac{5\sqrt{5}+11}{2^{6}}\right)^{\frac{1}{5}}\Rightarrow R\overline{R}=\frac{1}{4}\tag{1a,1b,1c}$$
2. Define an expression $r$ and its conjugate $\overline{r}$ such that each is linear in the radical $\sqrt{5}$:$$r\equiv a\sqrt{5}-b,\quad\overline{r}\equiv a\sqrt{5}+b\tag{2a,2b}$$
3. Motivated by eq.(1c), impose the condition:$$r\overline{r}=5a^2-b^2=\frac{1}{4}\tag{3}$$
4. Solve eq.(3) for $b$ and insert it into eq.(2) to find:$$r=a\sqrt{5}-\frac{1}{2}\sqrt{20a^{2}-1}\equiv\frac{\sqrt{5}}{n}-\frac{\sqrt{20-n^{2}}}{2n},\quad\overline{r}=\frac{\sqrt{5}}{n}+\frac{\sqrt{20-n^{2}}}{2n}\tag{4a,4b}$$where $n\equiv a^{-1}$.
5. Now demand that, when $r$ and $\overline{r}$ are raised to integer powers, the resulting expressions should contain no irreducible square-roots other than $\sqrt{5}$. The simplest way to guarantee this is to insist that the coefficients $n^{-1}$ and $(2n)^{-1}\sqrt{20-n^{2}}$ in eqs.(4) themselves contain no irreducible roots whatsoever. Then in particular, $\left|n\right|$ must fall in the range $0<\left|n\right|\leq\sqrt{20}$ so that $\sqrt{20-n^{2}}$ is real and $20-n^{2}$ must be a perfect square. The only two values that satisfy these criteria are $\left|n\right|=2$ and $\left|n\right|=4$, thus yielding the 4 values:$$r_{2}=\frac{\sqrt{5}}{2}-1,\quad\overline{r}_{2}=\frac{\sqrt{5}}{2}+1,\quad r_{4}=\frac{\sqrt{5}-1}{4},\quad\overline{r}_{4}=\frac{\sqrt{5}+1}{4}\tag{5a,5b,5c,5d}$$(Here I've ignored trivial variations involving multiplication by $-1$ on the right sides.)
6. Using Mathematica to raise the right sides of eqs.(5) to the fifth power, I verify the following 4 identities:$$\left(\frac{305\sqrt{5}-682}{2^{5}}\right)^{\frac{1}{5}}\equiv\frac{\sqrt{5}}{2}-1,\quad\left(\frac{305\sqrt{5}+682}{2^{5}}\right)^{\frac{1}{5}}\equiv\frac{\sqrt{5}}{2}+1\tag{6a,6b}$$and$$\left(\frac{5\sqrt{5}-11}{2^{6}}\right)^{\frac{1}{5}}\equiv\frac{\sqrt{5}-1}{4},\quad\left(\frac{5\sqrt{5}+11}{2^{6}}\right)^{\frac{1}{5}}\equiv\frac{\sqrt{5}+1}{4}\tag{6c,6d}$$Under the criteria set-out in step 5, these are the only 4 identities that satisfy $R=r,\overline{R}=\overline{r}, R\overline{R}=r\overline{r}=\frac{1}{4}$. Specifically, eq.(6c) establishes the identity in the OP's first post. QED

As far as I'm aware, I nowhere assume that $r=\frac{\sqrt{5}-1}{4}$. Rather, I derive it from the 3 criteria that: a) $r$ is linear in the radical $\sqrt{5}$, b) $r$ involves no other irreducible radicals, and c) $r\overline{r}=\frac{1}{4}$.

 

[votingmachine]

Proving the identity was never the problem. I'm sure what @FranzS meant was:
$$
0=x^5-\dfrac{5\sqrt{5}-11}{64} \Longrightarrow x=\dfrac{1}{4}\left(\sqrt{5}-1\right) \text{ is one (of 5) solution(s) }
$$
How can we figure out how this root is found?

I tried several methods (your conjugation trick, polynomial division, an ansatz, complex numbers, minimal polynomial ...) but couldn't find one that succeeded. I think meanwhile that @FranzS has taken $\dfrac{\sqrt{5}-1}{4}$ to WA, asked it for $\left(\dfrac{\sqrt{5}-1}{4}\right)^5,$ and posted this question. I finally gave up since I saw no general value in the question: "Does $x^5-\dfrac{5\sqrt{5}-11}{64}\in \mathbb{Q}(\sqrt{5})[x]$ has a root?" when there are technical tools like WA which easily answer that question.

Taking that as:
$$
x^5=\dfrac{5\sqrt{5}-11}{64}
$$

note that the product of the right side by it's conjugate is 2^-10
So:
X^5 times X^5 prime = 2^-10

by the product rule in comment #10, that is the same as:
(X times X prime)^5 = 2^-10

raise both sides to the 5th power:

X times X prime = 2^-2 = 1/4

X and X prime are two of the form:
(n(sqrt(a) + b)) and n((sqrt(a) -b))
and multiplied give:
n^2((a)-b^2)
which equals 1/4

Now it seems intuitive that the a-b^2 part has to be an exponent of 2, as does n. Since both sides seem to require a factorization of 2 ... might be flawed intuition, but ...

So intuitively ...
5-1=4 then times n=1/4 ... which is the solution arrived at.
It seems there are an infinite number of expressions that work though

My apologies for not using the dollars-sign formula stuff. I tried to copy it ... feel free to finish the final step, and to write it legibly.

 

[fresh_42]

My apologies, I introduced little-$r$ in post #23.
Let me present a revised compilation of my step-by-step reasoning:

1. Define the radical $R$, its conjugate $\overline{R}$, and find their product $R\overline{R}$:$$R\equiv\left(\frac{5\sqrt{5}-11}{2^{6}}\right)^{\frac{1}{5}},\quad\overline{R}\equiv\left(\frac{5\sqrt{5}+11}{2^{6}}\right)^{\frac{1}{5}}\Rightarrow R\overline{R}=\frac{1}{4}\tag{1a,1b,1c}$$
2. Define an expression $r$ and its conjugate $\overline{r}$ such that each is linear in the radical $\sqrt{5}$:$$r\equiv a\sqrt{5}-b,\quad\overline{r}\equiv a\sqrt{5}+b\tag{2a,2b}$$
3. Motivated by eq.(1c), impose the condition:$$r\overline{r}=5a^2-b^2=\frac{1}{4}\tag{3}$$
4. Solve eq.(3) for $b$ and insert it into eq.(2) to find:$$r=a\sqrt{5}-\frac{1}{2}\sqrt{20a^{2}-1}\equiv\frac{\sqrt{5}}{n}-\frac{\sqrt{20-n^{2}}}{2n},\quad\overline{r}=\frac{\sqrt{5}}{n}+\frac{\sqrt{20-n^{2}}}{2n}\tag{4a,4b}$$where $n\equiv a^{-1}$.
5. Now demand that, when $r$ and $\overline{r}$ are raised to integer powers, the resulting expressions should contain no irreducible square-roots other than $\sqrt{5}$. The simplest way to guarantee this is to insist that the coefficients $n^{-1}$ and $(2n)^{-1}\sqrt{20-n^{2}}$ in eqs.(4) themselves contain no irreducible roots whatsoever. Then in particular, $\left|n\right|$ must fall in the range $0<\left|n\right|\leq\sqrt{20}$ so that $\sqrt{20-n^{2}}$ is real and $20-n^{2}$ must be a perfect square. The only two values that satisfy these criteria are $\left|n\right|=2$ and $\left|n\right|=4$, thus yielding the 4 values:$$r_{2}=\frac{\sqrt{5}}{2}-1,\quad\overline{r}_{2}=\frac{\sqrt{5}}{2}+1,\quad r_{4}=\frac{\sqrt{5}-1}{4},\quad\overline{r}_{4}=\frac{\sqrt{5}+1}{4}\tag{5a,5b,5c,5d}$$(Here I've ignored trivial variations involving multiplication by $-1$ on the right sides.)
6. Using Mathematica to raise the right sides of eqs.(5) to the fifth power, I verify the following 4 identities:$$\left(\frac{305\sqrt{5}-682}{2^{5}}\right)^{\frac{1}{5}}\equiv\frac{\sqrt{5}}{2}-1,\quad\left(\frac{305\sqrt{5}+682}{2^{5}}\right)^{\frac{1}{5}}\equiv\frac{\sqrt{5}}{2}+1\tag{6a,6b}$$and$$\left(\frac{5\sqrt{5}-11}{2^{6}}\right)^{\frac{1}{5}}\equiv\frac{\sqrt{5}-1}{4},\quad\left(\frac{5\sqrt{5}+11}{2^{6}}\right)^{\frac{1}{5}}\equiv\frac{\sqrt{5}+1}{4}\tag{6c,6d}$$Under the criteria set-out in step 5, these are the only 4 identities that satisfy $R=r,\overline{R}=\overline{r}, R\overline{R}=r\overline{r}=\frac{1}{4}$. Specifically, eq.(6c) establishes the identity in the OP's first post. QED

As far as I'm aware, I nowhere assume that $r=\frac{\sqrt{5}-1}{4}$. Rather, I derive it from the 3 criteria that: a) $r$ is linear in the radical $\sqrt{5}$, b) $r$ involves no other irreducible radicals, and c) $r\overline{r}=\frac{1}{4}$.

This looks ok, but if you use Mathematica to raise the numbers to the fifth power, why not use it to calculate $\left(\dfrac{\sqrt{5}-1}{4}\right)^5$ and find $\dfrac{5\sqrt{5}-11}{2^6}$ or use the binomic formula to do it? I think the detour by the conjugates only makes things more complicated.

 

[FactChecker]

I think the detour by the conjugates only makes things more complicated.

Absolutely, (I did think that it was an interesting approach though.) Expanding the RHS using the binomial theorem and collection like terms is so methodical that I think I could write a computer program that would generate thousands of such identities.
It's so much easier to go in that direction than the other way. Consider how hard it would be to go from right to left in this step:
$\frac{1}{2^{10}} (25\sqrt{5}-5\cdot 25+10\cdot 5\sqrt{5}-10\cdot 5+5\cdot \sqrt{5}-1) = \frac{1}{2^{10}} (80\sqrt{5}-176)$

 

[renormalize]

This looks ok, but if you use Mathematica to raise the numbers to the fifth power, why not use it to calculate $\left(\dfrac{\sqrt{5}-1}{4}\right)^5$ and find $\dfrac{5\sqrt{5}-11}{2^6}$ or use the binomic formula to do it? I think the detour by the conjugates only makes things more complicated.

Because I am assuming that we haven't been given $\frac{\sqrt{5}-1}{4}$ in advance. The whole point of my exposition is to demonstrate that, knowing only the expression $\left(\frac{5\sqrt{5}-11}{2^{6}}\right)^{\frac{1}{5}}\equiv R$, you can derive the candidate expression $\frac{\sqrt{5}-1}{4}\equiv r$, and then demonstrate with Mathematica that indeed $r=R$.

 

[FactChecker]

Because I am assuming that we haven't been given $\frac{\sqrt{5}-1}{4}$ in advance. The whole point of my exposition is to demonstrate that, knowing only the expression $\left(\frac{5\sqrt{5}-11}{2^{6}}\right)^{\frac{1}{5}}\equiv R$, you can derive the candidate expression $\frac{\sqrt{5}-1}{4}\equiv r$, and then demonstrate with Mathematica that indeed $r=R$.

That's a good point. I wonder how often such an expression can be simplified that way and how to recognize which ones can be simplified. I would like to go through the steps one-by-one to see if it works often or is just a lucky shot in the dark.

 

[Petek]

That's a good point. I wonder how often such an expression can be simplified that way and how to recognize which ones can be simplified. I would like to go through the steps one-by-one to see if it works often or is just a lucky shot in the dark.

I might have an explanation for the theory behind this phenomenon. The explanation requires some knowledge about algebraic number theory, specifically of quadratic number fields. In the present example we're working in the quadratic number field $K = \mathbb{Q}(\sqrt{5})$ and its associated ring of algebraic integers $O_K$. Consider $\alpha = \frac{5\sqrt{5} - 11}{2}$. $\alpha$ is an element of $O_K$. In fact, $\alpha$ is a unit in $O_K$. That follows from $\alpha$'s norm equaling $1.$ (By definition, element's norm equals the product of the element by its conjugate.) Now let $\beta = \frac{\sqrt{5} - 1}{2}$. Then $\beta's$ norm also equals $1$, so $\beta$ is a unit in $O_K$. A key observation is that $\beta$ is what's called a fundamental unit: It's a generator of the cyclic group of all units in $O_K$. That implies that $\alpha$ is $\pm$ an integral power of $\beta$. In fact, $\alpha = \beta^{5}$. Now taking fifth roots and massaging the powers of $2$ gives the desired result. So, it seems to me that these "collapsing radicals" result from an interplay between quadratic number fields whose group of units is cyclic and a generator for such a group.

 

[FactChecker]

I might have an explanation for the theory behind this phenomenon. The explanation requires some knowledge about algebraic number theory, specifically of quadratic number fields. In the present example we're working in the quadratic number field $K = \mathbb{Q}(\sqrt{5})$ and its associated ring of algebraic integers $O_K$. Consider $\alpha = \frac{5\sqrt{5} - 11}{2}$. $\alpha$ is an element of $O_K$. In fact, $\alpha$ is a unit in $O_K$. That follows from $\alpha$'s norm equaling $1.$ (By definition, element's norm equals the product of the element by its conjugate.) Now let $\beta = \frac{\sqrt{5} - 1}{2}$. Then $\beta's$ norm also equals $1$, so $\beta$ is a unit in $O_K$. A key observation is that $\beta$ is what's called a fundamental unit: It's a generator of the cyclic group of all units in $O_K$. That implies that $\alpha$ is $\pm$ an integral power of $\beta$. In fact, $\alpha = \beta^{5}$. Now taking fifth roots and massaging the powers of $2$ gives the desired result. So, it seems to me that these "collapsing radicals" result from an interplay between quadratic number fields whose group of units is cyclic and a generator for such a group.

Thanks! That is really fascinating. I know nothing about that subject, but this is the first time I appreciated why people are interested in it.

 

[FranzS]

I'll explain the context which the initial fifth-root + nested-square-root expression comes from, in the remote case it can give any insights (and also to show that this wasn't a sort of exercise where the simplified expression involving just a square root was artificially manipulated to get the other).

So, as I wrote in a previous post, I was just playing around with the calculations of the fifth roots of unity (with no use of trigonometry). That is, finding the complex solutions to the equation:
$$
z^5=1 \iff z^5-1=0 \; , \; \; \; z \in \mathbb{C}
$$ Since no general quintic formula exists (in terms of radicals and elementary arithmetic operations), I adopted the following approach.
We can express $z$ in the form $\; z=a+ib \;$, where $\; i^2:=-1 \;$ and $\; a,b \in \mathbb{R} \;$ (without loss of generality, as this is of course the general Cartesian form of any possible complex number).
Then we can expand the initial equation in terms of the unknowns $a$ and $b \;$:
$$
z^5 - 1 = \left( a +ib \right)^5 - 1 = a^5 + i 5 a^4 b - 10 a^3 b^2 - i 10 a^2 b^3 + 5 a b^4 + i b^5 - 1 = 0
$$ Now, if we rearrange the terms by grouping the real and imaginary monomials separately (unnecessary parentheses are added for a possibly better visualization)...
$$
\left( a^5 - 10 a^3 b^2 + 5 a b^4 - 1 \right) + i \left( 5 a^4 b - 10 a^2 b^3 + b^5 \right) = 0
$$ ... it becomes clear that, when we look at $0$ (zero) on the RHS as the complex number $0 + 0i$, the real part and the imaginary part of the LHS must be both equal to $0$. As another way to look at it, one could simply rearrange the equation like this...
$$
a^5 - 10 a^3 b^2 + 5 a b^4 - 1 = i \left( - 5 a^4 b + 10 a^2 b^3 - b^5 \right)
$$ ... and ask the question: when is a real number (LHS) equal to an imaginary number (RHS)? Of course, this is only possible when both are equal to $0$, which is the only intersection point between the real and the imaginary axes.
Therefore, we end up with a system of two equations:
$$
\begin{cases}
\mathrm{I.} & a^5 - 10 a^3 b^2 + 5 a b^4 - 1 = 0
\\
\mathrm{II.} & 5 a^4 b - 10 a^2 b^3 + b^5 = 0
\end{cases}
$$ Now it's just a matter of carrying out the calculations carefully.
In equation $\mathrm{II.}$ we factor out the common term $b$ and we get:
$$
b \left( 5 a^4 - 10 a^2 b^2 + b^4 \right) = 0
$$ A first, trivial solution to this equation is $b=0$. We substitute in equation $\mathrm{I.}$, thus all terms containing $b$ vanish and we are left with:
$$
a^5 = 1 \Longrightarrow a = 1
$$ Please remember that $a, b \in \mathbb{R}$, thus $a=1$ is really the only solution!
The first and trivial solution to the original equation is then $1+0i=1$.
Proof that $1$ is indeed one of the fifth roots of unity: $1^5=1 \; \checkmark$ (who would have thought?).

Ok then, back to the other possible (and more interesting) solutions of equation $\mathrm{II.}$ by evaluating the case with $b \neq 0$ (at least, not necessarily):
$$
5 a^4 - 10 a^2 b^2 + b^4 = 0
$$ We can solve it for $b^2$ in terms of $a^2$ using the quadratic formula:
$$
\left( b^2 \right) ^2 - 10 a^2 b^2 + 5 a^4 = 0
$$ $$
\Longrightarrow b_{1,2}^2 =
\frac{ - \left( -10 a^2 \right) \pm \sqrt{ \left( - 10 a^2 \right) ^2 - 4 \cdot 1 \cdot 5a^4 }}{2 \cdot 1} =
a^2 \left( 5 \pm 2 \sqrt{5} \right)
$$
If we substitute one (to start with) of these expressions, namely $b^2 = a^2 \left( 5 + 2 \sqrt{5} \right)$, into equation $\mathrm{I.}$ we get:
$$
a^5 - 10 a^5 \left( 5 + \sqrt{5} \right) + 5 a^5 \left( 5 + \sqrt{5} \right) ^2 - 1 = 0
$$ $$
\Longrightarrow
a^5 = \frac{1}{2^4 \left( 11 + 5 \sqrt{5} \right) } = \frac{5 \sqrt{5} - 11}{2^6}
$$ $$
\Longrightarrow
a = \sqrt[5]{\frac{5 \sqrt{5} - 11}{2^6}}
$$ ... which is, finally, the infamous expression that is the subject of this thread.
At this point I stopped, since I had found the first explicit result (the real part $a$ of one of the non-trivial fifth roots for unity) and I wanted to check if it was correct before proceeding with further calculations.
So, I looked up on the internet and found that its expression was instead given as:
$$
\frac{\sqrt{5} - 1}{4}
$$ Finally (and honestly in disbelief at first) I inputted my result in WolframAlpha and, to my surprise, WA confirmed that it simplified down to the equivalent simpler expression with no nested radicals.
Eventually, because of all this, I started this thread in order to learn something about the simplification of nested radicals.

 

[FranzS]

By the way, we got the value of $\cos \left( \frac{2 \pi}{5} \right)$ "for free"