Find a closed form to the solution of $\sqrt{2-\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{2-x}}}}}=x$

Question

Hi I try to solve the following nested radical :

$$\sqrt{2-\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{2-x}}}}}=x$$

Miraculously the related polynomials is a quintic .More precisely :

$$ x^5 - x^4 - 4 x^3 + 3 x^2 + 3 x - 1=0$$

I know that we can reduce the quintic to a Bring quintic form and use Jacobi theta function .

My question :

Can we hope to see a closed form with radicals ?

Any helps is greatly appreciated

Thanks a lot for all your contributions.

Answer 1

The real valued fixpoint for the original problem is $$ 2 \cos \frac{3 \pi}{11} \approx 1.309721467890570128113850145 $$

parisize = 4000000, primelimit = 500000
? x = 3 * pi
%1 = 3*pi
? x = 3 * Pi
%2 = 9.424777960769379715387930150
? x /= 11
%3 = 0.8567979964335799741261754682
? x = 2 * cos(x)
%4 = 1.309721467890570128113850145
? 
? 
? p = sqrt(2-x)
%5 = 0.8308300260037728510585482985
? q = sqrt(2+p)
%6 = 1.682507065662362337723623298
? r = sqrt(2+q)
%7 = 1.918985947228994779780736114
? s = sqrt(2-r)
%8 = 0.2846296765465702808875853372
? t = sqrt(2 - s)
%9 = 1.309721467890570128113850145
? 
? t-x
%10 = 0.E-28

Here are all the roots of the full degree 32 item. All of these that are not integers (i.e. $-2,1$) are of one of the forms $$ 2 \cos \frac{n\pi}{11} \; , \; \; 2 \cos \frac{n\pi}{31} \; , \; \; 2 \cos \frac{n\pi}{33} \; . \; \; $$ Now that I think of it, we can also express $-2 = 2 \cos \pi$ and $1 = 2 \cos \frac{\pi}{3}$

---

  -2.000000000000000     1/1
  -1.963857394525413    31/33
  -1.959059882504989    29/31
  -1.856735866032145    29/33
  -1.837915623240461    27/31
  -1.682507065662362     9/11
  -1.641526882414553    25/31
  -1.44746807621014     25/33
  -1.377933838151373    23/31
  -1.160113819142396    23/33
  -1.057928020653925    21/31
  -0.8308300260037726    7/11
  -0.6946105056896403   19/31
  -0.4715178710188543   19/33
  -0.3028555550091532   17/31
  -0.09516383164748456  17/33
   0.1012983376774255   15/31
   0.2846296765465702    5/11
   0.5013050645174411   13/31
   0.6541359266348435   13/33
   0.8807883031152686   11/31
   1.000000000000000     1/3
   1.224211965095326     9/31
   1.30972146789057      3/11   +++fixpoint
   1.517516245385582     7/31
   1.572106189485575     7/33
   1.748693232289164     5/31
   1.777670897309847     5/33
   1.908278512800098     3/31
   1.918985947228995     1/11
   1.98973864678379      1/31
   1.990943845146169     1/33

---

n = x^32 - 32*x^30 + 464*x^28 - 4032*x^26 + 23400*x^24 - 95680*x^22 + 
283360*x^20 - 615296*x^18 + 980628*x^16 - 1136960*x^14 + 
940576*x^12 - 537472*x^10 + 201552*x^8 - 45696*x^6 + 
5440*x^4 - 256*x^2 + x + 2


? polroots(n)
=
[-2.000000000000000000000000000, 
-1.963857394525413400797348885, 
-1.959059882504988987876012886,  
-1.856735866032145220401177450, 
-1.837915623240461258254376347, 
-1.682507065662362337723623298, 
-1.641526882414552652727089123, 
-1.447468076210140323279715474, 
-1.377933838151373135601733608, 
-1.160113819142396358393962264, 
-1.057928020653924914730984788, 
-0.8308300260037728510585482985, 
-0.6946105056896405710837087110, 
-0.4715178710188544565010206406, 
-0.3028555550091533273149352935, 
-0.09516383164748459489957448806, 
0.1012983376774254245575037150, 
0.2846296765465702808875853372, 
0.5013050645174410786296040705, 
0.6541359266348432726834987403, 
0.8807883031152686190323430674, 
1.000000000000000000000000000, 
1.224211965095325688293411241, 
1.309721467890570128113850145,      ++++++ 2 cos (3 Pi/11) 
1.517516245385581803826509273, 
1.572106189485574939513592112, 
1.748693232289164237654969328, 
1.777670897309846932623197786, 
1.908278512800097702951793440, 
1.918985947228994779780736114, 
1.989738646783790292642706620, 
1.990943845146169209452510562]

The roots of $$ x^5 + x^4 - 4 x^3 - 3 x^2 + 3x + 1 $$ are $$ 2 \cos \frac{2 \pi}{11 } , \; \; 2 \cos \frac{4 \pi}{11 } , \; \; 2 \cos \frac{6 \pi}{11 } , \; \; 2 \cos \frac{8 \pi}{11 } , \; \; 2 \cos \frac{10 \pi}{11 } , \; \; $$

For your example, just negate these, to get $$ 2 \cos \frac{9 \pi}{11 } , \; \; 2 \cos \frac{7 \pi}{11 } , \; \; 2 \cos \frac{5 \pi}{11 } , \; \; 2 \cos \frac{3 \pi}{11 } , \; \; 2 \cos \frac{ \pi}{11 } , \; \; $$

%10 = x^32 - 32*x^30 + 464*x^28 - 4032*x^26 + 23400*x^24 - 95680*x^22 + 
283360*x^20 - 615296*x^18 + 980628*x^16 - 1136960*x^14 + 
940576*x^12 - 537472*x^10 + 201552*x^8 - 45696*x^6 + 
5440*x^4 - 256*x^2 + x + 2

? factor(n)
%11 = 
[x - 1 1]

[x + 2 1]

[x^5 - x^4 - 4*x^3 + 3*x^2 + 3*x - 1 1]

[x^10 + x^9 - 10*x^8 - 10*x^7 + 34*x^6 + 
34*x^5 - 43*x^4 - 43*x^3 + 12*x^2 + 12*x + 1 1]

[x^15 - x^14 - 14*x^13 + 13*x^12 + 78*x^11 - 66*x^10 - 220*x^9 + 
165*x^8 + 330*x^7 - 210*x^6 - 252*x^5 + 
126*x^4 + 84*x^3 - 28*x^2 - 8*x + 1 1]

Quite similar, the roots of $$ x^{15} + x^{14} - 14x^{13} - 13x^{12} + 78x^{11} + 66x^{10} - 220x^9 - 165x^8 + 330x^7 + 210x^6 - 252x^5 - 126x^4 + 84x^3 + 28x^2 - 8x - 1 $$ are all $$ 2 \cos \frac{2k \pi}{31} $$ with $1 \leq k \leq 15.$ Negate the even degree terms, the roots of $$ x^{15} - x^{14} - 14x^{13} + 13x^{12} + 78x^{11} - 66x^{10} - 220x^9 + 165x^8 + 330x^7 - 210x^6 - 252x^5 + 126x^4 + 84x^3 - 28x^2 - 8x + 1 $$ are all $$ 2 \cos \frac{31-2k \pi}{31} $$ with $1 \leq k \leq 15.$

A bit more complicated, the roots of $$ x^{10} - x^9 - 10 x^8 + 10 x^7 + 34 x^6 - 34 x^5 - 43 x^4 + 43 x^3 + 12 x^2 - 12 x + 1 $$ are $$ 2 \cos \frac{2k\pi}{33} $$ with $$ k = 1,2,4,5,7,8,10, 13,14,16 $$ Now negate the odd degree coefficients, the roots of $$ x^{10} + x^9 - 10 x^8 - 10 x^7 + 34 x^6 + 34 x^5 - 43 x^4 - 43 x^3 + 12 x^2 + 12 x + 1 $$ are $$ 2 \cos \frac{33 - 2k\pi}{33} $$ with $$ k = 1,2,4,5,7,8,10, 13,14,16. $$

Answer 2

Will Jagy's answer has led me to the following approach, which also makes clearer how these cosines come in.

Real solutions $x$ of the given equation satisfy $0\leq x\leq2$. We therefore put $x=2y$ with $0\leq y\leq1$, and obtain the new equation $$\sqrt{2-\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{2-2y}}}}}=2y\ .\tag{1}$$ Introduce the two auxiliary functions $$c(t):=\sqrt{{1\over2}(1+t)},\qquad s(t):=\sqrt{{1\over2}(1-t)}\ .$$ Equation $(1)$ can then be written as $$2\ s\circ s\circ c\circ c\circ s(y)=2\ y\ .\tag{2}$$ Let $y=\cos\eta$ with $\eta\in\bigl[0,{\pi\over2}\bigr]$. Then we get in turn $$\eqalign{ s(y)&=\sin{\eta\over2}=\cos{\pi-\eta\over2},\cr c\circ s(y)&=\cos{\pi-\eta\over4},\cr c\circ c\circ s(y)&=\cos{\pi-\eta\over8},\cr s\circ c\circ c\circ s(y)&=\sin{\pi-\eta\over16}=\cos{7\pi+\eta\over16},\cr s\circ s\circ c\circ c\circ s(y)&=\sin{7\pi+\eta\over32}=\cos{9\pi-\eta\over32},\cr}$$ whereby all angles appearing on the RHS are in $\bigl[0,{\pi\over2}\bigr]$. With $(2)$ we now have $$\cos{9\pi-\eta\over32}=\cos\eta\ ,$$ and this implies ${9\pi-\eta\over32}=\eta$, or $\eta={3\pi\over11}$. In this way we finally obtain $$x=2\cos\eta=2\cos{3\pi\over11}=1.30972\ .$$

Answer 3

One of the solution that can be obtained easily is by substituting $x = 2\cos\theta$

Steps as follows

$\sqrt{2-\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{2 -x}}}}} = x$ will become $\sqrt{2-\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{2 -2\cos\theta}}}}} = 2\cos\theta$

Now by applying Half angle cosine formula we can simplify as follows $\sqrt{2-\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{2\sin\frac{\theta}{2}}}}}} = 2\cos\theta$

$\sqrt{2-\sqrt{2-\sqrt{2+\sqrt{2+2\cos(\frac{\pi}{2} - \frac{\theta}{2})}}}} = 2\cos\theta$

$\sqrt{2-\sqrt{2-\sqrt{2+2\cos(\frac{\pi}{4} - \frac{\theta}{4})}}} = 2\cos\theta$

$\sqrt{2-\sqrt{2-2\cos(\frac{\pi}{8} - \frac{\theta}{8})}} = 2\cos\theta$

$\sqrt{2-2\sin(\frac{\pi}{16} - \frac{\theta}{16})} = 2\cos\theta$

$\sqrt{2-2\cos(\frac{\pi}{2} -\frac{\pi}{16} + \frac{\theta}{16})} = 2\cos\theta$

$2\sin(\frac{\pi}{4} -\frac{\pi}{32} + \frac{\theta}{32}) = 2\cos\theta$

$2\cos(\frac{\pi}{2} -\frac{\pi}{4} +\frac{\pi}{32} - \frac{\theta}{32}) = 2\cos\theta$

Simplifying $\theta$ on both sides

$\frac{(16-8+1)\pi}{32} = \theta + \frac{\theta}{32}$

$\theta = \frac{9\pi}{33} = \frac{3\pi}{11} $

Therefore $ x = 2\cos\frac{3\pi}{11}$