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$$\sqrt[3]{9+9\sqrt[3]{9+9\sqrt[3]{9+\cdots}}} - \sqrt{8-\sqrt{8-\sqrt{8+\sqrt{8-\sqrt{8-\sqrt{8+\cdots}}}}}} = 1$$ In the second nested radical, the repeating pattern is $(-,-,+)$. I approached this problem in a rather boring way. That is, the first expression satisfies the equation, $$y=\sqrt[3]{9+9y} \implies y^3=9+9y$$ and the second one satisfies, $$x=+\sqrt{8-\sqrt{8-\sqrt{8+x}}} \implies \left[(x^2-8)^2-8\right]^2=8+x$$ Now this magically factors into $$(x^2 - x - 8) (x^3 - 2 x^2 - 11 x + 23) (x^3 + 3 x^2 - 6 x - 17) = 0$$ It turns out that the $x$ we are looking is the solution of the third factor. And so, $$x^3 + 3 x^2 - 6 x - 17=0 \implies (x+1)^3-9(x+1)-9=0$$ And magically $x+1=y$. The problem here is that this does not give me much understanding of nested radicals, it is just plain bash.

Q. All in all, my question asks, if there's a way to prove the difference is 1 without (or a little less) bashing and a nice general way to deal with such expressions (like the examples below)?


Next I tried the following, Assume that we want to solve $$t=\sqrt{7-\sqrt{7+\sqrt{7-\sqrt{7+\sqrt{7-\sqrt{7+\cdots}}}}}}$$ What we can try is the following, $$t=\oplus\sqrt{7\ominus\sqrt{7+t}}$$ When we repeatedly square, we loose the information of the signs highlighted. That is, the equation $(t^2-7)^2=t+7$ has some nice roots corresponding to, $$t=+\sqrt{7+\sqrt{7+t}}\qquad \text{and}\qquad t=-\sqrt{7-\sqrt{7+t}}$$ But these can be reduced by their symmetry, $$t=+\sqrt{7+t}\qquad \text{and}\qquad t=-\sqrt{7+t}$$ Solution to both of these satisfy $t^2=t+7$ (which are the roots we don't want). And so we get the intuition why that huge equation has a quadratic factor. Now in this case, after long division we will end up with a quadratic equation whose solution (that we want) is $2$. If we use this in our original question, we get a 6th degree equation with no further intuition.

I also saw this on Wikipedia with no explanation. There is a typing error in the above image.

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    $\begingroup$ i think that whatever we do , in the end we have to solve a polynomial to get the value of the nested radical. $\endgroup$ – Subhajit Halder Sep 29 '18 at 8:03
  • $\begingroup$ In the nested radical with $7$s, what symmetry are you referring to? Anyways, since $t\ge 0$ you can neglect the case where $t$ is negative $\endgroup$ – AnotherJohnDoe Sep 29 '18 at 8:06
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    $\begingroup$ I'm not sure what's the question here. $\endgroup$ – Ennar Sep 29 '18 at 8:10
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    $\begingroup$ @Mark Define $x_0=0$ and then recursively define, $x_{n+1}=\sqrt{8-\sqrt{8-\sqrt{8+x_{n}}}}$, And so, $x=\lim_{n\to\infty} x_n$. Using a calc, you can see, $x\approx 2.41$. Now we had $x=\sqrt{8-\sqrt{8-\sqrt{8+x}}}$, converting this into an 8th degree equation will add 7 other roots which does not interest us. It then turns out the root of the equation near $2.41$ satisfies the third factor. $\endgroup$ – Devansh Sehta Sep 29 '18 at 8:26
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    $\begingroup$ Just also numerical calculation offer the difference is 1. Again also graphical image offer difference is 1. Neither exact method, though. $\endgroup$ – Takahiro Waki Sep 29 '18 at 16:40
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took me a while. The roots of your $$ u^3 - 9u - 9 = 0 $$ are $$ -2 \sqrt 3 \sin {\frac{2\pi}{9}} \; , $$ $$ -2 \sqrt 3 \sin {\frac{8\pi}{9}} \; = \; -2 \sqrt 3 \sin {\frac{\pi}{9}} , $$ $$ -2 \sqrt 3 \sin {\frac{14\pi}{9}} \; = \; 2 \sqrt 3 \sin {\frac{4\pi}{9}} . $$

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? poldisc( s^3 - 9 * s - 9)
%51 = 729
? polgalois( s^3 - 9 * s - 9)
%59 = [3, 1, 1, "A3"]

? polroots( s^3 - 9 * s - 9)
%52 = [-2.226681596905677465811651808 ,
       -1.184792530904095372701352048 , 
        3.411474127809772838513003856 ]~


? u = -2 * sqrt(3) * sin (2 * Pi / 9  )
%53 = -2.226681596905677465811651808
? v = -2 * sqrt(3) * sin (8 * Pi / 9  )
%54 = -1.184792530904095372701352048
? w = -2 * sqrt(3) * sin (14 * Pi / 9  )
%55 = 3.411474127809772838513003856
? 
? u^3 - 9 * u - 9
%56 = 0.E-27
? v^3 - 9 * v - 9
%57 = 0.E-27
? w^3 - 9 * w - 9
%58 = -8.07793567 E-28
? 

===========================

enter image description here

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  • $\begingroup$ Leaving aside how that answers the question, how can your initial quadratic polynomial have three different roots? $\endgroup$ – Henning Makholm Sep 29 '18 at 20:24
  • $\begingroup$ @HenningMakholm typo corrected. The sort of thing it answers is discussed here: math.stackexchange.com/questions/1065862/… $\endgroup$ – Will Jagy Sep 29 '18 at 20:28
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You also asked about $$x =\sqrt{ 5+ \sqrt{ 5 + \sqrt{ 5-x}}}$$ I figured out how to express $x$ in radicals.

Take $$ k = \frac{1 + \sqrt {13}}{2} $$ so that $$ k^2 - k + 2 = 5 \; . $$

Then $x$ above is the largest root of the cubic $$ x^3 + (k-1) x^2 - (k^2 + 2)x - (k^3 - k^2 + 2 k - 1) $$ which is the same as $$ x^3 + (k-1) x^2 - (k+5)x - ( 5 k - 1) $$

? k = (1 + sqrt(13))/2
%5 = 2.302775637731994646559610634
?  h = x^3 + (k-1) * x^2 - (k^2 + 2) * x - (k^3 - k^2 + 2 * k - 1 )
%6 = x^3 + 1.302775637731994646559610634*x^2 
    - 7.302775637731994646559610634*x - 10.51387818865997323279805317
? polroots(h)
%7 = [-2.549662363373543544157989866 + 0.E-28*I,
      -1.500778167203563600102045353 + 0.E-28*I,
       2.747664892845112497700424585 + 0.E-28*I]~
? w = x^3 +(k-1) * x^2 - (k+5) * x - (5 * k - 1) 
%8 = x^3 + 1.302775637731994646559610634*x^2 - 7.302775637731994646559610634*x 
       - 10.51387818865997323279805317
? polroots(w)
%9 = [-2.549662363373543544157989866 + 0.E-28*I, 
      -1.500778167203563600102045353 + 0.E-28*I, 
       2.747664892845112497700424585 + 0.E-28*I]~
? 

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In general, with positive integer $a \geq 2$ and real $k > 0$ such that $$ k^2 - k + 2 = a, $$ so that $$ k = \frac{1 + \sqrt{4a-7}}{2} \; , $$ with $$x =\sqrt{ a+ \sqrt{ a + \sqrt{ a-x}}},$$ we find $x$ is the largest root of $$ x^3 + (k-1) x^2 - (a+k)x - ( a k - 1) \; .$$ We can use Cardano at this point. The bad news is that the cube roots will be of complex numbers, we are in Casus Irreducibilis.

enter image description here

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    $\begingroup$ I just now realized that there is a typing error in the image I got from Wikipedia. They say that repeating pattern is ++-+ but actually in the image it is ++- . Maybe this is the problem... (Sorry for the trouble I caused you) the only reason I added the image was to know what will happen when repeating pattern has a length of 4 or more. $\endgroup$ – Devansh Sehta Sep 30 '18 at 22:05

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