Can anyone help me in finding a closed form of the infinite nested radical here $$\left({\sqrt {4+\sqrt {4+\sqrt {4-\sqrt {4+\sqrt {4+\sqrt {4- ......\infty}}}}}}}\right)$$ The signs are as "+,+,-,+,+,-" and so on

I tried using some trigonometric substitution but it didn't work out well. Can it be solved using some crazy trigonometric substitution. Does it have a closed form? Any hint or approach can be really helpful. Thanks!

  • 5
    $\begingroup$ Call the answer $x$, then $x=\sqrt{4+\sqrt{4+\sqrt{4-x}}}$. $\endgroup$
    – Steen82
    Commented May 10 at 18:56
  • $\begingroup$ @Steen82 that would give birth to an eight degree polynomial equation $\endgroup$ Commented May 10 at 19:01
  • $\begingroup$ wolframalpha.com/… Wolfram Alpha can solve it $\endgroup$ Commented May 10 at 19:03
  • $\begingroup$ @RiemanTieman approach0.xyz/search/… $\endgroup$
    – Etemon
    Commented May 10 at 19:07
  • $\begingroup$ @Steen82 would need to prove that the expression converges to some answer first $\endgroup$
    – Ma Ye
    Commented May 10 at 19:46

1 Answer 1



$$x={\sqrt {4+\sqrt {4+\sqrt {4-\sqrt {4+\sqrt {4+\sqrt {4- ......}}}}}}}$$ then

$x={\sqrt {4+\sqrt {4+\sqrt {4-x}}}}$ Now we just need to solve for $x$

Start by eliminating the square root of the RHS

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

We can expand this and move everything to one side and we get


This can be factored to


So $x^2+x-4=0$ or $x^3-2x^2-3x+5=0$ or $x^3+x^2-6x-7=0$

For $x^2+x-4=0 $ using the quadratic formula we get $ x=\frac{-1-\sqrt{17}}{2}$ or $x=\frac{-1+\sqrt{17}}{2}$

For $x^3-2x^2-3x+5=0 $ we can solve by substituting $y=x-\frac{2}{3}$ , write it as $5-3(y+\frac{2}{3})-2(y+\frac{2}{3})^2+(y+\frac{2}{3})^3=0$ expanding those terms we get

$$x^3-\frac{13x}{3}+\frac{65}{27}=0$$ Then we can make the beautiful substitution $y=t+\frac{λ}{κ}$ and get


Multiply both sides by $t^3$

$$t^6+t^4(3λ-\frac{13}{2})+\frac{65 t^2}{27}+t^2(3λ^2-\frac{13λ}{3})+lλ^3=0$$

We can now substitute $λ=\frac{13}{9}$ since this value makes the coefficients of $t^2$ and $t^4$ both zero. And then Substitute $z=t^3$ making it a quadratic equation in the variable $z$

$$z^2+ \frac{65z}{27}+\frac{2197}{729}=0$$

So $z=\frac{13}{54}i(5i+3\sqrt{3}) \implies t^3=\frac{13}{54}i(5i+3\sqrt{3}) \implies $



$$x=\frac{(-13)^{2/3}}{3\sqrt[3]{\frac{1}{2}i(3\sqrt{3}+5i}}-\frac{1}{3}\sqrt[3]{\frac{-13}{2}} \sqrt[3]{(5i+3\sqrt{3} )}$$ or

$$x=\ \frac{1}{3}(-1)^{2/3}\sqrt[3]{\frac{13i}{2}(5i+3\sqrt{3})}-\frac{\sqrt[3]{-2}13^{2/3}}{3\sqrt[3]{i(3\sqrt{3}+5i)}}$$

And for $x^3+x^2-6x-7$ We can substitute $y=x+\frac{1}{3}$ to eliminate the quadratic term


We expand

$$y^3-\frac{19y}{3}-\frac{133}{27}=0$$ We substitute $y=t+\frac{λ}{κ}$


Multiply both sides by $t^3$

$$t^6+t^4(3λ-\frac{19}{3})-\frac{133 t^3}{27}+t^2(3λ^2-\frac{19λ}{3})+λ^3=0$$

Then substitute $λ=\frac{19}{9}$ and $z=t^3$ we get


And we can easily find that $$z=\frac{19}{54}(7+3i\sqrt{3}) \implies t^3=\frac{19}{54}(7+3i\sqrt{3})$$

Finally we get

$$x=-\frac{1}{3}+\frac{19^{2/3}}{3\sqrt[3]{\frac{1}{2}(3i\sqrt{3}+7)}}+\frac{1}{3} \sqrt[3]\frac{19(3i\sqrt{3}+7)}{2}$$ or

$$x=-\frac{1}{3}+\frac{(-19)^{2/3}}{3\sqrt[3]{\frac{1}{2}(3i\sqrt{3}+7)}}-\frac{1}{3} \sqrt[3]\frac{19(3i\sqrt{3}+7)}{2}$$ or


And at the end we need to check if those solutions satisfy the original equation.

Long story short the only solution is

$$x=-\frac{1}{3}+\frac{19^{2/3} }{3\sqrt[3]{\frac{1}{2}(3i\sqrt{3}+7)}}+\ \frac{1}{3}\sqrt[3]{\frac{19(3i\sqrt{3}+7)}{2}}$$

Ans so we have :

$${\sqrt {4+\sqrt {4+\sqrt {4-\sqrt {4+\sqrt {4+\sqrt {4- ......}}}}}}}=x=-\frac{1}{3}+\frac{19^{2/3} }{3\sqrt[3]{\frac{1}{2}(3i\sqrt{3}+7)}}+\ \frac{1}{3}\sqrt[3]{\frac{19(3i\sqrt{3}+7)}{2}} \approx 2.507$$

  • $\begingroup$ If you take a cube root you should note which cube root should be chosen especially if you add two different cube roots together later (then it should be noted how they are chosen). $\endgroup$
    – user1318062
    Commented May 10 at 20:06

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