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!

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

Let

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

$$-x^8+16x^6-88x^4+192x^2-x-140=0$$

This can be factored to

$$-(x^2+x-4)(x^3-2x^2-3x+5)(x^3+x^2-6x-7)=0$$

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

$$\frac{65}{27}-\frac{13}{3}(t+\frac{λ}{κ})+(t+\frac{λ}{κ})^3=0$$

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

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

or

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

$$-7-6(y-\frac{1}{3})+(y-\frac{1}{3})^2+(y-\frac{1}{3})^3=0$$

We expand

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

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

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

$$z^2-\frac{133z}{27}+\frac{6859}{729}=0$$

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

$$x=-\frac{1}{3}+\frac{(-2)^{2/3}\sqrt[3]{19}(3i\sqrt{3}+7)^{2/3}-2\sqrt[3]{-2}19^{2/3}}{6\sqrt[3]{3i\sqrt{3}+7}}$$

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 :

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

• 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).
– user1318062
Commented May 10 at 20:06