# Does $\sqrt{a \sqrt{a \sqrt{a}}}…=\sqrt{b \sqrt{b \sqrt{b}}}… \implies a=b$?

I was solving an equation which states,

$$\sqrt{\cos{\theta} \sqrt{\cos{\theta} \sqrt{\cos{\theta}}}}...=1$$

I solved like this:

The given equation can be written as following,

$$\sqrt{\cos{\theta} \sqrt{\cos{\theta} \sqrt{\cos{\theta}}}}...= \sqrt{1 \sqrt{1 \sqrt{1}}}...$$

$$\implies \cos{\theta}=1$$

$$\implies \theta=\arccos {1}$$

So solution is $2n\pi, n \in \mathbb Z$.

Have I solved it wrong way?

• I see no particular reason for it to be true. After all, $x^2 = 1^2$ doesn't imply $x = 1$. – Patrick Stevens Feb 21 '16 at 10:08
• Take for example, $y=\sqrt{2\sqrt{2\sqrt{2\cdots}}}$ – GoodDeeds Feb 21 '16 at 10:09
• $\sqrt{a\sqrt{a\cdots }}=a^{1/2+1/4+\cdots}=a$ and $\sqrt{b\sqrt{b\cdots }}=b$, so the implication is true. – user236182 Feb 21 '16 at 10:17
• If anything the ellipses should be inside the square roots: $\sqrt{b \sqrt{b \sqrt{b\ldots}}}$. But "infinite expressions" are not expressions, and undefined unless some specific meaning is given to them using some appropriate limit (which of course must also converge). – Marc van Leeuwen Feb 23 '16 at 14:27

$$\sqrt{\cos{\theta} \sqrt{\cos{\theta} \sqrt{\cos{\theta}}}}...=1$$ $$(\cos{\theta})^{1/2+1/4+1/8+...}=1$$ $$(\cos{\theta})^{1}=1$$ $$\cos{\theta}=1$$ $$\theta=2k\pi,k\in\mathbb Z$$

To answer the question in the title, let $A=\sqrt{a \sqrt{a \sqrt{a\cdots}}}$ and $B=\sqrt{b \sqrt{b \sqrt{b\cdots}}}$.

Then $A=B$ implies $aA=A^2=B^2=bB$ and so $a=b$.

• Why would $aA=A^2$? – Man_Of_Wisdom Feb 21 '16 at 14:53
• @Man_Of_Wisdom, isn't $A=\lim a_n$ with $a_{n+1}=\sqrt{a\cdot a_n}$ and $a_0=1$? – lhf Feb 21 '16 at 15:38
• See for instance math.stackexchange.com/questions/589288/…. – lhf Feb 21 '16 at 15:41

suppose that $$\sqrt{\cos{\theta} \sqrt{\cos{\theta} \sqrt{\cos{\theta}}}}...=1$$ square both side you get $$cos{\theta} \sqrt{\cos{\theta} \sqrt{\cos{\theta}}}...=1$$ We know that $$\ \sqrt{\cos{\theta} \sqrt{\cos{\theta}}}...=1$$ thus $$\cos{\theta}*1=1$$ therefore $$\theta=2k\pi,k\in\mathbb Z$$
$$\sqrt{\cos{\theta} \sqrt{\cos{\theta} \sqrt{\cos{\theta}}}}...=1.$$
Multiply by $\cos(\theta)$ and take the square root. You get
$$\sqrt{\cos(\theta)}=\sqrt{\cos{\theta} \sqrt{\cos{\theta} \sqrt{\cos{\theta} \sqrt{\cos{\theta}}}}}...=1.$$