Does $\sqrt{\cos{\theta} \sqrt{\cos{\theta} \sqrt{\cos{\theta\dots}}}}=\sqrt{1 \sqrt{1 \sqrt{1\dots}}} \implies \cos{\theta}=1$?

Question

I was solving this equation:

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

I solved it like this:

The given equation can be written as:

\begin{align*} \sqrt{\cos{\theta} \sqrt{\cos{\theta} \sqrt{\cos{\theta\dots}}}}&=\sqrt{1 \sqrt{1 \sqrt{1\dots}}} \\ \cos{\theta}&=1 \\ \theta&=\arccos {1} \end{align*}

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

Have I solved it the wrong way?

(The title originally contained a more general question: Does $\sqrt{a \sqrt{a \sqrt{a\dots}}}=\sqrt{b \sqrt{b \sqrt{b\dots}}} \implies a=b$? The current title is consistent with the body and the accepted answer.)

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
$\sqrt{a\sqrt{a\cdots }}=a^{1/2+1/4+\cdots}=a$ 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

Adi Dani · Accepted Answer · 2016-02-21T10:56:49.237

16

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

edited Feb 21 '16 at 10:56

answered Feb 21 '16 at 10:16

Adi Dani

16,949

lhf · Answer 2 · 2016-02-23T14:07:52.270

4

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 $a=b$.

edited Feb 23 '16 at 14:07

answered Feb 21 '16 at 11:43

lhf

216,483

Why would $aA=A^2$? – Man_Of_Wisdom Feb 21 '16 at 14:53
2

@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 http://math.stackexchange.com/questions/589288/sqrt7-sqrt7-sqrt7-sqrt7-sqrt7-cdots-approximation. – lhf Feb 21 '16 at 15:41

score 3 · Answer 3 · answered Feb 23 '16 at 14:21

Another answer

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 $$\theta=2k\pi,k\in\mathbb Z$$

score 0 · Answer 4 · answered Feb 23 '16 at 14:21

Hint:

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

Multiply by $\cos(\theta)$ and take the square root. You get

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

Does $\sqrt{\cos{\theta} \sqrt{\cos{\theta} \sqrt{\cos{\theta\dots}}}}=\sqrt{1 \sqrt{1 \sqrt{1\dots}}} \implies \cos{\theta}=1$?

4 Answers4