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All real values of $x$ which satisfy the equation $\sqrt{1+\sqrt{1+\sqrt{1+x}}}=x$

$\bf{My\; Try::}$ Here $\sqrt{1+\sqrt{1+\sqrt{1+x}}} = x>0$

Now Let $f(x)=\sqrt{1+x}\;,$ Then equation convert into $f(f(f(x)))=x$

Now Here $f(x)=x$ be one function which satisfy above equation.

My question is how can we calculate other function which satisfy above functional equation.

Help required, Thanks

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    $\begingroup$ If You had $f(x)=x$, for some $x$, then $f^{\circ 3}(x)=x$ as well. So you can find one solution by solving $x=\sqrt{1+x}$. You are trying to solve a degree 8 equation, and finding the corresponding solutions will reduce it to a degree 4 equation. $\endgroup$ – QTHalfTau Jun 22 '16 at 5:24
  • $\begingroup$ This is a good question! $\endgroup$ – Ghartal Jun 22 '16 at 5:50
  • $\begingroup$ the title of the post should match the question in the post. $\endgroup$ – miracle173 Jun 22 '16 at 6:34
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    $\begingroup$ There is no substantial relationship between finding $x$ which satisfies $f(f(f(x)))=x$ where $f(x)=\sqrt{1+x}$ and finding a function $f(x)$ which satisfies $f(f(f(x)))=x$ for all $x$ $\endgroup$ – Henry Jun 22 '16 at 7:15
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    $\begingroup$ @Ghartal: I don't think so. If you read the comments to the question and the answer you can see that the question is very unclear. Actually the answer consists of two answers because Jennifer is not sure what the question is. The OP ignores that this question in unclear and does not try to improve this. I think the best is to vote it down until the OP improves the question. $\endgroup$ – miracle173 Jun 22 '16 at 7:56
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With your remark $x=\sqrt{1+x}$ should lead to a solution.

$$x=\sqrt{1+x} \iff x^2=1+x \iff x^2-x-1=0 \iff x=\frac{1\pm\sqrt5}{2}$$

But $x=\frac{1+\sqrt5}{2}$ is the only positive solution. We verify that it is indeed solution by using long division :

$$\sqrt{1+\sqrt{1+\sqrt{1+x}}} = x\iff \sqrt{1+\sqrt{1+x}}=x^2-1 \iff \sqrt{1+x}=x^4-2x^2\\\iff x+1=x^8-4x^6+4x^4 \iff x^8-4x^6+4x^4-x-1=0$$

$$x^8-4x^6+4x^4-x-1=(x^2-x-1)(x^6+x^5-2x^4-x^3+x^2+1)$$

Let $g(x)=x^6+x^5-2x^4-x^3+x^2+1$, notice that $\forall x>0, g'(x)>0$, so $g$ is strictly increasing and $g(0)=1>0$, so $g(x)=0$ has no solution on $\mathbb{R}^{*+}$.

Finally $x=\frac{1+\sqrt5}{2}$ is the only solution to the problem.


Notes : The functional equation $\;(f \circ f \circ f)(x) = x,\forall x>0$ only has one continuous solution : $f(x)=x$.

By studying the domain, you have $f : \mathbb{R}^{*+} \rightarrow \mathbb{R}^{*+}$.

$\forall x \in \mathbb{R}^{*+}$ the image by $f$ of $f(f(x))$ is $x$ so $f$ is onto. Also if $f(a)=f(b)$, then $a=f(f(f(a)))=f(f(f(b)))=b$, so $f$ is injective. So $f$ is one to one. If $f$ is decreasing $f\circ f$ is increasing and $f\circ f\circ f$ is decrasing, but $id$ is increasing, it is absurd, so $f$ is strictly increasing.

Suppose $f(x)>x$, since $f$ is strictly increasing $f(f(x))>f(x)>x$, so $x=f(f(f(x))>f(f(x))>f(x)>x$, it is impossible. In the same way $f(x)<x$ is impossible, so $\forall x, f(x)=x$. So $f=id$.

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    $\begingroup$ Your reasoning is incorrect. In the link, $f$ satisfies the property for ALL $x$. $\endgroup$ – Aravind Jun 22 '16 at 5:36
  • $\begingroup$ @Aravind I posted an adapted proof $\endgroup$ – Jennifer Jun 22 '16 at 5:59
  • $\begingroup$ Ah! I see now. Here, we already know that $f$ is strictly increasing in $x > 0$, so we could also use that. $\endgroup$ – Aravind Jun 22 '16 at 6:18
  • $\begingroup$ @EricWofsey I have to justify why only $x=\sqrt{1+x}$ is the only expression which leads to solutions, I don't see why it should be removed, can you explain me please ? $\endgroup$ – Jennifer Jun 22 '16 at 6:27
  • $\begingroup$ The problem statement does not tell you that $f(f(f(x)))=x$ for all $x$, it just asks you to find the values of $x$ for which this is true. $\endgroup$ – Eric Wofsey Jun 22 '16 at 6:30
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\begin{align}\sqrt{1+{\sqrt{1+\sqrt{1+x}}}}=x&\implies \sqrt{1+{\sqrt{1+\sqrt{1+x}}}}=x\\&\implies \left(1+{\sqrt{1+\sqrt{1+x}}}\right)=x^2\\&\implies\sqrt{1+\sqrt{1+x}}=x^2-1\\&\implies 1+\sqrt{1+x}=x^4-2x^2+1\\&\implies \sqrt{1+x}=x^4-2x^2\\&\implies 1+x=x^4(x^2-2)^2\\&\implies 1+x=x^4(x^4-4x^2+4)\\&\implies x^8-4x^6+4x^4-x-1=0\end{align}Now observe that , $$x^8-4x^6+4x^4-x-1=x^8\color{red}{-x^7}-x^6\color{red}{+x^7}-x^6\color{blue}{-x^5}-2x^6\color{blue}{+2x^5}+2x^4\color{blue}{-x^5}+x^4\color{green}{+x^3}+x^4\color{green}{-x^3}\color{violet}{-x^2}\color{violet}{+x^2}-x-1$$Consequently, $$x^8-4x^6+4x^4-x-1=0\implies (x^2-x-1)(x^6+x^5-2x^4-x^3+x^2+1)$$Now observe that the function $$g(x)=x^6+x^5-2x^4-x^3+x^2+1=\left(x^3-1\right)^2+\left(x-1\right)^2\,x^3+x^2 >0$$ for all $x\in \mathbb{R}^+$. So $(x^2-x-1)=0$ and hence $x=\dfrac{1+\sqrt{5}}{2}$ is the only positive solution.

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    $\begingroup$ How do you know $g(x)>0$ for all $x\in\mathbb{R}$? $\endgroup$ – Eric Wofsey Jun 22 '16 at 7:51
  • $\begingroup$ Isn't this answer already included in Jennifer's answer? $\endgroup$ – miracle173 Jun 22 '16 at 8:06
  • $\begingroup$ @miracle173: Actually. I didn't notice the edit that was being made in Jennifer's answer as I was writing my answer. So, the answer to your question is yes. $\endgroup$ – S. Das Jun 22 '16 at 8:13
  • $\begingroup$ @EricWofsey: edited the post so one can see that g(x)>0 $\endgroup$ – miracle173 Jun 22 '16 at 12:51
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Let $1+x=y^2$, where $y>0$ and $1+y=z^2$, where $z>0$.

Hence, $1+z=x^2$, where $x>0$.

Thus, $x-y=(y-z)(y+z)$ and $y-z=(z-x)(z+x)$.

  1. Let $x>y$. Hence, $y>z$, which says that $z>x$. It's contradiction.

  2. Let $x<y$. Hence, $y<z$, which says that $z<x$. It's contradiction again.

Id est, $x=y=z$ and $x=\frac{1+\sqrt5}{2}$.

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