Proof of Convergence: Nested Radicals with Constant Sum on May 8, 2019

[anemone]

Here is this week's POTW:

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Prove that $\sqrt[3]{9+9\sqrt[3]{9+9\sqrt[3]{9+\cdots}}} - \sqrt{8-\sqrt{8-\sqrt{8+\sqrt{8-\sqrt{8-\sqrt{8+\cdots}}}}}} = 1$.

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Remember to read the https://mathhelpboards.com/showthread.php?772-Problem-of-the-Week-%28POTW%29-Procedure-and-Guidelines to find out how to https://mathhelpboards.com/forms.php?do=form&fid=2!

 

[anemone]

Congratulations to Olinguito for his correct solution(Cool), which you can find below:

Spoiler

Let $x=\sqrt[3]{9+9\sqrt[3]{9+\cdots}}=\sqrt[3]{9+9x}$.

$\therefore\ x^3=9x+9\quad\ldots\fbox1$.

Then
$$(((x-1)^2-8)^2-8)^2$$
$=\ ((x^2-2x-7)^2-8)^2$

$=\ (x^4-4x^3-10x^2+28x+41)^2$

$=\ ((9x+9)(x-4)-10x^2+28x+41)^2\quad\text{(using}\ \fbox1)$

$=\ (-x^2+x+5)^2$

$=\ x^4-2x^3-9x^2+10x+25$

$=\ (9x+9)(x-2)-9x^2+10x+25\quad\text{(using}\ \fbox1\ \text{again)}$

$=\ x+7$.

$\therefore\ ((u^2-8)^2-8)^2\ =\ u+8$ where $u=x-1$.

It remains to show that $u=\sqrt{8-\sqrt{8-\sqrt{8+u}}}$.

Now, if $f(x)=x^3-9x-9$, then $f(3.41)=-0.038179<0$ and $f(3.42)=0.221688>0$. Hence $3.41<x<3.42$ $\implies$ $2.41<u<2.42$. Thus
$$\sqrt8>2.42>u>2.41$$
$\implies\ 8>u^2>5.8081$

$\implies\ 0<8-u^2<2.1919<\sqrt8$

$\implies\ (8-u^2)^2<8$

Hence
$$u+8\ =\ ((u^2-8)^2-8)^2$$
$\implies\ \sqrt{8+u}\ =\ |(u^2-8)^2-8|\ =\ 8-(u^2-8)^2$

$\implies\ 8-\sqrt{8+u}\ =\ (u^2-8)^2$

$\implies\ \sqrt{8-\sqrt{8+u}}\ =\ |u^2-8|\ =\ 8-u^2$

$\implies\ u^2\ =\ 8-\sqrt{8-\sqrt{8+u}}$

$\implies\ u\ =\ \sqrt{8-\sqrt{8-\sqrt{8+u}}}$

as required.