How to find $x$ in: $$ \sqrt{x^2 +\sqrt{4x^2 +\sqrt{16x^2+ \sqrt{64x^2+\dotsb} } } } =5 $$

  • $\begingroup$ Shouldn't the RHS be a function of $x$? $\endgroup$ – Bobson Dugnutt Feb 16 '16 at 0:10
  • $\begingroup$ I interpreted the question as showing that the two sides of the equation are equal. I think that @TitoPiezasIII's interpretation was more to the point of the question. My apologies $\endgroup$ – zz20s Feb 16 '16 at 0:12
  • $\begingroup$ @ClementC. Yeah, that makes more sense. $\endgroup$ – Bobson Dugnutt Feb 16 '16 at 0:13
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    $\begingroup$ It seems for any positive real number $x$, we have, $$ \sqrt{x^2 +\sqrt{4x^2 +\sqrt{16x^2+ \sqrt{64x^2+\dotsb} } } } =x+1\tag1 $$ Hence the solution to the OP's problem is $x=4$. However, it remains to prove the general case. $\endgroup$ – Tito Piezas III Feb 16 '16 at 0:44

Hint: $~x+1=\sqrt{x^2+2x+1}=\sqrt{x^2+\sqrt{4x^2+4x+1}}=\ldots~$ Can you take it from here ? :-$)$

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    $\begingroup$ @susy: It's not that difficult, $$\begin{aligned}x+1&=\sqrt{x^2+(2x+1)}\\&=\sqrt{x^2+\sqrt{(2x+1)^2}}\\& =\sqrt{x^2+\sqrt{4x^2+(4x+1)}}\\&=\sqrt{x^2+\sqrt{4x^2+\sqrt{(4x+1)^2}}}\\& =\sqrt{x^2+\sqrt{4x^2+\sqrt{16x^2+(8x+1)}}}\end{aligned}$$ and so on. $\endgroup$ – Tito Piezas III Feb 16 '16 at 0:57
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    $\begingroup$ Nice solution. However there seems to be some ambiguity the way the expression in the question can be interpreted (i.e. in what we take as the initial seed for the recursion). For example it looks like taking $x=0$ should give $0$ as the result and not $1$ as we get with this procedure. $\endgroup$ – Winther Feb 16 '16 at 1:01
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    $\begingroup$ @Winther: Notice that each nested radical ends in $1,$ not $0.$ So if we carefully and rigorously define the infinitely-nested radical in question as a limit of partially-nested radicals, the result will indeed be $1,$ and not $0.$ $\endgroup$ – Lucian Feb 16 '16 at 1:07
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    $\begingroup$ @Winther: The expression in question creates no problems for positive values of x or a. $\endgroup$ – Lucian Feb 16 '16 at 1:21
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    $\begingroup$ @Winther Yes, that is true here. But your comment, I believe, might help others on the site understand that there is a very important consideration (often ignored) when dealing with these types of "recursive" relationships. So, well done! - Mark $\endgroup$ – Mark Viola Feb 16 '16 at 2:11

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