How to find $x$ in: $$ \sqrt{x^2 +\sqrt{4x^2 +\sqrt{16x^2+ \sqrt{64x^2+\dotsb} } } } =5 $$
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Shouldn't the RHS be a function of $x$? – Bobson Dugnutt Feb 16 '16 at 00:10
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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 – zz20s Feb 16 '16 at 00:12
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@ClementC. Yeah, that makes more sense. – Bobson Dugnutt Feb 16 '16 at 00:13
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4It 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. – Tito Piezas III Feb 16 '16 at 00:44
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Hint: $~x+1=\sqrt{x^2+2x+1}=\sqrt{x^2+\sqrt{4x^2+4x+1}}=\ldots~$ Can you take it from here ? :-$)$
Lucian
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How is this any different from $x + 2 = \sqrt {x^2 + 4x+4} = \sqrt{x^2 + \sqrt{16x²+32x+16}} = \sqrt {x^2 + \sqrt{4x^2 + \sqrt{\ldots}}}$ ? – mercio Feb 16 '16 at 00:54
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How do you get from the third to the fourth, @mercio? (It is different as Lucian's solution does get the $1,4,16,64,\dots$, while yours gets $1,16,?,\dots$) – Clement C. Feb 16 '16 at 00:57
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2@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. – Tito Piezas III Feb 16 '16 at 00:57
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1@mercio Because the degree of the terms in your RHS will increase without bounds (e.g., the ldots there includes a term $144x^4$, since you're taking an expression that square roots to $12x^2+32x+16$). Showing convergence even in the 'right' case is definitely non-trivial, but your attempted matching expression should much more clearly diverge. – Steven Stadnicki Feb 16 '16 at 00:58
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7Nice 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. – Winther Feb 16 '16 at 01:01
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Sorry, I missed so explanation. I speak Spanish and there are things I do not know, what is LHS and RHS? – susy diaz Feb 16 '16 at 01:04
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left hand side, right hand side: https://en.wikipedia.org/wiki/Sides_of_an_equation – Jaume Oliver Lafont Feb 16 '16 at 01:04
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2@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 $0.$ – Lucian Feb 16 '16 at 01:07
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@susy: My explanation is as simple as it gets. Continue the procedure to $\sqrt{(8x+1)^2} = \sqrt{64x^2+16x+1}$ and you'll get the pattern. – Tito Piezas III Feb 16 '16 at 01:07
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@mercio: Maybe you can delete your comment? I think it is confusing Susy. :) – Tito Piezas III Feb 16 '16 at 01:11
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2@Winther: The expression in question creates no problems for positive values of x or a. – Lucian Feb 16 '16 at 01:21
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Now, I have answers and algebraic substitutions, until then it was fine but looked equation expressed recursively. With that solution should I be me, there are things that escape me, if I explain them in more detail please ... – susy diaz Feb 16 '16 at 01:21
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2It's just... puzzling. Like saying "I am vegan and lactose-intolerant but just ordered 500 lbs of cheese." – Clement C. Feb 16 '16 at 01:42
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@Lucian: I will be reviewing this later! This is not a big issue! – Mhenni Benghorbal Feb 16 '16 at 01:44
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1@Winther Exactly! There are so many of these posted questions that fall into the same category - ill-posed nested operations with no initial seed. Glad you posted your comment! +1 - Mark – Mark Viola Feb 16 '16 at 01:53
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@Dr.MV Yes I have also seen many questions where this issue is a real problem, but as Lucian cleared up above there is not a big problem in this particular case as long as the seed is stricktly positive. The only issue is at $x=0$ when the seed is $0$ which gives $f(0)=0$ instead of $f(0)=1$. – Winther Feb 16 '16 at 01:59
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2@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 – Mark Viola Feb 16 '16 at 02:11
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