Solving an infinitely long square root problem is easy but how to solve this one? The equation goes like this. $$\sqrt{4+\sqrt{4+\sqrt{4-\sqrt{4-x}}}} = x$$
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$\begingroup$ we get the solution as a polynomial of degree $12$ $\endgroup$– Dr. Sonnhard GraubnerOct 31, 2015 at 19:53
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1$\begingroup$ I'm not sure if it would work, but this could be solved as a fixed point iteration problem. If you take the limit to infinity of iterations it then becomes an infinite chain of nested radicals. $\endgroup$– OFRBGOct 31, 2015 at 19:56
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Its around 2.526062179 and Wolfram Alpha agrees.
You can come to the same result by visiting "Wolfram Alpha". You can also rearrange the term stepwise:
$$ x=\sqrt{4+\sqrt{4+\sqrt{4-\sqrt{4-x}}}} $$ $$ x^2-4=\sqrt{4+\sqrt{4-\sqrt{4-x}}} $$ $$ (x^2-4)^2-4=\sqrt{4-\sqrt{4-x}} $$ $$ 4-(4-((x^2-4)^2-4)^2)^2=x $$ When you now resolve the parentheses you get: $$ \small{x^{12}-24 x^{10}-x^9+228 x^8+16 x^7-1087 x^6-88 x^5+2720 x^4+191 x^3-3380 x^2-136 x+1633} = 0 $$ Which Wolfram Alpha resolves too: $$ x \approx 2.5260621791756334159 $$
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2$\begingroup$ I think a more constructive way to answer this question would be to talk about the technique in solving this question. $\endgroup$ Oct 31, 2015 at 20:13
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1$\begingroup$ @Antonios-AlexandrosRobotis : Do you like it better now? $\endgroup$ Oct 31, 2015 at 20:57
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$\begingroup$ Yes, thanks for revising your response! $\endgroup$ Oct 31, 2015 at 21:00
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$\begingroup$ @Antonios-AlexandrosRobotis Would you mind removing the delete flag? $\endgroup$ Oct 31, 2015 at 21:02
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$\begingroup$ I have not placed any delete flag on this post- if someone has, it wasn't me. $\endgroup$ Oct 31, 2015 at 21:08