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badreferences
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Difficult question. Find the value of\[\sqrt{1\sqrt{\frac{17}{16}\sqrt{1\sqrt{\frac{17}{16}\sqrt{1\cdots}}}}}\]
 2 years ago
 2 years ago
badreferences Group Title
Difficult question. Find the value of\[\sqrt{1\sqrt{\frac{17}{16}\sqrt{1\sqrt{\frac{17}{16}\sqrt{1\cdots}}}}}\]
 2 years ago
 2 years ago

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badreferences Group TitleBest ResponseYou've already chosen the best response.0
@KingGeorge @UnkleRhaukus @TuringTest @Zarkon @bahrom7893
 2 years ago

badreferences Group TitleBest ResponseYou've already chosen the best response.0
I have the solution, by the way. This is like a challenge. :)
 2 years ago

hartnn Group TitleBest ResponseYou've already chosen the best response.2
y^2 = 1sqrt{17/16y} (y^21)^2 = 17/16y solve...
 2 years ago

badreferences Group TitleBest ResponseYou've already chosen the best response.0
But \(y\) has one definite value as bounded by the equation before taking the square of both sides. So, which one is it?
 2 years ago

badreferences Group TitleBest ResponseYou've already chosen the best response.0
That's a good heuristic guess. It's correct. Why?
 2 years ago

UnkleRhaukus Group TitleBest ResponseYou've already chosen the best response.0
\[x=\sqrt{1\sqrt{\frac{17}{16}\sqrt{1\sqrt{\frac{17}{16}\sqrt{1\cdots}}}}}\] \[1x^2={\sqrt{\frac{17}{16}\sqrt{1\sqrt{\frac{17}{16}\sqrt{1\cdots}}}}}\] \[\left((1x^2)^2\frac{17}{16}\right)^2={{\sqrt{1\sqrt{\frac{17}{16}\sqrt{1\cdots}}}}}\] \[\left((1x^2)^2\frac{17}{16}\right)^2=x\]
 2 years ago

UnkleRhaukus Group TitleBest ResponseYou've already chosen the best response.0
am i on the right track?
 2 years ago

badreferences Group TitleBest ResponseYou've already chosen the best response.0
A polynomial equation that doesn't rigorously determine which of the \(x\)'s are correct. But yes, you and @hartnn are on the right track. From here it's about iteration behavior at certain values.
 2 years ago

UnkleRhaukus Group TitleBest ResponseYou've already chosen the best response.0
i think i accidentally dropped a negative sign
 2 years ago

badreferences Group TitleBest ResponseYou've already chosen the best response.0
Yes, you did. Haha, I didn't notice. I just saw the polynomials and went, "k"
 2 years ago

badreferences Group TitleBest ResponseYou've already chosen the best response.0
Not that it makes a difference. A mild algebraic rearrangement leads to the same problem.
 2 years ago

UnkleRhaukus Group TitleBest ResponseYou've already chosen the best response.0
\[\left(\frac{17}{16}(1x^2)^2\right)^2=x\]\[\left(\frac{1}{16}+2x^2x^4\right)^2=x\]\[\frac1{16^2}+\frac{x^2}4\frac{31x^4}{8}4x^6+x^8=x\]\[\frac1{256}x+\frac{x^2}4\frac{31x^4}{8}4x^6+x^8=0\]
 2 years ago

badreferences Group TitleBest ResponseYou've already chosen the best response.0
Ahaha, that's some unnecessary complexity, but ostensibly it seems right. The difficulty is in showing which one of the \(x\)'s it is, as bounded by the one value \(x\) in the original equation.
 2 years ago

sauravshakya Group TitleBest ResponseYou've already chosen the best response.1
cant we just substitute the value in y^2 = 1sqrt{17/16y} and see if it is TRUE OR NOT.
 2 years ago

sauravshakya Group TitleBest ResponseYou've already chosen the best response.1
As squaring gives some unnecessary roots.
 2 years ago

UnkleRhaukus Group TitleBest ResponseYou've already chosen the best response.0
\[1256x+64{x^2}992x^41024x^6+256x^8=0\] \[12^8x+2^6{x^2}31\times2^5x^42^{10}x^6+2^8x^8=0\] \[12^8x+2^4{(2x)^2}62\times(2x)^42^{4}(2x)^6+(2x)^8=0\]
 2 years ago

sauravshakya Group TitleBest ResponseYou've already chosen the best response.1
dw:1349505055344:dw
 2 years ago

sauravshakya Group TitleBest ResponseYou've already chosen the best response.1
Only 1/2 satisfies that loop
 2 years ago

badreferences Group TitleBest ResponseYou've already chosen the best response.0
That's correct, and answers the question, but I feel like I'm missing a technical detail here.
 2 years ago

sauravshakya Group TitleBest ResponseYou've already chosen the best response.1
dw:1349505240174:dw
 2 years ago

badreferences Group TitleBest ResponseYou've already chosen the best response.0
Hmm, you used the same solution as I did, but for some reason my solution was corrected to include rates of change around \(x=0.5\). Haha, I'm still trying to catch why.
 2 years ago

badreferences Group TitleBest ResponseYou've already chosen the best response.0
I think the idea is that asserting\[y=\sqrt{1\sqrt{\frac{17}{16}y}}\]assumes that \(y\) has the innermost radical as \(1\), and iteratively doing this will result in a technical approximation. So we had to determine that the values converge around \(x=0.5\) using straightforward slopes.
 2 years ago

badreferences Group TitleBest ResponseYou've already chosen the best response.0
But whatever, if there was subtlety here, I'm still missing it. You got the answer.
 2 years ago

UnkleRhaukus Group TitleBest ResponseYou've already chosen the best response.0
\[y=\sqrt{1\sqrt{\frac{17}{16}y}}\]\[1y^2=\sqrt{\frac{17}{16}y}\]\[(1y^2)^2={\frac{17}{16}y}\]\[y+(1y^2)^2{\frac{17}{16}}=0\]\[y+(12y^2+y^4){\frac{17}{16}}=0\]\[y2y^2+y^4{\frac{1}{16}}=0\]\[16y32y^216y^41=0\]\[2^3(2y)2^3(2y)^2(2y)^41=0\]
 2 years ago

UnkleRhaukus Group TitleBest ResponseYou've already chosen the best response.0
\[2^3(2y)2^3(2y)^2(2y)^4(2y)^0=0\]
 2 years ago
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