Here's the question you clicked on:
badreferences
Difficult question. Find the value of\[\sqrt{1-\sqrt{\frac{17}{16}-\sqrt{1-\sqrt{\frac{17}{16}-\sqrt{1-\cdots}}}}}\]
@KingGeorge @UnkleRhaukus @TuringTest @Zarkon @bahrom7893
I have the solution, by the way. This is like a challenge. :)
y^2 = 1-sqrt{17/16-y} (y^2-1)^2 = 17/16-y solve...
But \(y\) has one definite value as bounded by the equation before taking the square of both sides. So, which one is it?
That's a good heuristic guess. It's correct. Why?
\[x=\sqrt{1-\sqrt{\frac{17}{16}-\sqrt{1-\sqrt{\frac{17}{16}-\sqrt{1-\cdots}}}}}\] \[1-x^2={\sqrt{\frac{17}{16}-\sqrt{1-\sqrt{\frac{17}{16}-\sqrt{1-\cdots}}}}}\] \[\left((1-x^2)^2-\frac{17}{16}\right)^2={{\sqrt{1-\sqrt{\frac{17}{16}-\sqrt{1-\cdots}}}}}\] \[\left((1-x^2)^2-\frac{17}{16}\right)^2=x\]
am i on the right track?
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.
i think i accidentally dropped a negative sign
Yes, you did. Haha, I didn't notice. I just saw the polynomials and went, "k"
Not that it makes a difference. A mild algebraic rearrangement leads to the same problem.
\[\left(\frac{17}{16}-(1-x^2)^2\right)^2=x\]\[\left(\frac{1}{16}+2x^2-x^4\right)^2=x\]\[\frac1{16^2}+\frac{x^2}4-\frac{31x^4}{8}-4x^6+x^8=x\]\[\frac1{256}-x+\frac{x^2}4-\frac{31x^4}{8}-4x^6+x^8=0\]
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.
cant we just substitute the value in y^2 = 1-sqrt{17/16-y} and see if it is TRUE OR NOT.
As squaring gives some unnecessary roots.
\[1-256x+64{x^2}-992x^4-1024x^6+256x^8=0\] \[1-2^8x+2^6{x^2}-31\times2^5x^4-2^{10}x^6+2^8x^8=0\] \[1-2^8x+2^4{(2x)^2}-62\times(2x)^4-2^{4}(2x)^6+(2x)^8=0\]
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Only 1/2 satisfies that loop
That's correct, and answers the question, but I feel like I'm missing a technical detail here.
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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.
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.
But whatever, if there was subtlety here, I'm still missing it. You got the answer.
\[y=\sqrt{1-\sqrt{\frac{17}{16}-y}}\]\[1-y^2=\sqrt{\frac{17}{16}-y}\]\[(1-y^2)^2={\frac{17}{16}-y}\]\[y+(1-y^2)^2-{\frac{17}{16}}=0\]\[y+(1-2y^2+y^4)-{\frac{17}{16}}=0\]\[y-2y^2+y^4-{\frac{1}{16}}=0\]\[16y-32y^2-16y^4-1=0\]\[2^3(2y)-2^3(2y)^2-(2y)^4-1=0\]
\[2^3(2y)-2^3(2y)^2-(2y)^4-(2y)^0=0\]