anonymous
  • anonymous
Difficult question. Find the value of\[\sqrt{1-\sqrt{\frac{17}{16}-\sqrt{1-\sqrt{\frac{17}{16}-\sqrt{1-\cdots}}}}}\]
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
@KingGeorge @UnkleRhaukus @TuringTest @Zarkon @bahrom7893
anonymous
  • anonymous
I have the solution, by the way. This is like a challenge. :)
hartnn
  • hartnn
y^2 = 1-sqrt{17/16-y} (y^2-1)^2 = 17/16-y solve...

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
But \(y\) has one definite value as bounded by the equation before taking the square of both sides. So, which one is it?
hartnn
  • hartnn
1/2 ?
anonymous
  • anonymous
That's a good heuristic guess. It's correct. Why?
UnkleRhaukus
  • UnkleRhaukus
\[x=\sqrt{1-\sqrt{\frac{17}{16}-\sqrt{1-\sqrt{\frac{17}{16}-\sqrt{1-\cdots}}}}}\] \[1-x^2={\sqrt{\frac{17}{16}-\sqrt{1-\sqrt{\frac{17}{16}-\sqrt{1-\cdots}}}}}\] \[\left((1-x^2)^2-\frac{17}{16}\right)^2={{\sqrt{1-\sqrt{\frac{17}{16}-\sqrt{1-\cdots}}}}}\] \[\left((1-x^2)^2-\frac{17}{16}\right)^2=x\]
UnkleRhaukus
  • UnkleRhaukus
am i on the right track?
anonymous
  • anonymous
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.
UnkleRhaukus
  • UnkleRhaukus
i think i accidentally dropped a negative sign
anonymous
  • anonymous
Yes, you did. Haha, I didn't notice. I just saw the polynomials and went, "k"
anonymous
  • anonymous
Not that it makes a difference. A mild algebraic rearrangement leads to the same problem.
UnkleRhaukus
  • UnkleRhaukus
\[\left(\frac{17}{16}-(1-x^2)^2\right)^2=x\]\[\left(\frac{1}{16}+2x^2-x^4\right)^2=x\]\[\frac1{16^2}+\frac{x^2}4-\frac{31x^4}{8}-4x^6+x^8=x\]\[\frac1{256}-x+\frac{x^2}4-\frac{31x^4}{8}-4x^6+x^8=0\]
anonymous
  • anonymous
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.
anonymous
  • anonymous
cant we just substitute the value in y^2 = 1-sqrt{17/16-y} and see if it is TRUE OR NOT.
anonymous
  • anonymous
As squaring gives some unnecessary roots.
UnkleRhaukus
  • UnkleRhaukus
\[1-256x+64{x^2}-992x^4-1024x^6+256x^8=0\] \[1-2^8x+2^6{x^2}-31\times2^5x^4-2^{10}x^6+2^8x^8=0\] \[1-2^8x+2^4{(2x)^2}-62\times(2x)^4-2^{4}(2x)^6+(2x)^8=0\]
anonymous
  • anonymous
|dw:1349505055344:dw|
anonymous
  • anonymous
Only 1/2 satisfies that loop
anonymous
  • anonymous
That's correct, and answers the question, but I feel like I'm missing a technical detail here.
anonymous
  • anonymous
|dw:1349505240174:dw|
anonymous
  • anonymous
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.
anonymous
  • anonymous
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.
anonymous
  • anonymous
But whatever, if there was subtlety here, I'm still missing it. You got the answer.
UnkleRhaukus
  • UnkleRhaukus
\[y=\sqrt{1-\sqrt{\frac{17}{16}-y}}\]\[1-y^2=\sqrt{\frac{17}{16}-y}\]\[(1-y^2)^2={\frac{17}{16}-y}\]\[y+(1-y^2)^2-{\frac{17}{16}}=0\]\[y+(1-2y^2+y^4)-{\frac{17}{16}}=0\]\[y-2y^2+y^4-{\frac{1}{16}}=0\]\[16y-32y^2-16y^4-1=0\]\[2^3(2y)-2^3(2y)^2-(2y)^4-1=0\]
UnkleRhaukus
  • UnkleRhaukus
\[2^3(2y)-2^3(2y)^2-(2y)^4-(2y)^0=0\]

Looking for something else?

Not the answer you are looking for? Search for more explanations.