Open study

is now brainly

With Brainly you can:

  • Get homework help from millions of students and moderators
  • Learn how to solve problems with step-by-step explanations
  • Share your knowledge and earn points by helping other students
  • Learn anywhere, anytime with the Brainly app!

A community for students.

Difficult question. Find the value of\[\sqrt{1-\sqrt{\frac{17}{16}-\sqrt{1-\sqrt{\frac{17}{16}-\sqrt{1-\cdots}}}}}\]

Mathematics
See more answers at brainly.com
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.

Join Brainly to access

this expert answer

SEE EXPERT ANSWER

To see the expert answer you'll need to create a free account at Brainly

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...

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

But \(y\) has one definite value as bounded by the equation before taking the square of both sides. So, which one is it?
1/2 ?
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.
\[1-256x+64{x^2}-992x^4-1024x^6+256x^8=0\] \[1-2^8x+2^6{x^2}-31\times2^5x^4-2^{10}x^6+2^8x^8=0\] \[1-2^8x+2^4{(2x)^2}-62\times(2x)^4-2^{4}(2x)^6+(2x)^8=0\]
|dw:1349505055344:dw|
Only 1/2 satisfies that loop
That's correct, and answers the question, but I feel like I'm missing a technical detail here.
|dw:1349505240174:dw|
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.
\[y=\sqrt{1-\sqrt{\frac{17}{16}-y}}\]\[1-y^2=\sqrt{\frac{17}{16}-y}\]\[(1-y^2)^2={\frac{17}{16}-y}\]\[y+(1-y^2)^2-{\frac{17}{16}}=0\]\[y+(1-2y^2+y^4)-{\frac{17}{16}}=0\]\[y-2y^2+y^4-{\frac{1}{16}}=0\]\[16y-32y^2-16y^4-1=0\]\[2^3(2y)-2^3(2y)^2-(2y)^4-1=0\]
\[2^3(2y)-2^3(2y)^2-(2y)^4-(2y)^0=0\]

Not the answer you are looking for?

Search for more explanations.

Ask your own question