Question

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

Answer 1

@KingGeorge @UnkleRhaukus @TuringTest @Zarkon @bahrom7893

Answer 2

I have the solution, by the way. This is like a challenge. :)

Answer 3

y^2 = 1-sqrt{17/16-y}
(y^2-1)^2 = 17/16-y
solve...

Answer 4

But \(y\) has one definite value as bounded by the equation before taking the square of both sides. So, which one is it?

Answer 5

1/2 ?

Answer 6

That's a good heuristic guess. It's correct. Why?

Answer 7

\[x=\sqrt{1-\sqrt{\frac{17}{16}-\sqrt{1-\sqrt{\frac{17}{16}-\sqrt{1-\cdots}}}}}\]
\[1-x^2={\sqrt{\frac{17}{16}-\sqrt{1-\sqrt{\frac{17}{16}-\sqrt{1-\cdots}}}}}\]
\[\left((1-x^2)^2-\frac{17}{16}\right)^2={{\sqrt{1-\sqrt{\frac{17}{16}-\sqrt{1-\cdots}}}}}\]
\[\left((1-x^2)^2-\frac{17}{16}\right)^2=x\]

Answer 8

am i on the right track?

Answer 9

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.

Answer 10

i think i accidentally dropped a negative sign

Answer 11

Yes, you did. Haha, I didn't notice. I just saw the polynomials and went, "k"

Answer 12

Not that it makes a difference. A mild algebraic rearrangement leads to the same problem.

Answer 13

\[\left(\frac{17}{16}-(1-x^2)^2\right)^2=x\]\[\left(\frac{1}{16}+2x^2-x^4\right)^2=x\]\[\frac1{16^2}+\frac{x^2}4-\frac{31x^4}{8}-4x^6+x^8=x\]\[\frac1{256}-x+\frac{x^2}4-\frac{31x^4}{8}-4x^6+x^8=0\]

Answer 14

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.

Answer 15

cant we just substitute the value in y^2 = 1-sqrt{17/16-y} and see if it is TRUE OR NOT.

Answer 16

As squaring gives some unnecessary roots.

Answer 17

\[1-256x+64{x^2}-992x^4-1024x^6+256x^8=0\]
\[1-2^8x+2^6{x^2}-31\times2^5x^4-2^{10}x^6+2^8x^8=0\]
\[1-2^8x+2^4{(2x)^2}-62\times(2x)^4-2^{4}(2x)^6+(2x)^8=0\]