## badreferences Group Title Difficult question. Find the value of$\sqrt{1-\sqrt{\frac{17}{16}-\sqrt{1-\sqrt{\frac{17}{16}-\sqrt{1-\cdots}}}}}$ one year ago one year ago

@KingGeorge @UnkleRhaukus @TuringTest @Zarkon @bahrom7893

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

3. hartnn Group Title

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?

5. hartnn Group Title

1/2 ?

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

7. UnkleRhaukus Group Title

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

8. UnkleRhaukus Group Title

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.

10. UnkleRhaukus Group Title

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.

13. UnkleRhaukus Group Title

$\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.

15. sauravshakya Group Title

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

16. sauravshakya Group Title

As squaring gives some unnecessary roots.

17. UnkleRhaukus Group Title

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

18. sauravshakya Group Title

|dw:1349505055344:dw|

19. sauravshakya Group Title

Only 1/2 satisfies that loop

That's correct, and answers the question, but I feel like I'm missing a technical detail here.

21. sauravshakya Group Title

|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.

25. UnkleRhaukus Group Title

$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$

26. UnkleRhaukus Group Title

$2^3(2y)-2^3(2y)^2-(2y)^4-(2y)^0=0$