# (Serbia) Just squaring won't work

1. Jul 26, 2014

### Dinheiro

1. The problem statement, all variables and given/known data
Find the real solution $x = \sqrt{2+\sqrt{2-\sqrt{2+x}}}$

2. Relevant equations
Algebric equations

3. The attempt at a solution
I tried to find some kind of recurrence by expanding as follows:
x = $\sqrt{2+\sqrt{2-\sqrt{2+x}}}$
$2 - x = 2 -\sqrt{2+\sqrt{2-\sqrt{2+x}}}$
$\sqrt{2-x} = \sqrt{2-\sqrt{2+\sqrt{2-\sqrt{2+x}}}}$
then
$2+\sqrt{2-x} = 2 + \sqrt{2-\sqrt{2+\sqrt{2-\sqrt{2+x}}}}$
$\sqrt{2+\sqrt{2-x}} = \sqrt{2 + \sqrt{2-\sqrt{2+\sqrt{2-\sqrt{2+x}}}}}$
but I couldn't identify any, so I tried some substitutions and no solution yet :p
How should I do this?
Note: no calculators allowed

2. Jul 26, 2014

### phinds

Just out of curiosity, what does Serbia have to do with this?

3. Jul 26, 2014

### Dinheiro

It is a Serbian olympic question. I thought indicating its origin would help finding a solution though

4. Jul 26, 2014

### Ray Vickson

You can analyze some general properties of the solution, but getting the solution itself is challenging. Let $f(x) = \sqrt{2+\sqrt{2-\sqrt{2+x}}}$, so the equation is $x = f(x)$. In order for all square roots defining f(x) to be real, we need $2 \geq \sqrt{x+2}$, hence $x \leq 2$. Also, $f(x) \geq \sqrt{2}$ (because $f(x) = \sqrt{2 + \text{something} \geq 0}$) , so $x \geq \sqrt{2}$. Note that $f(\sqrt{2}) > \sqrt{2}$ and $f(2) = \sqrt{2} < 2$ so there IS a root between $\sqrt{2}$ and $2$.

Getting an exact root can be done, but I don't see how without using a computer algebra package. Squaring DOES work, but it introduces 7 extraneous roots that are not part of the original problem.
$$x = f(x) \: \Longrightarrow x^2-2 = \sqrt{2 - \sqrt{2+x}} \: \Longrightarrow (x^2-2)^2 = 2 - \sqrt{2+x} \: \longrightarrow x+2 = (2 - (x^2-2)^2)^2$$
This is a polynomial equation of degree 8. Packages like Maple or Mathematica can find all the roots, but they are complicated. Only one of the roots lies in the region $[\sqrt{2},2]$.

Last edited: Jul 26, 2014
5. Jul 26, 2014

### Infinitum

I have been at this problem for about half an hour now, and couldn't deduce much beyond what you said, Ray. But I am curious to know how you deduced there is only one root in the region $[\sqrt 2,2]$. I think there is at least one other real root(assuming the others are complex), which may be in that range, right?

6. Jul 27, 2014

### SammyS

Staff Emeritus
You can determine that any real root must be in that interval by considering the following.

You have $\ \sqrt{2+x} \$ and $\ \sqrt{2-\sqrt{2+x}\,}\,. \$ Right?

7. Jul 27, 2014

### Dinheiro

Dude, I have spent like 2 hours in it and I am really considering to explode this book now. The worst part is this question is at the easy exercises page of the chapter

Last edited: Jul 27, 2014
8. Jul 27, 2014

### Dinheiro

#%$&(@&*$, I got it! just take x = 2cos(y)

Last edited: Jul 27, 2014
9. Jul 27, 2014

### Dinheiro

What a relief! But this exercise is at factorizations chapter, I wonder if there is a solution by factoring this monster

10. Jul 27, 2014

### Mentallic

Would you mind sharing your solution? I'm not seeing it.

11. Jul 27, 2014

### Dinheiro

Ok
$x = \sqrt{2+\sqrt{2-\sqrt{2+2cosy}}}$
note that
2 + 2cosy = 2(1 + cosy)
and cosy = 2cos²(y/2) -1
so 2 + 2cosy = 4cos²(y/2)
Now, we have
$x = \sqrt{2+\sqrt{2-2cos(y/2)}}$
analogously 2 - 2 cos(y/2) = 4sin(y/4)
$x = \sqrt{2+2sin(y/4)}$
Finally, x = 2cos(y/8) <=> 2cos(y) = 2 cos(y/8)
y = 2pi/9
x = 2cos(2pi/9)

xD

12. Jul 27, 2014

### Ray Vickson

By finding all the roots and evaluating them numerically. Alternatively, plot f(x) and x in the range [√2,2]. All eight roots of the polynomial are real, but 7 of them are extraneous to the original problem. All work done in Maple.

13. Jul 27, 2014

### Infinitum

My point exactly. Which is why I was confused when Ray said only one real root lies in that interval, I believe there at least need to be two.

I tried that earlier in my attempt. The substitution does yield a 'neat' looking equation, but I don't see how to reach a numerical answer for x without the use of calculators/tables. Could you please share how?

14. Jul 27, 2014

### Ray Vickson

The function $f(x)$ is strictly decreasing on the interval $[\sqrt{2},2]$ (from the derivative), while the function $x$ is strictly increasing on that interval. There is only one root.

15. Jul 27, 2014

### Infinitum

That makes sense, thanks Ray.

As a side note, I should start using one of these softwares soon :tongue2:

The expressions of x in the two lines aren't equal. That, or I'm doing some trigonometric errors that are embarrassingly silly.

16. Jul 27, 2014

### ehild

After repeatedly squaring, you can simplify out a first-order factor and get an odd -number order polynomial equal to zero.

ehild

17. Jul 27, 2014

### Mentallic

At the step

$$2\cos y = \sqrt{2+2\sin\frac{y}{4}}$$

squaring and then using the double angle identity

$$\cos^2\alpha = \frac{1}{2}\left(1+\cos{2\alpha}\right)$$

yields

$$\cos{2y} = \sin{\frac{y}{4}}$$

And I can't proceed.

18. Jul 27, 2014

### ehild

$$\cos{2y} = \sin{(\pi/2\pm2y +2k\pi)}=\sin{\frac{y}{4}}$$

ehild

19. Jul 27, 2014

Of course!