(Serbia) Just squaring won't work

[Dinheiro]

Homework Statement

Find the real solution $x = \sqrt{2+\sqrt{2-\sqrt{2+x}}}$

Homework Equations

Algebric equations

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

 

[phinds]

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

 

[Dinheiro]

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

 

[Ray Vickson]

Homework Statement

Find the real solution $x = \sqrt{2+\sqrt{2-\sqrt{2+x}}}$

Homework Equations

Algebric equations

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

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 $[\sqrt{2},2]$.

 

[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?

 

[SammyS]

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?

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?

 

[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

 

[Dinheiro]

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

 

[Dinheiro]

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

 

[Mentallic]

Motherf**, I got it! just take x = 2cos(y)

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

 

[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

 

[Ray Vickson]

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?

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.

 

[Infinitum]

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?

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.

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 $[\sqrt{2},2]$.

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

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?

 

[Ray Vickson]

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?

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.

 

[Infinitum]

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.

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.

That makes sense, thanks Ray.

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

$x = \sqrt{2+2sin(y/4)}$
Finally, x = 2cos(y/8)

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

 

[ehild]

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?

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

ehild

 

[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.

 

[ehild]

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.

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


ehild

 

[Mentallic]

Of course!