Find the exact value of the solution to the equation √ 2+(√ 2-(√ 2+√( 2-x))))=x.

[anemone]

Find the exact value of the solution to the equation √ 2+(√ 2-(√ 2+√( 2-x))))=x.

Find the exact value of the solution to the equation
$ \sqrt {2 +\sqrt {2-\sqrt{2+\sqrt{ 2-x}}}}=x $

I wanted so much to share with you all this question because it's an easy yet the tricky one, imho.

 

[chisigma]

Re: Find the exact value of the solution to the equation √ 2+(√ 2-(√ 2+√( 2-x))))=x.

Find the exact value of the solution to the equation
$ \sqrt {2 +\sqrt {2-\sqrt{2+\sqrt{ 2-x}}}}=x $

I wanted so much to share with you all this question because it's an easy yet the tricky one, imho.

If You consider the first order difference equation...

$\displaystyle a_{n+1}= \sqrt{2-(-1)^{n}\ a_{n}}\ ;\ a_{0}=x$ (1)

... Your problem is the search of solutions of (1) with periodicity 4.In order to symplify the procedure You can start searching solutions of periodicity 2 that, of course, are also solutions with periodicity 4. The solutions with periodicity 2 are solutions of the equation...

$\displaystyle \sqrt{2+\sqrt{2-x}}=x$ (2)

... i.e. a non negative solution of the fourth order equation...

$x^{4} -4\ x^{2} + x + 2= (x-1)\ (x+2)\ (x^{2}-x-1)=0$ (3)

... that are $\displaystyle x_{1}=1$ and $\displaystyle x_{2}=\frac{1 + \sqrt{5}}{2}$. A simple test shows that $x_{2}$ is 'good solution'. Other solutions with periodicity 4 are to be found...

Kind regards

$\chi$ $\sigma$

 

[sbhatnagar]

Re: Find the exact value of the solution to the equation √ 2+(√ 2-(√ 2+√( 2-x))))=x.

$$ \sqrt {2 +\sqrt {2-\sqrt{2+\sqrt{ 2-x}}}}=x $$

Write this as:

$$\sqrt{2 +\sqrt {2-\sqrt{2+\sqrt{ 2-\sqrt {2 +\sqrt {2-\sqrt{2+\sqrt{ 2-x}}}}}}}}=x $$

Again write this as:

$$\sqrt{2 +\sqrt {2-\sqrt{2+\sqrt{ 2-\sqrt {2 +\sqrt {2-\sqrt{2+\sqrt{ 2-\sqrt {2 +\sqrt {2-\sqrt{2+\sqrt{ 2-x}}}}}}}}}}}}=x $$

Repeat this procedure till infinity.

$$\sqrt{2 +\sqrt {2-\sqrt{2+\sqrt{ 2-\sqrt {2 +\sqrt {2-\sqrt{2+\sqrt{ 2-\sqrt {2 +\sqrt {2-\sqrt{2+\sqrt{ 2-\cdots}}}}}}}}}}}}=x $$

This means

$$\sqrt{2+\sqrt{2-x}}=x$$

Take squares of both sides

$$2+\sqrt{2-x}=x^2 \quad \quad \cdots (1)$$

Multiply both sides by $2-\sqrt{2-x}$:

$$2-\sqrt{2-x}=\frac{2+x}{x^2} \quad \quad \cdots (2)$$

Add (1) and (2):

$$4=x^2+ \frac{2+x}{x^2}$$

$$x^4-4x^2+x+2=0$$

Factorize:

$$(x-1)(x-2)(2x-\sqrt{5}-1)(2x+\sqrt{5}-1)=0$$

This gives:

$$x=\frac{1}{2}(\sqrt{5}+1)$$

which is the desired solution.

 

[anemone]

Re: Find the exact value of the solution to the equation √ 2+(√ 2-(√ 2+√( 2-x))))=x.

If You consider the first order difference equation...

$\displaystyle a_{n+1}= \sqrt{2-(-1)^{n}\ a_{n}}\ ;\ a_{0}=x$ (1)

... Your problem is the search of solutions of (1) with periodicity 4.In order to symplify the procedure You can start searching solutions of periodicity 2 that, of course, are also solutions with periodicity 4. The solutions with periodicity 2 are solutions of the equation...

$\displaystyle \sqrt{2+\sqrt{2-x}}=x$ (2)

... i.e. a non negative solution of the fourth order equation...

$x^{4} -4\ x^{2} + x + 2= (x-1)\ (x+2)\ (x^{2}-x-1)=0$ (3)

... that are $\displaystyle x_{1}=1$ and $\displaystyle x_{2}=\frac{1 + \sqrt{5}}{2}$. A simple test shows that $x_{2}$ is 'good solution'. Other solutions with periodicity 4 are to be found...

Kind regards

$\chi$ $\sigma$

Hi chisigma, you're always capable of solving all kinds of maths-related questions using some sophisticated approachs, huh?
You're awesome!

$$ \sqrt {2 +\sqrt {2-\sqrt{2+\sqrt{ 2-x}}}}=x $$

Write this as:

$$\sqrt{2 +\sqrt {2-\sqrt{2+\sqrt{ 2-\sqrt {2 +\sqrt {2-\sqrt{2+\sqrt{ 2-x}}}}}}}}=x $$

Again write this as:

$$\sqrt{2 +\sqrt {2-\sqrt{2+\sqrt{ 2-\sqrt {2 +\sqrt {2-\sqrt{2+\sqrt{ 2-\sqrt {2 +\sqrt {2-\sqrt{2+\sqrt{ 2-x}}}}}}}}}}}}=x $$

Repeat this procedure till infinity.

$$\sqrt{2 +\sqrt {2-\sqrt{2+\sqrt{ 2-\sqrt {2 +\sqrt {2-\sqrt{2+\sqrt{ 2-\sqrt {2 +\sqrt {2-\sqrt{2+\sqrt{ 2-\cdots}}}}}}}}}}}}=x $$

This means

$$\sqrt{2+\sqrt{2-x}}=x$$

I see it this way:

Since $ f^{-1} (f(x))=x $, and we're given $ \sqrt {2 +\sqrt {2-\sqrt{2+\sqrt{ 2-x}}}}=x $,

then if I solve the following for x,
$\sqrt{2+\sqrt{ 2-x}}=x$
that means I solve the original equation for x too.:)
(By using your way or substitution method.)

 

[CellCoree]

I would approach this equation by first simplifying the nested square roots. We can start by simplifying the innermost square root, which is √(2-x). This can be rewritten as √2 - √x. Substituting this back into the original equation, we get:

√2 + (√2 - (√2 + √(√2 - √x))) = x

Next, we can simplify the nested square root again, which is √(√2 - √x). This can be rewritten as √√2 - √√x, or √√2 - x^(1/4). Substituting this back into the equation, we get:

√2 + (√2 - (√2 + √(√√2 - x^(1/4)))) = x

Now, we can simplify the nested square root one last time, which is √(√√2 - x^(1/4)). This can be rewritten as √√√2 - x^(1/8), or √√√2 - x^(1/8). Substituting this back into the equation, we get:

√2 + (√2 - (√2 + √√√2 - x^(1/8))) = x

At this point, we can see that the equation is becoming more and more complex, and it may be difficult to solve for x algebraically. However, we can use numerical methods, such as iteration or the Newton-Raphson method, to approximate the solution. This would involve repeatedly plugging in a value for x and solving for the resulting value until we get a solution that is close enough to the exact value.

In conclusion, the exact value of the solution to this equation may be difficult to find algebraically, but numerical methods can be used to approximate the solution.