Infinite Square Roots!

Let

$x = \sqrt {1 - \sqrt {\dfrac {17}{16} - \sqrt {1 - \sqrt {\dfrac {17}{16} - \sqrt {1 - \ldots}}}}}$

Then

$x = \sqrt {1 - \sqrt {\dfrac {17}{16} - x}}$ .

Simplifying, we get

$x^2 = 1 - \sqrt {\dfrac {17}{16} - x}$

or

$x^4 - 2x^2 + 1 = \dfrac {17}{16} - x$ .

Simplifying further, we yield

$x^4 - 2x^2 + x - \dfrac {1}{16} = 0$ .

Factorising this, we get

$(2x - 1)(8x^3 + 4x^2 - 14x + 1) = 0$ .

Therefore, either $x = \frac {1}{2}$ or $x$ is a root of $8x^3 + 4x^2 - 14x + 1 = 0$ . Solving this cubic equation, we get the roots $x \approx -1.62, 0.073$ and $1.05$ . Since the answer to the problem has to be positive and less than one, the first and third roots are ruled out.

The only real roots left are $\frac {1}{2}$ and $0.073$ . A double-check shows that neither was introduced in the squaring steps above. Both are correct solutions to

$x = \sqrt {1 - \sqrt {\dfrac {17}{16} - x}}$ .

But $x$ clearly has one definite value, so which is it? A few iterations on a calculator (Sorry, the only way I thought I could do easily), starting with a "1" under the innermost radical, shows that $x = \frac {1}{2}$ is definitely the answer. Why?

The answer lies in the behaviour of $f(x) = \sqrt {1 - \sqrt {\frac {17}{16} - x}}$ near the points $x = \frac {1}{2}$ and $x = \alpha \equiv 0.073 \ldots$ . The point $x = \frac {1}{2}$ is stable, in the sense that if we start with an $x$ value slightly different from $\frac {1}{2}$ , then $f(x)$ will be closer to $\frac {1}{2}$ than $x$ was. The point $x = \alpha$ , on the other hand is unstable, in the sense that if we start with an $x$ value slightly different from $\alpha$ , then $f(x)$ will be farther from $\alpha$ than $x$ was.

Said in another way, the slope of $f(x)$ at $x = \frac {1}{2}$ is less than 1 in absolute value (it is in fact $\frac {2}{3}$ ), while the slope at $x = \alpha$ is greater than 1 in absolute value (it is approximately 3.4). What this means is that points tend to head toward $\frac {1}{2}$ , but away from $\alpha$ , under iteration by $f(x)$ . In particular, if we start with the value 1, as we are supposed to do in this problem, then we will eventually get arbitrarily close to $\frac {1}{2}$ after many applications of $f$ . Therefore, $\frac {1}{2}$ is the correct answer.

There is more but I need to know how to upload pictures on solutions. This part makes it easier to understand the reasoning. I think this was my most hardcore solution I have made.

Let $x$ be equal to the expression.

We see that $x=\sqrt{1-\sqrt{\dfrac{17}{16}-x}}$ .

Simplifying: $x^2=1-\sqrt{\dfrac{17}{16}-x}$

Simplifying more: $x^4+2x^2+1=\dfrac{17}{16}-x$

Even more: $x^2+2x^2+x-\dfrac{1}{16}=0$

Using RRT: $x=\dfrac{1}{2}$ is a root.

Thus, we have that the answer is $\boxed{0.5}$ .

$x$ is the expression.

So therefore

$x^2=1-\sqrt{\frac{17}{16}-x}$ $\longrightarrow$

$16x^4-1+16x-32x^2=0$

Factor and grouping:

$(4x^2+1)(2x+1)(2x-1)+16x(1-2x)=0$ $\longrightarrow$

$(2x-1)((4x^2+1)(2x+1)-16x)=0$

One root is $\frac{1}{2}$ and the cubic I solved on calculator.

Now to choose the correct root is the last part. By close examination: the expression has to be greater than $0$ so that means that $1-\sqrt{\frac{17}{16}-x}>0$ which limits the values to $\frac{1}{16}<x<\frac{17}{16}$

This solution is not complete

So many square roots. We'll find the domain first and then proceed with the equation.

Let the given expression equal $x$ . Then we have

$\displaystyle\,x=\sqrt{1-\sqrt{\frac{17}{16}-x}}---(i)$ .

Keeping in mind that square root function is true for positive values of arguments, we have

$\displaystyle\frac{17}{16}-x>0$ and $\displaystyle\,1-\sqrt{\frac{17}{16}-x}>0$ . Both of them bound value of $x$ as

$\displaystyle\frac{1}{16}<x<\frac{17}{16}---(ii)$

Now taking $(i)$ we have

$\displaystyle\,1-x^2=\sqrt{\frac{17}{16}-x}$

Squaring both sides

$\displaystyle\,x^4-2x^2+x-\frac{1}{16}=0$

Now using Rational Root Theorem we factor the above biquadratic as

$\displaystyle\left(x-\frac{1}{2}\right)\left(x^3+\frac{x^2}{2}-\frac{7x}{4}+\frac{1}{8}\right)=0$ .

We see that the second polynomial doesn't have a positive root( $\textit{I couldn't help myself from using W|A for this}$ ), so we conclude that $x=\frac{1}{2}=\boxed{0.5}$ is a root which indeed satisfies $(i)$ and $(ii)$ .