What is the value of x in $2 = x + \sqrt{x + \sqrt{x + \dots}} $?

Question

If $x + \sqrt{x + \sqrt{x + \cdots}} = 2$, what is the value of $x$?

Answer 1

The only reasonable interpretation of $$ x+\sqrt{x+\sqrt{x+\cdots}} $$ is as the limit of a sequence $(y_{n})_{n}$ satisfying $$ y_{n+1}=x+\sqrt{y_{n}}. $$ If this sequence has a nonnegative limit $y$, then taking limits of both sides and using the fact that limits and continuous functions (namely, $\sqrt{\cdot}$) commute, we get $y=x+\sqrt{y}$. Equivalently, $$ y-\sqrt{y}-x=0. $$ Now, let $w=\sqrt{y}$ so that $$ w^{2}-w-x=0. $$ This is a quadratic equation in $w$ and we can solve it to get $$ w=\frac{1\pm\sqrt{1+4x}}{2}. $$ In your case, $w=\sqrt{2}$. However, to keep matters general, let's proceed without this assumption. Moving some terms around, $$ 2w-1=\pm\sqrt{1+4x}. $$ Squaring both sides, $$ x=\frac{\left(2w-1\right)^{2}-1}{4}. $$

Returning to the case of interest, we can plug in $w=\sqrt{2}$ to the above to get $$\boxed{x=2-\sqrt{2} \approx 0.58579.}$$ We can verify that this is indeed the correct answer by looking at the first "few" iterations of the sequence with $y_0 = x$, which can be seen converging to $2$:

y_0 = 0.585786437627
y_1 = 1.35115330236
y_2 = 1.74817763785
y_3 = 1.9079731256
y_4 = 1.96708044167
y_5 = 1.98831288717
y_6 = 1.99586192751
y_7 = 1.99853621289
y_8 = 1.99948237838
y_9 = 1.99981698128

Answer 2

In the wiki page for nested radicals you will find the formula (for $n>0$) $$ \sqrt{n+\sqrt{n+\sqrt{n+\ldots}}} = \frac{1+\sqrt{1+4n}}{2} $$ This implies $$ \frac{1+\sqrt{1+4x}}{2} = 2 - x, $$ showing $x = 2 - \sqrt{2}$.

This formula can be shown like this: Let $\phi := \sqrt{n+\sqrt{n+\sqrt{n+\ldots}}}$. Then $\phi^2 = n + \phi$. This is a quadratic in $\phi$ which one can easily solve with the quadratic formula. You have to choose the positive solution since roots of positive numbers are positive.