2018 Minuses

If there is a finite number of minuses in an infintite sequence of $+$ 's and $-$ 's, then after the last minus there is an embedded infinite nested radical consisting only of positive $x$ 's The value of this can be found by solving for $z$

$z = \sqrt{x+z}$

the resulting being $z = \dfrac{1}{2}\left(1+\sqrt{1+4x}\right)$

For the radical $\sqrt{x - z}$ to be real, $x \ge z$ , so that solving for $x$ here

$x = \dfrac{1}{2}\left(1+\sqrt{1+4x}\right)$

gets us $x \ge 2$ . Otherwise if $x<2$ , this radical becomes imaginary.

Suppose with the last minus in the sequence, the radical becomes imaginary. Thereafter, in going back outwards towards less embedded radicals, whenever a complex number is added or substracted from a real number ( $x$ in this case), the radical of the sum or difference is still imaginary. Thus, the entire infinite nested radical is imaginary, once any of its embedded radicals becomes imaginary. It may be possible that as a limit, some infinite nested radical beginning with an imaginary number in its most embedded radical "may approach a real number", but there is only a finite number of radicals from the last minus to the start.

If $x \ge 2$ , then the radical containing the last minus in the sequence has the value of $\ge 0$ . Since $x \ge \dfrac{1}{2}\left(1+\sqrt{1+4x}\right)$ for all $x \ge 2$ , regardless of the number of positive $x$ 's to the next minus, the radical containing that next minus will always be positive and yield a real value. Thus, all such infinite nested radicals are real for $x \ge 2$ .

In all this, we work from the most embedded radicals outward to the "largest" radical, the one that is not embedded at all.

Here plotted are some selected infinite nested radicals that have a small number of minuses to give one an idea. The plot that has real values for $0 \le x \le 2$ is the case where there are no minuses.

Note: If one tries to consider a case where there are "minuses at the end of of all pluses", that implies that there is an end to all pluses, when there is not. Hence, the most embedded radical which contains no minuses has the value $\dfrac{1}{2}\left(1+\sqrt{1+4x}\right)$ , and we then only need to consider a finite nested radical with all the finite number of minuses in it.

Given the expression $\sqrt{x \pm \sqrt{x \pm \sqrt{x \pm...}}}$ ...

The way I solved this is to imagine the worst-case scenario, which would be the case where the expression under the first radical...

$x \pm \sqrt{x \pm \sqrt{x \pm...}}$

...is negative for the most values of $x$ . This means we'd like to subtract the first term, $x$ , by the biggest number possible, because after the first term, the size of the number we can subtract by shrinks rapidly. Therefore, the first negative sign should be placed after the first term.

Our next goal is to maximize what follows the negative sign, the rest of the nested radical. If all of the signs could hypothetically be positive, which would yield a maximum value for the expression following $x$ , we would have $y=\sqrt{x+\sqrt{x+\sqrt{x+...}}}$ , where we can solve for y:

$\sqrt{x+\sqrt{x+\sqrt{x+...}}}=\sqrt{x+y}$

$y=\sqrt{x+y}$

$y^2-y-x=0$

$y=\frac{1}{2}(1+\sqrt{1+4x})$

The expression following the first term $x$ grows closer and closer to $y$ as we place the 2017 remaining negative signs, all consecutively, further and further down the nested radical, each place down maximizing the expression further, and therefore $y$ becoming a better and better approximation. Hence, the worst-case scenario is when the remaining negative signs are placed in the $n$ -through- $(n+2016)$ th positions as $n$ approaches infinity. This approaches $y$ , and thus letting $n$ approach infinity gives us the function representing the "worst-case scenario":

$f(x)=\sqrt{x-y}$

The smallest real value $f$ could give us would be 0, so solve for $x-y=0$ :

$x-\frac{1}{2}(1+\sqrt{1+4x})$

$2x-1=-\sqrt{1+4x}$

$4x^2-8x=0$

$x=0, x=2$

Since 0 isn't a positive value, 2 must be the answer.