What is the result of $\sqrt{-\frac{1}{4}+\sqrt{-\frac{1}{4}+\sqrt{-\frac{1}{4}+\sqrt{-\frac{1}{4}+...}}}}$

Given the golden ratio is equal to $\sqrt{1+\sqrt{1+\sqrt{1+\sqrt{1+...}}}}$ you get $f(x) = \sqrt{1+f(x)}$.

Solving for $f(x)$:

$f(x)^2 = x + f(x)$

$f(x)^2 - f(x) = x$

$f(x)^2 - f(x) + \frac{1}{4} = x + \frac{1}{4}$

$(f(x) - \frac{1}{2})^2 = x + \frac{1}{4}$

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

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

Checking values for $f(x)$:

I've checked all these between using the equation and the nested square roots and they all check out, but I have a hard time believing the nested radical $\sqrt{-\frac{1}{4}+\sqrt{-\frac{1}{4}+\sqrt{-\frac{1}{4}+\sqrt{-\frac{1}{4}+...}}}}$ would equal $\frac{1}{2}$

The trick is to see if there exists a sequence that would work. The trouble is that many times, recursive sequences can be sensitive to their starting seeds (exhibit chaotic behavior), thus these sequences can not be said to have well defined limits, only regularized values useful for specific applications. In this case though, there is an insensitivity to starting seed provided $x_0$ that $x_0\geq \frac{1}{2}$. Consider the sequence

$$x_{n+1} = \sqrt{x_n - \frac{1}{4}} \hspace{20 pt} x_0 \geq \frac{1}{2}$$

Heuristically, one can see from this graph of the "discrete derivative" $f(x_n)=x_{n+1}-x_n$ that $x=\frac{1}{2}$ is an unstable fixed point:

Points less than $\frac{1}{2}$ move away from $\frac{1}{2}$ whereas points greater than $\frac{1}{2}$ move closer toward it (that is the nature of this discrete derivative, it is nonpositive everywhere). All we have to prove now is that in iterating this dynamical system, our sequence never jumps to the left of our limit point. In this case,

$$x_n \geq \frac{1}{2} \implies x_n-\frac{1}{4} \geq \frac{1}{4} \implies x_{n+1} = \sqrt{x-\frac{1}{4}} \geq \frac{1}{2}$$

So we are in the clear and we have proven sufficient insensitivity to starting seed to say that this one-sided limit of the function you found does exist. Interestingly enough though, since the discrete derivative has a zero of multiplicity greater than $1$, the convergence of the sequence will be very slow ($\sim \frac{1}{n}$) as opposed to exponential convergence for a root with multiplicity $1$