1. Dec 12, 2011

### latentcorpse

We can see that if

$u=\sqrt{x+\sqrt{x+\sqrt{x+\dots}}}$

then $u^2=x+u$

so $u^2-u-x=0$

This has solution

$\left( u-\frac{1}{2} \right)^2 -\frac{1}{4}-x=0 \Rightarrow u=\frac{1}{2} \pm \sqrt{x + \frac{1}{4}}$

This means that $u \in \mathbb{R} \forall x \geq \frac{1}{4}$

In other words $\sqrt{ -\frac{1}{8} + \sqrt{ - \frac{1}{8} + \sqrt{-\frac{1}{8} + \dots}}}$ is real.

This is clearly true according to the above formula. However, I cannot get my head around it - to me it seems like it must be imaginary! Can anyone give an explanation of why this is turning out to be real?

2. Dec 12, 2011

### ehild

Writing in the opposite way, it will not look imaginary:

$$\sqrt {\dots \sqrt{ -\frac{1}{8} + \sqrt{ - \frac{1}{8} + \sqrt{-\frac{1}{8} + v}}}}$$

with v=1/4 for example.

Edit: I changed the last term to v>0 instead of 1/4,
ehild

Last edited: Dec 12, 2011
3. Dec 12, 2011

### latentcorpse

i dont see where you got the 1/4 from?

4. Dec 12, 2011

### ehild

Sorry, it was an example to show that the -1/8 terms do not lead necessarily to imaginary result.

ehild

Last edited: Dec 12, 2011
5. Dec 12, 2011

### latentcorpse

ok. but surely the original expression will have a negative under the first root giving an imaginary result and then the second root will be the root of an imaginary number giving a complex result. All subsequent roots will involve roots of complex numbers and will give complex results won't they?

6. Dec 12, 2011

### Office_Shredder

Staff Emeritus
The first thing you do is take the square root o
$$-1/8+\sqrt{-1/8+\sqrt{-1/8+....}}$$

As long as the set of nested square roots is bigger than -1/8 then we're OK. But how can we check if this nested set of square roots is OK? Well, the first thing we do is take the square root of
$$-1/8+\sqrt{-1/8+\sqrt{-1/8+....}}$$

So it's not at all obvious whether these things are positive or negative.

7. Dec 12, 2011

### SammyS

Staff Emeritus
Shouldn't that say: $u \in \mathbb{R} \forall x \geq -\frac{1}{4}$ rather than $x \geq \frac{1}{4}\ ?$

8. Dec 13, 2011

### ehild

When you write an expression you need to say if you mean it for real or for complex numbers.
Defining the expression for complex x with the use of the complex square root, the expression does not have a unique value for any x. It can corresponds to a set of complex numbers.

In case we require that x and u are real, and the expression has a defined value, there are some constraints. The square of a real number is non-negative real, the square-root function is defined only for non-negative numbers, and the square root of a number is non-negative.
The condition that the value of the nested radicals exist and real, implies that u=√(x+u). From this, it follows that u and x must satisfy the equation u2=x+u, or
(u-1/2)2=x+1/4, that is x≥-1/4, and u=1/2±√(x+1/4).
For x>0 only the + sign is valid as u ≥0. But there can be two values for u if -1/4<x<0. So the expression of the nestled radicals is not single-valued for x<0. The value of the nestled radicals does not exist.

So the statement that the value of the nestled radicals must be complex for a certain x, say for x=-1/8 does not hold, but we can say that it is undefined. But in case x>0 the expression returns a single real number u=1/2+√(x+1/4)

You can consider the nestled radicals as a sequence given by an+1=√(an+x), for x≥-1/4 and a0 given. If a0=u=1/2±√(x+1/4) all elements of the sequence are equal.

ehild