Nested Radicals: Why is u in \mathbb{R}?

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In summary: 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/
  • #1
latentcorpse
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We can see that if

[itex]u=\sqrt{x+\sqrt{x+\sqrt{x+\dots}}}[/itex]

then [itex]u^2=x+u[/itex]

so [itex]u^2-u-x=0[/itex]

This has solution

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

This means that [itex]u \in \mathbb{R} \forall x \geq \frac{1}{4}[/itex]

In other words [itex]\sqrt{ -\frac{1}{8} + \sqrt{ - \frac{1}{8} + \sqrt{-\frac{1}{8} + \dots}}}[/itex] 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?
 
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  • #2
Writing in the opposite way, it will not look imaginary:

[tex]\sqrt {\dots \sqrt{ -\frac{1}{8} + \sqrt{ - \frac{1}{8} + \sqrt{-\frac{1}{8} + v}}}}[/tex]

with v=1/4 for example.

Edit: I changed the last term to v>0 instead of 1/4,
ehild
 
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  • #3
ehild said:
Writing in the opposite way, it will not look imaginary:

[tex]\sqrt {\dots \sqrt{ -\frac{1}{8} + \sqrt{ - \frac{1}{8} + \sqrt{-\frac{1}{8} + \frac{1}{4}}}}}[/tex]

ehild

i don't see where you got the 1/4 from?
 
  • #4
latentcorpse said:
i don't see where you got the 1/4 from?

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

ehild
 
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  • #5
ehild said:
Sorry, it was an example to show that the -1/8 terms do not lead necessarily to imaginary result.

ehild

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
latentcorpse said:
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?

The first thing you do is take the square root o
[tex] -1/8+\sqrt{-1/8+\sqrt{-1/8+...}}[/tex]

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
[tex]-1/8+\sqrt{-1/8+\sqrt{-1/8+...}}[/tex]

So it's not at all obvious whether these things are positive or negative.
 
  • #7
latentcorpse said:
We can see that if

[itex]u=\sqrt{x+\sqrt{x+\sqrt{x+\dots}}}[/itex]

then [itex]u^2=x+u[/itex]

so [itex]u^2-u-x=0[/itex]

This has solution

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

This means that [itex]u \in \mathbb{R} \forall x \geq \frac{1}{4}[/itex]
Shouldn't that say: [itex]u \in \mathbb{R} \forall x \geq -\frac{1}{4}[/itex] rather than [itex]x \geq \frac{1}{4}\ ?[/itex]
 
  • #8
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
 

FAQ: Nested Radicals: Why is u in \mathbb{R}?

1. What is a nested radical?

A nested radical is an expression in mathematics that involves taking the square root (or any other root) of a number multiple times. For example, √2 + √(3 + √4) is a nested radical.

2. Why is it called "nested"?

It is called "nested" because the radical expressions are contained within each other, like a set of nesting dolls. In the example above, the √4 is nested within the √(3 + √4) expression.

3. What is the significance of u being in &mathbb;R?

The notation &mathbb;R represents the set of real numbers, which includes all rational and irrational numbers. In the context of nested radicals, it means that the expression is defined for all real numbers, and not just a specific subset.

4. Why is it important to consider the domain of u in nested radicals?

It is important to consider the domain of u in nested radicals because some values of u may result in an invalid or undefined expression. For example, if u is negative, then the expression √u will result in a complex number, which is not allowed in a nested radical.

5. How are nested radicals used in mathematics?

Nested radicals have various applications in mathematics, such as in solving equations, approximating irrational numbers, and in the proofs of certain theorems. They also have connections to other areas of mathematics, such as continued fractions and fractals.

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