Square root of the square root of the square root...

Relevant wiki: Nested Functions

An infinite number of square roots are applied on $x$ so we have $\cdots\sqrt{\sqrt{\sqrt{\cdots x}}}=\left(\left(\left(x^{1/2}\right)^{1/2}\right)^{1/2}\right)^{\cdots}=\lim_{n\to\infty} x^{1/2^n}=x^0=\fbox{1}.$

let a > 0 such that $\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{...x}}}}}}= a$

$\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{...x}}}}}}^2 = a^2$

$\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{...x}}}}} = a^2$

$a = a^2$

The equation, $a^2 - a = 0$ , has 0 and 1 as roots.

But since a > 0, therefore a = 1.

Basically, if you continuously square root any number, you tend to 1. Square root X = X^0.5 (square root of that = X^0.25, and so on....) X^0.5, X^0.25, X^0.125......tending towards X^0 which is 1

...√√√√√√...x can be written as (((((x^½ )^½ )^½ )^½ )^½ ...).

It can be simplified into:

(((((x^½ )^½ )^½ )^½ )^½ ...) = lim n -> ∞ (x^(1/(2^n))).

lim n -> ∞ (x^(1/(2^n))) = x^(1/∞) = x^0 = 1. Thus, 1 is the answer.

OR

Let ...√√√√√√...x = y

y² = ...√√√√√...x

But, ...√√√√√√...x = y

Thus, y²-y = 0, or y(y-1) = 0.

y = 0 or y = 1. But y = 0 cannot be the answer because,

lim n -> ∞ (x^(1/(2^n))) = 0 means that; x^0 = 0 or 1 = 0. IMPOSSIBLE!!!

Thus, 1 is the answer!

Easy trick: the problem asks for solution where x is ANY positive number. Just choose x = 1, the result is surely 1.

it's 1. Try it out on a calculator.

I found that this was able to be put into a simple formula

x=√(x)

(We substitute that function into "x" infinitely.)

Therefore, we would get

(X^2)=x

Which then can become a quadratic

(X^2)-x=0

We know that the roots of the quadratic are 1 and 0, but since it cannot be zero, it is One.

(This system can easily be used anywhere in any infinite summation, fraction, subtraction, or product. )

Let vvv.....x=S S²=S S=1

*vx:root x

Let us consider a positive number 'x' less than one first. Sq root of 'x' will be a number greater than itself but not equal to 1. But on repeating sq rooting the value will come closer to one and finally become one. After the value becomes 1 further sq rooting means nothing. Similarly let 'x' be a number greater than 1. Sq root of 'x' will be smaller than 'x' but not equal to 1. But repeated sq rooting will eventually provide us with 1.

If x is any positive real number, then just say that x is 1. Therefore, the value of the expression is 1.

Let X be any positive real number and taking root n times, i.e. (X)^(1/n), now as n tends to infinity , 1/n tends to 0 Hence (X)^0 =1.