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Infinite Square Roots

If you square root a number infinite amount of times, will it eventually become 1, or some sort of whole number? Can you please give proof as to why you got your answer? Thanks!

Note by Jonathan Hsu
2 years, 2 months ago

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Not really a proof, but if we look at \(\sqrt{x}\), we know the result will be less than \(x\) (unless of course \(x\) is less than or equal to 1.

When we square-root a number an infinite number of times, the number \(x\) must lower in value an infinite number of times, until 1 is reached.

Interestingly, when we square-root a number greater than -1 but less than 1 an infinite number of times, we get back to 1. When we square (not square-root) a similar number an infinite number of times, we get back to 0.

For negatives, squaring gets us infinitely high (or infinitely low), and square-rooting gets us imaginary numbers.

Clive Chen - 2 years, 2 months ago

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If the initially chosen number is \(N\geq1\) the sequence \(\{x_n\}\) defined as \[x_1=N, x_{n+1}=\sqrt{x_n}, \forall n\geq 1\] converges to \(1\). This follows from Banach's Fixed Point Theorem because the map \(f:[1,N] \to [1,N]\) defined by \(f(x)=\sqrt{x}\) is a contraction map, with the unique fixed point \(1\), in the metric space \([1,N]\), with the usual euclidean metric.

Abhishek Sinha - 2 years, 2 months ago

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It's some kind of elvish. I can't read it. :)

Vincent Miller Moral - 2 years, 2 months ago

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Let y = ... sqrt ( sqrt ( sqrt ( x

Square both sides:
y**2 = ... sqrt ( sqrt ( sqrt ( x 
OR
y**2 = y

y**2 - y = 0
y( y - 1 ) = 0
y = 0 or 1

y cannot be 0 since the input number (x) is >0
so y=1

Vincent Miller Moral - 2 years, 2 months ago

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You need to show that the sequence converges to \(1\). Finding a fixed point is not sufficient for convergence, e.g., instead of taking square roots, if we square the numbers, for any initially chosen number \(N>1\), the sequence diverges to \(\infty\). But you still have the equation \(y=y^2\), hence\(y=0\) or, \(y=1\).

Abhishek Sinha - 2 years, 2 months ago

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sqr(sqr(sqr(...sqr(n) = x , X^2= sqr(sqr(...sqr(n) , X^2=X , X= 1 or 0 . I think that would be it

Caio Garcez - 2 years, 2 months ago

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