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Square Roots and Limits

Date: 03/12/2003 at 21:39:19
From: Jonathan Coveney
Subject: Square Roots

Given a value of x, find the value of the expression sqrt(x sqrt(x
sqrt(x... and so on, and explain why.


Date: 03/13/2003 at 03:41:32
From: Doctor Ian
Subject: Re: Square Roots

Hi Jonathan,

The easiest way to deal with this is to make use of the fact that

  sqrt(x) 

and 

  x^(1/2)

are just two ways of saying the same thing. If that doesn't seem
familiar to you, take a look at 

   Properties of Exponents
   http://mathforum.org/library/drmath/view/57293.html 

So, 

    _______            
   |     __          1/2 1/2     
  \| x \| x  = (x * x   )         

                 3/2 1/2
             = (x   )

                (3/2 * 1/2)
             = x 

                3/4
             = x 
   
and 
    ____________   
   |     _______   
   |    |     __         3/4 1/2
  \| x \| x \| x = (x * x   )

                     7/4 1/2
                 = (x   )

                    (7/4 * 1/2)
                 = x   

                    7/8
                 = x   

At each step, the exponent changes from 

  (k-1)/k 

to 

  ((k + (k-1))/k)/2 = (2k - 1)/(2k)

More generally, for n square roots, the exponent is 

  (2^n - 1)/(2^n)
  
So you're right that as n approaches infinity, the exponent approaches
1, and the value of the expression approaches x.  

However, note that the denominator of the exponent will always be a
power of 2, so x^(19/20) isn't really part of the sequence.  

Now for a bit of terminology. The way that we express this situation
is to say

  Limit    (2^n - 1)/(2^n)  = 1
    n->inf 

  "The limit, as n goes to infinity, of [the expression] is
   equal to 1."

This is somewhat subtle.  Note that we _don't_ say that the expression
itself, 

  (2^n - 1)/(2^n)

is ever actually equal to 1, because that's never the case. And we
_don't_ say that n ever _becomes_ 'infinity', because infinity isn't a
number, so we can't use it in arithmetical expressions.  

We're saying that we can make the value of the expression as close to
1 as we wish, by choosing a suitably high value for n.  

You'll learn more about limits when you get into calculus.  

Does this help? 

- Doctor Ian, The Math Forum
  http://mathforum.org/dr.math/ 


Date: 03/13/2003 at 08:29:19
From: Jonathan Coveney
Subject: Thank you (Square Roots)

Thanks, I wondered what limits were, exactly. But now it makes plain 
sense.
Associated Topics:
High School Calculus
High School Polynomials
High School Square & Cube Roots

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