Infinite square root

Date: 6/4/96 at 13:29:14
From: Anonymous
Subject: Infinite square root

I know this is an easy question, but I do not know how to figure it 
out.

For instance, if y = sqrt(2+ sqrt(2+ sqrt(2+ sqrt(2+ ..., y = 2, and
this happens always, because if k = e+ sqrt(e+ sqrt(e+ ..., and it 
goes indefinitely, k = e. However, how can I prove that this is true, 
using normal properties of roots?

Date: 6/4/96 at 16:14:46
From: Doctor Darrin
Subject: Re: Infinite square root

If we think about the properties that y has to have, we can find its 
value.

What do we get if we square y?  We get 2+sqrt(2+sqrt(2+..., so we see 
that y^2-2=sqrt(2+sqrt(2+sqrt(2+... .  Thus, y^2-2=y, so y is a root 
of the quadratic equation y^2-y-2==0.  You can work out what the roots 
of this equation are, and this gives two possible values of y (one is 
positive and one is negative).  Since y is defined as a square root, 
it makes sense to take the positive value.  

I am not sure what you mean when you say that "this happens always".  
In the definition you give for y, if we replace 2 by another number 
(say n), we can use essentially the same procedure to find what y is, 
but in general, y does not equal n.  When n=3, y=2.302..., (the 
positive root of y^2-y-3=0) and when n=4, y=2.56...  The definition 
you give for k is different, and in fact k does not equal e in 
general.  For e=3, k=5.302... In fact, there is no value of e (except 
e=0) for which k=e.  

Thanks for asking the question, it is very interesting

-Doctor Darrin,  The Math Forum
 Check out our web site!  http://mathforum.org/dr.math/