Raising Sqrt(2)

Date: 10/14/97 at 18:49:15
From: Benno
Subject: Algebra

Dear Dr. Math,

What is the value of x in the following equation:

              ....
           x
        x
     x
  x
x                     = 2

Thank you.

Benno

Date: 10/14/97 at 19:15:09
From: Doctor Tom
Subject: Re: algebra

There's a trick to solving this problem instantly. Since the exponents 
go on forever, the exponent of the lowest and leftmost x is the same 
as x to that power, so replace the equation by

    2
   x  = 2

So x is the square root of 2, or 1.4142135...

You can see that it makes sense by repeatedly raising sqrt(2) to what 
you've got so far. To make typing simple, let's call sqrt(2) Q:

                 Q^Q = 1.63252691943
     Q^1.63252691943 = 1.76083955587
     Q^1.76083955587 = 1.84091086928
     Q^1.84091086928 = 1.89271269681

You can do this as long as you want, and the result will get closer 
and closer to 2, indicating that you've got the right answer.

A problem arises if you replace the "2" in the original equation
by 4 and try the same thing:

     x^4 = 4, so x + fourth-root(4) = sqrt(2).

That can't be right, since we just checked that the infinite tower
of sqrt(2)s converges to 2, not to 4.

It turns out you can use the trick only if the value on the right
of the equation is less than e = 2.718281828459045...  Above
that it gives bogus answers.

There are in fact some deep mathematical things going on here, and
to understand them you'll need to know a lot about convergence
of infinite series, among other things.

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