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Solving an Equation with Infinite Exponents

Date: 03/15/2007 at 16:30:56
From: Brenda
Subject: Infinite x^x problem

If x^x^x^x^x^x^x^x^x^x^x...... = 2, solve for x.

I have tried whole numbers -3 through 3 and nothing gives me the 
result of 2.  I have heard that this problem is fairly easy, which
makes me wonder, is the answer that the problem is impossible?  

On a calculator, I usually end up with syntax error or some 
outrageously negative decimal number.  For example, if x = -3 I get

  -3^-3^-3^-3= -0.348259388...

I got similar numbers for -2.  -1 gives me the result of -1.  0 is a 
domain error.  1 gives me the result of 1, and 2 goes on and on and 
on.  Please help me with this, as I would like to know if it in fact 
IS impossible.




Date: 03/15/2007 at 22:07:09
From: Doctor Luis
Subject: Re: Infinite x^x problem

Hi Brenda,

Assuming that x^x^x^... is indeed some finite number, you can just
call it y for now, so 

  y = x^x^x^x^....

Naturally you can also write that as y = x^(x^x^...).  Hmm, that
exponent looks familiar, doesn't it?  From that you can conclude that 

  y = x^y (do you see why?).

Now the original equation becomes much easier to solve

  x^x^x^... = 2

becomes

  y = 2

Great!  We solved for y, but we're really interested in x instead. 

That's ok, because y = x^y becomes a much easier equation after
substituting the known value of y:

  2 = x^2

And from that equation, you can now easily get x, right?

I actually didn't believe it at first, so I tried on a calculator 
with precision to 11 decimals, and it took about 70 exponentiations
of x for the result to remain stable at 2.  You should verify this 
for yourself.

I hope this helped!

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




Date: 06/15/2007 at 18:51:46
From: Doctor Tom
Subject: Re: Infinite x^x problem

Hi Brenda,

Here's a good way to think about it.  We don't know what's meant by
the infinite tower of exponents, but we could look at towers of size
2, 3, 4, and so on.  Let:

  x(1) = x
  x(2) = x^x
  x(3) = x^(x^x) = x^x(2)
  x(4) = x^(x^(x^x)) = x^x(3)

In general, x(n) = x^(x(n-1))

As n gets larger and larger, either the numbers converge, or they
don't.  If x were 2, for example, we'd obtain:

  x(1) = 2
  x(2) = 4
  x(3) = 16
  x(4) = 65536
  ... and they would get huge very rapidly.

If they DO converge, say to a value X, then this has to be true:

  x^X = X.  (Do you see why?)

Now, we want X = 2, so we need to solve:

  x^2 = 2, so x = sqrt(2) = 1.4142135...

You can check this answer, at least approximately, on a calculator:

Let's let r = sqrt(2) = 1.4142135...

  r^r = 1.6325...
  r^r^r = 1.7608...
  r^r^r^r = 1.8409
  r^...^r (20 times) = 1.9995...

This isn't a proof of convergence, but it's very suggestive.  It's
clear that the exponents can't ever be bigger than 2 since you have 
to raise sqrt(2) to the second power to get to 2.
  
In the examples above, we always raise to powers less than 2, and
hence each successive value is also less than 2.  But each time it's
clear that the value gets bigger, so the series has to converge 
(since it's increasing AND bounded above).

It's possible to obtain bogus results by this method.  If the value
had been 4 instead of 2, we'd solve:

  x^4 = 4, 

and this also yields x = sqrt(2), which is clearly wrong.  In fact,
the method WILL solve for any value of the constant up to 
e = 2.718281828459045...  Anything larger will not have a root.

- Doctor Tom, The Math Forum
  http://mathforum.org/dr.math/ 
Associated Topics:
High School Calculus
High School Sequences, Series

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