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What's 8 plus five (in digits only)?

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File: TwoEqualsFour

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My lastest posts can be
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Previous blog posts:
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********
!!! TwoEqualsFour - 2011/04/14

[[[>40
I'm amazed that I need to say this, but it appears
that I do ...

I */know/* this is fallacious, and I */do/* know
what's wrong with it.  Please don't assume that
I really do think that 2=4, or that 1=2.

Thanks.

This page has been TaggedAsMaths
]]]
Here's a cool puzzle.

Consider the equation $2=x^{x^{x^{x^{\ldots}}}}$
and suppose we want to solve it for /x./

Because the exponential tower is infinite, we can also write it as
$2=x^{\left({x^{x^{x^{\ldots}}}}\right)}$

But the part in brackets is the same as the whole, and hence is
equal to 2. Thus we have $2=x^2$.

So $x=\sqrt{2}$ and we have
$\sqrt{2}^{\sqrt{2}^{\sqrt{2}^{\sqrt{2}^{\ldots}}}}=2$

Now consider the equation $4=x^{x^{x^{x^{\ldots}}}}$
and again let's solve for /x./

As before, we can write it as
$4=x^{\left({x^{x^{x^{\ldots}}}}\right)}$
and again, the part in brackets is the same as the whole,
and so now we get $4=x^4$.

Take the square root of each side and we get $2=x^2$ and so
again $x=\sqrt{2}$.

Thus $\sqrt{2}^{\sqrt{2}^{\sqrt{2}^{\sqrt{2}^{\ldots}}}}=4$.

So
$\sqrt{2}^{\sqrt{2}^{\sqrt{2}^{\sqrt{2}^{\ldots}}}}$
is 2, and it's also 4.

Hence 2=4 (and halving it means 1=2).

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!! Comments

In an email, Iain Murray has pointed to exercise 4.20 on page 86
of the book:

* Information Theory, Inference, and Learning Algorithms
    * by David J.C. MacKay FRS

A copy of the book can be found online here:
  * http://www.inference.phy.cam.ac.uk/mackay/itila/book.html

The exercise itself is in this PDF:
  * http://www.inference.phy.cam.ac.uk/mackay/itprnn/ps/65.86.pdf

and the solution/discussion is on page 89:
  * http://www.inference.phy.cam.ac.uk/mackay/itprnn/ps/87.89.pdf

********<