Re: [math-fun] Iterated square root puzzle

No worries :)  The nested radicals are hard to see in a pure text format. 
But I combined a couple of things in that step without explanation, so
I'll try to break it down more clearly.

First, I multiplied both sides by 1/sqrt(2).  Then, on the right hand
side, I recursively propagated this value into the square roots, squaring
it each time.  So the factor within the first square root was 1/2, within
the second it was 1/4, then 1/16, and so on.  These happened to precisely
match the powers of 1/2 that formed the first term within each radical, so
the factor could be folded into those terms by squaring them, which
simplified to adding one to the exponents of the inner powers of two:

    1/2^(2^a) * 1/2^(2^a)

    1/2^(2^a + 2^a)

    1/2^(2*(2^a))

    1/2^(2^(a+1))

This allowed the original expression to be rewritten in terms of itself,
which was then easily solved.

Tom


On 8/14/10, Fred lunnon <fred.lunnon@gmail.com> wrote:
> On 8/14/10, Fred lunnon <fred.lunnon@gmail.com> wrote:
>> On 8/14/10, Tom Karzes <karzes@sonic.net> wrote:
>>  > Ok, this one I can do:
>>  >
>>  >     n = sqrt(1/2^(2^0) + sqrt(1/2^(2^1) + sqrt(1/2^(2^2) + ...)))
>>  >
>>  >     (1/sqrt(2))*n = sqrt(1/2^(2^1) + sqrt(1/2^(2^2) + sqrt(1/2^(2^3)
>> + ...)))
>>
>>
>> Er ... don't follow this step, I'm afraid ...    WFL
>>
>
> Apologies --- I had misread the parentheses.
>
> [And as a result wasted some time attempting to prove said number
> transcendental!]   WFL
>
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