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We know that $$e^{i \pi} = -1 .$$ We can transform all of the components of this identity into (generalized) continued fractions. When we start of with $\pi$, we see that $$ \Big(3+ \mathbf{K}_{i=1}^{\infty} \frac{(\sqrt{a_{i-1}}+2)^2}{6}\Big)^{e \cdot i} = -1 $$ When we take the $\frac{1}{i}$'th power, we see that $$ \Big(3 + \mathbf{K}_{i=1}^{\infty} \frac{(\sqrt{a_{i-1}}+2)^2}{6}\Big)^{e} = (-1)^{-i} ,$$

In which $a_1=1^2,a_2=3^2$. We could re-state this equality by writing $e, -1$ and the root the latter as continued fractions, but I think the question is quite clear without doing so (also: other GCF's of pi). What information about (generalized) continued fractions and calculating with them can we extract from Euler's identity?

Motivation: I was hoping that, if we could find a good explanation for this, we would have more insight in how continued fractions behave under arithmetic operations. With that insight, perhaps we could find ways to combine the continued fractions of certain constants and finding the values of some constants, like Apery's Constant and Catalan's, of which we do know the continued fraction representations but few more.

Thanks,

Max

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You don't have a general formula for the $a_i$? –  J. M. Dec 6 '10 at 15:58
    
@ J.M. the general formula is in the question ( $a_i = (\sqrt{a_{i-1}}+2)^2 $) –  Max Muller Dec 6 '10 at 15:59
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Continued fractions behave very badly under multiplication (as well as addition), so I don't see any reason that this would be a meaningful thing to do. –  Qiaochu Yuan Dec 6 '10 at 16:30
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@Max: I don't think you should be looking at continued fractions to explain Euler's identity, and I also don't think it's meaningful to state Euler's identity in the form you've given. –  Qiaochu Yuan Dec 6 '10 at 17:12
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Well, it's not "natural" to add, multiply, or exponentiate CFs... there's no neat way to construct a CF that is the result of operating over two completely different CFs. And no, you didn't mention that recursion formula you gave for your $a_i$ in the question. –  J. M. Dec 7 '10 at 0:11
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