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Prove that the continued fraction of $\tan(1)=[1;1,1,3,1,5,1,7,1,9,1,11,...]$. I tried using the same sort of trick used for finding continued fractions of quadratic irrationals and trying to find a recurrence relation, but that didn't seem to work.

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    according to Wikipedia, it seems that the trick is to use the continued fraction for tan(1/n), and set $n=1$. I don't immediately know how to find a continued fraction for tan(1/n) though.2010-11-02

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We use the formula given here: Gauss' continued fraction for $\tan z$ and see that

$\tan(1) = \cfrac{1}{1 - \cfrac{1}{3 - \cfrac{1}{5 -\dots}}}$

Now use the identity

$\cfrac{1}{a-\cfrac{1}{x}} = \cfrac{1}{a-1 + \cfrac{1}{1 + \cfrac{1}{x-1}}}$

To transform $\cfrac{1}{a - \cfrac{1}{b - \cfrac{1}{c - \dots}}}$ to

$\cfrac{1}{a-1 + \cfrac{1}{1 + \cfrac{1}{b-2 + \cfrac{1}{1 + \cfrac{1}{c-2 + \dots}}}}}$

to get the expansion for $\displaystyle \tan(1)$

The above expansion for $\tan(1)$ becomes

$ \cfrac{1}{1-1 + \cfrac{1}{1 + \cfrac{1}{3-2 + \cfrac{1}{1 + \cfrac{1}{5-2 + \dots}}}}}$

$ = 1 + \cfrac{1}{3-2 + \cfrac{1}{1 + \cfrac{1}{5-2 + \dots}}}$ $= 1 + \cfrac{1}{1 + \cfrac{1}{1 + \cfrac{1}{3 + \cfrac{1}{1 + \cfrac{1}{5 + \dots}}}}}$

To prove the transformation,

let $\displaystyle x = b - \cfrac{1}{c - \dots}$

Then

$ \cfrac{1}{a-\cfrac{1}{x}} = \cfrac{1}{a-1 + \cfrac{1}{1 + \cfrac{1}{x-1}}}$ $ = \cfrac{1}{a-1 + \cfrac{1}{1 + \cfrac{1}{b-1 + \cfrac{1}{c - \dots}}}}$

Applying the identity again to

$\cfrac{1}{b-1 + \cfrac{1}{c - \dots}}$

we see that

$\cfrac{1}{a-\cfrac{1}{x}} = \cfrac{1}{a-1 + \cfrac{1}{1 + \cfrac{1}{b-2 + \cfrac{1}{1 + \cfrac{1}{c-1 + \cfrac{1}{d - \dots}}}}}}$

Applying again to $\cfrac{1}{c-1 + \cfrac{1}{d - \dots}}$ etc gives the required CF.

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    Thanks Moron. I had been looking for a transform like the one you gave but couldn't seem to find it.2010-11-04