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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4Quadratic irrationals are not a good model for such proofs. A better one is here: http://arxiv.org/abs/math/0601660 – Alon Amit Nov 02 '10 at 07:00

1Letting $z=1$ in the continued fraction for $\tan z$ derived on Wikipedia (http://en.wikipedia.org/wiki/Gauss%27s_continued_fraction#Applications) gives $\tan 1=[0;1,3,5,7,9,11,\ldots]$. Maybe that can be transformed to your expression somehow (although I don't see how for the moment). – Hans Lundmark Nov 02 '10 at 09:36

@Hans: I went that route too, but I can't find the appropriate equivalence transformation also. FWIW, I tried to derive a recursion for the partial numerators and denominators, but that too was a blind alley (I used the formula in the OEIS entry for this sequence). – J. M. ain't a mathematician Nov 02 '10 at 10:09

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. – Nov 02 '10 at 16:33
1 Answers
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
$$\cfrac{1}{a\cfrac{1}{x}} = \cfrac{1}{a1 + \cfrac{1}{1 + \cfrac{1}{x1}}}$$
To transform $$\cfrac{1}{a  \cfrac{1}{b  \cfrac{1}{c  \dots}}}$$ to
$$\cfrac{1}{a1 + \cfrac{1}{1 + \cfrac{1}{b2 + \cfrac{1}{1 + \cfrac{1}{c2 + \dots}}}}}$$
to get the expansion for $\displaystyle \tan(1)$
The above expansion for $\tan(1)$ becomes
$$ \cfrac{1}{11 + \cfrac{1}{1 + \cfrac{1}{32 + \cfrac{1}{1 + \cfrac{1}{52 + \dots}}}}}$$
$$ = 1 + \cfrac{1}{32 + \cfrac{1}{1 + \cfrac{1}{52 + \dots}}}$$ $$= 1 + \cfrac{1}{1 + \cfrac{1}{1 + \cfrac{1}{3 + \cfrac{1}{1 + \cfrac{1}{5 + \dots}}}}}$$
To prove the transformation,
let $\displaystyle x = b  \cfrac{1}{c  \dots}$
Then
$$ \cfrac{1}{a\cfrac{1}{x}} = \cfrac{1}{a1 + \cfrac{1}{1 + \cfrac{1}{x1}}}$$ $$ = \cfrac{1}{a1 + \cfrac{1}{1 + \cfrac{1}{b1 + \cfrac{1}{c  \dots}}}}$$
Applying the identity again to
$$\cfrac{1}{b1 + \cfrac{1}{c  \dots}}$$
we see that
$$\cfrac{1}{a\cfrac{1}{x}} = \cfrac{1}{a1 + \cfrac{1}{1 + \cfrac{1}{b2 + \cfrac{1}{1 + \cfrac{1}{c1 + \cfrac{1}{d  \dots}}}}}}$$
Applying again to $\cfrac{1}{c1 + \cfrac{1}{d  \dots}}$ 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. – James Nov 04 '10 at 02:08