# Continued Fraction expansion of $tan(1)$

Prove that 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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Quadratic irrationals are not a good model for such proofs. A better one is here: arxiv.org/abs/math/0601660 – Alon Amit Nov 2 '10 at 7:00
Letting $z=1$ in the continued fraction for $\tan z$ derived on Wikipedia (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 2 '10 at 9:36
@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. Nov 2 '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. – user1709 Nov 2 '10 at 16:33

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. – James Nov 4 '10 at 2:08