I have been poring over many texts about continued fractions, but none of them seem to be helping me to prove the following beautiful continued-fraction identity (I am nowhere close): $$ \cfrac{1}{\dfrac{5}{2 \pi} - \cfrac{1}{\dfrac{15}{2 \pi} - \cfrac{1}{\dfrac{25}{2 \pi} - \cfrac{1}{\dfrac{35}{2 \pi} - \cfrac{1}{\dfrac{45}{2 \pi} - \ddots}}}}} = \sqrt{5 + 2 \sqrt{5}}. $$ I warmly welcome anyone in the community to offer any insights that he/she might have. Thank you very much!

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    $\begingroup$ Smells like Ramanujan. $\endgroup$ – anon Jan 6 '13 at 3:56
  • $\begingroup$ Same as always: where did you get this? $\endgroup$ – Will Jagy Jan 6 '13 at 3:57
  • $\begingroup$ Unfortunately, it comes from a friend whom I lost touch with many years ago. I don't even know where he is now, but he left some of his notebooks with me for safe-keeping. He definitely smells like Ramanujan, but he's not an Indian. Computing the first few truncations of the continued fraction strongly indicates the correctness of the identity, though. $\endgroup$ – Haskell Curry Jan 6 '13 at 4:01
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    $\begingroup$ Sometimes you can construct a continued fraction expansion for a number by knowing a quadratic with integer coefficients for which it is a zero. In this case, $\sqrt{5+2\sqrt{5}}$ is a zero of the integer coefficient quartic $x^4 -10x^2 + 5$. I don't think you can necessarily construct a continued fraction expansion from this, and even if you could, I don't know how $\pi$ would come into it. $\endgroup$ – Michael Albanese Jan 6 '13 at 4:33
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    $\begingroup$ Possibly useless observation: If you replace $\frac{5}{2 \pi}$, $\frac{15}{2 \pi}$, $\frac{25}{2 \pi}$,... by 1,3,5,..., the result appears to be $\tan 1$. $\endgroup$ – Ted Jan 6 '13 at 5:00

Consider the more general continued fraction

$$\cfrac{1}{x - \cfrac{1}{3x- \cfrac{1}{5x- \cfrac{1}{7x - \cfrac{1}{9x - \ddots}}}}}$$

Some numerical experimentation suggests that this is equal to $\tan (1/x)$. Then we can substitute $x = \frac{5}{2 \pi}$ to get the original identity.

Edit: This identity is the third continued fraction identity for tangent listed on this page.

Edit 2: A proof of a continued fraction for $\tan x$, taken from Chrystal's Algebra. To derive the identity above, substitute $1/x$ for $x$ in the identity on that page and then do some manipulations.

  • $\begingroup$ This is great! Thanks for the link! $\endgroup$ – Haskell Curry Jan 6 '13 at 5:48

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