In a chapter of Computational Algebra and Number Theory called Continued Fractions of Algebraic Numbers (Available at http://citeseerx.ist.psu.edu/viewdoc/download?rep=rep1&type=pdf&doi= , Bombieri and van der Poorten give conditions under which the continued fraction of some algebraic numbers may be calculated by several complicated recursive formulas.

Experimentation has led me to believe that these same formulas can be used to calculate the continued fraction representations of at least some transcendental numbers as well, although a proof is lacking.

The theorem giving the formulas for the continued fraction expansions is Theorem 3 on page 151 of the paper by Bombieri and van der Poorten.

Here I'll give the recursive formulas as they were used by me in a program written in Mathematica.

k = 1;
f[x_] := Cos[ k  x];
p[0] = 1; p[-1] = 1;
q[0] = 1; q[-1] = 0;
p[h_] := p[h] = c[h] p[h - 1] + p[h - 2];
q[h_] := q[h] = c[h] q[h - 1] + q[h - 2];
r[h_] := r[h] = p[h]/q[h];
c[h_] := c[h] = 
   Floor[((-1)^h ( f'[r[h - 1]])/((q[h - 1]^2) f[r[h - 1]])) - 
       q[h - 2]/q[h - 1] + ((-1)^(h - 1)/q[h - 1]^2)  
            Sum[(1/(r[h - 1] - (j + 1/ 2) Pi/k)) + (1/(r[h - 1] - (-j + 1/2) Pi/k)),  
                {j, 1, Infinity}]];

$$\begin{eqnarray}c(h+1) &=& \bigg\lfloor (-1)^{h+1} \frac{f'(r(h))}{q(h)^2 f(r(h))} - \frac{q(h-1)}{q(h)} \newline & &+ \frac{(-1)^h}{q(h)^2}\sum_{j=1}^\infty \left(\frac{1}{r(h) - (j+\frac{1}{2})\frac{\pi}{k}} + \frac{1}{r(h)-(-j+\frac{1}{2}) \frac{\pi}{k}} \right)\bigg\rfloor\end{eqnarray}$$

Letting the function f be cos(x) gives rise to the partial quotients for $2/(\pi-2)$. It is this that I'd like to prove.

  • $\begingroup$ What exactly are you asking? Why not state your proposed formula for $2/(\pi-2)$? Note that continued fraction expansions for $e$ and some related transcendental numbers are already well known, so you might want to make your target more specific. $\endgroup$ – user25199 May 14 '15 at 7:39
  • $\begingroup$ @Carl It would be a good idea to write out the formulas as you have suggested. I have a problem with entering formulas in the format used here that's why I haven't done so. If someone could help me out with the format, I'd be pleased to edit the problem. Have you read the paper and seen the formulas? $\endgroup$ – David S. Newman May 14 '15 at 13:56
  • $\begingroup$ If you edit your question to include explicit formulas for the coefficients of the simple continued fraction for $2/(\pi-2)$, then someone will probably typeset them for you. $\endgroup$ – Douglas Zare May 14 '15 at 13:59
  • $\begingroup$ @Douglas Zare I'm going to add the formulas as you've suggested using the format of the programming language Mathematica since I can, hopefully, just copy and paste that. Hopefully someone will typeset them. $\endgroup$ – David S. Newman May 14 '15 at 19:59
  • $\begingroup$ I wrote the recursion for $c(h+1)$ instead of $c(h)$. Please check it. $\endgroup$ – Douglas Zare May 14 '15 at 20:42

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