# Convert partial fraction to continued fraction?

Lets say you have a partial fraction of the form:

$$f(x) = a_0 + \sum_{n=0}^{\infty} \frac{a_n}{\lambda_n + x}$$

Can anyone explain to me, in mildly plain English, how to convert this partial fraction to a continued fraction of the form:

$$f(x)= b_0+\frac{b_1}{1+}\;\;\frac{b_2x}{1+}\;\;\frac{b_3x}{1+}\;\;...$$

I have tried to find good answers but have only been able to find very difficult academic papers detailing QD algorithms (Rutishauser, 1954), Jacobi matrices (de Boor & Golub, 1978), etc. I have also looked in the textbook Numerical Methods That Work (Acton) but he only has a small section on QD algorithms.

Any help or direction to a good resource is greatly appreciated.

• please check the solution here math.stackexchange.com/questions/6757/… it may help you – Nizar Oct 13 '15 at 19:58
• – Lucian Oct 13 '15 at 20:09
• Should it be $b_1 x$ instead of just $b_1$? – Hans Lundmark Oct 13 '15 at 22:01
• No, there, the form of the continued fractions which I have looked at are in the form with just $b_1$ at the top, not $b_1x$. – Darcy Oct 13 '15 at 22:48
• @Nizar The link you provided gives the solution for finding the partial sum given a continued fraction. I am looking for the reverse process whereby you arrive at a continued fraction given a partial sum. – Darcy Oct 13 '15 at 23:13

If you expand each term of your sum as a geometric series, then (provided that the order of summation can be changed) you can rewrite it as a Laurent series of the form $$f(x) = \frac{c_0}{x} - \frac{c_1}{x^2} + \frac{c_2}{x^3} - \dots$$ instead. Stieltjes explains in the first sections, especially in Section 7 (p. 419), how to rewrite a continued fraction as such a Laurent series (at least formally), and then in Section 11 (p. 425) he shows how to go the other way around (which is what you want to do); the outcome is that the coefficients of the continued fraction are given by formulas involving certain determinants where the coefficients $c_k$ appear (formula (7) on p. 427).