|The incremental profile of the series within a section|
Next to the factor, two successive terms of a series of the type U(a,b)F are needed to develop it. To develop all series involved in an approach, one needs two successive sections. Suppose a program peters out before completing the second section.
In the worst case scenario it just managed the first non-trivial sp-block:
Top- and middle row give the numerators and denominators of the successively better rational approaches of root 13. The bottom row gives the index. As far as root approach goes, we can simply develop the sp-series. But if, for some theoretical reason, we would want all series, it's good to know that all sections show the same 'incremental profile', so that it is possible to obtain the second section (and therewith all series) from the first one's profile.
Let's define the operations to 'increment' or 'decrement' one fraction with another as follows:
a/b (+) c/d = (a+c)/(b+d)An identical fraction in a different representation, renders a different result:
a/b (-) c/d = (a-c)/(b-d)
pa/pb (+) c/d = (pa+c)/(pb+d)p and q act as parameters of the 'weight' in the outcome, of a/b an c/d respectively.
a/b (-) qc/qd = (a-qc)/(b-qd)
A next term in a section can always be obtained by in- or decrementing the last term with previous ones, including the last one itself. All sections show the same pattern, called the 'incremental profile'.
To simplify things, we will indicate fractions by their index and increment the second section according to the first one's profile. The increment within the ab-block is disregarded, so there are 14 increments, starting on the Nb/a-fraction, of which the last one should render the next non-trivial a/b-fraction 842401/233640. Since we know that one already, it's a way to check the result.
The profile of the first section looks like this:
0/1: | 0 | 0 | 0 | 0 | 4 | 5 | 6 | 3*8,7 | 8 | 8 | 8 | 11 | 12 | 13 |
0/1: | 0 | 0 | 0 | 0 | 4 | 5 | 6 | 4*8,-6 | 8 | 8 | 8 | 11 | 12 | 13 |
To get the second section, take the indices modulo 15
2340/649: | 15 | 15 | 15 | 15 | 19 | 20 | 21 | 3*23,22 | 23 | 23 | 23 | 26 | 27 | 28 |
2340/649: | 15 | 15 | 15 | 15 | 19 | 20 | 21 | 4*23,-21 | 23 | 23 | 23 | 26 | 27 | 28 |
The result should look like this: