The notion of recursion extends to very advanced fields of MATHEMATICS: recursive set theory or recursive function theory. They principally articulate the notion of computability. A brief exposition articulates the notion of harmonic progression (HP).

RECURSIVE ENUMERABLE SET: A set, S such that an algorithm exists for computing every element of S, that is, enumerating it.

But sometimes the algorithm "throws in" every desired element, along with extraneous elements. This problem is dealt with by the next definition.

RECURSIVE SET: A recursive enumerable set, S such that an algorithm also exists for detecting nonmembers.

Yet, the problem is a little more complicated. Sometimes a set is recursive enumerable only in another set. This is the case with a harmonic progression: it is recursive enumerable in the AP set.

We've already seen a TEST for the set known as "an arithmetic progression": constant difference between adjoining members. And this TEST also detects non-AP's.

So, the set of all AP's is a RECURSIVE SET. But the only TEST for an HP is:

1. invert each member;
2. see if the reciprocals for an AP. If so, it's an HP. Otherwise, no.

Repeating, an HP is recursive in a recursive AP set.

(Yes, it might be better if HP were labeled "reciprocal progression (RP)", but traditionally it's called "a harmonic progression" with good reason. Pythagoras discovered that segmenting a connected string at 1/2, 1/3, ..., 1/n, intervals GENERATES THE OVERTONES OR HARMONICS OF THE DIATONIC SCALE.)

Thus, given AP 1, 2, 3, 4, 5, 6, 7, 8, 9, 20. Then, reciprocating, 1, 1/2, 1/3, 1/4, 1/5, 1/6, 1/7, 1/8, 1/9, 1/10, is an HP.

Again, if we START WITH 3 AND REPETITIVELY ADD 5, we have another AP: 3, 8, 13, 18, 23, 28, 33, 38, 43, 48. Then, RECIPROCALLY, we have the HP: 1/3, 1/8, 1/13, 1/18, 1/23, 1/28, 1/33, 1/38, 1/43, 1/48.

How do you, easily, SUM AN HP?

You resort to the BYPASS STRATEGY of

1. TURNING A DIFFICULT or IMPOSSIBLE PROBLEM BY TRANSFORMING TO A SIMPLER PROBLEM;
2. SOLVING THE SIMPLER PROBLEM;
3. TRANSFORM THE ANSWER BACK INTO THE LANGUAGE OF THE ORIGINAL PROBLEM, FOR THE FINAL SOLUTION.

Thus, to sum the AP, 1, 1/2, 1/3, 1/4, 1/5, 1/6, 1/7, 1/8, 1/9, 1/10:

1. We reciprocate the progression: 1, 2, 3, 4, 5, 6, 7, 8, 9 -- a SIMPLER ARITHMETIC PROGRESSION.

2. Elsewhere we found the simple FORMULA for this sum: n(n + 1)/2, for SUMMING n TERMS OF AN AP. 9(9 + 1)/2 = 9(10)/2 = 9(5) = 50.
3. To find the final answer, we RECIPROCATE (DOUBLE INVERSING) THE ANSWER: 1/50. This saves us the tedium of
   1. Finding the common denominator of the fractions;
   2. Converting to the equivalent fractions with this common denominator.
   3. Performing the addition to obtain the answer.

   How do we find the HARMONIC MEAN OF A HARMONIC PROGRESSION?

   The BYPASS turns this problem into finding the ARITHMETIC MEAN of a set of data.

                   harmonic mean (h.m.)
               ---------------------------->
   reciprocate |                           ^reciprocate
     data      |                           |answer for
               |                           |h.m. of data
               |                           |
               V--------------------------->
               a. m. of reciprocals of data

   I'll assume you understand how to do this in order to pass onto a PROBLEM which stumped me the first time I encountered it and has stumped every one I've tried it on, since. It has to do with "the average speed of a set of average speeds".

   Having learned about this, I suggest a return to BABYLONIAN ASTRONOMY AND ALGEBRA.