AN ELEMENTARY PROBLEM ABOUT CIRCLES.
INFINITE PRODUCTS, FILLING PROGRAMS,
AND INTEGRATION.

DRAFT COPY ONLY.
10/2/2005.
Raimond A. Struble, PhD.

Professor Emeritus
Department of Mathematics
North Carolina State University at Raleigh
Raleigh, NC.

http://www.medparse.com/strucirc.htm
© 2005, Raimond A. Struble, PhD.

Send comments and correspondence to: George.Moore4@med.va.gov
See also: http://www.medparse.com/stru0928.htm ............. http://www.medparse.com/struifpr.htm ............. http://www.medparse.com/struitgr.htm ............. http://www.medparse.com/struppma.htm ............. http://www.medparse.com/infnpapl.htm

INTRODUCTION.



Unlike good storytellers, in this article we give away the punch-line at the very beginning: the triple-spiked region defined by any three mutually tangent circles can be completely filled by non-overlapping circles. The real story concerns an unusual proof that the sum of the areas of these filling circles equals the area of the triple-spiked region. Instead of attacking the problem directly by adding up areas of circles, we shift the burden to one of establishing the divergence of an infinite series. The "shift" is a straightforward application of the theory of infinite products, while the delicate divergence proof becomes a pretty good story in itself. Its punch line, in turn concerns the somewhat mysterious emergence of a modified harmonic series. But getting to the end requires a considerable amount of analytical and visual argument, which is not so straightforward. However, any competent undergraduate mathematics student should be able to follow the steps with ease. We employ only a little bit of trigonometry and of algebraic manipulation, a little appreciation of the legitimate use of pictures and approximations, and the fact that a series dominating a divergent one also diverges.

What we do explicitly is to prove the result for a particular triple-spiked region defined by three circles of equal radii. The mechanics of the proof in this particular case suggests a simple extension to more general cases.

A sketch of the basic elements of infinite products, as formulated by the author, has been appended for those wanting to pursue the topic beyond this specialized problem. It is this formulation which puts the elementary problem about circles into proper perspective.
A. THE ARITHMETIC INVOLVED.


We start by envisioning three circles of (common) radius R tangent to each other and to a unit circle. The special filling process employed is a rather natural one, and consists of successively filling all unfilled regions by maximum-sized circles tangent to the earlier circles employed. The number of circles involved at each stage increases by the powers of 3. So following the single unit circle, there are 3 circles, then 9 circles, then 27 circles, etc., as illustrated in Figure 1.. Some of these circles are very small indeed, but all of them create, at their initial appearance, three new triple-spiked regions (much like the original one), requiring filling at subsequent stages. There always appear to be appreciable areas remaining to be filled in the same fashion. So then how can the process result in the complete filling of the original triple-spiked region?

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In order to understand the arithmetic of this complicated process, we first compute the radius R of the three large circles. As indicated in Figure 1, the 30o right triangle with side R and hypoteneuse R+1 requires that R = √3 / (2 - √3), approximately 6.464. We are also interested in the area of the original triple-spiked region. This is the difference between the area of the 2R equilateral triangle, √3 R2, and the combined area of the three 60o circular R-sectors, πR2/2 (a half-circle). For later reference, we also take note of the fraction π/R2(√3 - π/2), approximately 0.466, of this area occupied by the unit circle.

At the second stage, the three new circles required have a radius r determined by the 60o oblique triangle with sides R+1, R+r, and r+1. The cosine law detemines the value of r=(R+1)/(3R-1), approximately 0.406. These three circles occupy a fraction, 3πr2/R2(√3 - π/2) - π, approximately 0.432, of the area, R2(√3 - π/2) - π, of the then-unfilled region. At the third stage, there are 9 circles to contend with, and the three outer circles have a radius determined by another 60o oblique triangle as [R(1-2r) + 2r(1+2r)+1]/[3R-4r-1], approximately 0.220. We do not choose to determine the radius of the 6 inner circles now as the procedure has already become much too complicated. In fact, it seems "out of the question" to successively calculate the areas of all the circles in an attempt to settle the original problem. Imagine, in Figure 1., the complications involved in just determining the radii of the 27 circles at stage 4, or of the 81 circles at stage 5. Fortunately, an indirect approach is available, and one which turns out to provide a (verifiable) alternative procedure
B. ALTERNATIVE INDIRECT METHOD.


If one lets rn denote the fraction of the unfilled area actually filled at the nth filling stage, then (1 - rn) is the fraction remaining. Just above, we obtained r1 = 0.466 and r2 = 0.432, approximately, so that the fraction remaining at stage 3 becomes the product, (1 - r1)(1 - r2) = 0.303 approximately. It is clear, moreover, that the infinite product:
P = (1 - r1)(1 - r2)(1 - r3) ... (1 - rn) ...
gives the final remaining fraction, and the circles completely fill the triple-spiked region if (and only if) P=0. This happens exactly when the corresponding infinite series
∑ rn
diverges. (See Paragraph E, for clues leading to this simple fact.) Thus, our task now is to prove that this series does, in fact, diverge. The initial pictures of the process suggest that the sequence rn may not even tend to zero as n → ∞, but this we have been unable to prove.

Fortunately, we need not prove this in order to conclude that the series diverges. For notice that at stage 3, of the 9 circles to be placed, 3 are placed in the outer regions of the original spikes, and 6 are placed so as to initiate the filling of 6 new triple-spiked regions, bordering on the unit circle. The latter resemble (and closely approximate) the stage 1 process, so that the composite r3-value of these 9 circles may be only slightly less than the r2-value above. At stage 4, there are 27 circles to be placed, 3 in the outer region, 18 in the inner regions resembling (and closely approximating) the stage 2 process, and 6 in-between ones, initiating the filling of 6 new triple-spiked regions, bordering on the r-circles. These latter again resemble (and closely approximate) the stage 1 process. The composite r4-value of these 27 circles may be only slightly less than the r3-value. As stated, we do not know if the decreasing sequence rn actually tends to zero, or not, as n → ∞, but it is evident that at each stage, the most inefficient fillings, of their assigned regions, occurs with the 3 outer circles, as they advance toward the original spikes. The individual fractions for these outer circles (neglecting all others) do, in fact, tend to zero as n → ∞. But we establish below that they do so no faster than the terms of a harmonic series. Therefore, the series ∑rn (consisting of even larger terms) diverges, and the circles do completely fill the original triple-spiked region.
C. THE ESSENTIAL DIVERGENCE PROOF.


To establish this claim concerning the individual fractions for the outer circles, it is convenient to recast the calculation of the individual fraction of any outer circle, as that of a nominal sized circle squeezed in between two large circles. This, of course, is equivalent to the calculation of the individual fraction of a small outer circle squeezed in-between the two original R-circles. Using the standard Cartesian x,y coordinate system, we consider a (very) large circle of radius, (N2+1)/2 with center (N,(N2+1)/2) and passing through (0,1). See Figure 2, which, however, illustrates the situation for a relatively small value of N, in order to clarify the recast geometric picture.

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The large circle is tangent to the x-axis (and to a reflected circle) at x=N, and its coordinate equation simplifies to (N2+1)y = (x - N)2 + y2. Therefore, the relevant spike area (above and below) is given by:
AN = ∫0N 2y dx = 2/(N2+1) ∫0N (x - N)2 dx + 2/(N2+1) ∫0N y2 dx.
For large N, the second term is negligible (y2 < 1), and
AN ~ 2/3 N3/(N2+1) ~ 2/3 N.
Moreover, for very large N, a unit circle centered at the origin becomes effectively tangent at (0,±1) to the very large circles forming the spike. See Figure 3:

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which now illustrates the situation for some very large N. Reverting back to the original setup, this picture can be viewed as the case of a very small outer circle in the original triple-spiked region, and where the individual associated fraction (circle area/unfilled spike area) is
π/AN ~ π/(2/3)N = 3π/2N.
The exact scale factor (back to the original setup) is simply 2R/(N2+1), which, of course, is the actual radius of the small circle, since its image in Figure 3 is the unit circle at the origin.

We can choose a large N, and an integer K, so that this number agrees precisely with the radius of some small outer circle at some appropriate stage K. Then moving on to the next stage, K+1, we determine a new (increased) N-value say, N1, so that the quantity 2R/(N12+1) also matches the radius of the K+1-stage circle precisely. A new Figure 3 becomes apropos with the unit circle at the origin now representing the K+1-stage and a spike located at the new station, N1. To continue the program, a new N-value is determined so that the quantity 2R/(N22+1) matches the radius of the K+2-stage circle precisely. Thus, another new Figure 3 becomes apropos with the unit circle at the origin now representing the K+2-stage and a spike located at the station N2. This program, of matching the quantity 2R/(Nk2+1) to the radius of the (K+k)-stage circle precisely for k=1, 2, 3, ..., leads to the sequence of the very fractions (circle area/unfilled spike area) we seek (beyond K), and for which we then need to sum:
∑k=1∞ (π/ANk) ~ (3π/2) ∑k=1∞ 1/Nk.
We observe that the approximations have become ever more accurate with increasing N-values. We shall also require good estimates for these N-values. So to this end, we again refer back to Figure 3 and to its extended versions for k=1, 2, 3, .... Since at any (K+k)-stage, the unit circle at the origin always represents the (small) scaled-up (K+k)-circle, our filling process requires that the next scaled-up (K+k+1)-circle be represented in this figure by a circle tangent to the unit circle and to the sides of the spike, i.e., the dashed circle. (This is how the 3 small outer circles advance toward the original spikes.)

The radius ρ of the dashed circle is just the ratio (Nk2+1)/(Nk+12+1) of the radii of these two successive circles of the filling process, and must also satisfy the corresponding spike equation
(Nk2+1)2/4 = [(1+ρ) - Nk]2 + [ρ - (Nk2+1)/2]2
of Figure 3. However, ρ is very nearly 1 (for large Nk), and thus also very nearly y(2), which also satisfies this spike equation
(Nk2+1)2) / 4 = [2 - Nk]2 + [y(2) - (Nk2+1)/2]2.
When this latter equation is expanded, it can be re-expressed in the form
y(2)(Nk2+1) = Nk2 - 4Nk + 4 + y2(2), or,

y(2)(1 + 1/Nk2) = 1 - 4/Nk + [4 + y2(2)]/Nk2.
So for large Nk, we obtain the estimate
y(2) ~ 1 - 4/Nk.
This, in turn, readily leads from the approximation
(Nk2+1)/(Nk+12+1) = ρ ~ y(2)
to the viable estimate
Nk+1 ~ Nk (1 + 2/Nk) = Nk+2.
Since y(2) is slightly too small (slightly less than ρ), this estimate for Nk+1 is slightly too large, a fact which only enhances the subsequent arguments concerning divergence.

Iteration of the last estimate, starting from k=0 (with N0 = N) yields our sought-after (good) estimates:
Nk ~ N + 2k
for k=1, 2, 3, .... We conclude, therefore, that the following three series satisfy:
∑k=1∞ (π/ANk) ~ (3π/2) ∑k=1∞ 1/Nk ~ (3π/2) ∑k=1∞ 1/(N+2k).
All approximations used throughout our analysis become ever more accurate as k increases. Since the modified harmonic series on the right diverges, the two series on the left must also diverge, and, therefore, the original (dominating) series ∑ rn itself must diverge, as was claimed in Paragraph B. We have thus proved that the sum of the areas of the filling circles equals the area of the triple-spiked region.

However, there are numerous points of the region which are missed by the filling "discs". These are the limits of telescoping unfilled regions shrinking down to individual points. In this connection, suppose that one were to assign the countably many unfilled regions a numerical order. (Perhaps, more-or-less as they occur during the filling process of Figure 1). Then each missed point p can be associated with various subsequences of these assigned integers, corresponding to those sequences of unfilled regions which actually shrink down upon p. But there is a special subsequence, the telescoping one, which is uniquely associated with p. Thus each of the missed points become labeled by special subsequences of integers. All other points can be assigned a stage number, indicating when they are captured by the filling discs. It is very easy to visualize all the convergent sequences of all unfilled regions as they settle down upon particular points. But the set of all these missed points (which includes nearly all the boundary of the original triple-spiked region), of course, has measure zero.

Visualizing the missed points suggests an amusing fly-swatting game. Imagine that there are many, many flies, all attracted to the centers of our filling circles. A single fly lands on the center of the unit circle prior to our placing it there. The unit circle becomes our fly-swatter, but the fly escapes just before we are able to swat it. The fly lands on one of the second-stage centers, and is joined by 2 other flies at the other 2nd-stage centers. These 3 flies then escape just before we are able to swat them with the 3rd-stage fly-swatter circles. They land on the 4th-stage centers, and are joined by 6 other flies. The game continues by the addition of more and more escaping flies. Those flies with homing instincts, who never leave a spike once entered, chart direct paths to missed points. Other flies chart indirect paths, or else continually flit about without focusing anywhere.
D. THE GENERAL CASE.


The detailed examination of the filling process presented in Paragraph B demonstrates that the filling of the triple-spiked regions for starting circles with unequal radii, is tantamount to one "jumping in" upon a later stage of the process we did examine. Indeed, once any triple-spiked region arises, its subsequent filling process is independent of all other (external) ones, and the local process is characterized by a sequence of its own fractional numbers, rn. All the various and sundry intermediate triple-spiked regions resulting from our original problem must themselves be completely filled, and one can find so many varieties along the way so as to conclude that all such triple-spiked region schemes should result in successful filling processes. Furthermore, this suggests that any simple convex region of the plane would be completely filled upon placing, successively, the largest circles possible in all unfilled regions. With such a procedure, the unfilled regions ultimately become triple-spiked ones, as visualized and treated here.

One should note, however, that other filling programs are certainly available, and of interest. There is no necessity for successively employing the largest circles possible. In fact, any selection process characterized by filling fractions rn of unfilled portions, and which results in the divergence of the infinite series ∑ rn is a complete filling one. The "trick" is to select a process for which one can then prove that the series does diverge. We have been successful in this respect, precisely because we employed the special process of filling all unfilled regions by maximum-sized circles. Altenatively, one could avoid this nasty "trick" by first selecting any such sequence of fractions for which ∑ rn = ∞, and then requiring that some successive filling process conforms to these particular fractions. The filling program will then have to be completely successful. But can one expect to find such a conforming filling process? Perhaps yes, by employing many very small circles at each successive stage. How, for example, might this work out using the terms rn=1/n of the harmonic series for n > 3? Employing the original first two fractions, r1=0.466 and r2=0.432 (and the corresponding circles), we place the original 6 inner circles of stage 3 as before, and 3 reduced outer ones, with centers on the center-lines and tangent as before, but so as to match the composite fraction, 1/3 = r3. (One anticipates here that the original third composite fraction exceeds 1/3; it is perhaps more like 0.4). For stage 4, we first place again as many as possible of the original maximum-sized inner circles, as well as any others required for new in-between, unfilled regions. We then place 3 reduced outer circles with centers on the center-lines and tangent as before, but so as to match the composite fraction, 1/4 = r4. Such a program becomes complicated and somewhat confusing, and yet by filling as many unfilled inside-regions with non-overlapping maximum-sized circles as possible, and then using the 3 outer circles to establish the match, 1/n = rn, it appears that one should be able to produce a successful filling process, essentially conforming to the harmonic series. After all, the dominant original individual fractions for the outer circles decay approximately in this manner. See Paragraph F for other filling schemes where arbitrary choices of the fractions rn can be realized and readily exhibited.
E. RELATED FACTS.


As the filling scheme used here is universal in nature, and as this approach to filling problems may be relatively new to many readers, we state some rather simple related facts which might prove to be of interest. The most important fact is that for any decreasing infinite product
P = (1 - r1)(1 - r2) ... (1 - rn) ... (0 < rn < 1),
one can usefully associate a convergent infinite series
F = r1 + (1 - r1)r2 + (1 - r1)(1 - r2)r3 + ... + (1 - r1)(1 - r2) ... (1 - rn-1)rn + ...
satisfying the numerical relationship
F = 1 - P
with the product. The inevitable convergence of F has numerous significant implications. The numerical relationship is also shared by the partial products and partial sums in the same form:
Fn = 1 - Pn.
These relationships can be obtained in a straightforward fashion by simply expanding the products. But (as was done in this application to filling by circles), they can also be obtained by noting that if one successively fills "something" with "something" according to fractions rn of unfilled portions at each stage n, then a fraction (1 - rn) remains at each stage, and so the partial products P1, P2, ..., Pn, ..., become the, stepwise, remaining unfilled fractions, while P becomes the final remaining unfilled fraction. Of course, F = r1 + P1r2 + P2r3 + ... + Pn-1rn + ... becomes the final fraction actually attained. Many examples of the use of this fact can be found in [2] The second most important fact is that any increasing infinite product:
P = (1+r1)(1+r2) ... (1+rn) ...
becomes the reciprocal (P = 1/P*) of a decreasing infinite product
P* = (1-r1*)(1-r2*) ... (1-rn*) ... ,
whenever one replaces (1+rn) by its equal, 1/(1-rn*), where rn* = rn/(1+rn) for all rn > 0. Moreover, this type of reciprocal scheme can be employed for any (and all, since any rn might be zero) mixed increasing and decreasing products
P = (1-r1)(1+r2)(1-r3)(1+r4) ... (1-rn)(1+rn+1) ... .
These then become very revealing quotient products:
P = PQ = [(1-r1)/(1-r2*)] [(1-r3)/(1-r4*)] ... [(1-rn)/(1-rn+1*)]... ,
involving only decreasing factors above and below. Filling interpretations can then be immediately applied to these numerator and denominator fractions. For example, rather easy examinations of the convergence of the delicate indeterminate cases become possible, as well as very easy examinations of the absolutely convergent cases, where the numerical relationship P = PQ = PN/PD holds. Here the numerator and denominator products PN = (1-r1)(1-r3) ... (1-rn) ... and PD = (1-r2*) (1-r4*) ... (1-rn+1*) ... can be afforded separate filling interpretations. See: [3]. One of the most revealing facts about these real-valued infinite products is the close numerical connections they exhibit relative to corresponding infinite series. For if M = l.u.b.n rn< 1, then a decreasing product satisfies the inequalities
(1 - 1/M)S/M < P < e-S,
where S = r1 + r2 + ... + rn + ... (whether convergent or not). As M → 0, these actually become identities.

Similarly, for an increasing product, the revealing inequalities are:
(1 + 1/M)S/M < P < eS,
which also become identities as M → 0, but where the l.u.b.n rn = M values are unrestricted. These inequalities follow directly from simple comparisons of the graphs of straight lines and exponential curves (any use of the logarithm function can be completely avoided). An appropriate label for them is The Sandwich Theorem. See: [2]. The final fact we take note of here allows for the immediate application of these real-variable products to arbitrary complex-variable products. One simply expresses the latter in polar-coordinate form, with the magnitudes becoming a real-variable infinite product topic, and their arguments becoming an the auxiliary infinite series topic. In particular, this last aspect takes care of the real-variable situation involving mixtures of positive and negative factors.

We also note that there is little necessity for involving the concepts of analytic continuation (within the complex plane, or from the real line) at all, using this scheme. In fact, one doesn't even need i = √-1, since one can just as easily consider the "real representation", zR = ρ eθ(mod 2π), in lieu of z = ρ eiθ, and their products as real-valued functions. In this situation, a complex-valued infinite product,
P = z1 ∙ z2 ∙ z3 ∙ ...
(whatever its factors mean to the user) is realized in the form
PR = ρ1 ∙ ρ2 ∙ ρ3 ∙ ... eθ1 + θ2 + θ3 + ... (mod 2π)
of real factors and real sums. Of course, the factors do not conform to the distributive law with respect to addition, but this doesn't really matter, since the complex product P is simply recovered from the real product PR by re-inserting the i. Important examples of complex infinite products, like cos πz and sin πz, can be readily treated this way, but deriving them, in the first place, is another matter entirely.

See again: [3] for related details.
F. LINK TO INTEGRATION.


There is an interesting link from infinite (real) products (and filling schemes) to integration. In the present context, we first need to define (invent) a special function that we shall call a circle-function. If c denotes a circle in the plane lying above the horizontal x-axis, then c(x) will denote the circle-function, which for each x is the length, at x, of the vertical segment subtended by c. There is nothing fancy here, just a continuous function with compact support which rises in the middle (in the form of a Greek arch).

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Returning now to our original problem of filling by circles, we assume that the countably many circles used to fill the original triple-spiked region have been assigned a linear ordering, c1, c2, ..., cn ..., and that they all lie above the x-axis. The sequence of integrals, ∫ cn(x) dx, of the corresponding circle-functions is summable, and (since the filling circles are non-overlapping) sums to the quantity
(√3 - π/2)R2 = ∑ ∫ cn(x) dx,
the area of the original triple-spiked region. The sequence, cn(x), of circle-functions themselves, define an integrable function
L(x) = ∑ cn(x),
whose integral, ∫ L(x) dx = (√3 - π/2)R2. This is the direct filling approach we abandoned in Paragraph B, because we could not use it to prove that the circles actually fill the triple-spiked region. Now, of course, we know that they do, and so that is why we can claim that ∫ L(x) dx = (√3 - π/2)R2.

It seems appropriate, therefore, to rephrase the above integration link in the context of our indirect method, using fractions. The simplest way to do this is to recast the entire picture by scaling down the original triple-spiked region so as to have area 1. This requires a smaller R-value, √3 (√3 - π/2)1/2 ~ 2.490, approximately, so that all lengths are then reduced by the factor, √3 (√3 - π/2)1/2 / (2 - √3) ~ 0.385, approximately. This is just the ratio of these two R-values. Then the scaled-down filling circles cn are such that
F = ∫ L(x) dx = ∑ ∫ cn(x) dx = P-1,
where P is exactly the infinite product that we employed in Paragraph B. Of course, since the triple-spiked region is completely filled, P=0 and F=1, as they should be, and the integrable function L(x) is just the projection on to the x-axis of the (scaled down) original triple-spiked region.

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However, if some of the filling circles are further reduced in size, then the new product P will be greater than zero, and the new series F will be less than one. In this way, additional far more interesting (Lebesgue) integrable functions L(x) are produced by the circle-functions, and
F = ∫ L(x) dx = 1 - P < 1
will always yield the correct values of their integrals.

This last phenomenon is more visibly demonstrated in an alternative (simpler) setting, where one attempts to fill a rectangle with triangles. To demonstrate this, we successively place non-overlapping collections of isosceles triangles in a rectangle, as depicted in Figure 4:

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through the fourth stage. After the first stage, the number of triangles again increases according to the powers of 3. Just as in the circle problem, each new filling triangle (for n > 2) initiates 3 new unfilled (skewed) triangular regions to be filled at subsequent stages. Because the filling triangles occupy unfilled triangles, the maximum filling fractions rn can never exceed 1/4, and the unfilled fractions (1 - rn) must always equal or exceed 3/4.

But, subject to this limitation, all possibilities for the fractions rn can be realized and readily exhibited, upon employing slim or fat isosceles triangles.


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The final unfilled fraction of the rectangle is given (as always) by the infinite product:
P = (1 - r1)(1 - r2) ... (1 - rn) ... ,
where, of course, F = r1 + P1r2 + P2r3 + ... = 1 - P. Therefore, the filling of the rectangle by the isosceles triangles is completely successful if (and only if) P=0, i.e., if (and only if) the corresponding series
S = r1 + r2 + ... + rn ...
diverges.

However, upon employing slim or fat isosceles triangles, it is very easy to arrange for this sum to be finite, and when it is, the product P becomes positive, and F=1-P becomes less than 1. Lowering the filling triangles tn to the base of the rectangle (now the x-axis) leads to a sequence of triangle-functions, tn(x), (formerly called tent-functions because of the appearance of their graphs) satisfying
F = ∑ ∫ tn(x) dx = 1 - P,
and defining a (Lebesgue) integrable function
L(x) = ∑ tn(x),
provided that we again normalize the picture and make the area of the rectangle equal to 1. When the series ∑ rn diverges, and the rectangle is completely filled, then L(x) is a constant function. But when the series converges, and the rectangle is incompletely filled by the triangles, then L(x) may become another very interesting (Lebesgue) integrable function.

A somewhat different type of geometric example concerns the attempted filling of a circle by rectangles. Following the insertion of a maximum square, non-overlapping collections of rectangles are successively inserted into all unfilled regions, so as to be in contact with the circle, and to share sides with the previously placed rectangles. Figure 5 illustrates the process through the third stage.

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Simple geometric considerations show that filling fractions rn can be arranged for so as to exceed 1/4 for all n. In such a case, the infinite series ∑ rn certainly diverges, and the rectangles completely fill the circle. However, by using slim or fat rectangles, one can arrange for the convergence of the series ∑ rn. (Actually, all possibilities for the fractions rn < 1/4 can be realized and readily exhibited.) In this situation, the circle is incompletely filled by the rectangles, and rather fascinating contours, interior to the circle, can arise. When the filling rectangles, bn, are lowered to an x-axis below, they define a sequence, bn(x), of rectangle-functions (formerly called brick-functions, because of the appearance of their graphs). If we specify that the area of the circle is 1, then these rectangle-functions satisfy
F = ∑ ∫ bn(x) dx = 1 - P,
and also define the (Lebesgue) integrable functions
L(x) = ∑ bn(x).
Whenever the series ∑ rn diverges, these functions are one and the same, and just the projection onto the x-axis of the circle (the induced circle-function). However, in the circumstance where the series ∑ rn converges, the integrable functions L(x) can again be expected to be spectacular.

For both of these latter two examples, it is shown in [3] that one can obtain a reversal of this situation, and find infinite product representatives of a filling process leading to any given positive, bounded (Lebesgue) integrable function (on an interval). The additional arguments require the employment of a development [1] in the theory of integration due to the late Prof. Jan Mikusinski. In [3], it is further shown how the filling of double-rectangles (above and below the x-axis) correlates with the quotient product form PQ mentioned in Paragraph E. This allows for the elimination of the above restriction to only positive functions.
G. THE RIEMANN ZETA-FUNCTION.


We conclude the paper with a brief mention of an historically significant example of an infinite product [4]. In 1859, Bernhard Riemann dealt with the function defined by the simple infinite series
ζ(s) = ∑ 1/ns (all positive integers n),
in a celebrated investigation of the prime number theorem. He started with Euler's familiar infinite product formula (of 1737)
P(s) = (1 - 1/2s) (1 - 1/3s) (1 - 1/5s) (1 - 1/7s) ... = 1 / ζ(s)
for s > 1, which involves only the prime numbers. He then studied the analytic continuation of ζ(s) into the complex plane, sans the simple pole at s=1, and its significance for the prime number theorem. He called the extended function the zeta-function, a name by which it has forever since been known.

The series itself converges only for s > 1 (real part of s > 1), and exhibits the expected divergent behavior as s → 1. So in our special symbolism, we can write
F(s) = 1 - P(s) = 1 - 1/ζ(s)
for s > 1, because of the Euler formula, but we can write more generally,
F(s) = 1/2s + (1 - 1/2s)1/3s + (1 - 1/2s)(1 - 1/3s)1/5s + (1 - 1/2s)(1 - 1/3s)(1 - 1/5s)1/7s + ...
throughout the extended range, s > 0. While the ζ-series diverges for 0 < s < 1, the F-series converges (as always) and to the value 1 here. Figure 6 illustrates the interesting graph of F(s) for the extended range.

676.
As a practical matter, either this F-series, or the infinite product, P(s), become useful tools for approximating the sum ζ(s) of the series ∑ 1/ns, which converges very slowly as s → 1+.

The most significant theoretical matter is the Riemann Hypothesis, which states that all of the non-trivial zeros of the zeta-function lie on the line, real part of s=1/2. It has infinitely many zeros along the negative real line (all of them considered trivial), but the validity of the hypothesis continues to remain unsettled.
H. REFERENCES.


1. Mikusiński J.
The Bochner Integral.
New York, San Francisco: Academic Press. Harcourt Brace Jovanovich, Publishers. 1978;:. Pure and Applied Mathematics.
Basel: Birkhäuser. Lehrbücher und Monographien aus dem Gebiete der exakten Wissenschaften: Mathematische Reihe. [Textbooks and monographs from the area of exact sciences: mathematical series.] 1978;55:.
ISBN: 3764308656, 233 pages.

2. Struble RA.
Infinite Products Rescued.
http://www.medparse.com/struifpr.htm

3. Struble RA.
Infinite Products and Integration.
http://www.medparse.com/struitgr.htm

4. Derbyshire J.
Prime Obsession: Bernhard Riemann and the Greatest Unsolved Problem in Mathematics.
New York: Plume Books. 2004. ISBN: 0452285259.


Last updated: 10/2/2005, by Raimond A. Struble, PhD.