INFINITE PRODUCTS RESCUED.
Raimond A. Struble, PhD.

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

http://www.infiniteproduct.info/struifpr.htm
© 2005, Raimond A. Struble, PhD.



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ABSTRACT.

A straightforward treatment of the elementary mathematical analysis˚ of infinite products˚ is expounded. This approach goes well beyond the typical treatment, that simply links the convergence˚ of an infinite product with the convergence of a related infinite series˚. The numerical/arithmetical link between the two, exploited here through the employment of inequalities, might, perhaps, be viewed as the genuine mathematical theory of infinite products. The technical background needed for a thorough understanding of the present treatment is extremely limited: the knowledge of, and routine experience with, the exponential function˚, the logarithmic function˚, and the most rudimentary notions of convergence. (Some auxiliary mathematical material is included throughout to interest the advanced reader.)

Following this, many facets of the workings of mathematics in a variety of applications are illustrated through the mechanism of specific examples, involving computational, graphical, and analytical aspects of infinite products. There, one encounters the Riemann zeta function˚, routinely intertwining many of the examples as a known numerical entity. Also, there is an extensive treatment of ordinary installment loans˚, such as home purchases, and the rudiments of investment plans subjected to compound interest payments˚. These are concrete applications, given specific mathematical formulations in relation to the concepts of infinite products˚, as developed in the basic theoretical portions of the work.

TABLE OF CONTENTS.


Abstract.
Table of Contents.
Introduction.
Chapter 1: Decreasing Products.
       Section 1A: Definition of Infinite Product.
       Section 1B: Special Topic: Finite Product.
Chapter 2: Increasing Products.
       Section 2A: Increasing Infinite Product.
       Section 2B: Increasing Product Theorem.
       Section 2C: Decreasing Product Theorem.
       Section 2D: Arithmetical Example of Increasing Products.
Chapter 3: Increasing And/Or Decreasing Products.
       Section 3A: Combined Infinite Products.
       Section 3B: Example of Increasing and Decreasing Products (for the Advanced Reader).
       Section 3C: Footnote: Zero and Negative Factors.
Chapter 4: Applications of Decreasing Products.
       Section 4A: Rectangle-filling.
       Section 4B: Processes analoguous to rectangle-filling.
       Section 4C: Installment Purchases.
       Section 4D: Finite Products for Realism.
       Section 4E: Further Finite Products.
       Section 4F: Some Divergence-Convergence Considerations.
       Section 4G: Footnote for the Advanced Reader.
       Section 4H: The Cosine Function (for the Advanced Reader).
       Section 4I: The Sine Function (for the Advanced Reader).
       Section 4J: Square-Filling by Circles and Circle-Filling by Rectangles.
       Section 4K: Turning Mixed Products into Decreasing Products.
       Section 4L: Effectiveness of Decreasing Infinite Products.
Chapter 5: Applications of Increasing Products.
       Section 5A: Compound-Interest.
       Section 5B: Computable Increasing Products.
       Section 5C: Additional convergence considerations.
       Section 5D: Finding the square-root of a number.
       Section 5E: Expanding a number as an infinite product and summation processes for an infinite series.
       Section 5F: An exponential limit.
       Section 5G: Turning Mixed Products into Increasing Products.
       Section 5H: Triangle-Filling by Increasing Products.
       Section 5I: Triangle-Filling by Mixed Products.
       Section 5J: Absolute Convergence.
Chapter 6: Applications of Mixed Increasing and Decreasing Products.
       Section 6A. Compound-interest with Finance Charges.
       Section 6B. Rectangle-filling with a growing rectangle.
       Section 6C. Modified processes analogous to rectangle-filling.
       Section 6D. Comprehensive view of products in applications.
       Section 6E. Infinite products of functions (for the Advanced Reader).
       Section 6F. Infinite convolution products (for the Advanced Reader).
Chapter 7. References and Additional Reading.
Chapter 8. R. A. Struble. Bibliography.
Chapter 9. Acknowledgments.

INTRODUCTION.


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It has recently come to the attention of the author [7.1] that the consideration of infinite products˚ in mathematics courses (pre-calculus, advanced calculus, calculus, post-graduate) is largely ignored. Perhaps the lack of real-life applications is one reason for this, but in any event, it seems a pity, since the topic affords a nice block of elementary and interesting analytical experiences, which could enrich a student's mathematical life. We hope to demonstrate some of these experiences, and supply a sampling of possible applications to boot! (According to the internet, there remains a healthy interest in infinite products as research topics. This writer is essentially unaware of it all. He ventured into the present work solely because of [7.1].

The first step will be to give a complete, elementary (hopefully, easy-to-follow) mathematical treatment, generally missing from textbooks. One exception, perhaps, in Apostol [1957], where, in a three-page treatment in a 543-page book on Mathematical Analysis (subtitled: A Modern Approach to Advanced Calculus, 1957), the necessary mathematics concerning convergence is succinctly presented.

In the following mathematical development, the only prerequisites are elementary arithmetic˚ and algebra˚, some very primitive notions of convergence˚, and a knowledge of some simple properties of exponential functions˚, such as ex, and its inverse˚, the natural logarithm˚, denoted loge x, or simply, log x. One exception is the more "advanced" fact that of the limit˚, limn → ∞ (1 + 1/n)n = e (in various guises), which might be accepted as an analytical or numerical fact (See (5.3)). It is the writer's belief that the entire development is very appropriate for a calculus course˚ at any level. The second step (and principal objective) will be to discuss a number of applications to fill the apparent void. Some of these applications are rather fanciful˚ (such as infinite-product convergence tests˚ for infinite series˚), but others have real-life content (such as the compounding of interest on savings accounts or installment loan payments on home purchases, and growth of surface cancers˚). Not every undertaking has an obvious reason for inclusion, beyond the fact that one can just do it. Others, however, lead to notable and/or intriguing results of interest. The ˚Riemann zeta-function, for instance, arises naturally in many situations as a known numerical function, throughout many examples. As an infinite series, the ˚Riemann zeta-function is also the premiere example used throughout this work. The first example of decreasing products in Chapter 4: Applications of Decreasing Products is certainly worthy of the special attention afforded here, as well as earlier in [7.1]. It is paramount, as a "visible" interpretation of any decreasing product, and all applications are readily mirrored in this simple geometric example. But our first objective now is the presentation of a mathematical foundation of the elementary theory of infinite products (in the ˚real-variable domain).

CHAPTER 1. DECREASING PRODUCTS.


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SECTION 1A. DEFINITION OF INFINITE PRODUCT.

We consider a sequence of numbers satisfying 0 < rk < 1, for all k = 1, 2, 3, ..., and the corresponding n-products:
(1.1)        Pn = (1-r1)(1-r2)(1-r3) ... (1-rn),        for n = 1, 2, 3, ....
Since each factor, (1-rk), is positive but less than 1; the sequence of numbers, Pn, is decreasing with increasing n; and the Pn possess a limit, P, as n → ∞, lying between 0 and 1, or possibly at 0 itself. In any event, P is called the infinite product (i.e., the value of the infinite product), and is written:
(1.2)        P = (1-r1)(1-r2)(1-r3) ...
The obvious question to ask is: Does P=0 or is P>0? A related question to ask is: What is the limit, S, of the increasing sequence of partial sums:
(1.3)        Sn = r1 + r2 + r3 + ... + rn.
of the same numbers as n→∞? Does S=∞ or is S<∞? In any event, S is called the corresponding infinite series, and is written:
(1.4)        S = r1 + r2 + r3 + ....
Before pursuing answers to these questions, we show that the infinite product, P, can be expressed as an infinite series. To this end, we note that:
P2 = (1-r1)(1-r2) = (1-r1) - (1-r1)r2 = 1 - [r1 + P1r2],
and that:
P3 = (1-r1)(1-r2)(1-r3)
= 1 - [r1 + P1r2] - (1 - r1)(1-r2)r3
= 1 - [r1 + P1r2 + P2r3],
and more generally that:
Pn = Pn-1(1-rn) = Pn-1 - Pn-1rn.
So:
Pn = (1-r1)(1-r2)(1-r3)...(1-rn) = 1 - [r1 + P1r2 + P2r3 ... + Pn-1rn] = 1 - Fn
holds for all n. Thus, the infinite product satisfies the equation:
(1.5)              P = 1-F,
where:
F = limn → ∞ Fn
is the infinite series:
(1.6)        F = r1 + P1r2 + P2r3 ... + Pn-1rn + ....
This infinite series always converges and, of course, equals the total decrease from one of the infinite product. Since P<Pn for all n, we conclude from (1.6) that F>PS, and so P = 1-F < 1-PS. This leads to the inequality:
P<1/(1+S).
which implies that P=0 if S=∞. We can improve upon this inequality by noting that 1-x < e-x holds for all x.
301.

Thus the inequality 1-rk < e-rk holds for all k, and upon multiplying n of these inequalities together, we conclude that:
Pn = (1-r1)(1-r2)(1-r3)...(1-rn) < e-r1 e-r2 e-r3 ... e-rn = e-(r1+r2+r3+...+rn).
holds for all n. Consequently:
(1.7)        P < e-S
Using a similar exponential inequality, we can derive a useful lower bound for P, whenever S<∞. In this circumstance, since rk<1 for all k, and rk→0, it follows that:
R = max1<k<∞ rk<1.
Now the inequality:
(1 - R)x/R < 1-x
holds for 0 < x < R.
302.

We conclude, therefore, that the inequality:
(1-R)(1/R)rk < 1 - rk
holds for all k, and so:
(1-R)(1/R)(r1 + r2 + r3 + ... rn) < (1-r1)(1-r2)(1-r3) ... (1-rn) = Pn
holds for all n. Consequently, we have:
(1-R)(1/R)S < P.
It is useful to recast this inequality in the form:
(1.8)        a-S < P < e-S,
where a = 1/(1-R)(1/R) is a number greater than or equal to e. In fact, a→e as R→0. In the limit as n → ∞, let R = 1/(n+1), so that 1/(1-R)(1/R) = (1 + (1/n))n+1. Although (1+1/n)n tends to e from below as n → ∞, (1+1/n)n+1 actually exceeds e, and:
limR → 0 1/(1-R)(1/R) = limn → ∞ (1 + (1/n))n+1 = limn → ∞ (1 + (1/n))n(1 + (1/n)) = e.
In any event, P is sandwiched in between two decaying exponential functions of S:
(1.9)        a-S < P < e-S
The sandwiching is sharp (i.e., a is nearly e) whenever R = maxk rk is small. However, for each series S, there exists a number, bS, such that the equality:
(1.10)        P = bS-S              (e < bS < a)
actually holds. Thus P is essentially a decaying exponential function of S. In particular:
(1.11)        P>0, F<1, if S<∞.
                  P=0, F=1, if S=∞.
It turns out that in many applications, the P=0, F=1 case is really the most desirable one, where:
(1.12)        1 = r1 + (1-r1)r2 + (1-r1)(1-r2)r3 + (1-r1)(1-r2)(1-r3)r4 + ...
generally reflects the success of some process. These are discrete, successive, step-by-step processes, such as might occur yearly, monthly, daily, or even secondly, etc. Therefore, in some cases, it is of interest to consider a limiting situation of instantaneous multiplication, where the process proceeds continuously in time t. If we let P(t) denote the instantaneously decreasing product, which is decreasing at the rate -P(t)r(t) at each instant (recall that Pn = Pn-1 - Pn-1rn), then we have:
dP(t)/dt = -P(t)r(t),
or:
dlogP(t)/dt = -r(t),
which yields:
(1.13)        P(t) = e -0∫tr(u)du              (P(0)=1)
Here:
0∫tr(u)du
is the continuous analogue of the partial sum:
Sn = r1 + r2 + r3 + ... + rn
of a discrete process. Indeed, if r(u) is a step function with constant values rk over discrete unit intervals of time, then (with t → ∞):
0∫∞r(u)du = r1 + r2 + r3 + ... = S.
Similarly:
(1.14)        P(∞) = e -0∫∞r(u)du
is the continuous analogue of the (final) completed discrete product:
P = (1-r1)(1-r2)(1-r3)....
So we can now rewrite (1.14) in the simple form:
(1.15)        P = e -S,
for the completed result of a continuous product, where P = P(∞) and S = 0∫∞r(u)du. This is the limiting form of (1.10), where bS becomes e. In particular, as in discrete processes:
(1.16)        P>0, F<1 if S<∞.
                   P=0, F=1 if S=∞.


SECTION 1B. SPECIAL TOPIC: FINITE PRODUCT.

Because the treatment of loan payments on a home purchase, for example, does not actually conform to an infinite product situation (i.e.,the loan is required to be completely paid off after finitely many, say 360, monthly payments), we conclude this chapter with some special mathematical details appropriate to this type of problem. Practical considerations will be postponed to Chapter 4: Applications of Decreasing Products. This special circumstance requires that fractions rk be selected in a special manner, which we now describe.

Recalling from (1.6) that F = limn→∞ Fn, where:
(1.17)        Fn = r1 + P1r2 + P2r3 + ... + Pk-1rk + ... + Pn-1rn,
we choose each rk to satisfy the equation:
(1.18)        Pk-1rk = p - jPk-1.
Here, p represents the constant (fractional) monthly house payment, and jPk-1 represents the (fractional) interest charge paid to the lender at month k. Indeed, Pk-1 is the current loan balance following month k-1, and j is the constant monthly interest rate. Since Pk-1 = 1-Fk-1, (1.18) can be re-expressed in the form:
Pk-1rk = (p - j) + jFk-1,
so that:
(1.19)        Fk = Fk-1 + Pk-1rk = Fk-1(1+j)+(p-j)
holds for all k. Starting with F0 = 0, we then have, in turn:
F1 = p-j
F2 = F1(1+j) + p-j = (p-j)[(1+j)+1)]
F3 = F2(1+j) + p-j = (p-j)[(1+j)2+(1+j)+1)],
and ultimately:
(1.20)        Fn = (p-j)[(1+j)n-1 + (1+j)n-2 + ... + (1+j)+1)] = (p-j)[(1+j)n-1)/j].
(Recall that 1 + x + x2 + ... + xn-1 = (xn-1)/(x-1).)

Now the loan is completely paid off for n=360, if F360=1 i.e., P360 = 1-F360=0. In this circumstance, the required monthly (fractional) payment, p, is then given by (1.20) as:
(1.21)        p = j[1 + 1/(1+j)n-1)] = j(1+j)n / [(1+j)n - 1]
for n=360. Multiplying this expression by the initial loan value is what your calculator gives you when you punch in j and n. A typical value of p, for j=0.005 (6% annually) with n=360 is 0.00599. This requires a monthly payment of $599 on a $100,000 loan. In this type of process (installment loan process), the factors (1 - rk) are continually decreasing, until finally (1 - rN)=0 for some N. In our perspective here, we will still consider the corresponding infinite product:
P = (1 - r1)(1 - r2)(1 - r3) ... (1 - rN)(1 - rN+1) ... = 0,
where the rk for k>N are just irrelevant (0 < rk < 1). This artifice allows for the application of all the results of this chapter relating P to the infinite series, in a similar manner, by declaring that the partial sums:
Sn = r1 + r2 + r3 + ... + rn
tend to ∞ as n → ∞. In effect, the "irrelevant" rk do add up to ∞. However, as a practical matter, the fact that the partial sums, Sn, are actually finite numbers, and the fact that the Pk are greater than zero for k<N, allows for some very relevant use of the various inequalities and identities of this chapter, when S and P are replaced by Sk and Pk, for k<N. Thus the partial products can be estimated and expressed in terms of the corresponding partial sums, as is the situation in any infinite product case; it is just that the results become trivial, and are ignored for k>N. Also, by this artifice, we will be able to convert some rather fanciful applications into some more realistic ones.

CHAPTER 2. INCREASING PRODUCTS.


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SECTION 2A. INCREASING INFINITE PRODUCT.

We now consider a sequence of positive numbers, rk, without further restrictions. The corresponding n products:
(2.1)        Pn = (1+r1)(1+r2)(1+r3) ... (1+rn), for n = 1,2,3,...
form a sequence of increasing numbers which possesses a limit P as n→∞, which is either finite or ∞. In either case, we call P the infinite product and write:
(2.2)        P = (1+r1)(1+r2)(1+r3) ....
Once again, we consider the partial sums:
Sn = r1 + r2 + r3 + ... + rn
of the same numbers, and denote by S (finite or infinite) the limit of the Sn as n → ∞. Also, once again, we can express P in the form of an infinite series:
(2.3)        P = 1+ F,
where again:
(2.4)        F = r1 + P1r2 + P2r3 + ...
In this case, it is useful to observe that (since Pk>1 for all k), F>S, and so the inequality, 1+S<P, holds for all S. Thus, P = ∞ if S = ∞; and S < ∞ if P < ∞. (An obvious conclusion also from (2.2).) On the other hand, since (1+x)<ex for all x,
303.

we conclude that 1+rk<erk holds for all k. Therefore:
Pn = (1 + r1)(1 + r2)(1 + r3) ... (1 + rn)
< er1+r2+r3+...+rn = eSn
holds for all n, and so:
(2.5)        P < eS.
To obtain an improved lower bound for P, we note that for any R>0, the inequality, (1+R)x/R < 1+x holds for 0<x<R.
304.

Therefore, if S<∞, then max rk = R < ∞, and so necessarily:
(1+R)rk/R < 1 + rk
holds for all k. We conclude that:
(1+R)Sn/R < Pn
holds for all n. Thus P satisfies the inequality:
(2.6)        (1+R)(1/R)S < P.
Letting a = (1+R)(1/R), which lies between 1 and e, inequalities (2.5) and (2.6) demonstrate that P is sandwiched in between two increasing exponential functions of S:
(2.7)        aS < P < eS,
which is sharp (a nearly equal to e) whenever R = max rk is small. In any event, there is a number, bS, such that P satisfies the equality:
(2.8)        P = bSS (bS depends upon the series S, of course.
where 1<bS< e. Thus P is now essentially an increasing exponential function of S and:
(2.9)        P < ∞ if S < ∞.
             P = ∞ if S = ∞.
In applications, generally the divergent case, P = 1+F = ∞, F = ∞ retains some meaning, despite the lack of mathematical content, ∞ = 1 + ∞.

For the limiting situation of instantaneously increasing multiplicative processes, P(t), we have d(P(t)/dt = P(t)r(t), which results in:
(2.10)        P(t) = e   0∫tr(u)du.
With t → ∞, the completed process results in:
(2.11)        P = eS,
where P=P(∞) and S=0∫∞   r(u)du.

This is the limiting form of (2.8) for continuous, instantaneously increasing products.

SECTION 2B. INCREASING PRODUCTS THEOREM.

It seems useful at this point to summarize the essence of Chapter 1: Decreasing Products and Chapter 2: Increasing Products, as formal theorems:

INCREASING PRODUCTS THEOREM:
Let r1, r2, r3, ..., be a sequence of non-negative numbers, and employ the following notation:
(i). Pn = (1 + r1)(1 + r2)(1 + r3) ... (1 + rn)
(ii). P = limn → ∞ Pn (finite or infinite)
(iii). Sn = r1 + r2 + r3 + ... + rn
(iv). S = limn → ∞ Sn (finite or infinite)
(v). R = max1<k<∞ rk (least upper bound)
If R < ∞, then the following inequalities hold:
(1+R)Sn/R < Pn < eSn

(1+R)S/R < P < eS, and P = ∞, if and only if S = ∞.
If R = ∞, then P = S = ∞

SECTION 2C. DECREASING PRODUCTS THEOREM.

DECREASING PRODUCTS THEOREM:
Let (ii) through (v) be , as above, and replace (i) by:
(i). Pn = (1 - r1)(1 - r2)(1 - r3) ... (1 - rn),
with 0 < rk < 1 for all k. Then if R < 1, the following inequalities hold:
(1 - R)Sn/R < Pn < e-Sn

(1 - R)S/R < P < e-S

and P = 0 if and only if S = ∞.
If R = 1, then P = 0, and S = ∞. (The circumstance in which R > 1 is not permitted.)

SECTION 2D. ARITHMETICAL EXAMPLE OF INCREASING PRODUCTS.

We will conclude this chapter with a simple, arithmetical example of increasing products. For each x>1, let:
(2.12)        P0(x) = (1 + 1/x)(1 + 1/x2)(1 + 1/x3) ... (1 + 1/xn) ...
The corresponding infinite series:
S0(x) = 1/x + 1/x2 + 1/x3 ... + 1/xn ... = 1/(x-1)
converges, and thus by (2.5) and (2.6), P0(x) satisfies the inequalities:
(2.13)        (1 + 1/x)x/(x-1) < P0(x) < e1/(x-1),
since here R = 1/x. Now in the left member, we have:
2 < (1 + 1/x)x < e,
and so (2.13) can be rephrased as the inequalities:
(2.14)        21/(x-1) < P0(x) < e1/(x-1).
The left member of (2.14) is an accurate approximation for x near 1, while the right member is an accurate approximation for x → ∞. The graph of P0(x) is illustrated below. Equation (2.12) yields rather interesting rational products whenever x is an integer. For x=2, we obtain:
(3/2)(5/4)(9/8)(17/16)(33/32)(65/64)(129/128) ... = 2.3842... = bS0(2)
Now if 0 < x < 1, then the infinite product, (2.12) diverges, but the inequalities (2.5) and (2.6) (modified for partial products and series) can still be used to obtain estimates as to how fast divergence takes place. Indeed, if one desires estimates of the nth partial product:
(2.12)n        Pn(x) = (1 + 1/x)(1 + 1/x2)(1 + 1/x3) ... (1 + 1/xn)
for 0 < x < 1, then these modified inequalities imply that:
(2.15)n        (1 + 1/xn)(1 - xn)/(1-x) < Pn(x) < e(1 - xn)/(1-x)xn)
hold for all n and 0 < x < 1. This is because R is now 1/xn, and the partial sum:
Sn(x) = 1/x + 1/x2 + 1/x3 + ... + 1/xn = (1 + x + x2 + ... + xn-1)1/xn = (1 - xn)/(1-x)xn.
For x close to zero, (2.15)n becomes essentially:
(2.16)n        1 + 1/xn < Pn(x) < e1/xn,
while for x close to one, (2.15)n becomes essentially:
(2.17)n        2(1-xn)/(1-x) < Pn(x) < e(1-xn)/(1-x).
These inequalities supply considerable information above and beyond the simple conclusion that (2.12) diverges for 0 < x < 1.

342.


CHAPTER 3. INCREASING AND/OR DECREASING PRODUCTS.


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SECTION 3A. COMBINED INFINITE PRODUCTS.

We now combine the results of the previous two chapters to consider situations wherein some factors of an infinite product may be increasing (1 + rk) and/or some factors may be decreasing (1 - rk), with rk < 1.

We will again let S = S+ + S- denote the infinite series, r1 + r2 + r3 + ... consisting of all these numbers, rk, where, however:
(3.1)        S+ = sum of all the rk coming from factors of the form, (1+rk),
and:
(3.2)        S- = sum of all the rk coming from factors of the form, (1-rk).
Whenever S <∞, the order of the terms of the infinite series and the order of the factors in the infinite product is immaterial (absolute convergence. See Section 5J: Absolute Convergence). So we denote by P+ the product of all factors of the form (1+rk), and we denote by P- the product of all factors of the form (1-rk). Employing the results of Chapter 1: Decreasing Products and Chapter 2: Increasing Products, we conclude that the inequalities:
a+S+ < P+ < eS+
and:
a--S- < P- < e-S-
hold, and that P = P+P-, so long as S < ∞. Therefore, P satisfies the inequalities:
(3.3)        a+S+a--S- < P < eS+-S-,
so long as S = S+ + S- < ∞. Recall that:
a+ = (1+R+)1/R+,
where R+ = maxrkCS+ rk and:
a- = 1/(1-R-)1/R-,
where R- = maxrkCS- rk < 1. If R = max(R+,R-) is nearly zero, then both a+ and a- are nearly e, so that P is closely sandwiched in between two exponentials:
aS+-S- < P < eS+-S-
where a = [(1+R)/(1-R)]1/R is nearly e. In any event, we conclude from (3.3) that if S < ∞, then:
0 < P < 1 if -∞ < S+-S- < 0
and
1 < P < ∞ if 0 < S+-S- < ∞
Briefly, we have, for S = S++S- < +∞:
0 < P = P+P- < +∞ if and only if -∞<S++S-<∞
We will employ the symbolism -∞ = S+ - S- to mean that S+ < ∞ and S- = ∞, while S+ - S- = ∞ means that S+ = ∞ and S-<∞. This symbolism makes sense, since each partial product Pn always factors conveniently into two partial factors of P+ and P-, so that Pn → P+P- equals either zero or infinity in the above two circumstances. The same thing happens with each partial sum, Sn, so that the sum of two partial sums of S+ and S- lead to S+ - S- = ±∞, with the indicated results. Then we can extend the above to:
(3.4)        0 < P = P+P- < ∞ if and only if S+ + S- < ∞,
and
              P=0 if -∞ = S+ - S- and P=∞ if S+ - S- = ∞.
The circumstances S+ = S- = ∞, i.e., P+ = ∞, P- = 0, when S = ∞, are indeterminate, of course, and include cases wherein conditional convergence can take place for S and for P with suitable rearrangements of terms and factors, but without implications between the two of them. Generally speaking, our applications retain some meaning in these circumstances as well as in the determinate ones.

We will illustrate the workings of inequalities (3.3) with the following simple arithmetical example. For each x > 1, consider the mixed increasing and decreasing product:
(3.5)        Pm(x) = (1 - 1/x)(1 + 1/x2)(1 - 1/x3)(1 + 1/x4) ...
for which:
S+ = 1/x2 + 1/x4 + 1/x6 + ... = 1/x2[1 + 1/x2 + 1/x4 + ...] = 1/x2[1/(1-1/x2)] = 1/(x2-1),

S- = 1/x + 1/x3 + 1/x5 + ... = 1/x[1 + 1/x2 + 1/x4 + ...] = 1/x[1/(1-1/x2)] = x/(x2-1),

and

S+ - S- = (1-x)/(x2-1) = -1/(x+1).
Since, in this case, R+ = 1/x2, R- = 1/x, and a+ = [1 + 1/x2]x2, a- = 1/[1 - 1/x]x the inequalities (3.3) become explicitly:
[(1 + 1/x2)(1 - 1/x)]x2/(x2-1) < Pm(x) < e-1/(x+1),
which demonstrate that 0 < Pm(x) < 1 for 1 < x < ∞. Clearly, both sides of these inequalities (consequently Pm(x) itself) tend to 1 as x → ∞, while the left-hand member tends to 0 as x → 1+, and the right-hand member tends to e-1/2. These latter limits, of course, are not definitive, but since (1-1/xn) < (1-1/xn+1), one sees that the product (3.5) also satisfies the inequality:
Pm(x) < (1 - 1/x2)(1 + 1/x2)(1 - 1/x4)(1 + 1/x4) ... = (1 - 1/x4)(1 - 1/x8) ...,
which guarantees that P(x) → 0 as x → 1+. For subsequent use, we extend this product to all real x by defining Pm(x) = 0 for x < 1. It is interesting to note that if the plus and minus signs in (3.5) are interchanged, to yield:
PM(x) = (1 + 1/x)(1 - 1/x2)(1 + 1/x3)(1 - 1/x4) ...
then the inequalities (3.3) become explicitly:
[(1 + 1/x)(1 - 1/x2)]x2/(x2-1)] < PM(x) < e1/(x+1),
and:
PM(x) < (1 - 1/x4)(1 - 1/x8) ...
because:
(1+1/xn) < (1+1/xn-1)
Thus again, PM(x) → 1 as x → ∞ and PM(x) → 0 as x → 1+. The functions Pm(x) and PM(x) have rather simple (but rather interesting) graphs as illustrated below:

353.




SECTION 3B. ARITHMETICAL EXAMPLE OF INCREASING AND DECREASING PRODUCTS (FOR THE "ADVANCED" READER):

We will now examine the more challenging mixed infinite product (depending upon the real number x):
(3.6)        P(x) = (1 - x2)(1 + (x2)2/2!)(1 - (x2)3/3!)(1 + (x2)4/4!) ...,
first for 0 < x < 1. All factors are positive, and the combined infinite series:
(3.7)        S(x) = -x2 + (x2)2/2! - (x2)3/3! + (x2)4/4! - ...,
is clearly convergent for such x, and might be recognized as the power series expansion (about zero) of
e-x2-1.
Since R = x2, the infinite product is thus given approximately by:
P(x) ~ eS(x) = e(e-x2-1),
for small x. Therefore, P(x) → 1 as x → 0, and is initially decreasing with x increasing from zero. More precisely, we note that each of the infinite series (see (3.1) and (3.2)):
S+(x) = (x2)2/2! + (x2)4/4! + (x2)6/6! + ...
and:
S-(x) = x2 + (x2)3/3! + (x2)5/5! - ...
is convergent, and so according to (1.7) and (2.5), the inequalities:
0 < P+(x) < eS+(x)
and:
0 < P-(x) < e-S-(x),
hold. Therefore, by (3.3), we have:
0 < P(x) = P+(x)P-(x) < e[S+(x)-S-(x)].
But:
[S+(x) - S-(x)]
= [(x2)2/2! - x2] + [(x2)4/4! - (x2)3/3!] + [(x2)6/6! - (x2)5/5!] + ...
= (x2)/2! [x2 - 2] + (x2)3/4! [x2 - 4] + (x2)5/6! [x2 - 6] + ...,
which is certainly negative.

On the other hand, since:
1 - 1/N! < 1 - (x2)N/N!
and:
1 + (x2)N+1/(N+1)! < 1 + 1/(N+1)!
hold (so long as 0 < x < 1), the inequalities :
(3.8)        1 - 1/N! < (1 - (x2)N/N!) (1 + (x2)N+1/(N+1)!) < 1 + 1/(N+1)!
hold for N = 1, 3, 5, 7, 9, ..., we can take advantage of these factorial divisors to accurately estimate the infinite product, (3.6). If we employ only the first two factors, then we obtain (using (3.8) for N = 3, 5, 7, 9, ...):
(1 - x2)(1 + (x2)2/2!) (1 - 1/3!)(1 - 1/5!) ... < P(x) < (1 - x2)(1 + (x2)2/2!) (1 + 1/4!)(1 + 1/6!) ...
i.e.,
(1 - x2)(1 + (x2)2/2!)(0.82625) < P(x) < (1 - x2)(1 + (x2)2/2!) (1.04314),
for 0 < x < 1 (accurate to the degree exhibited). If greater accuracy is desired, we can employ the first four factors to obtain:
(3.9)        (1 - (x2)(1 + (x2)2/2!)(1 - (x2)3/3!)(1 + (x2)4/4!)(0.99150) < P(x) < (1 - (x2)(1 + (x2)2/2!)(1 - (x2)3/3!)(1 + (x2)4/4!)(1.00141)
for 0 < x < 1 (accurate to the degree exhibited). Clearly, therefore, P(x) → 0 as x → 1, because of the first factor, (1 - x2). If this factor is removed, then the truncated infinite product, P(x)/(1 - x2) becomes (1+1/2!)(1-1/3!)(1+1/4!)(1-1/5!) ... = 1.29123 for x=1, and the truncated infinite series,
S(x)-1 + x2 = e-x2-1 + x2
becomes 1/e. Moreover, the remaining factors of P(x)/(1 - x2) are all positive for 0 < x < 61/6 (>1), and so P(x) converges there because the truncated series does. (It converges to e-x2-1 + x2). This number, 61/6 (>1), is the value of the next larger zero of P(x), when (1 - x2)3/3!)=0. Quite generally, P(x) has zero factors for x=1, 61/6, 1201/10, ... (N!)1/2N, ..., whenever (x2)N/N! = 1. These zeros, xN = (N!)1/2N → ∞, as N → ∞, (odd N), but very slowly. If the second, potentially vanishing factor, (1 - (x2)3/3!) is also removed from the infinite product, then the truncated product, P(x)/(1 - (x2)(1 - (x2)3/3!) of positive terms converges for x in the extended range, 0 < x < (120)1/10 = x5 , because the truncated corresponding series, S(x) - x + x2 - (x2)3/3!) does. (It converges to e-x2-1 + x2 - (x2)3/3!)). The value of P(x) / (1 - x2)(1 - (x2)3/3!) becomes (1 + 62/3/2!)(1 + 64/3/4!)(1 - 65/3/5!) ... = 3.20... for x = 61/6 = x3. In this way, we conclude:
(a) that the infinite product (3.6), P(x), converges for all positive x;
(b) that P(x) becomes zero for each of the special numbers, x1, x3, x5,..., (N!)1/2N,...;
(c) that P(x) oscillates from positive to negative values between these numbers; and
(d) that the magnitudes of the oscillations increases as x increases to ∞.
The somewhat startling graph of P(x) is illustrated below. For future reference, we note that P(x) extends naturally to an even function of x for all real x:

343.



SECTION 3C. ZERO AND NEGATIVE FACTORS.:

In most cases considered in the above chapters, our factors in an infinite product are assumed to be positive numbers. Zero factors come into play rather naturally for installment loan processes, and can be treated by the artifice discussed in Chapter 1: Decreasing Products. However, the inclusion of infinitely many negative and positive factors allows only for a convergent product P which is zero (since then there would be infinitely many changes of signs with the Pn).

This situation can be the case if and only if the relevant series, S, is divergent. In fact, if we remove all the minus signs, so that all the factors are positive, then the conclusion stems from the results of this chapter above: P→0 if and only if S+ - S- → -∞. If there are only finitely many negative factors, these can be removed, the remaining factors treated as above, and then the negative factors can be applied to the result, yielding a final conclusion as to the value and the sign of P. We know of no real-life applications of such products, and will not pursue the mathematical aspects any further at this point, except to finish this chapter with an intriguing example, exemplifying these auxiliary concepts.

Let r1, r2, r3,... be a sequence of monotonically decreasing, positive numbers, with limit zero. We then define a function by the infinite product:
(3.10)        P(x) = (1 - r1/x)(1 - r2/x)(1 - r3/x) ...,
which converges (perhaps to zero) for each x>0, since rk/x < 1 for all k sufficiently large. For each x>0, there are at most finitely many of these factors which are negative, and there is a change of sign of P(x) (when nonzero) as x transverses each rk. Of course, P(rk)=0 for every k, since one factor in (3.10) vanishes. Moreover, if the infinite series, r1 + r2 + r3 + ... = ∞, then by (1.11), P(x)=0 for all x>0. However, if r1 + r2 + r3 + ... < ∞, then |P(x)|>0, so long as x ≠ r1, r2, r3, .... In this latter case, P(x) oscillates from positive to negative values, with x varying between the rk. Clearly, P(x)>0, when x is greater than all rk, and P(x) → 1 as x → ∞, by (1.9), since S(x) = (r1 + r2 + r3 + ...)/x → 0 and R = r1/x → 0(i.e., a → e) as x → ∞. However, it is also clear that the individual factors in (3.10) tend to -∞ as x → 0. Thus, the values of |P(x)| for x lying between the rk "tend" to be large for small values of x, while P(x) actually vanishes at each rk. The spectacular nature of the graph of y=P(x) is critically affected not only by the positions of the rk, but also by their spacings along the x-axis. Recall that P=0 if r1 + r2 + r3 + ... = ∞, so that even for rn = 1/n, this spectacular nature simply vanishes as every infinite product, (3.10) is zero.

344.


CHAPTER 4. APPLICATIONS OF DECREASING PRODUCTS.


Next Chapter.
Previous Chapter.
Return to Table of Contents.

SECTION 4A. RECTANGLE-FILLING.

The following example provides a visible illustration of the overall meaning of decreasing products. The Filling lemma obtained is a universal maxim, and all applications of such INFINITE products can be viewed in the geometry of this concrete example.

Suppose we ask: Can a succession of symmetrically-placed isosceles triangles (pointing upward) and non-overlapping ultimately fill a rectangle? A sequence of steps might appear as follows:

STEP 1.
31.


STEP 2.
32.


STEP 3.
33.


STEP 4.
34.


STEP 5.
28.



  • If the first triangle fills an r1 fraction of the area of the rectangle (r1 = 1/2?), leaving a (1-r1) fraction of the area;

  • the second two triangles fill an r2 fraction of the remaining area, leaving a (1-r1)(1-r2) fraction of the area;

  • the next six triangles fill an r3 fraction of the remaining area, leaving a (1-r1)(1-r2)(1-r3) fraction of the area;

  • the next 18 triangles fill an r4 fraction of the remaining area, leaving a (1-r1)(1-r2)(1-r3)(1-r4) fraction of the area;

  • and if this process is continued indefinitely, then the infinite product:
    P = (1-r1)(1-r2)(1-r3)(1-r4) ....
    gives the final fraction of area remaining.
    • If the final fraction is zero, then the rectangle has been completely filled with isosceles triangles. If P>0, then that number is the fraction of the area of the rectangle that has been omitted, and:
    F = 1 - P = r1 + (1-r1)r2 + (1-r1)(1-r2)r3 + ...
    is the fraction of the area of the rectangle that has actually been filled. Using (1.11) from Chapter 1: Decreasing Products, we can formulate these results as follows:

    FILLING LEMMA. A sequence of non-overlapping isosceles triangles, conforming to the fractions rn of unfilled portions of a rectangle, will completely fill the rectangle, if and only if the corresponding infinite series:
    S = r1 + r2 + r3 + r4 + ...
    diverges (S=∞). Whenever S converges (S<∞), the corresponding infinite product P is positive, and equals the fraction of the area of the rectangle that is unfilled.

    For example, if, beyond r1 = 1/2, all rn = 1/4 (the maximum allowed), then:
    S = 1/2 + 1/4 + 1/4 + 1/4 + 1/4 + 1/4 + ... = ∞,
    and the corresponding isosceles triangles completely fill the rectangle (P=0). This is the case pictured previously. We illustrate other possibilities below, but first give another view of the lemma. Indeed, using (1.9) with R = 1/4, we can rephrase the Filling lemma quantitatively through the simple inequalities:
    (256/81)-S < P < e-S
    for the unfilled portion, P. Curiously, (256/81) = 3.16... ~ π = 3.14.... Clearly, the conclusions of the lemma stem directly from these inequalities, which, in turn, impose restrictions on all such rectangle-filling processes.

    However, we need to clarify what happens with the isosceles triangles when the fractions are less than 1/4, and especially when they are small. As suggested, no fractions rn (except r1 = 1/2) can be greater than 1/4. This is because each subsequent isosceles triangle pointing upward must be placed somewhere in a skewed triangle pointing downward. As shown below, this results in either a fat triangle or a slim triangle with a reduced area, or a maximum-area isosceles triangle with 1/2 the dimensions of the skewed triangle.

    FAT:
    36.


    MAX (1/4):
    73.


    SLIM:
    38.



    So if an isosceles triangle corresponds to a fraction greatly less than 1/4, then it must be either a very fat one or a very slim one, and the resulting pictures can be much different in appearance than those shown above.

    To illustrate this fact, we will examine what takes place if we use a sequence of decreasing fractions represented by the terms of the harmonic series, rn=1/n (for n > 4). Since the series diverges, the isoceles triangles completely fill the rectangle. We show below only the SLIM triangles when n = 1/4, 1/5, 1/6, and 1/7, where some already appear merely as vertical line-segments, and are barely visible. The subsequent ones will all begin to appear as vertical line-segments as 1/n decreases.
    355.

    A more extreme case is illustrated by choosing the rapidly decreasing fractions rn satisfying (for a fixed number, s>1):
    r1 = 1/2s,
    r2 = 1/3s,
    r3 = 1/5s,
    r4 = 1/7s,
    r5 = 1/11s, ...,
    where rn = 1/ps, and p is the nth prime number. We refer to these as Riemann zeta-function numbers. Furthermore, for convenience, we disregard the first (big) triangle with area equal to 1/2 (now labeled r0), and consider the subsequent isosceles triangles as filling (if so) the remaining area of the rectangle. Actually in this particular case, the relevant series:
    S = ∑ rn = ∑ 1/ps = 1/2s + 1/3s + 1/5s + 1/7s + 1/11s + ....
    converges when s > 1, and so the corresponding product:
    (4.1)        P = (1 - 1/2s)(1 - 1/3s)(1 - 1/5s) (1 - 1/7s)(1 - 1/11s) ...
    is greater than zero. Thus, according to the Filling lemma, not all the rectangle is filled. A fraction, F=1-P, of the area of the rectangle remains unfilled by triangles. The following picture depicts a select few of the slim isosceles triangles prescribed by the Riemann zeta-function numbers as the (unsuccessful) filling process proceeds, when s=2.
    356.



    All the triangles occurring during the first couple of steps are shown. We here exhibit only some of the more visible ones for the next three steps. Beyond these, all triangles, if exhibited, would be indistinguishable from vertical spiked segments. The picture (beyond the initial triangles), if completed, would appear throughout, mostly as a dense forest of vertical segments. Notice how even the visible triangles shown climb up the big triangle in ever-decreasing increments. The last ones exhibited here correspond to the fraction, (1/49). They do not reach to the top, so that there are two empty spaces there. We have chosen not to show the forest throughout, but suggest what will happen in a portion of the rectangle. This dense forest pattern is omnipresent in the upper regions of all areas between the visible, or nearly-visible triangles. It's a startling picture contemplated here, particularly realizing that approximately 60% of this area of the rectangle is not filled by the isosceles triangles. Of course, fat triangles could be substituted for the slim ones, or a mixture of the two, to obtain other startling pictures. The following picture depicts some of the FAT isosceles triangles represented by the Riemann zeta numbers for s=2. These are labeled to indicate the steps of the process. Again, these become indistinguishable from (horizontal) line-segments as n increases. They do not fill up the lower areas. Finally, we illustrate what the picture becomes when slim and fat triangles are alternated during the first four steps of the process. The triangles rapidly morph into vertical and horizontal line-segments.
    357.


    375.


    376.



    The quantities P and F could be estimated directly using (1.9), in terms of S, but in this case we revert to some convenient historical events for assistance. For in 1859, Bernhard Riemann˚ defined a function of s>1 by an infinite series:
    (4.2)        ζ(s) = 1 + 1/2s + 1/3s + 1/4s + 1/5s + 1/6s + ...        (all integers)
    and chose to call it the zeta-function˚. Ever since that time, the function has been called this, and is considered "well known". Some hundred years earlier, Leonhard Euler˚ had shown that the reciprocal of this series (4.2) could be written as an infinite product:
    (4.3)        1/ζ(s) = (1 - 1/2s)(1 - 1/3s)(1 - 1/5s)(1 - 1/7s)(1 - 1/11s) ...        (all primes)
    Therefore, according to (4.1):
    P = 1/ζ(s)
    gives the final fraction remaining when rn=1/ps. The final fraction of the rectangle actually filled is, of course:
    (4.4)        F = 1-P = 1 - 1/ζ(s) = 1/2s + (1 - 1/2s)1/3s + (1 - 1/2s)(1 - 1/3s)1/5s + ...
    for s>1, according to (1.5) and (1.6). If, on the other hand, we fill the rectangle with triangles conforming to the Riemann zeta-function numbers, with s lying between 0 and 1 (we have to drop some early terms greater than 1/4), then the corresponding series ∑ rn = ∑ 1/ps diverges, and so P=0, and the final fraction filled becomes, simply:
    (4.5)        F = 1.
    Thus the rectangle is completely filled only when 0<s<1. This situation is depicted below where we give a plot of F versus s for all s>0. (See Ref. [7.8] for information concerning the convergence or divergence of ζ(s) in (4.2).)
    306.

    In this very special Riemann zeta-function case, we have S = ∑ 1/ps, which means that big S becomes a function S(s) of little s, as we vary little s. From (1.10), we have further:
    P = bS-S.
    So in this case, P also becomes a function P(s) of little s. This function, of course, is none other than 1/ζ(s) for s>1. Using the results from Chapter 1: Decreasing Products, we know that the number bS is nearly equal to e if R = max rk = 1/2s is small. Hence for large s (both R and S →0 as s → ∞), we have approximately:
    P(s) ~ e-S(s) = e-∑1/ps
    with S(s) → 0 as s → ∞. Thus P(s) → 1 exponentially as s → ∞. For 0<s<1, P(s)=0. Elsewhere:
    P(s) = bS(s)-S(s) < e-S(s)
    holds, since e<bS(s). Thus P(s)→0 exponentially as s→1+, since S(s)→∞ as s → 1+. All these facts are illustrated in the the accompanying figure:

    305.

    A rather interesting consequence here is that the equality:
    ζ(s) = bS(s)S(s)
    holds for all s>1. So we have here an exact manifestation of (1.10) over a complete range of values of s. This gives us a means of computing the nefarious number, bS, namely:
    (4.6)        bS(s) = elogζ(s)/S(s) = ζ(s)1/S(s),
    where S(s) = ∑ 1/ps (all primes) and ζ(s) = ∑ 1/ns (all integers). Equation (4.6) confirms the fact that bS(s) → e as s → ∞, since:
    ζ(s) = 1 + (1/2s + 1/3s + 1/4s + ...) = 1 + θ
    and:
    S(s) = 1/2s + 1/3s + 1/5s + ... = φ (<θ)
    Thus:
    ζ(s)1/S(s) = (1+θ)1/φ → e
    with both φ, θ→0 as s→∞ and φ/θ → 1. (See: Chapter 5: Applications of Increasing Products, Section 5B: Computable Increasing Products, for additional arithmetical details concerning these two infinite series.)

    This rectangle-filling application of decreasing products illustrates two aspects of a certain process and just what may be considered a successful conclusion. The filling is found to be successful when P=0 (Pn decreases to 0), and this success is actually manifested in the equation, F=1, i.e., the rectangle is completely filled (Fn increases to 1). However, the non-filling example (using the Riemann zeta-function˚ numbers) can be thought of as furnishing a more interesting mathematical and intriguing picture of this process. The failed filling illustration, for example, leads to interesting Lebesgue-integrable functions˚ if the filling triangles are lowered to the base of the rectangle, and the fraction F is actually the value of the Lebesgue integrals˚ This seems something of a success. We will encounter this dual interpretation of success in other applications.

    SECTION 4B. PROCESSES ANALOGUOUS TO RECTANGLE-FILLING.

    Here we supply very brief descriptions of the examples, as the finer details have been covered in Section 4A: Rectangle-filling. One could envision the rectangle-filling process for both qualitative and quantitative details.

    SECTION 4B(i). PERSON DISPOSES PERIODICALLY OF FRACTIONAL WEALTH.

    Suppose a wealthy individual wishes to dispose, periodically, of a fraction of his wealth, and makes the decision as to what fraction, rk to dispose of at each period, k, based upon the current balance in his bank account. (No other funds are involved). Then after n periods, his current balance becomes (in proportion to his initial wealth):
    Pn = (1-r1)(1-r2)(1-r3) ... (1-rn),
    and he has dispersed the portion:
    Fn = 1 - Pn = r1 + P1r2 + P2r3 + ... + Pn-1rn
    of his wealth. If he ultimately disposes all of his funds, then Fn → F = 1 and Pn → P = 0. On the other hand, he retains some of his wealth, F=1-P, if P>0. What counts as success here is up to one's point-of-view. Either the starving public gets it all, or his deserving (selfish) heirs may inherit some of it.

    SECTION 4B(ii). GOVERNMENT TAXES A CITIZEN PERIODICALLY OF WEALTH.

    We can reverse this dispersal idea completely by imagining a government (greedy political system) which taxes each of its citizens, periodically, a fraction, rk, of one's current wealth. If a citizen has some wealth, but no income, he eventually loses it all if:
    P = (1-r1)(1-r2)(1-r3) ... = 0.
    His heirs may inherit something, F=1-P, if P>0.

    SECTION 4B(iii). MOMENT-TO-MOMENT JOB EVENTUALLY COMPLETED.

    Any day-to-day (or moment-to-moment) job is eventually completed if a fraction, rk, of an unfinished job at each day (or moment) k is secured, and if P=0. If P>0, then the job is never completed. Imagine a preacher trying to save the souls of his congregation, and at each k (Sunday, of course), he is successful with an rn fraction of the unsaved. If P=0, then the preacher wins over all the souls. One could imagine many such (silly?) situations, such as painting a house over many days, successively spraying to kill weeds, inoculating or vaccinating a population, in fact, just any process which takes into account the previous successes, and rk becomes the kth fractional advance based upon the previous success. Other processes more reasonably modeled by such products are discretely or continually decaying-type physical processes.

    SECTION 4B(iv). CONTINUALLY DECAYING PROCESSES.

    For continually decaying processes, see (1.13). It seems traditional to accept the model wherein the fractional rate of decay, r(t), of an isotope is assumed to be constant. This assumption forces the discussion to revert to the so-called half-life concept (where F(t)=1/2), in order to avoid the (nonrealistic) complete vanishing of the isotope as t → ∞. A more realistic model is one in which the rate of decay, r(t), decreases with increasing t, and leads to a final fractional product:
    (1.14)        P(∞) = e -0∫∞r(u)du
    greater than zero for the residue. Since these decaying processes are not actually continuous, the use of a discrete product model:
    P = (1-r1)(1-r2)(1-r3) ...
    might even be more appropriate. Here, rn might reflect and incorporate some statistical aspects of the true physical process.

    SECTION 4C. INSTALLMENT PURCHASES.

    Installment purchases, as commonly understood and explained previously, become infinite products only through the use of an artifice. These are instances where a complete payoff occurs after a finite number of payments, and the representing product becomes zero, due to a zero factor, (1-rN). However, since in many cases, the finite number, N, may be large, and the treatment requires a careful examination of a process whereby credit for a payment is based upon the unpaid balance, the inclusion of installment purchases and similar processes seem appropriate for this presentation. Also, it will allow for, perhaps, greater realism, when we later revisit some of the above (fanciful) applications. In Chapter 1: Decreasing Products, we derived the mathematical framework for installment purchases (phrased as a home purchase situation), where the fraction of the loans paid after payment n, is given by:
    (1.20)        Fn = (p-j)[(1+j)n-1)/j].
    Here, p represents the constant payment, and j the constant interest rate charged. We recall that this conclusion is based on the facts that (in general):
    Fn = r1 + P1r2 + P2r3 + ... + Pk-1rk + ... + Pn-1rn,
    and that the kth contribution to the principal, namely, Pk-1rk, is the payment, p, minus the interest payment, jPk-1, that goes to the lender. The simplest application of (1.20) is to an installment loan for which the total number of payments, N, and the interest rate, j, are specified. Then one calculates the payment, p, by setting FN = 1, which means that the loan is completely paid off at payment N. This results in:
    (1.21)        p = j(1+j)N / [(1+j)N-1-1]
    for the fractional payment, which, in real life, is multiplied by the original loan amount. This formula represents the traditional home-buying, car-buying, or other installment-buying schemes. We might examine one of the simplest of these schemes, where the loan is refinanced at some stage k<N. In this situation, the balance of the loan is obtained from (1.20) with n=k and Pk = 1-Fk, At this point, the lender and borrower agree to another set of conditions: a new N, a new j, and, consequently, a new p, given by (1.21), but most importantly, a new starting-balance which could be Pk, multiplied by the original starting balance. However, often times this amount is increased or decreased, as the borrower either takes out extra cash or pays down something on the principal. Such refinancing schemes are fairly common, and are often repeated numerous times during the course of the loan contract. One item of concern may be the amount of interest paid. For a non-refinanced loan, this amount is simply (Np-1) multiplied by th original loan amount. Upon refinancing at k, the interest paid up to that point is, similarly, kp - Fk, multiplied by the original loan balance. Often times, the consideration of increased or extra auxiliary payments along the way are contemplated. We might take note of just how such payments affect the mathematics, and, more importantly, the borrower's fortunes. The precise mathematics is not very revealing, but an alternative perspective is. For example, if an auxiliary (fractional) payment, q, is made just after the kth step of an installment loan, then the value of such a payment is reflected in an increase, q, of the paid-off portion, Fk. The lender just reduces the loan balance fraction by the amount, q. Thereafter, (n>k), the fraction paid off becomes:
    Fn = ((p-j)/j)[(1+j)n-1] + q(1+j)n-k
    Clearly, the earlier the auxiliary payment is made, the sooner the loan is paid off (i.e., FN = 1). It is more revealing to determine approximately how many extra payments the auxiliary payment of q is worth to the borrower. To this end, one determines an ℓ>k such that Fℓ = Fk + q.

    Using the values for Fk and Fℓ given by (1.20), one obtains such an ℓ. After some manipulation (recall that log xy = y log x) we get:
    (4.7)        ℓ = k + log[1 + jq/(p-j)(1+j)k]/log(1+j)    ~    k + q/(p-j)(1+j)k
    Of course, it is just the nearest integer in ℓ that matters here. In Chapter 1: Decreasing Products, we noted a home purchase case where j=0.005 and p=0.00599, for example. With these data, and with one auxiliary payment of q=p=0.00599 at k=10, we obtain from (4.7), ℓ=15.7. So, one extra regular payment after ten months is worth about 6 payments that show up at the end. If the extra payment is made after 120 months (10 years), then the extra payment is worth only about 3 regular payments at the end.

    Other installment loan schemes might be those requiring an early lump sum payment (same as when refinancing), or, variable regular loan payments, and/or variable interest rates (quite common nowadays). For these, one might determine the successive fractions, rk, from the requirement that Pk-1rk = p - jPk-1 holds (where p and j may change with k), and obtain the developing loan balance from the product:
    Pn = (1-r1)(1-r2)(1-r3) ... (1-rn).
    This requires the strange choice of fractions:
    rk = j(p-j)(1+j)k-1/[p - (p-j)(1+j)k-1]
    where the p and j may change with k. In lieu of this procedure, one could employ the formula (1.20) stepwise, as the parameters p and j change, and obtain a succession of values for the paid-off fraction, Fk. Then the sum of these, in turn, depicts the developing stepwise loan balance, Pk = 1 - Fk.

    The paid portion, Fn, is always subjected to the limitation, Fn<1. A simple no-interest loan, for example, corresponds to Fn=np, so that Np=1, simply says that the fractional payment, p, is just 1 divided by the number N of payments.

    SECTION 4D. FINITE PRODUCTS FOR REALISM.

    We now examine some situations wherein a product, PN, terminates in a zero value with a factor (1-rN)=0. Installment loans are straightforward examples, but other scenarios are possible. In the usual installment loan case, the factors, (1-rn), decrease in a systematic fashion, dictated by the somewhat strange requirement that the two quantities, Pk-1(rk-j)=p and j remain constant throughout. This requirement leads to equation (1.20), for Fn = 1-Pn, which, in turn, is required always to be less than or equal to 1. The process terminates when FN=1, rN=1, and:
    PN = (1-r1)(1-r2)(1-r3) ... (1-rN) = 0,
    for some N.

    Perhaps, most of the fanciful real-life applications considered in Section 4B: Processes analogous to rectangle-filling. become more acceptable when a finite product model, terminating as above, is used in lieu of the infinite product model. For example, the wealthy philanthropist in Section 4B(i) might wish to select his fractional payment, p, retaining a fraction, j, for his expenses, and then calculate from (1.20), how many steps:
    (4.8)        N = log[1+j/(p-j)] / log(1+j) = log[p/(p-j)] / log(1+j)
    it takes to dispose of the whole package. The philanthropist could stop at k<N to leave something, Pk=1-Fk, with Fk, given by (1.20), as inheritance for his heirs.

    Other scenarios are possible, and he might, after choosing j for his expenses and anticipating a subsequent life-expectancy of N years, use (1.21), with n=N to calculate the constant withdrawal fraction p from his bank account, and leave nothing for his heirs. In Section 4B(ii) , a more "benevolent" government might let its citizens retain a small portion, jPk, of current wealth for expenses, but still take a fraction, q=p-jPk, of that current wealth. Here, p=q+jPk then becomes the constant periodic drain on one's bank account, and (4.8) gives the number of years, N, after which there is nothing left to tax.

    For most of real-life situations, coming to a final result after a finite number of steps, N, equations (1.20), (1.21), and (4.8) are useful for reinterpreting otherwise infinite processes. Instead of choosing fractions rk, to accomplish a certain result after infinitely many steps, one could seek interpretations of the parameter p and j to fit a given application, and then follow the progress according to (1.20), so long as Fn<1, with (4.8) giving the terminal step. If, on the other hand, a terminal step, N, is prescribed, then the parameters p and j are restricted by equation (1.21), with n=N. In all cases, p must exceed j but cannot exceed 1. It is possible for j to be negative (but greater than -1), where, for example, one receives assistance in paying down a loan.

    We will briefly illustrate the reinterpreting technique, by considering again the interesting problem of Section 4A: Rectangle-filling. Previous considerations were given to filling a rectangle in infinitely many steps by the use of isosceles triangles. These isosceles triangles put a limitation rk<1/4 (after r1=1/2) on the possible fractions that can be used. If we were to allow other filling geometries, then these restrictions can be relaxed to rk<1. One need only select figures Δk (any shape) within the unfilled portions, in order to proceed with any rk sequence. If, in addition, after some selected step, k, a terminating finite sequence rn is then chosen to yield a zero factor, (1-rN)=0, then much of the original character can be preserved, while terminating the process after finitely many steps. For example, if we fill the rectangle with isosceles triangles through step 11, using the Riemann zeta-function numbers for triangles with s=2, then we retain the interesting picture shown, but we can terminate the filling process, starting with the fraction remaining at step 12.

    SECTION 4E. FURTHER FINITE PRODUCTS.

    Following the model of installment loans, we now consider a less-realistic problem, in which a payment (instead of being reduced by an interest payment to the lender) is actually enhanced somehow by some mechanism. Perhaps some relative contributes to the payment. In any case, the extra contribution made is proportional to the current loan balance. This means that the loan is paid off much sooner. To this end, we suppose that the fractions, rk, are selected to satisfy the equation, rkPk-1 = p + jPk-1, with p and j between 0 and 1, but with p+j<1. Then since Pk = 1 - Fk, and Fk = Fk + rkPk-1 ... , we obtain the recursion formula:
    Fk = Fk-1(1-j)+(p+j).
    Starting with F0=0, we obtain (recall the analogous steps leading to (1.20) in Chapter 1: Decreasing Products):
    (4.9)        Fn = (p+j)[1-(1-j)n]/j,
    which holds so long as Fn <1. If FN=1 for some N, then p satisfies:
    (4.10)        p = j(1-j)N / [1 - (1-j)N].
    This is the assisted payer's contribution at each step. If p and j are prescribed, then the process terminates at step N (greatest integer in N) given by:
    (4.11)        N = log(p/p+j) / log(1-j).


    These various formulas convey the mathematical content of an installment loan being paid off, with a borrower's contribution being enhanced by a contribution proportional to the compound balance. Lenders don't do this sort of thing, so the extra payments must come from some other source.

    If the home-purchasing data p=0.00599 and j=0.005 are employed in the last formula, (4.11), then it is interesting to observe that the loan is paid off in approximately N=125 months, rather than the usual 360 months. Alternatively, if the next-to-last formula, (4.10), is used, then one finds that the reduced payment of approximately p=0.001 per month is required for a standard 360-payment loan.

    A wierd application could be furnished by an army eliminating its enemy, a p-fraction-a-day while getting assistance from some of the survivors just committing suicide. Here, Pn = 1 - Fn represents the fraction of survivors, while jPn represents the fraction which commit suicide. Of course, PN=0 represents the complete annihilation of the enemy. It takes N, given by (4.11), days to do the job. If p=1/100 and j=1/1000 the war is over in N=95 days, rather than 100 days. One wonders if there are physical processes with such strange characteristics? A decaying process getting assistance during the decay from the remaining elements? In the medical world?

    In the political world, a political party could receive assistance in eliminating its opposition, by having some of the unpersuaded citizens simply drop out politics. The plan would be to convert a fraction, p, of citizens each year, and hope that of the remaining Pn citizens, each year a fraction, jPn, of citizens become discouraged. Equation (4.9) gives the fraction of citizens converted at year n, and equation (4.11) gives the year at which the opposition party is completely eliminated. Politics should be so simple? Two-parameter politics? If p=10-3 and j=10-4, it takes a long time, N=950 years, to become a one-party country (like the old Soviet Union or Saddam Hussein's Iraq), using these fanciful numbers.

    Suppose we wish to eradicate a pest, by successively spaying the population. If we eliminate a fraction, p, at each spaying, k; and of the remaining fraction, Pk-1, a fraction, jPk-1, dies anyway from other causes; then the fraction eliminated after n spayings is given by (4.9), and the number of spayings required to completely eliminate the pests is given by (4.11).

    SECTION 4F. SOME DIVERGENCE-CONVERGENCE CONSIDERATIONS.

    In order to emphasize the central role played by rectangle-filling schemes for understanding decreasing products, we consider a modified version of the Riemann zeta-function example (See: Section 4A: Rectangle-filling). For this demonstration, suppose that we now extend our definition of the Riemann zeta-function:
    (4.2)        ζ(s) = 1 + 1/2s + 1/3s + 1/4s + 1/5s + 1/6s + ...        (all integers)
    to encompass (symbolically, at least) the divergent range, 0<s<1, as well as the usual convergent range, 1<s < ∞. In this application, we will reconsider the problem of filling a rectangle with triangles, corresponding to the fractions, 1/ns, from (4.2), rather than the subset of the Riemann zeta-function numbers, 1/ps (primes only), used previously in Section 4A: Rectangle-filling. The modified process is now governed by the infinite product:
    (4.12)        P(s) = (1 - 1/2s)(1 - 1/3s)(1 - 1/4s) ...,
    with S = ζ(s) - 1 in (4.2), the corresponding governing infinite series (as treated in Chapter 1: Decreasing Products). So, this modified process comes about simply by interchanging the roles played by the two series, ∑1/ps and ∑1/ns - 1 = ζ(s) - 1. Thus according to the Filling lemma, the rectangle is completely filled if and only if P(s) = 0, which is the divergent range, 0<s <1 of the series ζ(s) in (4.2). As always, the equation:
    (4.13)        P(s) = 1 - F(s)
    holds in all cases (convergent or divergent), where now:
    (4.14)        F(s) = 1/2s + (1 - 1/2s)1/3s + (1 - 1/2s)(1 - 1/3s)1/4s + ....


    This series converges regardless of the value of s, and gives the final fraction of the rectangle filled by the isosceles triangles. In fact, for 0<s <1, the series converges simply to the number one:
    (4.15)        1 = 1/2s + (1 - 1/2s)1/3s + (1 - 1/2s)(1 - 1/3s)1/4s + ...,
    which is exactly when the filling process is successful. This equality affords a meaning even as the series ζ(s) in (4.2) diverges. In particular, for s=1, we obtain the interesting arithmetical formula:
    1 = 1/2 + (1/2)(1/3) + (1/3)(1/4) + (1/4)(1/5) + ...,
    which is just the simplest case of (4.15). Quite generally, the convergent infinite series, F of (1.6) in Chapter 1: Decreasing Products, is always of interest, regardless of the outcome of the infinite product, P of (1.1). It seems likely that it could be a useful tool in the study of divergent series.

    However, continuing our examination of the special infinite product, (4.12), we will have P(s)>0 and F(s)<1 whenever s>1. This is the convergent range of the series ζ(s) in (4.2), where the filling process is unsuccessful. The numerical value of F(s), given by (4.14), can be estimated using the general formula of Chapter 1: Decreasing Products. Inequalities (1.9), for example, with S = ζ(s)-1 and P = 1-F(s), lead to the inequalities:
    (4.16)        1 - 1/eζ(s)-1< F(s) < 1 - 1/aζ(s)-1
    where a = 1/(1-R)(1/R) → e as R = 1/2s → 0.

    Thus we obtain an excellent approximation:
    (4.17)        F(s) ~ 1 - 1/eζ(s)-1
    for large values of s. Also, as s → 1+, we have R → 1/2 and a → 4, so that we obtain the approximation:
    (4.18)        F(s) ~ 1 - 1/4ζ(s)-1
    for s near 1+. In between (because of (4.16)), there is a number, bs, lying between e and 4 and satisfying the equality:
    (4.19)        F(s) = 1 - 1/bsζ(s)-1
    for all s>1. The graph of (4.14), shown below, is quantitatively similar to the graph encountered previously ((4.4) in Section 4A: Rectangle-filling), using the Riemann zeta-function numbers˚, 1/ps, but there are quantitative differences 1 - 1/bsζ(s)-1 versus 1 - 1/ζ(s), for s>1.


    320.



    As observed earlier in this chapter, the Riemann zeta-function ˚, ζ(s), which here and elsewhere plays such a dominant role, is considered "well-known", and the source of numerical results. Using the ordinary integral comparison test for series (with the integral ∫0∞ dt/(1+t)s = (1/(s-1), one can obtain the inequalities 1/(s-1) < ζ(s) < s/(s-1), which are useful for rough estimates. (For these and very extensive analytical facts, see Ref. [7.8]). Indeed, its value, ζ(s), when s is an even integer, can be expressed in closed form. For example, one has ζ(2) = π2/6, which is why we claimed that P=1/ζ(2), the unfilled fraction of the interesting picture in Section 4A: Rectangle-filling, is approximately 60%. This particular case is of passing interest for still another reason, because of the special factoring:
    (1 - 1/k2) = (1 - 1/k)(1 + 1/k)
    possible. Thus the convergent product, (4.3) with s=2:
    P=1/ζ(2) = (1 - 1/22)(1 - 1/32)(1 - 1/52) ... = 6/π2
    formally factors, using the (invalid) rearrangement into two divergent products:
    P+ = (1+1/2)(1+1/3)(1+1/5) ... = ∞
    P- = (1-1/2)(1-1/3)(1-1/5) ... = 0.
    So that in this last form, we obtain an indeterminate case, with P = P+P- = ∞ × 0, and where S = S+ = S- = ∞ for the corresponding sequences in Chapter 3: Increasing And/Or Decreasing Products. On the other hand, the valid rearrangement of the factors of the convergent product:
    1/ζ(4) = (1 - 1/24)(1 - 1/34)(1 - 1/54) ...
    into the two convergent products:
    P+ = (1 + 1/22)(1 + 1/32)(1 + 1/52) = ?
    P- = (1 - 1/22)(1 - 1/32)(1 - 1/52) = 1/ζ(2) = 6/π2
    yields the correct value, P = P+P- = 90/π4. Thus, P+= 15/π2 = 15.19817753..... (Note: π2 = 9.869604401...; 1/π2 = 0.101321184...; 15/π2 = 15.19817753..... We will generalize this situation in the following Chapter 5B: Computable Increasing Products.

    However, the main point to be made here in this section is that (4.14) reduces to (4.15), and (4.4) reduces to (4.5), for the divergent range, 0<s<1, as shown in the two graphs, and should not be dismissed out-of-hand, simply because a series such as (4.2) diverges. Indeed, these are the only instances when the rectangle is actually completely filled; ostensibly, the "objective" in the first place: to complete the picture. This axiom is dramatically borne out in the following chapter, where only divergent cases seemingly give meaningful results (at least in real applications).



    SECTION 4G. FOOTNOTE FOR THE "ADVANCED" READER.

    As suggested earlier, the rectangle-filling process by isosceles triangles has a connection with Lebesgue integration, which was extensively pursued in: Struble RA. Can one do serious Mathematics using pictures and calculus? (See: http://www.infiniteproduct.info/stru0928.htm ). Briefly, the connection comes about in each case (4.1), by translating the corresponding triangles to the base of the rectangle, and interpreting them as simple functions (called tent-functions in [7.1]), t1(x), t2(x), t3(x), ..., of x along that base. It turns out that their pointwise sum:
    L(x) = t1(x) + t2(x) + t3(x) + ...
    is necessarily a Lebesgue-integrable function, and its integral:
    ∫L(x) = ∫t1(x) + ∫t2(x) + ∫t3(x) + ...
    is the number, F=1-P. Note that the right-hand member is simply the sum of the areas of the triangles. When F=1 (P=0), L(x) is a constant function and ∫L(x)dx = area of the rectangle, reflecting a successful filling of the rectangle. However, if F<1, then ∫L(x)dx is something less (P>0) than the area of the rectangle, and the filling process has been unsuccessful. The examples in Section 4A: Rectangle-filling, and in Section 4F: Some Divergence Considerations are interesting illustrations of this phenomenon, where the variable, s, is merely a parameter meandering across families of Lebesgue-integrable functions, L(x), for which we have obtained quantitative results for F = ∫L(x). The associated Lebesgue-integrable functions, L(x), for s>1 are likely to be rather bizarre-looking, coming as they are from those extreme-type triangles encountered earlier in Section 4A: Rectangle-filling, with all those dense forests, etc. When using slim triangles exclusively, the resulting function L(x) fails to be differentiable in those dense forest regions. When using fat triangles exclusively, the resulting L(x) is continuous. However, when using slim and fat triangles alternatively, the resulting function L(x) might be expected to be continuous and non-differentiable. The reason for this is suggested by the sketch below, demonstrating the mixing of some slim isosceles triangles with some fat ones. The fat ones flatten down the vertical variations, while the slim ones accentuate them. Thus when the triangles are lowered to the base of the rectangle, there is some blending of the two effects.


    358.





    SECTION 4H. THE COSINE FUNCTION. FOR THE "ADVANCED" READER.

    As a standard analytical application of a decreasing product, we will take note of a well-known product-formula for the cosine function:
    (4.20)        cos πx = (1 - 4 x2)(1 - 4 x2/32)(1 - 4 x2/52) ... (1 - 4 x2/(2n-1)2) ...
    for 0 < x < 1/2. What is of particular interest here for us, is the fact that we can sum the corresponding infinite series:
    (4.21)        S(x) = 4x2(1 + 1/32 + 1/52 + 1/72 + 1/92 + 1/112 + ...),
    and are then able to derive a surprisingly good approximation for cos πx. For from (1.10) in Chapter 1: Decreasing Products, we have:
    (4.22)        S(x) = -log P / log bS = -log (cos πx) / log bS.
    Since, in this case, R = max rn = 4x2 and:
    e < bS < (1/(1-R))1/R = (1/(1 - 4x2)1/4x2,
    we obtain (from their logs):
    (4.23)        1 < log bS < - 1/4x2 log(1-4x2) = 1 + (4x2)/2 + (4x2)2/3 + ...,
    using the logarithmic expansion. Since:
    (4.24)        1 + (4x2)/2 + (4x2)2/3 + ... < 1 + 4x2 + (4x2)2 + (4x2)3 + ... = 1/(1-4x2),
    using a geometric expansion, we obtain finally from (4.22), (4.23), and (4.24), the simple inequalities:
    (4.25)        - (1 - 4x2) log (cos πx) < S(x) < - log (cos πx)
    . These expressions would yield an excellent approximation to S(x) for small x. However, we can do much better, by evaluating the limit:
    limx → 0 S(x) / 4x2 = limx → 0 - log (cos πx) / 4x2.
    To accomplish this objective, we also need the power series expansion of the cosine function:
    (4.26)        cos πx = 1 - (πx)2/2! + (πx)4/4! - (πx)6/6! + ... ,
    so that, upon using the logarithmic expansion one more time, we obtain:
    - log(cos πx) = [(πx)2/2! - (πx)4/4! + ...] + 1/2 [(πx)2/2! - (πx)4/4! + ...]2 + ...
    Using this expression, we obtain from (4.21):
    (4.27)        1 + 1/32 + 1/52 + 1/72 + ... = limx → 0 S(x) / 4x2
    = limx → 0 - log (cos πx) / 4x2 = limx → 0 (πx)2/2! / 4x2 = π2/8,
    as the sum of an interesting series. Moreover, by (4.21), finally:
    (4.28)        S(x) = (π2/2) x2
    for the exact sum. We can also obtain a really interesting approximation to the cosine function, using inequality (1.7) of Chapter 1: Decreasing Products:
    (4.29)        cos πx = P(x) < e-S(x) = e-(πx)2/2,
    which is bound to be good for small x. Upon using the exponential expansion on the right, this expression becomes:
    (4.30)        cos πx ~ 1 - (πx)2/2 + [(πx)2/2]2)(1/2!) - [(πx)2/2]3)(1/3!) + ...
    = 1 - (πx)2/2 + (πx)4/8 + ....,
    By comparing the quartic terms, it can be seen that (4.30) over-estimates (4.26), as is should, for small x, and suggests the actual