DRAFT COPY ONLY.
7/4/2009.
© Raimond A. Struble, PhD.
http://www.infiniteproduct.info/struifpr.htm
Professor Emeritus
Department of Mathematics
North Carolina State University
Raleigh, NC.
(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)and more generally that:
= 1 - [r1 + P1r2] - (1 - r1)(1-r2)r3
= 1 - [r1 + P1r2 + P2r3],
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 - Fnholds for all n. Thus, the infinite product satisfies the equation:
(1.5) P = 1-F,where:
F = limn → ∞ Fnis 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.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-SUsing 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-xholds for 0 < x < R.
302.(1-R)(1/R)rk < 1 - rkholds for all k, and so:
(1-R)(1/R)(r1 + r2 + r3 + ... rn) < (1-r1)(1-r2)(1-r3) ... (1-rn) = Pnholds 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-SThe 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<∞.It turns out that in many applications, the P=0, F=1 case is really the most desirable one, where:
P=0, F=1, if S=∞.
(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)duis the continuous analogue of the partial sum:
Sn = r1 + r2 + r3 + ... + rnof 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)duis 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=∞.
(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-jand ultimately:
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)],
(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).)
(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 + ... + rntend 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.
(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 + ... + rnof 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.Pn = (1 + r1)(1 + r2)(1 + r3) ... (1 + rn)holds for all n, and so:
< er1+r2+r3+...+rn = eSn
(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.(1+R)rk/R < 1 + rkholds for all k. We conclude that:
(1+R)Sn/R < Pnholds 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 < ∞.In applications, generally the divergent case, P = 1+F = ∞, F = ∞ retains some meaning, despite the lack of mathematical content, ∞ = 1 + ∞.
P = ∞ if S = ∞.
(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.
(i). Pn = (1 + r1)(1 + r2)(1 + r3) ... (1 + rn)If R < ∞, then the following inequalities hold:
(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)
(1+R)Sn/R < Pn < eSnIf R = ∞, then P = S = ∞
(1+R)S/R < P < eS, and P = ∞, if and only if S = ∞.
(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-SnIf R = 1, then P = 0, and S = ∞. (The circumstance in which R > 1 is not permitted.)
(1 - R)S/R < P < e-S
and P = 0 if and only if S = ∞.
(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.(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+ = maxrk
a- = 1/(1-R-)1/R-,where R- = maxrk
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- < 0Briefly, we have, for S = S++S- < +∞:
and
1 < P < ∞ if 0 < 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- < ∞,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.
and
P=0 if -∞ = S+ - S- and P=∞ if S+ - S- = ∞.
(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),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:
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).
[(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.(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)]which is certainly negative.
= [(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] + ...,
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!) ...for 0 < x < 1 (accurate to the degree exhibited). If greater accuracy is desired, we can employ the first four factors to obtain:
i.e.,
(1 - x2)(1 + (x2)2/2!)(0.82625) < P(x) < (1 - x2)(1 + (x2)2/2!) (1.04314),
(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 + x2becomes 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;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:
(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 ∞.
343.(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.
31.
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33.
34.
28.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: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.
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:
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.
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-Sfor 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.
36.
73.
38.
355.r1 = 1/2s,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:
r2 = 1/3s,
r3 = 1/5s,
r4 = 1/7s,
r5 = 1/11s, ...,
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.
357.
375.
376.(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.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/pswith 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.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 + θand:
S(s) = 1/2s + 1/3s + 1/5s + ... = φ (<θ)Thus:
ζ(s)1/S(s) = (1+θ)1/φ → ewith 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.)
Pn = (1-r1)(1-r2)(1-r3) ... (1-rn),and he has dispersed the portion:
Fn = 1 - Pn = r1 + P1r2 + P2r3 + ... + Pn-1rnof 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.
P = (1-r1)(1-r2)(1-r3) ... = 0.His heirs may inherit something, F=1-P, if P>0.
(1.14) P(∞) = e -0∫∞r(u)dugreater 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.
(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-kClearly, 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.
(4.7) ℓ = k + log[1 + jq/(p-j)(1+j)k]/log(1+j) ~ k + q/(p-j)(1+j)kOf 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.
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.
PN = (1-r1)(1-r2)(1-r3) ... (1-rN) = 0,for some N.
(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.
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).
(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 + ....
(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.
(4.16) 1 - 1/eζ(s)-1< F(s) < 1 - 1/aζ(s)-1where a = 1/(1-R)(1/R) → e as R = 1/2s → 0.
(4.17) F(s) ~ 1 - 1/eζ(s)-1for 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)-1for 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)-1for 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.(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/π2formally factors, using the (invalid) rearrangement into two divergent products:
P+ = (1+1/2)(1+1/3)(1+1/5) ... = ∞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:
P- = (1-1/2)(1-1/3)(1-1/5) ... = 0.
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) = ?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.
P- = (1 - 1/22)(1 - 1/32)(1 - 1/52) = 1/ζ(2) = 6/π2
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.(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) / 4x2as the sum of an interesting series. Moreover, by (4.21), finally:
= limx → 0 - log (cos πx) / 4x2 = limx → 0 (πx)2/2! / 4x2 = π2/8,
(4.28) S(x) = (π2/2) x2for 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!) + ...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 error in the approximation. For x=1/4, it is but 0.0317..., while (4.26) becomes 1 / √ 2 = 0.707... ~ 0.737..., both surprisingly good!. We complete the picture by an illustration of inequality (4.29) and the lower bound:
= 1 - (πx)2/2 + (πx)4/8 + ....,
a-S(x) = (1 - 4x2)π2/8 < cos πxgiven by (1.7), (with R = 4x2)) as the dashed curve.
332.(4.31) sin πx / πx = (1 - x2) (1 - x2/22)(1 - x2/32) (1 - x2/42) ...for 0 < x < 1. Again, our particular interest is in the fact that we can obtain the sum of the corresponding infinite series:
(4.32) S(x) = x2 ( 1 + 1/22 + 1/32 + 1/42 + ...),and are then able to derive a good approximation to sin πx / πx. We observe that the sum of this series is proportional to ζ(2) = π2/6. As a sidelight, we observe that using the sum in Section 4H: The sine function, for the odd terms in (4.32) of π2/8, we can obtain the sum of the even terms in (4.32), simply as ζ(2) - π2/8 = π2/24. But our main interest concerns the use of inequality (1.7) of Chapter 1: Decreasing Products, which leads to:
(4.33) sin πx / πx < e-S(x) = e -(πx)2/6,and which is bound to yield a good estimate for small x. For x=1/2, sin πx / πx becomes 2 / π = 0.6766..., while e -(πx)2/6 becomes 0.6613..., (not bad at all for such a large value of x). The inequality (4.33) is illustrated by the interesting picture below, and demonstrates an even better approximation than was obtained for cos πx. The approximation for the sine function itself becomes:
(4.34) sin πx < πx e -(πx)2/6,and is also illustrated below. To complete the picture, the lower bound:
(4.35) πx a-S(x) = πx (1-x2)π2/6 < sin πxgiven by (1.7) (with R = x2) is included in the illustration as the dashed curve. It matches the accuracy of the upper bound (4.34) up to x=1/2, and is qualitatively better for 1/2 < x < 1.
334.
335.
359.(4.36) P = (1-r1)(1-r2)(1-r3) ... (1-rn) ...yields the untlimate unfilled portion of the square, and the Filling lemma now applies: the square is completely filled by the succession of circles, if and only if the infinite series:
(4.37) S = r1 + r2 + r3 + ... + rn + ...diverges. An examination of the above geometric process suggests that the terms of this series may not even tend to zero as n → ∞. Indeed, the early steps suggest rn values in the neighborhood of 1/4 or more. As the process progresses, and smaller and smaller circles come into play, the contact picture becomes increasingly complex, but the suggestion remains that the apparent rn values (fractions of the multiple unfilled portions filled by the circles at each step) continue to approximate something of the order of 1/4 or so.
363.
364.
403.2Ry = (x-T)2 + y2,where R = (T2+1)/2. This graphic is particularly appropriate for illustrating the effect on the area of a spiked region compared to that of a small contact circle, as the contact point recedes. This is essentially what happens if subsequent small circles are inserted in contact with (between) two much larger circles. Since y = [(x - T)2 + y2]/2R (for the upper circle), the area AT of the spiked region is given by:
AT = 0∫T 2y dx = (1/R) 0∫T (x-T)2 dx + (1/R) 0∫T y2 dx.For large T, the second term on the right is actually negligible, and so:
AT > 1/3 T3/R = 2/3 T3 / (T2 + 1) ~ 2/3 T.Thus the area of a small contact circle becomes a tiny fraction of the area of the spiked region, whenever T is very large, and nothing at all like 1/4. (The fraction here is less than 1/4 only if T > 19 and R > 180.) However, for a given value of n, (the nth value of the filling process) most new filling circles are not sandwiched in between much larger ones, but rather fill regions more representative of the 1/4 fraction. Indeed, the process does not generally lead to the extreme cases discussed above. One way to see this is to associate with each new inserted circle the acute (or right) triangle defined by the three tangent lines drawn through the contact points. Most triangles do not become extreme as this process continues.
∑2∞ rn* = ∑2∞ rn Pn/Pn*,where Pn and Pn* are the corresponding partial products, with Pn* > P* > 0. The question then becomes, does Pn → 0 as n → ∞ (sufficiently rapidly), so that ∑2∞ rn* < ∞? If not, then P = limn → ∞ Pn cannot be zero and ∑2∞ rn must converge.
P * ∑2∞ rn* < ∑2∞ rn Pn = F - r1 = 1 - P - π/4.So, in the limit, as r1* → r1, we obtain P ∑2∞ rn < 1 - P - π/4, where the left member is consistent with either an indeterminate form, 0 × ∞, or else an ordinary product form, P > 0 × ∑2∞ rn < ∞, as required by the finite right member. (The writer is unaware whether or not this square-filling problem has been previously investigated and, of course, what possible conclusions might have been drawn.)
540.S = r1 + r2 + r3 + ... + rn + ...diverges, and the infinite product,
P = (1-r1)(1-r2)(1-r3) ... (1-rn) ...is zero. Moreover, the filling lemma applies to other possible filling schemes: the circle is completely filled if and only if P=0, and if S does converge, then P>0 is the fraction of the area of the circle that remains unfilled. So less-efficient schemes may still completely fill the circle, yet too-efficient schemes will not.
(4.38) P = (1 - r1)(1 + r2)(1 - r3) (1 + r4) ... (1 - r2n+1)(1 + r2n+2) ...one assumes that the numbers:
r1' = (r1-r2) + r1r2, r3' = (r3-r4) + r3r4, r5' = (r5-r6) + r5r6,i.e.:
(4.39) r2n+1' = (r2n+1-r2n+2) + r2n+1r2n+2,satisfy the inequalities:
0 < (r2n+1-r2n+2) + r2n+1r2n+2 < 1/4for all n. This requires only that 0 < r2n+2 < r2n+1/(1 - r2n+1), resulting in 0 < r2n+1' < r2n+1 < 1/4. Then P becomes the simple decreasing product:
(4.40) P = P' = (1 - r1')(1 - r3')(1 - r5') ... (1 - r2n+1') ...,subjected to the Filling lemma, since indeed:
(1 - r2n+1)(1 + r2n+2) = 1 - [(r2n+1 - r2n+2) + r2n+1r2n+2] = 1 - r2n+1'.Thus, by (1.9) (using odd indices), the product (4.40) (i.e., (4.38)) satisfies the inequalities:
(4.41) a-S' < P < e-S'with corresponding series:
(4.41) S' = r1' + r3' + r5' + ... = (r1 - r2 + r1r2) + (r3 - r4 + r3r4) + (r5 - r6 + r5r6) + ...and a = 1/(1-R)1/R, where R = max r2n+1' < max r2n+1. Therefore, P=0 if and only if S' = ∞. This situation needs to be compared with the formulation for mixed products (3.3), which in this case concerns the two series:
S+ = r2 + r4 + r6 + ...and (the negation of) their differences:
and
S- = r1 + r3 + r5 + ...
S- - S+ = (r1 - r2) + (r3 - r4) + (r5 - r6) + ....The extra products, r1r2, r3r4, r5r6, ... in (4.41) will be interpreted subsequently, but for now, we just observe that the presence of increasing factors (1 + r2n+2) in (4.38) merely retard the decreasing aspects somewhat. Of course, they can retard the process sufficiently to convert an otherwise apparently zero value for (4.38) (i.e., S- = ∞) into an actual positive one for (4.40) (i.e., S' < ∞). In any event, this formal scheme turns mixed products (4.38) into decreasing products (4.40). We now change perspective and consider a more visible one.
371.(1 - r1) + r1r2of the rectangle-yet-to-be-filled. Then step 3 consists of filling an r3 fraction of the rectangle reflected by this last quantity with suitable right-side-up isosceles triangles, followed by the remanding of a fraction r4 of each, using upside-down isosceles triangles, at step 4. This rather messy process is illustrated below for steps 1 through 4.
372.[(1 - r1) + r1r2] [(1 - r3) + r3r4]gives the fraction of the rectangle-yet-to-be-filled, beginning with step 5. Thus if the "trick" of using suitable upside-down triangles remanded back to unfilled ones is maintained throughout the infinitely many steps (as n → ∞), then the ultimate unfilled portion of the rectangle is given by the infinite decreasing product:
(4.43) P' = [(1 - r1) + r1r2] [(1 - r3) + r3r4] [(1 - r5) + r5r6] ... = [(1 - r1(1- r2)] [(1 - r3(1- r4)] [(1 - r5(1- r6)] ... [(1 - r2n+1(1- r2n+2)] ...,which also retards the filling process somewhat (in an obvious numerical fashion). Though this is just another different scheme for turning mixed products (see (4.44) below) into decreasing ones, it has a visible interpretation, while (4.40) does not. Also, one now has a possible interpretation of the extra products, r2n+1r2n+2 in S': they are the analogues of the areas of the remained, upside-down isosceles triangles used above for (4.43). They also enter into the process of converting (4.43) back into a mixed product:
(4.44) P' = (1 - r1)(1 + r2')(1 - r3) (1 + r4') ...,where r2' = r1r2/(1-r1), r4' = r3r4/(1-r3), ..., i.e.:
(4.45) r2n+2' = r2n+1r2n+2/(1-r2n+1)Thus if one starts with the mixed increasing and decreasing product (4.44) (with given values of r2n+1 < 1/4 and r2n+2' < r2n+1/4), one can invert the formulas (4.45) to obtain:
(4.46) r2n+2 = (1 - r2n+1) r2n+2' / r2n+1 < 1/4,and, thereby, turn the mixed product into the simple decreasing product, (4.43). As emphasized before, this scheme has a visible intrepretation using the upside-down triangles, while (4.40) does not. However, it is of considerable interest to observe that if :
(4.47) r2n+1 << 1,(meaning that r2n+1 is negligible with respect to 1) holds for all n, then the two processes are essentially one and the same. To see this, one sets r2n+1 = 0 in the above divisors 1/(1-r2n+1) in (4.45) and substitutes the resulting expressions r2n+2' = r2n+1r2n+2 into (4.38) in place of the original r2n+2. Then P and P' become nearly equal, since in (4.40), we have:
(1 - r2n+1')approximately (since r2n+12 << r2n+1). This conforms with (4.43), and the two decreasing product schemes essentially amount to the same thing, and where the extra products r2n+1r2n+2 ARE the same in both. In the convergent cases, r2n+1 → 0 as n → ∞, so that always in the end, they certainly amount to the same thing, although they could differ numerically due to the early factors. In ref. [7.1], numerous visible filling schemes are discussed ((4.43), however, is an example not considered previously) and all are subjected to a universal Filling lemma, in which the restriction, rn < 1/4 can be relaxed to rn < 1. It's the employment of the isosceles triangles that creates this special restriction here.
= 1 - (r2n+1 - r2n+2') + r2n+1r2n+2'
= 1 - (r2n+1 - r2n+2r2n+2') + r2n+12r2n+2'
= 1 - r2n+1(1 - r2n+2),
(4.48) T = x1 + x2 + x3 + ... + xn + ...of positive terms. The statement simply is: the related infinite series
(4.49) S = x1 / (x1 + x2 + x3 + ... ) + x2 / (x2 + x3 + ... ) + ... + xn / (xn + xn+1 + ... ) + ...DIVERGES. To prove the result, we can recast this series in a more suggestive form:
(4.50) S = x1/T + x2/(T - x1) + x3/(T - x1 - x2) + ... + xn/(T - x1 - x2 ... - xn-1) + ...where:
(4.51) rn = xn/(T - x1 - x2 ... - xn-1)is just the fraction that the term xn is of the remaining sum in (4.48), having removed the first (n-1) terms. Therefore, the quantity (1 - rn) is the fraction of the remaining sum, (T - x1 - x2 - ... - xn), which actually remains after the removal of xn from the sum (4.48). It follows that the infinite product:
(4.52) P = (1 - r1)(1 - r2)(1 - r3) ... (1 - rn) ...is the fraction remaining of the sum (4.48) after the removal of all the terms on the right-hand side, which, of course, has to be zero. Since the infinite product (4.52) is zero, according to the filling lemma, the corresponding infinite series (4.49), consisting of the same rn-fractions, must diverge, as claimed.
Pn = (xn+1 + xn+2 + ...)/T.On the other hand, when xn = 1/n2, then:
rn = 1/[1 + (n/(n+1))2 + (n/(n+2))2 + ... ]so that the divergence conclusion in (4.49) is at least a subtle one.
< 1 / [1 + (n/(n+1))2 + (n/(n+2))2 + ... + (n/(n+n))2 ]
< 4 / [n + 1],
rn = 1 / [1 + (n/(n+1))s + (n/(n+2))s + ... + (n/(n+n))s]is seen to hold for s>1, where ζ(s) is the Riemann zeta-function. (The same scheme shows that 1/n[ζ(s)] < rn.) Therefore, while the partial sums:
+ [(n/(2n+1))s + (n/(2n+2))s + ... + (n/(2n+n))s ]
+ ... + [(n/(3n+1))s + (n/(3n+2))s + ... + (n/(3n+n))s ] + ...
< 1 / [n (1/2)s] + [ n (1/3)s] + [ n (1/4)s] + ...
= 1 / n[ζ(s) - 1]
∑1m xn = 1/2s + 1/3s + 1/4s + ... + 1/msof the convergent series, T = [ζ(s) - 1] = 1/2s + 1/3s + 1/4s + ... increase as s → 1+, those of the related, divergent series, S,
∑1m rn < ( 1/2 + 1/3 + 1/4 + ... + 1/m ) / [ζ(s) - 1],tend to zero as s → 1+. The first suggesting divergence, the second suggesting convergence.
(4.53) rn(u) = xn / (u + xn + xn+1 + ... ) = xn / (T + u - x1 - x2 - ... - xn-1)for u > 0. Of course, the modified series:
(4.54) S(u) = ∑1∞ xn / (u + xn + xn+1 + ... ) = ∑1∞ xn / (T + u - x1 - x2 - ... - xn-1)is now convergent, and satisfies the (obvious) inequalities:
(4.55) T/(u + T) < S(u) < T/u.These formulas are not very revealing as to the effect of u, but a "closed form expression" for the corresponding infinite product is:
(4.56) P(u) = (1 - r1(u))(1 - r2(u)) ... (1 - rn(u)) ... = u/(T + u)In this case, each rn(u)-value is the fraction the term xn(u) is of the modified series:
(4.57) T + u = u + x1 + x2 + x3 + ... + xn + ...remaining, having removed the x1, x2, x3, ..., xn-1 terms. Therefore, the quantity (1 -rn(u)) is the fraction of the remaining sum (T + u - x1 - x2 - x3 - ... - xn), which actually remains after the removal of xn from the sum (4.57).
(4.58) (1 - R(u))S(u)/R(u) < P(u) = u/(T+u) < e-S(u),where R(u) = maxk rk(u). We recall that these inequalities are sharp whenever R(u) is small, and then
P(u) ~ e-S(u).Hence:
(4.59) S(u) ~ log((T+u)/u)is a good approximation for small values of R(u). If the original series is a decreasing one, then R(u) = x1 / (T+u) < x1/T. For a geometric series, T = θ + θ2 + θ3 + ..., one obtains for (4.49), S = (1 - θ) + (1 - θ) + (1 - θ) + ..., and for (4.52), P = θθθ..., while (4.53) becomes rn(u) = (1 - θ) θn/[θn + (1 - θ)u]; and (4.56) becomes P(u) = (1 - θ)u / [θ + (1 - θ)u].
(4.60) x1/x1 + x2/(x1 + x2) + x3/(x1 + x2 + x3) + ... + xn/(x1 + x2 + x3 + ... + xn)when all the divisors x1 + x2 + x3 + ... + xn are non-zero. In this case, it turns out simply that (with one minor caveat), (4.60) converges if and only if the series:
(4.61) x1 + x2 + x3 + ... + xn + ...converges. This follows by considering the fact that
1 - rn = 1 - xn/(x1 + x2 + x3 + ... + xn)and so the (collapsing) partial products:
= (x1 + x2 + x3 + ... + xn-1) / (x1 + x2 + x3 + ... + xn),
(4.62) Pn = (1 - r1) (1 - r2) (1 - r3) ... (1 - rn)reflect the simultaneous convergence (or not) of the two series, (4.60) and (4.61), as n → ∞. The one caveat concerns the possibility that (4.61) may converge to zero, in which case (4.60) diverges.
= x1 /(x1 + x2 + x3 + ... + xn)
(4.63) ∑1∞ cos-1(an / an+1)necessarily diverges. This follows by observing that the similarly (collapsing) partial product:
(4.64) Pn = (a1/a2) (a2/a3) (a3/a4) ... (an/an+1)tends to zero (as n → ∞), so that the corresponding infinite series:
= (1 - r1) (1 - r2) (1 - r3) ... (1 - rn) = (a1/an+1) (rk = 1 - (ak/ak+1)
(4.65) ∑1∞ rn = ∑1∞ (1 - (an / an+1))necessarily diverges. But a glance at the geometric graph of the inverse cosine function shows that cos-1(an / an+1) > (1 - (an / an+1)) holds, and so (4.63) diverges.
541.P1 = M(1+j),for yearly compounding.
Pk = M(1+j)(1+j) ...(1+j)= M(1+j)k.However, if the compounding is done m times a year, these formulas modulate into the forms:
(5.1) P1 = M(1+j/m)m (compounding interest rate is j/m)and:
(5.2) Pk = M(1+j/m)km.Here, m=12 for monthly compounding and m=365 for daily compounding. The latter now seems to be almost universal, and savers and especially lenders, now speak of an a.p.r. (annual percentage rate). This is a fictitious value, j' = (1+j/m)m -1, representing the accumulated interest payment, just as if that payment were made at the end of each year.
(5.3) P1 = Mej.In fact, the expression, (1+j/365)365 = [(1+j/365)365/j]j is only slightly less than ej, because:
limn → ∞(1+1/n)n = e.For example, 365/j>3650 and (1+1/3650)3650/e>0.9998 for j<0.1. The exact return to the investor, of course, results from applying (5.1), while (5.3) is the continuous compounding upper limit to any real compounding scheme, as given by (2.10), in Chapter 2: Increasing Products. Should the interest rate change during this first year (a common occurrence now-a-days), then the exact formula, (5.1), needs to be replaced by a product of m (generally different) factors:
(5.4) P1 = M(1+j1/m)(1+j2/m)(1+j3/m) ... (1+jm/m)reflecting the annual interest rates, j1, j2, j3, ... jm, involved each day.
(5.5) Pk = Mejk
(5.6) Pk = M(1+j1/m1) (1+j2/m2)(1+j3/m3) ... (1+jk/mk),where the total number of factors depends upon the corresponding fractions employed. Upper and lower bounds for Pk are readily available from (2.7), with P and S replaced by the partial products, Pk, and the partial sums:
Sk = j1/m1 + j2/m2 + j3/m3 + ... jk/mk.Thus, the inequalities:
MaSk < Pk < MeSkhold, where a = (1+R)1/R and R = max jh/mh. As observed above, R is likely to be extremely small, and a is well-approximated by e. So (5.6) becomes essentially:
(5.7) Pk= MeSk.Again, this is the continuous compounding upper limit to any real compounding scheme. The formulas, (5.3), (5.5), and (5.7), constitute the practical, capsulized, compound-interest story that investors and lenders might possibly use.
P = (1+r1)(1+r2)(1+r3) ... (1+rk-1)eSkaffords a realistic method for accurately computing P; the accuracy only becomes greater with increasing k. Reversing the process, this scheme also leads to a method for computing (approximately) the sum of a convergent infinite series, S = r1 + r2 + r3 + .... In this scheme, one first obtains a very accurate estimate of the value of the corresponding infinite product, P = (1+r1)(1+r2)(1+r3) ...., one way or another. Then one determines an index, h, large enough (based upon the necessary "smallness" of the "tail-end" terms rk for k > h, so that the tail-end product, Qk = (1+rk)(1+rk+1)(1+rk+2) ...., must be very well-approximated by the exponential, eSk, where Sk = rk + rk+1 + rk+2 +...., Then, with P = (1+r1)(1+r2)...(1+rk-1) eSk, the "tail-end" sum can be computed accurately from the equation:
Sk = log[P/(1+r1)(1+r2)...(1+rk-1)].So finally, of course, the accurate value for the infinite series, S, is obtained by adding Sk to the finite sum, r1 + r2 + ... + rk-1:
S = r1 + r2 + ... + rk-1 + log[P/(1+r1)(1+r2)...(1+rk-1)].This computing scheme can also be employed using decreasing products. Let us see how it works out in the special case of the interesting series:
∑ 1/ps = 1/2s + 1/3s + 1/5s + ...of the Riemann zeta-function numbers. Here the corresponding decreasing infinite product is:
(4.3) P(s) = (1 - 1/2s)(1 - 1/3s)(1 - 1/5s)(1 - 1/7s)(1 - 1/11s) ... 1/ζ(s)So, in this case, we can obtain an exact value for this product. Suppose we choose, for example, s=6, so that P(s)=P(6)=945/π6. Then the terms 1/ps of the series can be considered small beyond, say, the prime p=5: 1/ps < 1/56 = 1/15625. Thence, the "tail-end" product becomes, (1 - 1/56)(1- 1/76) ... = e-(1/56 + 1/76 + ... ), and so the "tail-end" series becomes:
1/56 + 1/76 + ... = -log[ 945/π6) / (1 - 1/26(1 - 1/36)] = 0.000294....Thus:
∑ 1/ps = ∑ 1/p6 = 1/26 + 1/36 + 0.000294 = 0.017291....accurate to the extent displayed.
S = j1/m1 + j2/m2 + j3/m3 + ....In such a (strange) case, the ultimate return, namely:
P = M(1+j1/m1)(1+j2/m2) (1+j3/m3) ...will be finite, of course, but any practical concern is in the partial return, (5.6), after k steps, just as above. The practical result is again given by (5.7). As an example, suppose that jk/mk = pk, where 0<p<1. Then:
P = M(1+p)(1+p2)(1+p3) ...(Recall example (2.12) with p = 1/x.) While:
S = p + p2 + p3 + ... = p/(1-p), and P = ep/(1-p) exactly,and (5.7) gives, as an approximation, simply:
Pk = MeS(1-pk).Note:
p + p2 + p3 + ... + pk = S - pk(p + p2 + p3 + ...) = S(1 - pk).This example illustrates a possible (?) real-life case where the interest rate decreases systematically, perhaps annually.
Pn = P0(1+r1)(1+r2)(1+r3) ... (1+rn)where P0 is the initial population.
P0aS < P < P0 eS,with a very nearly e for R = max rk small.
(5.8) P(s) = (1 + 1/2s)(1 + 1/3s)(1 + 1/5s) ...where s>0. Again, we note that the corresponding infinite series:
(5.9) S(s) = 1/2s + 1/3s + 1/5s + ...is finite if s>1, and P(s) then is finite. Subsequently, we will obtain an exact formula for this P(s), and so our first concern is to derive reasonable estimates for the series, S(s). Using the inequalities (2.7) in Chapter 2: Increasing Products, we have:
(5.10) aS(s) < P(s) < eS(s),and, as noted in Chapter 4: Applications of Decreasing Products, these bounds are sharp (a is nearly e) for large values of s. Specifically, we have a = (1 + 1/2s)2s = (1 + 1/n)n, with n = 2s. In the worst case, s=1 and a=9/4=2.25, and so (rather remarkably, since e = 2.718281828...), the explicit inequalities:
(2.25)S(s) < P(s) < eS(s),hold for s > 1. These, in turn, lead to the fixed bounds:
log(2.25) < S(s)/ log P(s) < 1,for all s > 1.
(5.11) S(s) = [ζ(s)-1] - [ 1/4s + 1/6s + 1/8s + 1/9s + ... ].Indeed, ζ(s) = ∑ 1/ns summed over all integers, n, while S(s) = ∑ 1/ps summed over all primes, p, so that their difference (sans 1) is the above-indicated sum:
(5.12) E(s) = 1/4s + 1/6s + 1/8s + 1/9s + ...summed over all composite integers, 4, 6, 8, 9, ....
ζ(2) = π2/6 = 1.6449.... S(2) = 0.4564.... E(2) = 0.1885....So the approximation: S(s) = ζ(s) - 1 is beginning to appear reasonable for increasing s, and is probably all that is ever needed for s > 20. Note that the relative error is E(s)/S(s) ~ 1/2s for large s, and that 220 exceeds one million.
ζ(4) = π4/90 = 1.0966.... S(4) = 0.0901.... E(4) = 0.0065....
ζ(6) = π6/945 = 1.0173.... S(6) = 0.0172.... E(6) = 0.0001....
(5.13) S(s) = ∑ 1/pnswhen the kth prime, pk, is large enough to render the approximation:
= 1/2s + 1/3s + 1/5s + ... + 1/pks - log[ζ(s)(1 - 1/2s)(1 - 1/3s)(1 - 1/5s)...(1 - 1/pks)],
(1 - 1/pk+1s)(1 - 1/pk+2s)(1 - 1/pk+3s) ... = e-[1/pk+1s + 1/pk+2s + 1/pk+3s + ...]acceptable. (Note that the reciprocal of the bracketed quantity in the log is, by Euler's formula, this last "tail-end" infinite product.) The numerical values of S(s) for s=2, 4, 6, ... above were obtained using this summation process, (5.13). With s=2 (the most difficult case), we were able to use, simply:
S(2) = 1/22 + 1/32 + 1/52 + 1/72 - log[π2/6(1 - 1/22)(1 - 1/32)(1 - 1/52)(1 - 1/72)].To receive the same degree of accuracy by dirrect summation, it takes the addition of the terms, 1/p2 of the series up to the prime, 211. This certainly demonstrates the overwhelming superiority of this summation process using products. For s=6, we were able to use:
1/26 + 1/36 - log[π6/945(1 - 1/26)(1 - 1/36)],because 56 = 15625.
(5.14) S = r1 + r2 + r3 + ... + rk - log[(1 + r1)(1 + r2)(1 + r3) ... (1 + rk)/P+]for a convergent infinite series, hinges upon the inevitable validity of one of the approximations:
S = r1 + r2 + r3 + ... + rk + log[(1 - r1)(1 - r2)(1 - r3) ... (1 - rk)/P-]
where:
P+ = (1 + r1)(1 + r2)(1 + r3) ... (0 < rn)
P- = (1 - r1)(1 - r2)(1 - r3) ... (0 < rn < 1)
(1 + rk)(1 + rk+1)(1 + rk+2) ... = e[rk + rk+1 + rk+2 + ...]with increasing k. The inevitability, of course, stems from the fact that rk→0 as k→∞. To use these approximations effectively, however, one must be able to obtain sufficiently accurate values for the infinite products, P+ or P-, so that the quotients in the bracketed quantities in the logs lead to acceptable approximations. In (5.13), the infinite product is known exactly, but in other cases, one could conceivably evaluate these quotients by other means. But for now, we will derive the promised formula for the infinite product for P(s) of (5.8), which furnishes the final population that we envision.
or
(1 - rk)(1 - rk+1)(1 - rk+2) ... = e-[rk + rk+1 + rk+2 + ...]
P+ = (1 + 1/2s)(1 + 1/3s)(1 + 1/5s) ... = ?with:
P- = (1 - 1/2s)(1 - 1/3s)(1 - 1/5s) ... = 1/ζ(s)
S+ = 1/2s + 1/3s + 1/5s + ...the corresponding sequences for s>1. In this special case, S+ and S- are the same sequence S(s) as in (5.9). Since they are convergent, both P+ and P- converge (absolutely), and so the product:
S- = 1/2s + 1/3s + 1/5s + ...
P = P+P- = (1 + 1/2s)(1 + 1/3s)(1 + 1/5s) ... (1 - 1/2s)(1 - 1/3s)(1 - 1/5s) ...can be rearranged to obtain the expansion:
(5.15) P = P+P- = (1 - 1/22s)(1 - 1/32s)(1 - 1/52s) ... = 1/ζ(2s)(by Euler's formula, (4.3)). Note that (1 + 1/ks)(1 - 1/ks) = (1 - 1/k2s), and we have formed these products pairwise from P+P-. Hence the product, P+, is obtained from (5.15) as:
(5.16) P+ = P/P- = ζ(s)/ζ(2s),which is the promised exact value of P(s) in (5.8). The invalid rearrangement shown above in Chapter 4: Applications of Decreasing Products, and leading to indeterminacy, is the divergent case, s=1 here. Now since P(s) = bsS(s), we have exactly:
S(s) = log [ζ(s)/ζ(2s)] / log bS(where bS → e as s → ∞), which is to be compared with the exact formula, S(s) = [ζ(s) - 1] - E(s) in (5.11). For large values of s, these lead to an interesting (approximate) formula:
ζ(2s) = ζ(s) e[1 - ζ(s)],when bS = e, E(s)=0. The principle source of any inaccuracies stem from setting E(s)=0, and the much more precise formulation:
ζ(2s) = ζ(s) e[1 - ζ(s)]+E(s) = ζ(s) e-S(s),is accurate to within 5%, even for the small value s=2.
(1 - 1/2s)(1 - 1/3s)(1 - 1/5s) ... = 1/ζ(s), (s > 1)an analogous increasing infinite product formula:
(5.17) (1 + 1/2s)(1 + 1/3s)(1 + 1/5s) ... = ζ(s)/ζ(2s),involving the prime numbers. In historical jargon, we reformulate this expression in explicit form:
χ(s) = [1 + 1/4s + 1/9s + 1/16s + ... ] / [1 + 1/2s + 1/3s + 1/4s + ...Could the analytic extension into the complex plane of χ(s) lead to an alternative approach to the prime number theorem, π(N) ~ N/log(N)? One has an alternative "Golden Key" calculus version:
= 1 / (1 + 1/2s)(1 + 1/3s)(1 + 1/5s) ....
1/s log χ(s) = 0∫∞ JA(x) x-s-1 dx (= 0∫∞ J(x) x-s-1 (1- x-1) dx,with:
JA(x) = π(x) - 1/2π(√x) + 1/3π(3√x) - 1/4π(4√x) + 1/5π(5√x) - ...,having alternating signs, as contrasted with J(x) (See: pages 299-309 in Ref. [7.8]).
P+ = (1 + 1/2s)(1 + 1/3s)(1 + 1/4s) ...and:
P- = (1 - 1/2s)(1 - 1/3s)(1 - 1/4s) ... (all integers)
S = S+ = S- = ζ(s) - 1.Then for s > 1, these products converge (absolutely), and so the following rearrangement of their product:
P+P- = (1 - 1/22s)(1 - 1/32s)(1 - 1/42s) ...using (1 + 1/ks)(1 - 1/ks) = (1 - 1/k2s) is valid. But this is just the product, (4.12), with s replaced by 2s. Since P-(s) = P(s) by (4.12), we obtain P+(s) = P(2s)/P(s). So the increasing product:
= (1 - 1/4s)(1 - 1/9s)(1 - 1/16s) ...,
(5.18) P+(s) = (1 + 1/2s)(1 + 1/3s)(1 + 1/4s) ... = P(2s)/P(s)can be obtained directly from the decreasing product:
(4.12) P(s) = (1 - 1/2s)(1 - 1/3s)(1 - 1/4s) ...,The estimates (4.16), for F(s) = 1 - P(s), now become:
a[1 - ζ(s)] < P(s) < e[1 - ζ(s)],and are useful for estimating P+(s). One obtains from (5.18):
a = 1/ (1 - 1/2s)2s
a[1 - ζ(2s)]e[ζ(s)-1] < P+(s) < a[ζ(s)-1]e[1-ζ(2s)],leading to P+(s) ~ e[ζ(s)-ζ(2s)], since a ~ e for large s. This result is to be compared with P-(s) ~ e[1-ζ(s)] from (4.17). It turns out ([7.10] MATHSOFT. Infinite product constants) that the values of the increasing infinite product:
(5.18) P+(s) = (1 + 1/2s)(1 + 1/3s)(1 + 1/4s) ...are known in (simple) closed forms for s = 2, 3, 4. Evidently, these values are:
P+(2) = sinh(π)/π.According to the same source, the values of certain decreasing infinite products, analogous to (4.12) for s = 2, 3, 4, are also known in closed forms:
P+(3) = cosh(√3/2π)/π.
P+(4) = [cosh(√2π - cos(√2π] / π2.
(1 - 4/32)(1 - 4/42)(1 - 4/52) ... (1 - 4/n2) ... = 1/6.
(1 - 8/33)(1 - 8/43)(1 - 8/53) ... (1 - 8/n3) ... = sinh(√3π) / 42 √3π.
(1 - 16/34)(1 - 16/44)(1 - 16/54) ... (1 - 16/n4) ... = sinh(2π) / 120π.
(2.7) aS < P < eS,where:
(2.2) P = (1+r1)(1+r2)(1+r3) ....and:
(5.19) S = r1 + r2 + r3 + ...Here, there are no restrictions on the sizes of the positive numbers, r1, r2, r3, so all series with positive terms are in play. One immediately concludes from (2.7) that S<∞ if and only if P<∞. In fact, more adroitly, we have:
(2.8) P = bSS.This shows that the infinite series and the infinite product are intimately (numerically) connected, and in cases of real interest, the number, bS, satisfies 1 < bS<e. Recall that a = (1+1/R)1/R, whenever rk < R for all k, and that a → e as R → 0.
log P = S log bS,particularly when bS is very nearly e.
(1.1) Pn = (1-r1)(1-r2)(1-r3) ... (1-rn), for n = 1, 2, 3, ....in Chapter 1: Decreasing Products., where 0 < rk < 1. We then expand the reciprocal, 1/Pn, as a product of n (convergent) geometric series:
1/(1-rk) = 1 + rk + rk2 + rk3 + ...This results in a complicated convergent series:
1/Pn = (1 + r1 + r12 + ...) (1 + r2 + r22 + ...) ... (1 + rn + rn2 + ...)consisting of numerous products of elements, rkh for 0< h < ∞ and k<n.
(5.20) P(t) = (1 - 1/t)(1 - 1/t2)(1 - 1/t3) ...which holds for t > 1. This equation gives the sum of the infinite series:
= e-[1/(1 - t) + 1/2 × 1/(t2-1) + 1/3 × 1/(t3-1) + ...] = e-S*(t),
S*(t) = 1/(t-1) + 1/2 × 1/(t2-1) + 1/3 × 1/(t3-1) + ... = - log (1 - 1/t)(1 - 1/t2)(1 - 1/t3) ...in terms of an infinite product, or rather, its logarithm. Of course, the right-hand member here is also the sum of an infinite series:
- log (1 - 1/t) - log(1 - 1/t2) - log (1 - 1/t3) - ...,which suggests a method for extablishing the result in the first place (using geometric and logarithmic expansions). Rather interestingly, we note that the corresponding infinite series:
S(t) = 1/t + 1/t2 + 1/t3 + ... = 1/(t - 1),is the first term of S*(t). Therefore, by (1.9), (5.20) satisfies:
(1-1/t)t/(t-1) < P(t) < e-1/(t-1),and so tends to 1 as t → ∞, and to 0 as t → 1+.
log P = log (1 - r1) (1 - r2) ...with S = r1 + r2 + r3 + ... and Š = 1/2 [ r12 + r22 + r32 + ... ] + 1/3 [ r13 + r23 + r33 + ... ] + ... > 0. Thence:
= log (1 - r1) + log (1 - r2) + ...
= - [r1 + 1/2 r12 + 1/3 r13 + ... ] - [r2 + 1/2 r22 + 1/3 r23 + ... ] + ...
= - [r1 + r2 + r3 + ... ] - 1/2 [r12 + r22 + r32 + ... ] + ....
= - S - Š,
P = (1 - r1)(1 - r2)(1 - r3) ... = e-S - Š = e-S e-Š < e-S,which is an alternative verification of (1.7). A similar formula applies to an increasing product:
(5.21) P = (1 + r1)(1 + r2)(1 + r3) ... = eS - Š = eS e-Š < eS,(See (2.5)), where
(5.22) P1/2 = (1 + r1)(1 + r2)(1 + r3) ....We will do this by first finding such a sequence satisfying:
(5.23) P = (1 + r1)2 (1 + r2)2 (1 + r3)2 ....For this purpose, we employ a simple numerical (geometrical?) fact: The inequality:
(5.24) (1 + r)2 < 1 + (P+2)rholds for 0 < r < P. Indeed, (1 + P)2 = 1 + (P + 2)P.
368.(5.25) 1+(P+2) r1 = P > 1.So:
(5.26) r1 = (P-1)/(P+2) < 1 < P.Therefore:
P1 = P / (1 + r1)2 = (1+(P+2) r1)/(1 + r1)2 > 1,because of (5.24). We then select r2 to satisfy:
(5.27) 1+(P+2) r2 = P1.So:
(5.28) r2 = (P1 - 1) / (P+2) < (P-1)/(P+2) = r1 < P.Therefore:
P2 = P1 / (1 + r2)2 = (1+(P+2) r2) / (1 + r2)2 > 1,again, because of (5.24). We then select r3 to satisfy:
(5.29) 1+(P+2) r3 = P2.So:
(5.30) r3 = (P2-1)/(P+2) < (P1-1)/(P+2) = r2 < P.Therefore:
P3 = P2/(1+r3)2 = (1+(P+2)r3)/ (1+r3)2 > 1,once again by (5.24). By an induction argument, we can define the decreasing sequence, r1, r2, r3, ..., rn, ..., for each n, to satisfy:
(5.31) rn = (Pn-1-1)/(P+2),where:
Pn-1 = P / (1 + r1)2 (1 + r2)2 ... (1 + rn-1)2 > 1.Since this is a decreasing sequence, rn → r > 0 as n → ∞. But if r > 0, then the series:
S = 2 r1 + 2 r2 + 2 r3 + ... 2 rn + ...diverges, and so does the corresponding infinite product:
(1 + r1)2 (1 + r2)2 (1 + r3)2 ... (1 + rn)2 ...,which contradicts (5.31). Hence, rn → 0 as n → ∞. Finally, since:
P / (1 + r1)2 (1 + r2)2 (1 + r3)2 ... (1 + rn-1)2 = Pn-1 = 1+(P+2) rnwe conclude that, indeed:
(5.23) P = (1 + r1)2 (1 + r2)2 (1 + r3)2 ....Since S < ∞:
(5.22) P1/2 = (1 + r1)(1 + r2)(1 + r3) ...,as desired. To obtain the square-root of a positive number, Q < 1, let P = 1/Q, and invert the infinite product (5.22), using tk = rk/(1+rk) to obtain:
Q1/2 = (1 - t1)(1 - t2)(1 - t3) ...as a decreasing infinite product. This example appears to exhibit some "nice" mathematics, and not much else. The next example is even more inane/esoteric, but one can turn it into some effective infinite series-infinite product interaction, as shown.
(5.32) P = I.d1d2d3...dn...where I is the whole-number-part of P, and 0.d1d2d3...dn... is the decimal expansion of the fractional-part of P. One then selects the sequence of numbers, r1, r2, r3, ..., rn, ..., to satisfy:
(1 + r0) = I;and more generally:
(1 + r0)(1 + r1) = I.d1;
(1 + r0)(1 + r1)(1 + r2) = I.d1d2;
(1 + r0)(1 + r1)(1 + r2) (1 + r3) = I.d1d2d3;
(5.33) (1 + r0)(1 + r1)(1 + r2) (1 + r3)...(1 + rn) = I.d1d2d3...dn = Pn.If d1 ≠ 0, then one obtains explicitly:
r0 = I-1;and more generally:
r1 = [I.d1/I - 1] = [0.d1/I]
r2 = [I.d1d2/I.d1 - 1] = [0.0d2 / I.d1]
r3 = [I.d1d2d3/I.d1d2 - 1] = [0.00d3/I.d1d2]
(5.34) rn = [ I.d1d2d3...dn / I.d1d2d3...dn-1 - 1] = [0.00 ... 0.dn / [I.d1d2d3...dn-1]. = [0.00...0dn/Pn-1]Clearly, as n → ∞, (5.33) yields our sought-after expression:
(5.35) (1 + r0)(1 + r1)(1 + r2) (1 + r3)... = I.d1d2d3...dn... = P.If dk is the first non-vanishing digit, then these formulas are modified as follows:
(5.36) (1 + r0)(1 + rk)(1 + rk+1)... = I.00...0dkdk+1... = P.and:
(5.37) rn = [ 0.00 ... 0dn / I.00...0dkdk+1...dn-1 ], for n > k+1.
r0 = 2;or, in decimal form:
r1 = 0.1/3;
r2 = 0.04/3.1;
r3 = 0.001/3.14;
r4 = 0.0005/3.141;
r5 = 0.00009/3.1415,....
r0 = 2;Thus:
r1 = 0.0333333....
r2 = 0.0129032....
r3 = 0.0003184....
r4 = 0.0001591....
r5 = 0.0000286....
π = 3(1.0333333...)(1.0129032...)(1.0003184...)(1.0001591...)(1.0000286...) ...Round-off errors prevent the exact result, 3.14159..., as required by (5.33). The famous WALLIS PRODUCT for π/2 uses rn = 1 / 4n2 - 1 for n > 1, and gives:
= (3.141589...)(1.0000?...)....
π = 2(1.0333333...)(1.0666666...)(1.0285714...)(1.0153846...)(1.0101010...)...a much slower process. However, π = P/S in the Wallis case, since:
= (3.0007319...)(1.00?...)...
S = ∑ 1 / 4n2 - 1Historical Note: Wallis (whose surname means: `man from Wales') is the person who gave us the symbol ∞ for infinity (see [7.52].)
= 1/2 ∑ (1 /(2n - 1) - 1 /(2n + 1))
= 1/2[(1/1-1/3) + (1/3-1/5) + (1/5-1/7) + ...] = 1/2(1/1) = 1/2.
(5.38) S = r0 + r1 + r2 + r3 + ... rn + ...,in this special situation. These terms decrease very rapidly, as seen by (5.34) and (5.37). Indeed, rn becomes approximately [ 0.00...0dn / P] for large n, and so upon adding these terms up for n > k+1, results in the excellent approximation (for large k):
(5.39) S ~ r0 + r1 + r2 + r3 + ... rk + [0.00...dk+1dk+2.../P]which becomes exact as k → ∞. Note that this approximation underestimates the exact sum, S, since P exceeds every Pk. A useful, alternative form, for arbitrary convergent infinite series, is obtained by subtracting (5.33) from (5.35) (P - Pn = 0.00...0dn+1dn+2...):
(5.40) S ~ r0 + r1 + r2 + r3 + ... rk + [1-Pk/P],where:
Pk = (1 + r0)(1 + r1)(1 + r2)(1 + r3) ... (1 + rk).In this form, where:
Pk/P = 1 / (1 + rk+1)(1 + rk+2)...,it can be compared with the earlier version, (5.14), of a summation process for a general convergent infinite series. The validity of the earlier process, leading to the summation term:
(5.41) -log(Pk/P) = -log[1-(1-Pk/P)]depended upon the appropriateness of the exponential approximation, P ~ eS for small rn (n > k+1), while the validity of the summation process leading to the summation term [1-Pk/P] in (5.40), depended upon the appropriateness of replacing Pn-1 by P in (5.34) or (5.37), for n > k+1, relative to this special expansion in (5.35) or (5.36). The use of (5.41) also leads to an underestimate of the exact sum, S, because:
= log[1/1-(1-Pk/P)]
= log[1 + (1-Pk/P) + (1-Pk/P)2 + (1-Pk/P)3 + ... ]
= [(1-Pk/P) + (1-Pk/P)2 + (1-Pk/P)3 + ... ]
- 1/2[(1-Pk/P) + (1-Pk/P)2 + (1-Pk/P)3 + ... ]2 + ....,
-log(Pk/P) = log(P/Pk) = log[(1+rk+1)(1+rk+2)...]However, the expansion on the right in (5.41) (using first a geometric series, followed by a logarithmic series), indicates that a few terms of the geometric series:
= log(1 + rk+1) + log(1 + rk+2) + ...
= rk+1 - (1/2) rk+12 + ... + rk+2 - (1/2) rk+22 + ...
< rk+1 + rk+2 + ....
(1-Pk/P) + (1-Pk/P)2 + (1-Pk/P)3 + ...,whose total sum:
1 / 1 - (1-Pk/P) = P / Pkactually exceeds the exact "tail-end" sum, rk+1 + rk+2 + rk+3 ... should yield an improved estimate for the latter. Therefore, a practical compromise between simplicity and accuracy might be the simple approximation:
= (1 + rk+1)(1 + rk+2)(1 + rk+3) ...
(5.42) S ~ r0 + r1 + r2 + r3 + ... rk + [1-Pk/P] + [1-Pk/P]2which gives a workable summation process for any case where the value of the corresponding infinite product:
P = (1 + r0)(1 + r1)(1 + r2)(1 + r3) ... > 1is known, a-priori. It is interesting to note that the full quadratic term on the right in (5.41) is:
1/2 [1 - Pk/P]2,which seems to support the appropriateness of the approximation in (5.42).
S = r0 + r1 + r2 + r3 + ...,with terms limited by 0 < rn < 1, and for which the value of the corresponding decreasing infinite product:
P = (1 - r0)(1 - r1)(1 - r2)(1 - r3) ... ( < 1 )is known, the analogous approximation:
(5.43) S ~ r0 + r1 + r2 + r3 + ... rk + [P/Pk -1] - 1/2 [P/Pk-1]2might be expected to exhibit an accuracy similar to that of (5.42). Here we have used the first two terms of the expansion of:
log (Pk/P) = log [1 - (Pk/P - 1)]These two terms alone underestimate the "tail-end" series, whereas log(Pk/P) overestimates it.
= (Pk/P - 1) - 1/2 (Pk/P - 1)2 + 1/3 (Pk/P - 1)3 - ...
P = (1 - 4/32)(1 - 4/42)(1 - 4/52) ... (1 - 4/n2) ... = 1/6,we can obtain the exact sum of the corresponding infinite series:
S = 4/32 + 4/42 + 4/52 + ... + 4/n2 + ...in terms of the Riemann zeta-function for s=2:
S = 4 [ ζ(2) - 1 - 1/22 ] = 4 [ π2/6 - 5/4 ] = 1.57972....When we choose k=7 in this case, and directly sum the finite series:
S7 = 4/32 + 4/42 + 4/52 + 4/62 + 4/72 = 1.047184...,we obtain S - S7 = 0.532... (exact value), while [P7/P - 1] - 1/2[P7/P - 1]2 = 0.459... (approximate low value), where we have computed the finite product:
P7 = (1- 4/32)(1- 4/42)(1- 4/52)(1- 4/62)(1- 4/72) = 0.2857142...also directly. The summation term log(P7/P) = 0.591... (approximate high value) appears to be the more accurate one.
e-S = e-4 (π2/6 - 5/4) = e-1.57972....and P = 1/6. So how close is 1.57972... to log 6 ~ 1.8... ? Not very, but the number, bS, for which P = bS-S holds exactly, is simply their ratio, 6/1.57972... = 3.79814... = bS . Therefore:
P = 1/6 = (3.79814...)-1.57972....Recall the interesting Wallis product situation, where P/S = π, then bS = (πS)1/S = (π/2)2 = 2.467396..., which is much closer to e. For the interesting case cited in Section 5C: Convergence tests for Infinite Products, one has for the number
bS = e-S*(t)/S(t)where:
S*(t) = 1/(t-1) - 1/2 (1/t2 - 1) - 1/3 (1/t3 - 1) + ...,and:
S(t) = 1/t - 1/t2 - 1/t3 + ... = 1/(t-1),
P(t) = (1 - 1/t) - (1 - 1/t2) - (1 - 1/t3) - ...for t > 1. So one could explore numerical circumstances here, using the equation:
bS = e-[1 - 1/2(1/(t+1)) - 1/3(1/(t2 - t - 1) + ...]For example, bS converges to e-ζ(2) = e-6/π2 as t → 1+.
(5.44) ex = lim n → ∞ (1 + xn/n)n, xn > 0,whenever lim n → ∞ xn = x. To do this, we can simply interpret the quantity:
Pn = (1 + xn/n)n = (1 + xn/n)(1 + xn/n) ... (1 + xn/n)as a product of n factors, with identical rn = xn/n values. So by (2.7), with Sn = (xn/n)n = xn, the inequalities:
(5.45) axn < (1 + xn/n)n < exnhold, where a = (1 + R)1/R , provided that rn = xn/n < R. Then for any R > 0, the limit in (5.45), yields the inequalities:
(5.46) (1 + R)x/R < lim n → ∞ (1 + xn/n)n < ex.Letting R → 0 in (5.46) establishes (5.44).
lim m → ∞ M (1 + jm/m)m = M ejis an exact outcome for the limiting continuous formula in (5.3), for the annual return.
(4.38) P = (1 - r1)(1 + r2)(1 - r3) (1 + r4) ... (1 - r2n+1)(1 + r2n+2) ...where now our objective is to render this into an ultimately increasing product. To do this, we require that the inequalities:
(5.47) 0 < (r2n+2 - r2n+1) - r2n+1 r2n+2 = r2n+2'hold for all n sufficiently large. (This requires r2n+1 < r2n+2/(1 + r2n+2)). For such n, we have:
(1 - r2n+1)(1 + r2n+2) = (1 + [(r2n+2 - r2n+1) - r2n+1r2n+2])and so under (5.47), these two factors do contribute a simple, increasing one to the product. Thus (4.38), can be rewritten in the form:
= (1 + r2n+2')
(5.48) P = (1 - r1)(1 + r2) ... (1 + r2k+2')(1 + r2k+4') ...,for some k, where:
r2k+2j' = (r2k+2j r2k+2j-1) - r2k+2j-1 r2k+2j > 0 for all j > 1.So the tail-end increasing product:
(5.49) Pte' = (1 + r2k+2')(1 + r2k+4') (1 + r2k+6') ...is governed by the corresponding series:
(5.50) Ste' = r2k+2' + r2k+4' + r2k+6' ...and according to (2.7), by:
(5.51) aSte' < Pte < eSte'where a < e. In particular, Pte and P are finite if and only if Ste' < ∞. The inequality (5.47) shows the bias toward decreasing factors, when r2n+2' < 0 as in Section 4K. Turning Mixed Products into Decreasing Products. Again, we point out that (4.38) is not as special as it appears to be, since some r2n+1 values can be allowed to be zero. In this case, the presence of the decreasing factors (1 - r2n+1) in (4.38) merely retard the otherwise increasing aspects of the process, but can convert an otherwise divergent one into a convergent one.
(2.2) P = (1+r1)(1+r2)(1+r3) ... (1+rn) ...we associate an infinite series:
(2.4) F = r1 + P1r2 + P2r3 + ...for which P = 1+F. It turns out that the partial sums:
(5.52) Fn = r1 + P1r2 + P2r3 + P3r4 + ... Pn-1rnof this series F are just what we need in our display. Here, Pn = (1 + r1)(1 + r2) (1 + r3) ... (1 + rn). If we begin with an isosceles triangle of unit altitude (proportional to its area), and successively increase the altitude according to the values of Fn, then one "sees" the partial products, Pn of (2.1) as the corresponding areas. In particular, the altitudes of the isosceles triangles are bounded by the value of P=1+F. As shown, the incremental increases of the altitudes are just the terms:
(5.53) Pn-1 rnof the infinite series, F:
373.(4.38) P = (1 - r1)(1 + r2)(1 - r3) (1 + r4) ... (0 < r2n+1 < 1).with, in this case (recall the derivation of P=1-F in Chapter 1: Decreasing Products.):
(5.55) F = P-1 = - r1 + P1r2 - P2r3 + P3r4 + ...where:
Pn = (1 - r1)(1 + r2)(1 - r3) ... (1 ± rn)Then the partial sums:
(5.56) Fn = - r1 + P1r2 - P2r3 + ... ± Pn-1rndepict the rise AND FALL of the altitudes of our isosceles triangles as the quantities ± Pn-1rn depict these changes. The altitudes for even-numbered steps always increase, while those for the odd-numbered steps always decrease. Therefore, the ultimate altitude (value P) will vary from 0 to ∞ from case to case (if convergent). If the inequalities 0 < (r2n+2 - r2n+1) - r2n+1r2n+2 hold for all n as illustrated below, then the limiting value of P exceeds 1 (may be equal to ∞). However, all possibilities (including indeterminate ones with continual up-and-down fluctuations) are easily imagined, using this scheme of increasing and decreasing isosceles triangles. In this visual scheme, the entire rectangle-filling process of Chapter 4: Applications of Decreasing Products takes place within the original isosceles triangle of altitude 1, with decreasing isosceles triangles converging down to the altitude of P < 1.
374.(5.57) PA = (1 + r1)(1 + r2)(1 + r3)(1 + r4) ...,obtained by replacing all ±rn values with their corresponding absolute-values, |±rn| = rn, is convergent. In such a case, the series (5.55) is replaced by the companion series:
(5.58) FA = PA - 1 = r1 + P1r2 + P2r3 + P3r4 + ...,obtained by replacing all ±rn values with their corresponding absolute values, rn. The Pk in (5.58) are not the same ones as those in (5.55), but definitely exceed the latter. Thus, if the series (5.58) converges, then so does the series (5.55), because for infinite series, absolute convergence implies convergence. Because P = 1 + F, the same process takes place in the case of infinite products, namely:
(5.59) AN INFINITE PRODUCT CONVERGES IF IT IS ABSOLUTELY CONVERGENT.This maxim is needed repeatedly in Section 6E: Infinite products of functions and in Section 6F: Special infinite products of operators of the following chapter. We note also that since rearrangements of factors in (4.38) are mirrored in the same rearrangements of the terms of the series in (5.55), all rearrangements are allowed whenever absolute convergence is present.
(5.1) P1 = M(1+j/m)mSuppose that the lender then imposes a small finance-charge, fraction i, of the year-end balance. The principal then is reduced to the amount:
(6.1) P1 = M(1+j/m)m(1-i)If this arrangement is continued for k years, then the principal becomes:
(6.2) Pk = M(1+j/m)km(1-i)k,or essentially, Mejk(1-i)k (as explained in Chapter 5: Applications of Increasing Products). In the symbolism of Chapter 3: Increasing and/or Decreasing Products, this modified investment scheme corresponds to an indeterminate situation, namely:
S+ = limk→∞ (j/m)km = ∞ (rn = j/m)As indicated above, this application manifests a real-life meaning, though the mathematical meaning in the limit is at best vague. However, the practical outcome is generally predictable, as indicated below.
S- = limk→∞ ik = ∞ (rn = i).
(6.3) a+jka--ik < Pk/M < e(j-i)k,where a+= (1+j/m)m/j and a-= 1/(1-i)1/i are nearly equal to e, for small values of i and j. Thus (6.3) means, in effect, that:
Pk = M e(j-i)k,the continuous compounding upper bound to any real-life scheme. Should the interest rate j and the finance charge rate i change periodically, then (6.2) must be reformulated in the form:
(6.4) Pk = M(1+j1/m)(1+j2/m)(1+j3/m) ... (1+jk/m) × (1-i1)(1-i2)(1-i3) ... (1-ik),and (6.3) becomes:
(6.5) a+(j1/m + j2/m + j3/m ... + jk/m) a--(i1 + i2 + i3 + ... + ik) < Pk/M < e(j1/m + j2/m + j3/m ... + jk/m) - (i1 + i2 + i3 + ... + ik)In effect:
(6.6) Pk ~ M e(Sk-S'k),where:
Sk = (j1 + j2 + ... + jk)/mand:
S'k = (i1 + i2 + ... + ik)
S = limk → ∞ Skand:
S' = limk → ∞ S'k,then, of course, we are actually dealing with a determinate case of Chapter 3: Increasing And/Or Decreasing Products, and (6.5) yields:
a+S a--S' < P/M < eS e-S', as k → ∞Also, (6.6) yields: P ~ M eS e-S' = M eS-S'.
Pk = [(1-r1)(1-r2) ... (1-rk)][(1+t1)(1+t2) ... (1+tk)].The ultimate result as k → ∞ determines the outcome, and the rectangle gets completely filled if and only if that limit is zero. In this situation, we have a contest between filling and expanding, with:
Sk = r1 + r2 + ... + rkand:
Tk = t1 + t2 + ... + tkplaying the corresponding roles as described in Section 6A: Compound-interest with finance charges., above.
(6.7) a+Tk a--Sk < Pk < eTk-SkIf Tk < ∞ and Sk → ∞ as k → ∞, then the rectangle is ultimately completely filled. Other (interesting) outcomes are certainly possible. The circumstance Tk → ∞ depicts the case of a rectangle increasing in size without limit, regardless of any technical mathematical interpretation or what happens to the limits limk → ∞ Sk or limk → ∞ Tk - Sk. If the latter two are both ∞, then the infinitely increasing rectangle nonetheless gets filled. (Recall paragraph B of Chapter 4). (For other interpretations of the rectangle-filling process using mixed increasing and decreasing products, see ref. [7.1], and Sections 4K, 5G, and 5I).
(6.8) P(x) = (1 - r1(x)(1 - r2(x)(1 - r3(x) ... (1 - rn(x) ...is then governed by the inequalities (following (1.9)):
(6.9) a-S(x) < P(x) < e-S(x),where:
S(x) = r1(x) + r2(x) + r3(x) + ...Presently, we shall lift the restrictions on the rk(x) values. We have previously considered examples of (6.8) in Chapter 4: Applications of Decreasing Products:
a = 1/[1 - R(x)]1/R(x),
R(x) = max1<k<∞ rk(x)
(4.20) cos πx = (1 - 4 x2)(1 - 4 x2/32)(1 - 4 x2/52) ... (1 - 4 x2/(2n-1)2) ...and:
(4.31) sin πx / πx = (1 - x2)(1 - x2/22)(1 - x2/32)(1 - x2/42) ...where, however, in order to avoid negative factors, the ranges for the quantity, x, was restricted to 0 < x < 1/2 and 0 < x < 1, respectively. Should we let x vary over the entire real line, these formulas still apply, and we obtain, not only the correct negative values for the trigonometric functions, but the infinite products are generally of the mixed type of increasing and decreasing products, which converge according to (5.59). Then these become analytical applications appropriate to this chapter. The interesting examples from Chapter 3: Increasing and/or Decreasing Products:
(3.6) P(x) = (1 - x2)(1 + (x2)2/2!)(1 - (x2)3/3!)(1 + (x2)4/4!) ...,with zeros and negative factors, and involving both increasing and decreasing factors, are still other appropriate examples here. (The latter one requires some kind of extension to the entire real line, see: Section 6F: Infinite convolution products.)
and
(3.10) P(x) = (1 - r1/x)(1 - r2/x)(1 - r3/x) ...,
(6.10) P-(x) = (1 - x)(1 - x/2)(1 - x/3) ... (1 - x/n) ...in order to obtain a convergent product:
(6.11) Pw(x) = (1-x)e-x (1-x/2)e-x/2 (1-x/3)e-x/3 ... (1-x/n)e-x/n ...of functions that has zeros at all positive integers. We can obtain the same result here by constructing the formal product of the decreasing product of (6.10) with that of the increasing product:
P+(x) = (1 + x)(1 + x/2)(1 + x/3) ... (1 + x/n) ...to obtain:
(6.12) P+(x)P-(x) = (1-x2) (1-(x/2)2) (1-(x/3)2) ... (1-(x/n)2) ....The related infinite series, S(x) = x2 + (x/2)2 + (x/3)2 + ..., guarantees the convergence for all x, according to (6.9).
(6.13) P*(x) = (1 - e-x)(1 + e-2x)(1 - e-3x)(1 + e-4x) ... (1-e-nx)(1+e-(n+1)x) ...whose related infinite series:
S(x) = - e-x + e-2x - e-3x + e-4x + ...guarantees the convergence for x > 0, according to (6.9). Here, P*(0) = 0 by convention, and P*(x) → 1 as x → ∞, according to (3.3), since R(x) = e-x. (It is essentially given by the first factor alone for large values of x). Furthermore, P*(x) can be factored into the product, P+(x)P-(x), where:
P+(x) = (1 + e-2x)(1 + e-4x)(1 + e-6x) ...are, respectively, a convergent increasing product and a convergent decreasing product with corresponding infinite series:
and
P-(x) = (1 - e-x)(1 - e-3x)(1 - e-5x) ...
S+ = e-2x + e-4x + e-6x + ...Therefore:
and
S- = e-x + e-3x + e-5x - ....
[S+ - S-] = - e-x + e-2x - e-3x + e-4x - e-5x + e-6x - ...which tends to 0 as x → ∞. In fact, the second factor here can be summed as a geometric series:
= - e-x [ 1 - e-x] - e-3x [ 1 - e-x] - e-5x [ 1 - e-x] - ...
= - [ 1 - e-x] × [ e-x + e-3x + e-5x + e-7x + ...],
e-x [ 1 + (e-x)2 + (e-x)4 + (e-x)6 + ... ]So that finally we have:
= e-x [ 1 + (e-x)2 + ((e-x)2)2 + ((e-x)2)3 + ... ]
= e-x / [ 1 - (e-x)2].
[S+ - S-] = - e-x(1 - e-x)/ (1 - (e-x)2)This tends to -1/2 as x → 0, and therefore the upper bound of (6.13), e[S+ - S-] = e-1/(ex+1) tends to 1/e1/2 as x → 0. This function is illustrated below. It is amazingly close to a straight line segment for 0<x<1. In fact:
= - e-x / (1 + e-x)
= -1 / (ex + 1).
e-1/(1+ex) = 0.6065 + 0.162 x(shown as the dashed curve) is accurate to within 1/4% for 0<x<1. We might pursue this strange circumstance by expanding the exponential as a power series:
e1/(1+ex) = 1 + (1/(1+ex)) + 1/2! (1/(1+ex))2 + 1/3! (1/(1+ex))3 + ...The bracketed quantity here is largest, 1.296..., when x=0, and smallest, 1.147..., when x=1. Hence the inequalities:
= 1 + (1/(1+ex)) [1 + 1/2 (1/(1+ex)) + 1/6 (1/(1+ex))2)) + 1/24 (1/(1+ex))3)) + ... ].
(ex + 1) / (ex + 2.3...) < e-(1/(1+ex)) < (ex + 1) / (ex + 2.1...)hold approximately for 0 < x <1. The left member is nearly exact for x=0, and the right member is nearly exact for x=1. The middle term, of course, traverses from left to right as x traverses from 0 to 1. From these, one might speculate that something like:
(ex + 1)/(ex + 2.3 - 0.2x)might be a reasonably good approximation to e-1/(1+ex), but it is not nearly as accurate nor as simple as the above straight-line segment. Also shown in the illustration, of course, is the infinite product itself:
P*(x) = P-(x)P+(x) = (1 - e-x)(1 + e-2x)(1 - e-3x)(1 + e-4x) ...which tends to zero as x → 0. The lower bound, (1 - e-x)ex/(ex+1)) (here, R = e-x)) is the dotted curve in the illustration. For the purposes of the following Section 6F: Infinite convolution products, we extend P*(x) to the real line as an even function of x.
< (1 - e-2x)(1 + e-2x)(1 - e-4x)(1 + e-4x) ...
= (1 - e-4x)(1 - e-8x) ...,
339.(6.14) P**(x) = (1 - e-x)(1 + e-2x)(1 - 3e-3x)(1 + 3e-4x) ... (1 - ne-nx)(1 + ne-(n+1)x) ...As above, P**(x) factors into increasing products, P+(x) and decreasing products P-(x), where:
P+(x) = (1 + e-2x)(1 + 3e-4x)(1 + 5e-6x) ....Following precisely the same procedure used above, we can obtain the expression:
P-(x) = (1 - e-x)(1 - 3e-3x)(1 - 5e-5x) ....
(6.15) [S+(x) - S-(x)] = - [1 - e-x] e-x [1 + (e-x)2 + 3(e-x)4 + 5(e-x)6 + ... ]for the difference of the corresponding infinite series. Except for the odd multiples in the last bracketed factor on the right in (6.15), this factor would sum to 1/(1 - (e-x)2), as above, and the limit of [S+(x) - S-(x)], would be -1/2, as above. However, the presence of these odd multiples will clearly force this limit to be -∞ in this case. Thus P**(x) → 0 as x → 0. We observe that the negative factors in (6.14) arise as x traverses the zeros xn of P**(x). Interestingly enough, these are given by:
(6.16) xn = log n / n for odd nwhich eventually tend monotonically to zero, as n → ∞, just as in the case of (3.10).
406.(a). cos iπs = (I + 4s2)(I + 4s2/32)(I + 4s2/52) ....We emphasized that juxtaposition of operators means convolution, and convergence is just inferred from that of the corresponding arithmetical products in Section 6E: Infinite products of functions. Incidentally, 1/s is the integration operator, usually symbolized by H(t), the Heaviside step-function˚, considered as an operator.
(b). sin iπs = iπs(I + s2)(I + s2/22)(I + s2/32) ....
(c). P(s/i) = (1+s2)(1+(s2)2/2!) (1+(s2)3/3!) ... (see (3.6)))
(d). P+(s/i)P-(s/i) = (1+(s/2)2)(1+(s/3)2)(1+(s/4)2) ... (see (6.12)))
(e). hP(s/i) = h(1 - ir1/s)(I - ir2/s)(I - ir3/s) .... (see (3.10)))
(f). hPw(s/i) = (I + is)eis)(I + is/2)eis/2)(I + is/3)eis/3) ....
(g). P*(s/i) = (I - eis)(I + e2is)(I - e3is)(I + e4is) ....
(h). P** (s/i) = (I - eis)(I + e2is)(I - 3e3is)(I + 3e4is) ....
6F(vi)(a). The convolution product˚ (I - e-s)(I - e-2s/22)(1 - e-3s/32) ..., where e-ns is the real translation operator whose F.T. is einx, becomes the arithmetical infinite product:(1 - eix)(1 - e2ix/22)(1 - e3ix/32) ...by the F.T. This product converges for all x, because of the numerical divisors. Without them, arithmetical convergence fails.
6F(vi)(b). The infinite convolution product˚ of operators, (I - s)(I - (s/2)2)(I - (s/3)2)..., with F.T. becomes (1 - ix)(1 + (x/2)2)(1 - i(x/3)2).... This product is a mixed infinite product, with decreasing (complex) factors and increasing (real) factors, converging for all x, because of (5.59) (extended to complex-valued products).
6F(vi)(c). Let φ be a non-zero c∞-function with compact support˚. If φ is considered an operator, then its F.T. is the classical one, given by ψ(x) = ∞ ∫ ∞ e-ixt φ(t)dt. The infinite convolution product:(I - φ)(I - φ2)(I - φ3) ...,becomes the arithmetical infinite product:(1 - ψ(x))(1 - ψ2(x))(1 - ψ3(x)) ...by the F.T. This product converges for all |x| sufficiently large, since ψ(x) is a rapidly decreasing (entire-analytic) function of x. A scalar factor multiplying φ can insure convergence for all x, to salvage this example.
6F(vi)(d). The convergence˚ of the convolution infinite product˚:(I - 2is)(I - 3is)(I - 5is)(I - 7is) ...might be in doubt, but the F.T. gives (surprise):(1 - 2-x)(1 - 3-x)(1 - 5-x) ...,which converges to 1/ζ(x), for x > 1, and to zero otherwise. This is the reciprocal of the Riemann zeta-function˚, as treated here, and so is zero for all x < 1.
6F(vi)(e). Similarly, the F.T. sends the infinite convolution product˚(I - 2is)(I - 3is)(I - 4is) ...to the arithmetical infinite product (4.12):(1 - 2-x)(1 - 3-x)(1 - 4-x) ...,converging for x as above, and which we encountered in the modified filling process.
6F(vi)(f). The F.T. of the infinite convolution operator product with mixed "increasing" and "decreasing" factors:Pm(s/i) = (I + i/s)(I - i/s2)(I + i/s3)(I - i/s4) ...produces the mixed arithmetical product in (3.5):Pm(x) = (1 - 1/x)(1 + 1/x2)(1 - 1/x3)(1 + 1/x4) ...with increasing and decreasing factors. It tends to 0 as x → 1+, and has been extended to all x by defining Pm(x) = 0 for x < 1. One could premultiply Pm(s/i) by h, for greater clarity.
6F(vi)(g). By the F.T., the convolution infinite product of "increasing" factors:(I + 2is)(I + 3is)(I + 5is) ...becomes the arithmetical infinite product (5.8) of increasing factors:(1 + 2-x)(1 + 3-x)(1 + 5-x) ...,converging for x > 1, to ζ(x)/ζ(2x) (5.16). In this case, we need to premultiply by H(x-1) and preconvolute by e-sh to salvage this example.
6F(vi)(h). Finally, for the "increasing" infinite convolution product of operators:(I + se-s)(I + s2e-s)(I + s3e-s)(I + s4e-s)(I + s5e-s)...,the F.T. gives:(1 + ixe-ix)(1 - x2e-2ix)(1 - ix3e-3ix)(1 + x4e-4ix)(1 + ix5e-5ix)...,which converges only if -1 < x < 1. There seems to be no simple way to salvage this example for operators, except to insert some kind of convergence factors, like e-x2/n, such as Weierstrass did. The corresponding convolution infinite product becomes:(I + se-s)es2 (I + s2e-s)es2/2 (I + s3e-s)es2/3 ...,which now qualifies for consideration here, since its F.T.:(1 + ie-ix)e-x2 (1 - x2 e-2ix)e-x2/2 (1 - ix3 e-3ix)e-x2/3 ...converges for all x.
(6.17) P(x) = (1 - cψ(x)) (1 - c2ψ2(x)) (1 - c3ψ3(x)) ...,which converges for all x. This is because c is chosen to be so small that R = c maxx |ψ(x)| < 1. Since ψ(x)→0 (rapidly) as x→∞, the infinite product (6.17) is given approximately as:
P(x) ~ e-S(x)for large |x|. The corresponding infinite series:
S(x) = cψ(x) + c2ψ2(x) + c3ψ3(x) + ....tends (rapidly) to zero as |x|→∞, which implies that P(x)→1 (rapidly) as |x|→∞. Therefore (see (1.5) and (1.6)):
(6.18) 1 - P(x) = F(x) = cψ(x) + [1 - cψ(x)] c2ψ2(x) + [(1 - cψ(x))(1-c2ψ2(x)] c3ψ3(x) + ... + [(1 - cψ(x))(1 - c2ψ2(x)) ... cn-1ψn-1(x))] cnψn(x) + ...not only converges, as it always does, but is integrable˚ with respect to x, since it converges to zero (rapidly) as |x|→∞. Thus, F(x) possesses an ordinary numerical I.F.T.:
(6.19) F(t) = -∞∫∞ eixtF(x)dx/2π,which, here, is considered to be an ˚operator. Because of (6.18), it is expressible in the interesting operator form (recall that φ = I.F.T. ψ is a c∞-function with compact support˚):
(6.20) F = cφ + [I - cφ] c2φ2 + [(I - cφ)(I - c2φ2] c3φ3 + ... + [(I - cφ)(I - c2φ2) ... (I - cn-1φn-1)] cnφn + ...where I, of course, is the singular (non-function) identity operator˚. All other symbols in (6.20), are regular (function) operators˚. On the other hand, (6.20) can also be expressed as a infinite convolution product˚:
(6.21) F = I - P(s/i) = I - (I - cφ)(I - c2φ2)(I - c3φ3) ...,(see (6.17)), where again, I is the only singular operator involved.
(6.22) S = r1 + r2 + r3 + ... + rn + ...converges in F, then the convolution product:
(6.23) P = (1 + r1)(1 + r2)(1 + r3) ... (1 + rn) ...converges in F.
(6.24) F = r1 + (1 + r1)r2 + (1 + r1)(1 + r2)r3 + ...guarantees success, regardless of the nature of the F.T. rn in F. (See Section 5J: Absolute Convergence.).