Based on the infinite geometric sum 1 = 1/2 + 1/4 + 1/8 +... we have the trivial infinite product 1/2 1/4 1/8 a = a a a .... for any number a. Of course, dividing each term in the geometric sum by 2 gives the equivalent identity 1/2 = 1/4 + 1/8 + 1/16 + ..., so without altering the product we can introduce an arbitrary number b to give / a \ 1/2 1/4 1/8 a = ( --- ) (ab) (ab) .... \ b / Likewise we can introduce a sequence of arbitrary quantities c,d,e,... and we have the identity / a \1/2 / ab \1/4 / abc \1/8 / abcd \1/16 a = ( --- ) ( ---- ) ( ----- ) ( ------ ) .... \ b / \ c / \ d / \ e / Now, suppose we set a = 1-x b = 1+x c = 1+x^2 d = 1+x^4 e = 1+x^8 and so on. Then the above identity becomes / 1 - x \1/2 / 1 - x^2 \1/4 / 1 - x^4 \1/8 (1 - x) = ( ----- ) ( ------- ) ( ------- ) ... \ 1 + x / \ 1 + x^2 / \ 1 + x^4 / Naturally there are infinitely many variations on this approach. For example, if we set a = 1-x^4 b = 1+x^4 c = 1+x^8 d = 1+x^16 e = 1+x^32 we'll end up with an expression for (1-x) that looks like this 1 / 1 - x^4 \1/2 / 1 - x^8 \1/4 / 1 - x^16 \1/8 (1-x) = ------------ ( ------- ) ( ------- ) ( -------- ) ... (1+x)(1+x^2) \ 1 + x^4 / \ 1 + x^8 / \ 1 + x^16 / On the other hand, if we set a = 1-x b = (1+x)(1+x^2) c = (1+x^4)(1+x^8) ... we can define another infinite family of identities, based on / 1 - x \1/2 / 1 - x^4 \1/4 / 1 - x^16 \1/8 (1-x) = ( ------------ ) ( -------------- ) ( ---------------- ) ... \(1+x)(1+x^2)/ \(1+x^4)(1+x^8)/ \(1+x^16)(1+x^32)/ All of these are based somewhat loosely on "binary" exponents. Of course, the infinite product 1 ------ = (1 + x)(1 + x^2)(1 + x^4)(1 + x^8) ... 1 - x is sort of a standard example of an infinite product. It's even given as the example under the definition of "infinite product" in "The Harper Collins Dictionary of Mathematics" by Borowski and Borwein. (However, I notice they have a typo, because they list the exponent on x as 2n instead of 2^n.) This is based directly on the geometric, since, as Euclid showed, 1/(1-x) equals the geometric series 1 + x + x^2 + x^3 + ... so there are many different infinite products, corresponding to the possible complete numeration systems. In other words, we want a set of sets of numbers such that every positive integer has a unique representation as a sum of precisely one element of each set, and each such representation corresponds uniquely to a positive integer. The "standard" example noted above corresponds to the binary number system, which has the sets {0,1}, {0,2}, {0,4}, {0,8}, ... whereas the base 3 number system has the basis sets {0,1,2}, {0,3,6}, {0,9,18},... which suggests the infinite product oo 1/(1-x) = PROD [1 + x^(3^n) + x^(2*3^n)] n=0 = (1 + x + x^2)(1 + x^3 + x^6)(1 + x^9 + x^18)... and similarly for any other base. Likewise, the factorial number system suggests the product oo 1/(1-x) = PROD [ 1 + x^(1*(n!)) + x^(2*(n!)) + ... + x^(n*(n!)) ] n=1 = (1 + x) (1 + x^2 + x^4) (1 + x^6 + x^12 + x^18) (1 + x^24 + x^48 + x^72 + x^96)... Since the nth partial sum of the geometric series is (1-x^n)/(1-x), these products all just correspond to cyclotomic factorizations, although not necessarily primitive factorizations. Incidentally, one simple way of "seeing" why these products work is to notice how the terms "telescope". For example, beginning with the identity /1 + x^2\1/2 /1 + x^4\1/4 /1 + x^8\1/8 1 = (1 - x^2)( ------- ) ( ------- ) ( ------- ) ... \1 - x^2/ \1 - x^4/ \1 - x^8/ which we can instantly see is true, because the factors "telescope" into each other, i.e., if we combine the first two terms on the right side we get the equivalent relation 1/2 /1 + x^4\1/4 /1 + x^8\1/8 1 = (1 - x^4) ( ------- ) ( ------- ) ... \1 - x^4/ \1 - x^8/ Then we combine the first two terms on the right side of THIS expression to give the equivalent relation 1/4 /1 + x^8\1/8 1 = (1 - x^8) ( ------- ) ... \1 - x^8/ and so on. Thus it's clear that if |x| is less than 1 the right hand infinite product equals 1. We can also see how such products are related to power series expansions of the natural log. If we write the original relation as 1 /1 + x^2\1/2 /1 + x^4\1/4 /1 + x^8\1/8 --------- = ( ------- ) ( ------- ) ( ------- ) ... (1 - x^2) \1 - x^2/ \1 - x^4/ \1 - x^8/ and take the natural log of both sides, we have the infinite sum 1 /1 + x^2\ 1 /1 + x^4\ 1 /1 + x^8\ -ln(1 - x^2) = -ln( ------- ) + -ln( ------- ) + -ln( ------- ) + ... 2 \1 - x^2/ 4 \1 - x^4/ 8 \1 - x^8/ Now, the two most common power series expansions involving the natural log function are u^2 u^3 u^4 -ln(1 - u) = u + --- + --- + --- + ... 2 3 4 and 1 /1+u \ u^3 u^5 u^7 -ln( ---- ) = u + --- + --- + --- + ... 2 \1-u / 3 5 7 so the logs on the right side of the preceding equation give the overall sum x^6 u^10 u^14 x^2 + --- + ---- + ---- + ... 3 5 7 x^4 x^12 --- + ---- + ... 2 6 x^8 --- + ... 4 which of course is the expansion of -ln(1 - x^2). We could also express the above relations between logs in terms of the respective continued fractions, noting that u ln(1 + u) = ---------- u 1 + ---------- u 2 + ---------- u 3 + --------- ... and 1 /1+u\ u -ln( --- ) = ----------- 2 \1-u/ u^2 1 - ---------- 4u^2 3 - --------- 9u^2 5 - ---------- 16u^2 7 - ---------- ... Anyway, it's clear that all the infinite products above are based on a sequence of functions that is multiplicatively cumulative in a nice way. For example, the cumulative products of [1-x] [1+x] [1+x^2] [1+x^4] [1+x^8] ... are all of the form 1 - x^n. Another sequence of functions with a similar property is [b^x - b^(-x)] [b^x + b^-x] [b^(2x) + b^(-2x)] ... Thus we can construct infinite products based on these functions, which of course are essentially just hyperbolic trig functions. The arch- typicaly identity of this family is / 1 - exp(-z) \1/2 / 1 - exp(-2z) \1/4 1 - exp(-z) = ( ------------- ) ( -------------- ) ... \ 1 + exp(-z) / \ 1 + exp(-2z) / More interestingly, we can apply the same approach to the ordinary trigonometric functions, noting the multuiplicatively cumulative sequence of functions sin(z) 2cos(z) 2cos(2z) 2cos(4z) 2cos(8z) ... From this we immediately arrive at the intesting result 1 | / 1/2 1/4 1/8 \ | |sin(z)| = --- |( tan(z) tan(2z) tan(4z) ... )| 2 | \ / | This relates the magnitudes (norms) of the two sides, but in terms of the real and imaginary components, the identity can be explicitly written as 1/2 1/4 1/8 sin(2z) + i[1 - cos(2z)] = tan(z) tan(2z) tan(4z) ... The left hand side is as shown below: The "identity" of this simple circular locus with the infinite product on the right is non-trivial, because the right-hand side is evidently discontinuous at every real value of z. Notice that the right side vanishes when any of the arguments z, 2z, 4z,... is an integer multiple of PI, so it vanishes for z = n PI/2^k for any integers n and k. Also, the right side is infinite if any of the arguments 2^k z is of the form n PI + PI/2, which implies it is infinite for z = (n PI + PI/2)/2^k for any integers n and k. Thus every z is arbitrarily close to a value for which the function is zero, and also to a value for which the function is infinite. To see how this evolves into the circular locus, consider the partial product N 1/2^(k+1) f(z) = PROD tan(2^k z) k=0 If we plot the real and imaginary parts of f(z) as z ranges from 0 to PI, with N=0, we see that f(z) = tan(z)^(1/2), so the result is purely real, approaching positive infinity as z approaches PI/2. Then when z exceeds PI/2 the value of tan(z) branches to negative infinity, so the function f(z) becomes purely imaginary, coming in from inf*i to 0. This is illustrated in the left-hand figure below. In the right-hand figure below we have set N=1, so we get more branches of the tangent function, and each branch gives a straight line locus in the complex plane, with the density of points determined by the slope of the tangent and the effect of the respective root extractions. Increasing the number of factors to N=2, and then to N=3, increases the number of branches and focuses the density even more, as shown in the figures below. It should be noted that the lines for the branches has non-zero density at all distances from the origin, but since the plot only has finite resolution, we see only the points at discrete intervals, which truncates below some threshold density. With N=4 and N=5 the points are clearly arrayed in a "fan" that begins to outline the circular locus. Going on for just a couple of more steps, to N=6 and N=7, we see that nearly all the density in on the simple circular locus described previously. Nevertheless, the range of f(z) for any integer N includes the entire upper half of the complex plane, even though the density away from the circular locus can be made arbitrarly small by increasing N. Thus we argue that the LIMIT of f(z) as N goes to infinity IS the circular locus, but this must be understood in the sense of densities.

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