Infinite products of Bernoulli and Euler
In 1720, Christian Goldbach's experiments with infinite series was published in the German journal Acta Eruditorum, he titled his paper Specimen methodi ad summas serierum and proceeded to furnish a number of worked examplesThe following examples are given in Goldbach's 1720 paper: \begin{align}\tfrac{1}{2}+\tfrac{1}{6}+\tfrac{1}{12}+\tfrac{1}{20}+\ldots\\ \tfrac{2}{64}+\tfrac{11}{544}+\tfrac{26}{1972}+\tfrac{47}{5104}+\ldots \\ \tfrac{1}{2}+\tfrac{1}{4}+\tfrac{1}{8}+\tfrac{1}{16}+\ldots \\ \tfrac{1}{3}+\tfrac{1}{81}+\tfrac{1}{19683}+\tfrac{1}{43046721} + \ldots\end{align}whose general terms are \(\frac{1}{x^2+x}\), \(\frac{3x^2-1}{9x^4+36x^3+33x^2-6x-8}\), \(\frac{1}{2x}\), and \(\frac{1}{2x^x-x}\). of his finite difference technique for relocating the general term of a given sequence. The general principle is that in the general case, the \(n\)-th term, \(a_n\), of any number sequence $$a_1, a_2, a_3, \ldots$$ can be expressed by using an infinite series involving differences of known terms: $$a_n = a_1 + \frac{n-1}{1!}\Delta a_1 + \frac{(n-1)(n-2)}{2!}\Delta^2 a_1 + \ldots$$ where the delta's are iteratively defined by \(\Delta a_1 = a_{n+1} - a_n\) and $$\Delta^n a_1 = \Delta^{n-1}a_{n+1} - \Delta^{n-1}a_n=\sum_{k=0}^{n-1}\binom{n-1}{k}a_{n+1-k}$$ This recurrence is actually related to the backward finite difference scheme, since the modern rewrite of the equation involves binomial coefficients. Construction of the general term is curiously important because you can then use the formula to inject an arbitary members between terms in the original number sequence. For example, we can ask: what should be the value assigned to \(\color{brown}{a_{\frac{1}{2}}}\), \(\color{brown}{a_{1\frac{1}{2}}}\), \(\color{brown}{a_{2\frac{1}{2}}}\), etc? $$\phantom{.}^{\searrow} \color{brown}{a_{\frac{1}{2}}} , \quad a_1, \phantom{.}^{\searrow} \color{brown}{a_{1\frac{1}{2}}}, \quad a_2, \phantom{.}^{\searrow}\color{brown}{a_{2\frac{1}{2}}}, \quad a_3, \ldots$$ Many years later in 1728, Goldbach polished the same idea and wrote a paper titled De terminus generalibus serierum. For example, when \(a_n = \frac{1}{n!}\), the equation reduced to $$\tfrac{1}{n!} = 1 - \tfrac{1}{2!}\tfrac{n-1}{1!} + \tfrac{1}{3!}\tfrac{(n-1)(n-2)}{2!} + \tfrac{1}{4!}\tfrac{(n-1)(n-2)(n-3)}{3!} + \ldots$$ and when \(n = 1\frac{1}{2}\), $${1\tfrac{1}{2}!} = \frac{1}{1 - \tfrac{1}{2!}\tfrac{1}{2} - \tfrac{1}{3!}\tfrac{1}{2 \cdot 4} + \tfrac{1}{4!} \tfrac{1}{2\cdot 4 \cdot 6} + \ldots}$$ Daniel BernoulliGoldbach met Daniel's brother, Nikolaus, in Venice, and at Nikolaus's suggestion, Goldbach, in April 1723, initiated a series of mail exchanges with Daniel, which lasted until 1730. Daniel's father, Johann, was Euler's teacher at the University of Basel, Switzerland. apparently has a copy of this paper and he presented the Goldbach's idea in a seminar at St. Petersburg Academy, on 23 September 1729. We suppose the young Euler had attended the seminar and he realized that Goldbach's construct of computing the general term of a number sequence using another infinite sequence is not the most elegant solution. Euler must have mentioned this to Daniel, and so Daniel told him to talk to Goldbach directly. Subsequently in 13This is a date in the old Julian calendar, it is proleptically equivalent to 24 October 1729 (Monday). Later when Euler moved to Berlin, he began to mark his letter with Gregorian dates. However, the dates on Goldbach replies are consistently marked using the Gregorian system. October 1729, the 22-year-old Euler wrote his first mail to Goldbach. This historic mail travelled 700 kilometers southeast from St. Peterburg to Moscow and it carried the following equation: $$m! = \frac{1\cdot 2^m}{\color{brown}{1+m}}\frac{2^{1-m}\cdot3^m}{\color{brown}{2+m}}\frac{3^{1-m}\cdot4^m}{\color{brown}{3+m}}\frac{4^{1-m}\cdot5^m}{\color{brown}{4+m}}\cdots$$ This equation announced to Goldbach that Euler had found a solution to the interpolation problem, mentioned by Daniel Bernoulli in a seminar three weeks earlier. After giving some mundane numerical examples on how to use his infinite product formula, e.g. \(\frac{4}{3}\frac{9}{8}\frac{16}{15}\frac{25}{24}\frac{36}{35}\cdots = 2\) and \(\frac{8}{4}\frac{27}{20}\frac{64}{54}\frac{125}{112} \cdots = 6\), Euler went on to explain that by reframing the problem geometrically as quadrature, he can actually obtainActually what Euler wrote in the letter was: Terminem autem exponentis \(\frac{1}{2}\) aequalis inventus est huic \(\frac{1}{2}\sqrt{}(\sqrt{-1} . l\, -1)\), seu quod huic aequale est, lateri quadrati aequalis circulo, cujus diameter \(= 1\). The formula essentially means $$\tfrac{1}{2}\sqrt{\sqrt{-1} \ln (-1)}$$ in modern notation. Since \(\ln(-1) = i\pi + 2i\pi k, k \in \mathbb{Z}\), the expression \(\tfrac{1}{2}\sqrt{\sqrt{-1} \ln (-1)}\) can be reduced to \(\tfrac{1}{2}\sqrt{i(i\pi+2i\pi k)} = \frac{1}{2}\sqrt{\pi}\), when \(k = -1\).: $$\tfrac{1}{2}! = \tfrac{1}{2}\sqrt{\sqrt{-1} \ln (-1)}$$ and proceeded to say that the result is equivalent to \(\frac{1}{2}\sqrt{\pi} = 0.8862269\). And with this he showed that he can easily calculate \(1\frac{1}{2}!\) as $$1\tfrac{1}{2}! = 1\tfrac{1}{2} \times \tfrac{1}{2}! = \tfrac{3}{4}\sqrt{\pi} = 1.3293403\ldots$$
Interestingly, another letter was addressed to Goldbach, at about the same time, by Daniel Bernoulli himself to talk about a certain cycloid problem. However, in the postscript of the letter, he briefly mentioned the use of the following infinite product to compute \(x!\): $$x! = \left(A + \tfrac{1}{2}x\right)^{x-1} \frac{2}{1+x}\frac{3}{2+x}\frac{4}{3+x}\cdots\frac{A}{A-1+x}$$ It is not known which letter reached Goldbach first. It could be well that both of them reach Moscow on the same day if they are transported on the same mail carrier, even though the dates mentioned in the letter are gapped by one week.
You can roughly convince yourself that Euler's expression for \(m!\) is true by the following mental process: Bring all of the factors in the denominator of the right hand side of the equation and combine it with \(m!\) to form $$m!\color{brown}{(m+1)(m+2)(m+3)(m+4)\cdots}$$ which is really \(1\cdot 2 \cdot 3 \cdot 4 \cdots\) since the dot-dot-dot continues indefinitely. On the other hand, the numerator can be rewritten as: $$\require{cancel}1 \cdot \bcancel{\color{brown}{2^m}} \left(\frac{2}{\bcancel{\color{brown}{2^m}}}\cdot \bcancel{\color{brown}{3^m}}\right)\left(\frac{3}{\bcancel{\color{brown}{3^m}}}\cdot \bcancel{\color{brown}{4^m}}\right)\left(\frac{4}{\bcancel{\color{brown}{4^m}}}\cdot \bcancel{\color{brown}{5^m}}\right)\cdots$$ which is apparently equivalent to \(1\cdot 2 \cdot 3 \cdot 4 \cdots\) since the dots extend ad infinitum.
Let us now turn our attention to Bernoulli's infinite product. In modern notation, his formula can be compressed to \(b(\infty, x, 2)\), where $$b(A, x, u) = \left(A+\tfrac{1}{u}x\right)^{x-1}\prod_{k=1}^{A-1}\frac{k+1}{k+x} $$ Using this, Bernoulli cited two numerical examples, one of which is \(3! \approx b(16, 3, 2) = 6\frac{1}{204}\). Bernoulli's expression is correct in general since it can be transformedFirst we move all the factors in the denominator to the left and multiply them with \(x!\) to form: $$x!(x+1)(x+2)\cdots(x+A-1) = (x+A-1)!$$Then we bring \(A! = 2\cdot 3\cdots A\) on the numerator to the left hand side of the equation, so that we have $${\rm LHS} = \frac{(x+A-1)!}{A!} = \frac{\bcancel{\color{white}{A!}}(A+1)(A+2)\cdots(A+x-1)}{\bcancel{\color{white}{A!}}}$$ Then we bring \((A+\frac{1}{u}x)^{x-1}\) to the denominator of LHS:$$\frac{A+1}{A+\frac{1}{u}x}\frac{A+2}{A+\frac{1}{u}x}\cdots \frac{A+x-1}{A+\frac{1}{u}x}$$ to $$\lim_{A \to \infty} \prod_{k=1}^{x-1} \frac{1+\frac{k}{A}}{1+\frac{x/u}{A}} = 1$$ Note that this relation is true for all values of \(u\) so the choice of \(u = 2\) in Bernoulli's infinite product is curiously interesting, for if we define $$g(A, x, u) = \left| x! - b(A, x, u)\right|$$ It is immediately clear that \(u =2\) is the value associated with the best asymptotic behavior.
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