# Exponential Function as an Infinite Product

Is there any representation of the exponential function as an infinite product (where there is no maximal factor in the series of terms which essentially contributes)? I.e.

$$\mathrm e^x=\prod_{n=0}^\infty a_n,$$

and by the sentence in brackets I mean that the $a_n$'s are not just mostly equal to $1$ or pairwise canceling away. The product is infinite but its factors don't contain a subseqeunce of $1$, if that makes sense.

There is of course the limit definition as powers of $(1+x/n)$., but these are no definite $a_n$'s, which one could e.g. divide out.

• cheating: let $a_n=e^{b_nx}$ where $\sum_nb_n=1$ – user8268 Sep 23 '13 at 6:56
• @user8268: Okay, I like your idea. So for example $a_n=\mathrm e^{2^{-(n+1)}}$ In my mind, the $a_n$'s were of course simpler to compute than $\mathrm e^x$ itself. Like like rationals. – Nikolaj-K Sep 23 '13 at 6:58
• Unfortunately you can't get it really simpler. That is, if $a_n$'s are entire functions of $x$ then they must be non-0 everywhere (as their product is), so each of $a_n$ is exp(some entire function). Perhaps there is som econtrived formula with non-entire functions. – user8268 Sep 23 '13 at 7:02
• While in the lhs there is a (variable) $x$ I see only constants on the rhs. Where shall the variability be encoded in the rhs? – Gottfried Helms Jan 27 '14 at 9:21
• @GottfriedHelms: Is the letter $a$ a constant by default? I intended those to be functions of $x$ - of course, solutions $a_n(x)=c_n^x$ where $c_n$ is an complex number are a slight cop-out, but work too. I came to ask the question because I generally have no idea how one does come up with product representations, and am baffled when I then see things like the Weierstrass factorization theorem. – Nikolaj-K Jan 27 '14 at 9:31

If $x\geqslant0$ (or $x\ne-2^n$ for every $n\geqslant0$), one can use $$a_0=1+x,\qquad a_{n+1}=\left(1+\frac{x^2}{2^{n+2}(x+2^n)}\right)^{2^n}$$ If $x\leqslant0$ (or $x\ne2^n$ for every $n\geqslant0$), one can use $$a_0=\frac1{1-x},\qquad a_{n+1}=\left(1-\frac{x^2}{(2^{n+1}-x)^2}\right)^{2^n}$$ Where does this come from? From the identity, valid for every $n\geqslant0$, $$\prod_{k=0}^na_k=\left(1\pm\frac{x}{2^n}\right)^{\pm2^n}.$$ The first identity (when $\pm=+$) yields a nondecreasing sequence of partial products. The second identity (when $\pm=-$) yields a nonincreasing sequence of partial products.

• +1, Spontanously I don't see how to check the convergence, but I see your approach via approaching $(1+x/n)$. You "skip" the entire function thing, but making the coefficients case dependend. I wonder, can one get rid of the $2$'s for an arbitrary rational $r>1$? – Nikolaj-K Sep 23 '13 at 7:16

Not sure if this satisfies your assumptions, but this is an interesting infinite product for $|z|<1$, $$e^z=\prod_{k=1}^\infty (1-z^k)^{-\frac{\mu(k)}{k}},$$ where $\mu(k)$ is the Möbius function. See here.

• Why is the right hand side not zero at $z=1$? – Nikolaj-K Feb 6 '17 at 19:53
• $z=1$ is not permitted, i.e. $|z|<1$. – Pixel Feb 6 '17 at 21:00

There exists an infinite product for $e$ as follows:

If we define a sequence $\lbrace e_n\rbrace$ by $e_1=1$ and $e_{n+1}=(n+1)(e_n+1)$ for $n=1,2,3,...;$ e.g. $$e_1=1,e_2=4,e_3=15,e_4=64,e_5=325,e_6=1956,...$$ then $$e=\prod_{n=1}^\infty\frac{e_n+1}{e_n}=\frac{2}{1}.\frac{5}{4}.\frac{16}{15}.\frac{65}{64}.\frac{326}{325}.\frac{1957}{1956}. ...$$ For proof, first by induction we can show that if $s_n=\sum_{k=0}^n\frac1 {k!}$, then $e_n=n!s_{n-1}$,for $n\in\mathbb N$. And this immediately follows that $s_n/s_{n-1}=(e_n+1)/e_n$ and $s_n=\prod_{k=1}^n\frac{e_k+1}{e_k}$. Hence, $$e^x=\prod_{n=1}^\infty\left(\frac{e_n+1}{e_n}\right)^x.$$

• Very cool thanks. I translated it to Mathematica: e[n_] = If[n == 1, 1, n (1 + e[n - 1])]; Table[Product[1 + 1/e[n], {n, 1, N}], {N, 1, 7}] – Nikolaj-K Jan 27 '14 at 8:31

Amazingly, the exponential function can be represented as an infinite product of a product! That result was shown in the 2006 paper "Double Integrals and Infinite Products For Some Classical Constants Via Analytic Continuations of Lerch's Transendent" by Jesus Guillera and Jonathan Sondow.

It is proven in Theorem 5.3 that $$e^x=\prod_{n=1}^\infty \left(\prod_{k=1}^n (kx+1)^{(-1)^{k+1} {{n}\choose{k}}}\right) ^{1/n}$$

I dunno, this was too cool not to show you.

• That the sign of the exponent jumps in $(-1)^{k+1}$ is very odd. – Nikolaj-K May 1 '16 at 13:07

From $$\cos x= \prod_{k=1}^\infty1-\frac{x^2}{(2k-1)^2 \pi^2}$$ and $$\sin x=x \prod_{k=1}^\infty1-\frac{x^2}{k^2 \pi^2}$$ we have $$e^{ix}=\cos x + i \sin x=\prod_{k=1}^\infty1-\frac{x^2}{(2k-1)^2 \pi^2}+ i x \prod_{k=1}^\infty1-\frac{x^2}{k^2 \pi^2}$$ even if not a single infinite product. Maybe can we factor this expression in some way?

This function is related to the one Antinous posted, but it is not the same: $$e=\prod\limits_{n=1}^\infty\left(1-\frac{1}{\tau^n}\right)^{\frac{\mu(n)-\phi(n)}{n}}$$ where $\tau$ denotes the golden ratio, $\mu(n)$ denotes the Möbius Function and $\phi(n)$ denotes Euler's Totient Function.
("A Golden Product Identity for e" by Robert P. Schneider)

• I'ma give you a +1, but this is so odd I don't even want to check ^^ – Nikolaj-K Mar 3 '17 at 23:59

OP here. I just realized that the following should hold in general:

$$\lim_{n\to\infty}a_n=a_1+\sum_{n=1}^\infty(a_{n+1}-{a_n}),$$

and for finite $a_n$, similarly

$$\lim_{n\to\infty}a_n=a_1\cdot\prod_{n=1}^\infty\frac{a_{n+1}}{a_n}.$$

Hence, with $a_1=\prod_{n=1}^\infty (a_1)^{2^{-n}}$ and for $x$ that aren't negative integers,

$$\mathrm {exp}(x)=\prod_{n=1}^\infty\ (1+x)^{2^{-n}}\left(1+\frac{x}{n+1}\right)^{n+1}\left(1+\frac{x}{n}\right)^{-n}.$$

For any $$x\in \mathbb{C}/{\mathbb{N}^{-}}$$, we have : $$e^{x}=(1+x)\prod_{n=1}^{\infty}\left(1+\frac{x}{n}\right)^{-n}\left(1+\frac{x}{n+1}\right)^{n+1}$$

We have$$\mathrm e^{-x}=\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad$$$$(1-x)(1+\tfrac12x^2)(1+\tfrac13x^3)(1+\tfrac38x^4)(1+\tfrac15x^5)(1+\tfrac{13}{72}x^6)(1+\tfrac17x^7)(1+\tfrac{27}{128}x^8)(1+\tfrac{8}{81}x^9)\cdots$$for $$|x|<1$$. Some coefficients for these product terms are listed in OEIS as A170910 and A170911. See the paper on power product expansions by Gingold et al. (1988). The result is due to O. Kolberg (1960), who showed that the coefficient of $$x^n$$ in the $$n$$th factor is $$1/n$$ if and only if $$n$$ is prime.