# 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 $\sum_nb_n=1$ Commented Sep 23, 2013 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. Commented Sep 23, 2013 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. Commented Sep 23, 2013 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? Commented Jan 27, 2014 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. Commented Jan 27, 2014 at 9:31

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.

Proof

Let's start with the logarithm of the product, $$-\sum_{k=1}^\infty\frac{\mu(k)}{k}\log\left(1-z^k\right)=\sum_{k=1}^\infty\frac{\mu(k)}{k}\sum_{\ell=1}^\infty\frac{z^{k\ell}}{\ell}.$$ We can rewrite this $$\sum_{n=1}^\infty\frac{z^n}{n}\sum_{d\mid n}\mu(d)=z+\sum_{n=2}^\infty\frac{z^n}{n}\sum_{d\mid n}\mu(d)=z$$ since $$\sum_{d\mid n}\mu(d)=0$$ for $$n\geq 2$$.

• Why is the right hand side not zero at $z=1$? Commented Feb 6, 2017 at 19:53
• $z=1$ is not permitted, i.e. $|z|<1$. Commented Feb 6, 2017 at 21:00
• Could you please provide a reference to the proof? Commented Feb 24, 2021 at 13:03
• @JKDASF I don't have a reference but I will try to prove it. I've already made a start but don't have much time right now so will return to it later. Commented Feb 24, 2021 at 15:51
• @Pixel Thanks! Actually I've found a reference for that. See page 410 here: Bellman, Richard, and R. C. Buck. “4072.” The American Mathematical Monthly, vol. 51, no. 7, 1944, pp. 409–410. Your starting proof looks promising and we can just proceed by considering series expansion of log(1-x) and the compare the coefficients. Commented Feb 25, 2021 at 8:24

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}] Commented Jan 27, 2014 at 8:31
• Question math.stackexchange.com/q/4279145/442 asks where this product expansion first appeared. Commented Oct 17, 2021 at 14:16
• Commented Oct 17, 2021 at 15:17
• Note: @user91500 is the one who posted this here. That HTML link seems faulty, but the postscript version comes out fine for me. numbers.computation.free.fr/Constants/E/e.ps This product is at the end of § 4. Commented Oct 17, 2021 at 18:22

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 $r>1$? Commented Sep 23, 2013 at 7:16

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.

This function is related to the one Pixel 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)

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

• Beautiful! How does the ⁻ transform ℕ and how does the /ℕ⁻ transform ℂ? Commented Jul 29, 2021 at 22:05
• If you start at n=0 then that will produce your (1+x) implicitly. Commented Jul 31, 2021 at 3:27
• i finally found your joke: Ex nihilo quod libet! For all elements x of an empty set (like all the zero complex negative nonnegatives) it is true that all their infinitely many factors cancel out and exponentiate to 1! Commented Aug 2, 2021 at 20:52
• Haha! Your product is one. None negative natural exists to equal ex nihilo quod libet. Commented Aug 2, 2021 at 21:33

From Euler's $$\sin[2x]=\Im[e^{xi+xi}]=\Im[[\cos x+i\sin x][\cos x+i\sin x]]=2\sin[x]\cos[x]$$ $$\frac{\sin[πx]}{πx} =\prod_{n=1}^\infty\left[1^2-\left[\frac{x}n\right]^2\right]$$ we get $$\cos\left[\frac\pi2x\right]=\frac{\sin[πx]}{2\sin[\frac\pi2x]} =\frac{πx\prod_{n=1}^\infty\left[1^2-[\frac{x}{n}]^2\right]}{2\frac\pi2x\prod_{n=1}^\infty\left[1^2-[\frac{x}{2n}]^2\right]} \\=\frac{ [\frac{1-x²}{1}][\frac{4-x²}{4}] [\frac{9-x²}{9}][\frac{16-x²}{16}]\cdots} {[\frac{4-x²}{4}][\frac{16-x²}{16}] [\frac{36-x²}{36}][\frac{64-x²}{64}]\cdots} =\prod_{n=1,3,5}^\infty\left[1^2-\left[\frac{x}{n}\right]^2 \right]$$

which together with$$\sin\left[\fracπ2x\right]=\cos\left[\fracπ2[x-1]\right]$$

yields $$i^x=\exp\left[i\fracπ2x\right] =\cos\left[\frac\pi2x\right]+i\sin\left[\frac\pi2x\right] \\=\left[\prod_{n=1,3,5}^\infty\left[1^2-\left[\frac{x}{n}\right]^2 \right]\right] +i\left[\prod_{n=1,3,5}^\infty\left[1^2-\left[\frac{x-1}{n}\right]^2 \right]\right]$$ consisting of two infinite products. Maybe can we factor this expression in some way?

• (i^)=transpose.map(scanl(*)1.forM[1,3..].(1-).(^2).(%)).zipWith(+)[0,-1].replicate 2 flowz outta dat! Commented Jul 30, 2021 at 7:50

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.

• That's a really cool paper. Commented Apr 13, 2020 at 22:23

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,

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

• Good to know that for $z\in(-1,1)$, we have $${\mathrm e}^z = \prod_{n=1}^\infty (1-z^n)^{-\mu(n)/n}$$ Series[Product[(1-z^n)^(-MoebiusMu[n]/n), {n,1,5}], {z,0,5}] Commented Dec 11, 2020 at 22:44
• I just added a proof in my old answer above. Commented Feb 25, 2021 at 21:39