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Is there any representation of the exponentil function as 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.

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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. –  NikolajK 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 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. –  NikolajK Jan 27 at 9:31

3 Answers 3

up vote 4 down vote accepted

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.

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+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 $r>1$? –  NikolajK Sep 23 '13 at 7:16

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, $$e^x=\prod_{n=1}^\infty\left(\frac{e_n+1}{e_n}\right)^x.$$

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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}] –  NikolajK Jan 27 at 8:31

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?

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