MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Some power series factorize; $1+\sum_{n=1}^\infty x^n=\prod_{n=1}^\infty (1+x^{2^n})$ and $1+\sum_{n=1}^\infty x^{2n}/(2n+1)!=\prod_{x=1}^\infty (1+x^2/n^2\pi^2)$ for example; while others do not----in particular, $1+\sum_{n=1}^\infty x^n/n!$. Specifically the question is: what is known about necessary and sufficient conditions on the coefficients of the power series $1+\sum_{n=1}^\infty a_nx^n$ (assuming a positive radius of convergence) for it to factorize as $\prod_{n=1}^\infty (1+ b_nx+c_nx^2)$, where the coefficients $a_n$, $b_n$, and $c_n$ are real constants?

Remarks The Weierstrass factorization theorem doesn't directly answer the question, since here we allow Taylor series of non-entire functions (e.g. the first example above) and exclude non-polynomial factors. The roots of the question are algebraic: loosely stated, under what conditions can the fundamental theorem of algebra be pushed to infinity? A positive radius of convergence is supposed because formal power series with no functional meaning make me uncomfortable, and admitting them might complicate the answer. I previously posted an unanswered version of this question on Math.StackExchange.

Edit Thanks to juan and Douglas Zare for showing my error in the preamble of the question by citing the interesting factorization of the exponential series by Gingold et al. As Aaron Meyerowitz indicates, restriction to linear and quadratic factors leads respectively to quite different situations. Alexandre Eremenko answered the linear case. Allowing polynomial factors of unrestricted degree opens a wide vista. The rather simple power series $1$ includes among its polynomial product representations, valid for $|x| \lt r_0$, such forms as $$\left(1-\dfrac{p(x)}{r} \right) \prod_{n=0}^\infty \left[1+\left(\dfrac{p(x)}{r} \right)^{2^n} \right],$$ where $r_0$ is an arbitrary positive constant, $p(x)$ is an arbitrary polynomial, and $r$ is any real constant strictly exceeding $|p(x)|$ whenever $|x|\leqslant r_0$. Myriad expansions like these may be woven into any polynomial product expansion of any power series. I find this vista daunting and so will stick to the specified quadratic case, which remains unanswered so far.

share|cite|improve this question
Why are you so sure that the exponential has no factorization on polynomials convergent on a certain disk $|z|<r$ ? Specially as if in your first example you admit polynomial of arbitrary degree. – juan Aug 13 '12 at 10:54
Perhaps because the exponential never vanishes... – Feldmann Denis Aug 13 '12 at 12:26
See the OEIS A170910 and the references there. – juan Aug 13 '12 at 12:57
According to the OEIS entry juan mentioned, $\exp(-x) = $ $(1-x)(1+\frac12x^2)(1+\frac13x^3)(1+\frac38x^4)(1+\frac15x^5)(1+\frac{13}{72}x^‌​6)(1+\frac17x^7)(1+\frac{27}{128}x^8)(1+\frac{8}{81}x^9)...$ – Douglas Zare Aug 13 '12 at 14:31

I agree with John Bentin, that the question has little sense in the formal category. Let me restrict myself to the simpler question of factorization into linear factors (so everyting will be over complex numbers).

A formal infinite product $(1+a_1x)(1+a_2x)(1+a_3x)...$ makes no sense, because we have to sum infinite series to obtain the coefficient at $x$.

For this reason I disagree with solution of Will Sawin. The product $(1+a_1x)(1+a_2x^2)...$ does make sense, formally, but after you factor each multile into linear factors, it does not, even formally!

So let us restrict to convergent series and products. Then the next question, of course, is "where are they supposed to converge?"

If we want convergence in the whole plane, a satisfactory answer can be given. This is not exactly Weierstrass, but Weierstrass combined with Hadamard:

$$\rho:=\limsup\frac{n\log n}{-\log|a_n|}\; \; \; < 1,$$ then $$1+a_1x+a_2x^2+\ldots=(1+c_1x)(1+c_2x)\ldots,$$ and both sides are convergent in the ordinary sense in the whole plane. Notice that some convergence of $$\sum c_n$$ is necessary just to make sense of the RHS. If this series of $c_n$ is ABSOLUTELY convergent, then then the right hand side is convergent in the whole plane, so it defines the left hand side (the power series, and for this power series $\rho\leq 1$ must hold.

So we see that the condition $\rho<1$ is almost best possible. The case $\rho=1$ can be also completely studied, but this is somewhat more complicated, and for this I refer to B. Levin, Distribution of zeros of entire functions, AMS, 1964.

Let me repeat the main point. For the product $(1+c_1x)(1+c_2x)\ldots$ to make sense, even FORMALLY, the series of $c_n$ must converge. Now, of course one can discuss various meanings of this convergence. But if it converges in the most usual sense, that is absolutely, the product converges in the whole plane, and we are within the subject of entire functions. Exact condition on the coefficients of the power series can be written, and for this I refer to Levin's book. So a necessary and sufficient condition on the series which guarantees that it has a product expansion with linear factors can be given. Passing from complex to real product is routine, of course.

share|cite|improve this answer
Yes, but as I said before, the first example in the question refers to factorization in polynomials of arbitrary degree. This is other question, for which I suspect the only condition must be on the value at $z=0$ and the order of the possible zero there. In this sense the exponential admits convergent factorizations possibly not unique. But this requires some proofs. – juan Aug 13 '12 at 17:00
Formal factorization into a product of polynomials of high degree is very non-unique, as the examples above show. That's why it seems reasonable to me to consider only convergent products. For linear polynomial factors, this is almost the same (see my "answer"). – Alexandre Eremenko Aug 13 '12 at 19:29
@Alexandre: formal factorization into a product of polynomials is unique in the form specified in Will Sawin's answer. – Qiaochu Yuan Aug 13 '12 at 19:43

Write the polynomial in the form $(1+a_1x)(1+a_2x^2)\dots$ for the unique choice of $a_n$ such that after $n$ factors the first $n+1$ terms of the product are the first $n+1$ terms of the power series. Then factor $1+a_nx^n$ into real quadratics. I don't know what form of convergence you want for the product so I'm not sure if/when this qualifies.

share|cite|improve this answer
After you expand each factor into linear or quadratic factors the formal infinite product will make no sense. (Formal product of factors $(1+a_jx)$ makes no sense!) – Alexandre Eremenko Aug 13 '12 at 14:43
This product makes non-formal sense sometimes though. – Will Sawin Aug 13 '12 at 19:54

Here is an observation which shows the delicacy of factorization: Your second product $$\prod_{n=1}^\infty (1+x^2/n^2\pi^2)=\sum_{n=0}^\infty x^{2n}/(2n+1)!=\frac{\sinh{x}}{x}$$ is absolutely convergent but is only conditionally convergent if it is written as $$\prod_{n=1}^\infty (1+ix/n\pi)(1-ix/n\pi).$$ So an order dependent factorization does not sound like the fundamental theorem of algebra.

We can leave complex numbers out entirely by shifting to a famous product of Euler (which is not an Euler product.) $$\prod_{n=1}^\infty (1-x^2/n^2\pi^2)=\sin{x}/x=\sum_{n=0}^\infty (-x^2)^{n}/(2n+1)!$$ Of course this was not rigorously justified at the time and I think Euler was perhaps content to leave it as $$\prod_{n=1}^\infty (1+x/n\pi)(1-x/n\pi).$$ The heuristic justification (one treatment is page 3 of this article) was that the product gives a function $f(x)$ with $f(0)=1$ and a single root of $f(n\pi)=0$ for non-zero integers $n$. Hence it should be $\sin{x}/x$ for which we do know the power series.

In volume I of Polya's Mathematics and Plausible Reasoning it is related that the motivation was to find the exact (and unknown) value of $\sum 1/n^2$ which does converge to about $1.644934.$ Of course comparing the coefficient of $x^2$ gives $\sum 1/n^2=\pi^2/6.$ The conclusion is "(extremely) daring", says Polya, but it does agree with the known value and the method yields other similarly valid looking results, so it seems acceptable.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.