[Math] known about the polynomial factorization of power series

factorizationpower series

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.

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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:

If
$$\rho:=\limsup\frac{n\log n}{-\log|a_n|}\; \; \; < 1,$$ 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.

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[Math] construct a power series with infinitely many zeros in the complex plane, bounded coefficients??

It's been noted already in the comments that the problem is still too easy as stated, because one can easily find functions $f(z) = \sum_{n=0}^\infty c_n z^n$ such as $f(z) = \sin(1 / (r-z))$, with infinitely many zeros inside the circle of convergence $|z|<r$, and as long as $r>1$ the coefficients $c_n$ decay exponentially, and thus satisfy constraints such as $\sum_n n|c_n| < \infty$ or $\left|c_n-c_{n-1}\right| \ll n^{-3}$ with plenty of room to spare.

[It's apparently not been noted that the "nonpolynomial" condition is automatic unless $f$ is identically zero, since a nonzero polynomial has finitely many roots anywhere in ${\bf C}$.]

However, the problem becomes somewhat more interesting if we require (as Johan Andersson suggested, and as the proposer may have intended) that $f$ have infinitely many roots in the unit disc $|z|<1$, which is the largest disc on which the power series is guaranteed to converge by the constraints on $c_n$.

Still, even with that interpretation there are power series whose coefficients that satisfy either condition, or even $|c_n - c_{n-1}| \ll n^{-k}$ for any constant $k$, and nevertheless have infinitely many real zeros in $\left|z\right|<1$. I'll deal with $\left|c_n - c_{n-1}\right| \ll n^{-3}$, though the same technique applies for all $k$. Since $c_n - c_{n-1}$ is the $z^n$ coefficient of $(1-z) \phantom. f(z)$, it is enough to construct $g(z) = \sum_{n=0}^\infty a_n z^n$ with $|a_n| \ll n^{-3}$ and infinitely many real zeros, and then set $f(z) = g(z) / (1-z)$.

Define $g(z) = S(z) - P(z)$, where $$ S(z) = z - \frac{z^2}{2^3} + \frac{z^4}{4^3} - \frac{z^8}{8^3} + \frac{z^{16}}{16^3} - + \cdots = \sum_{m=0}^\infty (-1)^m \frac{z^{2^m}}{2^{3m}}, $$ and $P(z)$ is the cubic polynomial $$ P(z) = \frac89 - \frac45(1-z) - \frac1{15} (1-z)^2 - \frac1{60}(1-z)^3. $$ Clearly $S$ satisfies the functional equation $S(z) = z - \frac18 S(z^2)$. The cubic $P$ was chosen to satisfy the same equation within $O((1-z)^4)$: we have $$ P(z) = z - \frac18 P(z^2) + (1-z)^4 \frac{(1+z)(3+z)}{480}. $$ Hence $g(z) = -\frac18 g(z^2) + O((1-z)^4)$. It follows that as $z \rightarrow 1$ the ratio $f(z) / (\log(1/z))^3$ approaches some function $\phi(z)$ that satisfies $\phi(z^2) = -\phi(z)$ exactly. Unless this function is identically zero, it must have at least one zero in each interval $(z,z^{1/2}]$, and thus an infinite sequence of zeros approaching $1$. Now there are several ways to show that $\phi$ is not identically zero, but possibly the simplest is numerical computation, which shows $\phi$ oscillating with amplitude about $2.2 \cdot 10^{-5}$. The first zero of $f(z)$ appears near $z=0.998131$, then $0.998891$, $0.999493$, $0.999736$, $0.999871$, $0.999935$, $0.999967$, etc. with each zero about twice as close to $1$ as the previous one.

Here's some gp code that gives a rough plot showing $f$ first crossing zero after not quite reaching it around $0.9945$ (so it's 99.44% certain to be negative?...), and then locates the first seven zeros exhibited above.

f(z) = suminf(n=0,(-1)^n*z^2^n/8^n)
P(z) = 8/9 - (1-z)*4/5 - (1-z)^2/15 - (1-z)^3/60
R(z) = (f(z)-P(z)) / log(1/z)^3

plot(x=.99,.9993,R(x))    
for(i=11,17,print(solve(x=1-3/(2^(i-1)),1-3/2^i,R(x))))

[Math] Rearrangements of a power series at the boundary of convergence

This question is not very well defined, so

there is a general science on rearragment of series in Banach spaces (see e.g. here). Now there are many ways to understand your question. For example, you can think about the function on the boundary as an element of $L_1(\mathbb{T})$ . Then because the functionals of scalar products with $z^n$ form a total system, the set of possible resulting functions is single point. (Note, that single point in $L_1$ is not a single function in ``pointwise'' sense)

However I don't know how to approach the problem when there is no natural Banach space of functions in which you want your resulting sum to lie in.

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[Math] construct a power series with infinitely many zeros in the complex plane, bounded coefficients??

[Math] Rearrangements of a power series at the boundary of convergence

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