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In complex field, assume $f(x)=\sum_{i=1}^{\infty} a_i x^i$ where $a_i \in {\bf N}$ or $a_i = 0$, and $f(x)$ converges in an area.

Question 1: are there $$f(x)=p(x)+\sum_{i=1}^{\infty}r_i(x), $$ or $$f(x)=p(x)+\sum_{i=1}^{n}r_i(x),$$ where $p(x)$ is a polynomial with all coefficients which are natural numbers, and $r_i(x)$ is a quotient of polynomials with at least one pole(which means denominator is polynomial with at least one zero point),$r_i(x)$ can be expanded as $$r_i(x) =\sum_{j=1}^{\infty} b_{ij} x^j, \text{ }b_{ij} \in N\text{ or } b_{ij} = 0 ?$$

Question 2: If they do exist, how to compute them?

Question 3: Under what condition $$f(x)=p(x)+\sum_{i=1}^{n}r_i(x),n <\infty?$$

Question 4: Are there finite number of $$T_i(x) =\sum_{j=1}^{\infty} b_{ij} x^j, \text{ }b_{ij} \in N\text{ or } b_{ij} = 0 $$,where $$r_i(x) =\sum_{j=1}^{\infty} b_{ij} x^j, \text{ }b_{ij} \in N\text{ or } b_{ij} = 0 $$, are all "finitely generated" by $T_i(x)$ by multiply and addition?

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Seems to me part of the question is trivial --- if $r_i(x)=a_ix^i$ then $f(x)=\sum^{\infty}r_i(x)$. –  Gerry Myerson Jul 20 '13 at 23:56
    
@Gerry,maybe,but $r_i(x)$is rational quotient of polynomials,thank you for your comment, –  XL _at_China Jul 21 '13 at 0:21
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@GerryMyerson You are a better man than I, since I can't even parse the question. –  Igor Rivin Jul 21 '13 at 1:15
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@XL_at_China, the function $a_ix^i$ is a quotient of polynomials. It is the quotient of $a_ix^i$ and $1$, for example. –  Mariano Suárez-Alvarez Jul 21 '13 at 1:55
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@GerryMyerson,sorry, the original sentence of "$r_i(x) is a quotient of polynomials with at least one pole(which means denominator is polynomial with at least one zero point)," is " ri(x) is a quotient of polynomials ",it is my fault to ignore it,your comment does not make nonsense –  XL _at_China Jul 21 '13 at 2:22
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1 Answer

up vote 2 down vote accepted

Consider the sequence $1,0,0,1,0,0,0,0,1,\dots$ given by $a_n=1$ if $n$ is a square, otherwise $a_n=0$. I claim that no nonzero sequence $b_n$ of non-negative integers with $b_n\le a_n$ for all $n$ can satisfy a constant coefficient homogeneous linear recurrence relation. For any sequence satisfying such a relation of order $d$ with $d$ consecutive terms equal to zero must be the zero sequence, and any sequence of non-negative integers dominated by $a_n$ must have arbitrarily long runs of zeros.

Now if $r(x)$ is a rational function, then the coefficients of its power series expansion must satisfy a constant coefficient linear recurrence. So if $f(x)=x+x^4+x^9+\cdots$, it can't be written as a sum, finite or infinite, of rational functions with poles and with power series with non-negative integer coefficients.

EDIT. More generally, a sum of a finite number of rational functions is a rational function, and a power series represents a rational function if and only if its coefficients satisfy a constant coefficient homogenous linear recurrence, so the question of representing $f$ as a finite sum boils down to the question of determining whether the coefficients of $f$ satisfy such a relation. How to do this will depend on how $f$ is known.

I don't know what to say about infinite sums, other than to give an example. If we write $d(n)$ for the number of divisors of $n$, then $$f(x)=\sum d(n)x^n={x\over1-x}+{x^2\over1-x^2}+{x^3\over1-x^3}+\cdots$$ while $f$ is certainly not a rational function.

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Why does the argument apply to infinite sums of rational functions? The degrees of the denominators might increase, no? –  Mariano Suárez-Alvarez Jul 21 '13 at 3:00
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@MarianoSuárez-Alvarez It seems to me that, even in the case of an infinite sum of rational functions of increasing degrees, each individual one of the summands would, as Gerry says, be coefficient-wise majorized by the original sequence (of coefficients of $f$) and would therefore have arbitrarily long runs of zero coefficients. So, because the coefficients eventually satisfy a linear recurrence, such a rational function would be a polynomial, contrary to the stipulation in the question. –  Andreas Blass Jul 21 '13 at 3:48
    
Good,another related question is when can such a function $f(x)$ be expanded as $$f(x)=p(x)+\sum_{i=1}^{\infty}r_i(x), $$ or $$f(x)=p(x)+\sum_{i=1}^{n}r_i(x),$$ with $p(x)$,$r_i(x)$ like the above. –  XL _at_China Jul 21 '13 at 4:54
    
@AndreasBlass, ah! that works :-) –  Mariano Suárez-Alvarez Jul 21 '13 at 5:11
    
Yes,$$f(x)=\sum d(n)x^n={x\over1-x}+{x^2\over1-x^2}+{x^3\over1-x^3}+\cdots$$ is transcendental function,and I remember that it is a formula of sieving in number theory.I am wondering when transcendental or algebraic non-rational function can be expanded as infinite sum of $r_i(x)$ defined in the post –  XL _at_China Jul 22 '13 at 2:17
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