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EDIT: This is a case of being too wrapped up in a formulation ($e_j,p_i,$ and the like) to try something simple. It did not occur to me to pull exp to the outside in the weeks I have stared at this. Thanks to Aaron Meyerowitz and Brendan McKay for a humbling revelation. I leave it to others to decide if this is an object lesson in mathematical myopia or a question to be removed. END EDIT

If the following turns out to be a homework question, great! Just tell me what course, textbook (or online accessible .pdf) and page number has the problem, and I'll go check it out.

Let $m$ and $k$ be integers with $k \gt e^m \geq 1$. I am considering two expressions in variables $x_i$, and looking at the partial sums with monomials of degree at most $m$, which I note by $\cong_m$. (So if any term like, say, $x_hx_i^{m-1}x_j$ occurs in my sum, I toss that term and consider what is left with $\cong_m$.) The question in brief is: $$\prod_{0 \leq i \leq k}(1 - x_i) \cong_m \prod_{1 \leq j}(\sum_{0 \leq l} [ - \sum_{0 \leq i \leq k}(x_i)^j/j ]^l/l! ) ?$$

Here is some background and motivation. I rewrite the above question in terms of elementary symmetric polynomials in the $x_i, 0\leq i \leq k$.
In order to have enough terms to make sense, assume $k$ is as large as needed. (Likely I just need $k > e^{e^m}$, but it would be nice to know if, say, just $k \gt m^2$ will do.) Let us write $q_j$ for $ \sum_{0 \leq i \leq k} (x_i)^j/j $. This is $p_j/j$, or the $jth$ power symmetric polynomial divided by $j$. Let us also write $XP_m(t)$ for some polynomial $t$ to be the formal expression (which we will chop off after the terms get too big) $\sum_{0\leq j \leq m} (-t)^j/j!$, after the series expansion for $e^{-t}$. For the left hand side, we rewrite it in terms of the elementary symmetric polynomials $e_j$, which is the sum over all $j$-sets (not $j$-tuples) $\{i_1, \ldots , i_j\}$ of $\{0,\ldots,k\}$ of the monomials $x_{i_1}\ldots x_{i_j}$. I also declare $e_0=1$. Now the left hand side is written as a sum, and as I am concerned only with that part containing monomials of total degree at most $m$, I can limit the summation indices and the question now becomes:

$$\sum_{ 0 \leq i \leq m} (-1)^ie_i \cong_m \prod_{1\leq j \leq m} XP_m(q_j) ?$$

If I had facility with any computer algebra system, I would have tried it there first before asking here. I have verified the equality for small $m$, and would like a reference to a proof, or disproof. Also, Newton's identities $je_j = \sum_{1\leq i \leq j}(-1)^{i-1}p_ie_{j-i}$ give me some hope that the above is true, but I have not found a derivation.

My motivation is in exploring the limits of an argument I learned in a paper of Harlan Stevens. He uses the left hand side with prime reciprocal values for the $x_i$, and looks for the smallest odd integer m such that the sum is positive, but all partial sums for smaller odd positive integers are negative, so in particular $q_1=p_1\gt 1$ for interesting examples.
If it is true, I hope to show that the smallest odd positive $m$ that gives a positive partial sum implies $p_1 \gt Cm$ for some real constant $C$, which would then tell me that this argument is limited to showing upper bounds no tighter than $Ak^{B\log\log k}$, and no smaller.

Gerhard "Also, I Find It Pretty" Paseman, 2014.01.19

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  • $\begingroup$ It occurs to me that this might be in a text like Enumerative Combinatorics or generatingfunctionality. If someone knows where this is treated, a text and section number would be appreciated. Gerhard "Lacks Good Web Search Terms" Paseman, 2014.01.19 $\endgroup$ – Gerhard Paseman Jan 19 '14 at 22:49
  • $\begingroup$ DO you know that $k=m$ is not enough? $\endgroup$ – Aaron Meyerowitz Jan 19 '14 at 23:16
  • $\begingroup$ Gerhard, unclear, do you have the two books you mention? I think I have both as pdfs. I found the second one already.. $\endgroup$ – Will Jagy Jan 20 '14 at 0:05
  • $\begingroup$ Your first display has "[" once and "]" twice. Please fix it. $\endgroup$ – Brendan McKay Jan 20 '14 at 1:04
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    $\begingroup$ I don't really get the question. The right side of your first display is just a rearrangement of $\exp(\sum_i \ln(1-x_i))$ after applying Taylor expansion to the exp and ln functions. When it is formally expanded out, all terms with total degree more than $k$ vanish, as they should by Taylor's theorem. $\endgroup$ – Brendan McKay Jan 20 '14 at 1:21
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Try this: $$\prod_{1 \leq j}^{\infty}(\sum_{0 \leq l}^{\infty} [ - \sum_{i=0}^k\frac{x_i^j}{j} ]^l/l! ])=$$ $$\prod_{1 \leq j}^{\infty}\mathop{exp}\left( - \sum_{i=0}^k\frac{x_i^j}{j}\right)=$$ $$\mathop{exp}\left(\sum_{j=1}^{\infty}\left( - \sum_{i=0}^k\frac{x_i^j}{j}\right)\right)=$$ $$\mathop{exp}\left(\sum_{i=0}^k\left( -\sum_{j=1}^{\infty} \frac{x_i^j}{j}\right)\right)=$$ $$\mathop{exp}\left(\sum_{i=0}^k\ln({1-x_i)}\right)=$$ $$=\prod_{i=0}^k(1-x_i )$$

So I think $m=k$ would be just enough.

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  • $\begingroup$ Darn, beat by 3 seconds! $\endgroup$ – Aaron Meyerowitz Jan 20 '14 at 1:21
  • $\begingroup$ Oh, it's only that Brendan types faster. Gerhard "You Still Get The Upvote" Paseman, 2014.01.20 $\endgroup$ – Gerhard Paseman Jan 20 '14 at 22:41

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