Limit of Product of Monomials, Specifically $\prod_{i=1}^{n}(x+i)$

Question

If $$p_n(x)=\prod_{i=1}^{n}(x+i)$$ then what is the order of the power of $x$ (with respect to $n$) with the greatest coefficient for the following polynomial? $$\lim_{n \to \infty} p_n(x)$$

I plugged in some small values of $n$ and saw that the desired power tends to hang within the smaller powers of $x$ (for example, $p_4(x)=x^4 + 10 x^3 + 35 x^2 + 50 x + 24$, so the power of $x$ with the greatest coefficient is $1$), but I don't know enough about infinite polynomials to make any sort of generalization.

Also, I know that the power goes to infinity as $n$ goes to infinity, what I want to know is the order of the power relative to $n$- is it $\sqrt{n}$? $\ln(n)$? $n^{\frac{1}{e}}$? That sort of thing.

Answer 1

$\displaystyle \prod\limits_{k=1}^n (x+k) = \sum\limits_{v=0}^n a_{n,v}x^v \approx\frac{n!n^x}{\Gamma(1+x)}=\frac{1}{\Gamma(1+x)}\sum\limits_{v=0}^\infty\frac{n!(\ln n)^v}{v!}x^v$

For $~v\ll n~$ we can write $\displaystyle~a_{n,v}\approx \frac{n!(\ln n)^v}{v!}~$ .

So, the question is about $\displaystyle~\max\frac{(\ln n)^v}{v!}~$ .

With the Stirling formula $\displaystyle~v!\approx \left(\frac{v}{e}\right)^v\sqrt{2\pi v}~$ we get $\displaystyle~a_{n,v}\approx \frac{n!}{\sqrt{2\pi v}}\left(\frac{e\ln n}{v}\right)^v~$ .

We have $\displaystyle~\max\left(\frac{e\ln n}{v}\right)^v \leq\left(e^{1/e}\right)^{e\ln n} = n =\left(\frac{e\ln n}{v}\right)^v|_{v=\ln n}~$ .

It follows $~v\approx \ln n~$ .

Answer 2

Let $$p_n(x) = x^n + a_1x^{n-1}+a_2x^{n-2}+\ldots +a_{n-1}x+a_n$$

Then, $$-a_1 = 1+2+3+\ldots +n =\displaystyle\sum_{i=1}^n i = {n(n+1)\over 2}$$ $$a_2 = 1\cdot 2 + 1\cdot 3 + 1\cdot 4 +\ldots +n\cdot n-1 = 1(2+3+ \ldots + n) + 2(3+ 4+ \ldots n) +\ldots+ n $$ $$ = \displaystyle\sum_{i=1}^n\left[ i\displaystyle\sum_{j=i+1}^n j\right] =\displaystyle\sum_{i=1}^n\left[ i\left(\displaystyle\sum_{j=1}^n j - \sum_{k=1}^i k\right)\right] =\displaystyle\sum_{i=1}^n\left[i\left({n(n+1)\over 2}-{i(i+1)\over 2}\right)\right]$$ $$ = {n(n+1)\over 2}\sum i -\frac 12\sum(i^3+i) = {n^2(n+1)^2\over 4}+\frac 12 \left({n^2(n+1)^2\over 4} + {n(n+1)\over 2}\right)$$

$$=\frac 38 n^2(n+1)^2 + \frac 12 n(n+1)$$

To be continued...

Answer 3

Since this is the product of n linear polynomial in x, the degree is n. The coefficients are, I believe, unsigned Stirling numbers of either the first or second kind. One of my answers gave an asymptotic expansion in powers of $x+(k+1)/2$.

Answer 4

In terms of the unsigned Stirling numbers of the first kind $$ p_n (x) = \sum\limits_{k = 0}^n {\left[ {\begin{array}{*{20}c} {n + 1} \\ {k + 1} \\ \end{array}} \right]x^k } . $$ By a result of Hammersley, the index $k(n)$ of the maximal coefficient (which is unique) is then given by $$ k(n) = \left\lfloor {\log (n + 2) + \gamma - 1 + \frac{{\zeta (2) - \zeta (3)}}{{\log (n + 2) + \gamma - 3/2}} + \frac{{\theta _n }}{{(\log (n + 2) + \gamma - 3/2)^2 }}} \right\rfloor - 1, $$ with a suitable number $\theta_n$ satisfying $-1.1 < \theta _n < 1.5$. Here $\zeta$ is the Riemann zeta function and $\gamma = 0.5772\ldots$ is the Euler–Mascheroni constant. For more information, see, e.g., this paper.