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If $$p_n(x)=\prod_{i=1}^{n}(x+i)$$ then what is the order of the power of $x$ (with regards 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 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.

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  • $\begingroup$ Not sure you are aware, but formally you can write $$a_k(n)=\frac{1}{k!} \, \frac{{\rm d}^k}{{\rm d}x^k} \frac{\Gamma(x+n+1)}{\Gamma(x+1)} \Bigg|_{x=0}$$ where $a_k(n)$ is defined by $p_n(x)=\sum_{k=0}^n a_k(n) x^k$. $\endgroup$
    – Diger
    Nov 20 '19 at 18:08
  • $\begingroup$ Nope, definitely was not aware of that. Now I need to look into this and justify it. $\endgroup$ Nov 20 '19 at 19:54
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Let $$p_n(x) = x^n + a_1x^{n-1}+a_2x^{n-2}\dots a_{n-1}+a_n$$

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

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

To be continued...

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  • $\begingroup$ I see what you’ve done so far, but I don’t understand how what’s been written generalizes. Could you please continue your solution? $\endgroup$ Dec 26 '19 at 18:23
  • $\begingroup$ I will do it if i get time. I was trying to go the inductive route. $\endgroup$ Dec 27 '19 at 4:14
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$\displaystyle \prod\limits_{k=1}^n (x+k) = \sum\limits_{v=0}^n a_{n,v}x^v \approx\frac{n!n^x}{\Gamma(1+x)}=\frac{1}{\Gamma(1+x)}\sum\limits_{v=0}^\infty\frac{n!(\ln n)^v}{v!}x^v$

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 $~v\approx \ln n~$ .

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    $\begingroup$ Why do you identify $a_{n,v}\approx \frac{n!(\ln n)^v}{v!}$? What about the denominator $\Gamma(x+1)$? $\endgroup$
    – Diger
    Nov 20 '19 at 22:38
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    $\begingroup$ @Diger : It's an approximation for $n\to\infty$, where $x$ doesn't play a role $(a_{n,v}$ is independend of $x$). And of course: It's not exact. It must be $v\ll n$, that's why we can write $a_{n,v}\approx\frac{n!(\ln n)^v}{v!}$ by comparing the coefficients of $x^v$ . $\endgroup$
    – user90369
    Nov 21 '19 at 6:06
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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 $x+(k+1)/2$.

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