# Probability generating function.

I have this collection of generating functions in $x$ parametrized by $n$:

$$f_n(x) = \frac{1}{n!}(1-x)^{n+1}\sum_{k \geq 1} k^n x^k,$$

so that $n$ is a constant. For instance

$$f_6(x) = (1/720)x+(19/240)x^2+(151/360)x^3+(151/360)x^4+(19/240)x^5 + (1/720)x^6.$$

I am interested in $f'_n(x)|_{x =1}$. Why? Because these coefficients are probabilities and I want to determine the expectation.

I should get $1/2$ for my answer because of the symmetry of these polynomials. However, I am getting stuck because somewhere in my algebraic manipulations; I must be dividing by zero and really causing some trouble.

On my own, I would like to be able to calculate the variance (which means I'll need the second derivative); so a solution that relies on showing the symmetry of these expressions isn't something that I'm interested in.

Thank you for any help you may be able to offer.

Note: To anyone who may find this, I realized that this is a generating function for the Eulerian polynomials. It has even been shown that these coefficients (as n goes to infinity) follow a normal distribution.

• First show that $$(n+1)f_{n+1}(x)=x(1-x)f'_n(x)+(n+1)xf_n(x)$$ then consider expansions at $x=1$. First order expansios show that $m_n=f'_n(1)$ solves $$(n+1)m_{n+1}=n+1+nm_n$$ which, together with $m_1=1$, yields $$m_n=\frac{n+1}2$$ Likewise, expansions up to order $(x-1)^2$ yield a recursion on $$\sigma_n^2=f''_n(1)+f'_n(1)-f'_n(1)^2$$ – Did Feb 10 '17 at 7:32
• Thank you! This was just enough to get me going. Thank you for your help. – curiousaboutthings Feb 10 '17 at 15:57
• "This was just enough to get me going" Great. Post what you got as an answer below? – Did Feb 11 '17 at 8:37

$$f_n(x) = \frac{1}{n!}(1-x)^{n+1}\sum_{k=1}^\infty k^n x^k=\frac{(1-x)^{n+1} }{n!}\text{Li}_{-n}(x)$$ where appears the polylogarithm function.$$f'_n(x) =\frac{(1-x)^{n+1} }{x n!}\text{Li}_{-n-1}(x)-\frac{(n+1) (1-x)^n }{n!}\text{Li}_{-n}(x)$$ I must confess that I have not been able to compute the limits of $f'_n(x)$ when $x\to 1$.
I just used brute force with the very first terms and a very clear pattern appears $$f'_n(1)=\frac{n+1}2$$