Let $P_0(y)=2y-3$ and define recursively $$P_{n+1}(y)=4y\cdot P_n'(y)+(5-4y)\cdot P_n(y).$$ I would like to know as many properties of $P_n$ as I can. For example, it can be shown that each $P_n$ has only real simple positive zeros and that $P_n$ and $P_{n+1}$ strictly interlace for every $n$. It can also be shown that the recursive strict Turan Inequality is satisfied, for $y>0$, $$T_n(y):=P_{n+1}(y)^2-P_n(y)\cdot P_{n+2}(y)>0.$$ Empirical evidence (Mathematica) indicates that $T_n(y)$ is an increasing polynomial with only one real zero at $0$. My goal is to find a good estimate of $T_n(\pi)$ and show that $T_n(y)>T_n(\pi)$ for $y>\pi$. In an effort to establish that goal it would be nice to find a generating function for the $\{P_n\}$.


In an attempt to find such generating function I have made a terrible error, but I can't seem to find what my error is. Please show me the error of my ways.

Define the formal power series, $$f(y,t):= \sum_{n=0}^\infty \frac{P_n(y)}{n!}t^n.$$ we see then that $$f_t=\frac{d}{dt}f(y,t)=\sum_{n=1}^\infty \frac{P_n(y)}{(n-1)!}t^{n-1}=\sum_{n=0}^\infty \frac{P_{n+1}(y)}{n!}t^n.$$ Using the differential equation above we have, $$\sum_{n=0}^\infty \frac{P_{n+1}(y)}{n!}t^n=\sum_{n=0}^\infty \frac{4y\cdot P_n'(y)+(5-4y)\cdot P_n(y)}{n!}t^n.$$ So we arrive at $$f_t=4y\cdot f_y+(5-4y)\cdot f.$$ A quick check verifies that $f(y,t)=e^{5t+y}$. And so we have the generating function $$e^{5t+y}=\sum_{n=0}^\infty \frac{P_n(y)}{n!}t^n,$$ but this implies, differentiate with respect to $y$, that $$e^{5t+y}=\sum_{n=0}^\infty \frac{P'_n(y)}{n!}t^n,$$ and so $$P_n(y)=P'_n(y)$$ for every $n$ and $y$. How can that be? $\{P_n\}$ is a sequence of polynomials???

I was motivated to try generating functions by reading "Rainville - Special Functions - Page 188" and the derivations for Hermite polynomials.


If you expand $e^{5t+y}$ in function of $t$ you get : $$e^{5t+y}=\sum_{n=0}^\infty e^y\frac{(5\,t)^n}{n!}$$ And your $\ P_n(y)=5^n\,e^{y}\;$ verifies indeed $\;P_n(y)=P'_n(y)\;$ while $\;P_0(y)=2y-3\,$ is not verified !

Your error is in the resolution of the P.D.E. $$f_t-4y\cdot f_y=(5-4\,y)\cdot f$$ which admits more general solutions than $\;f(y,t)=e^{5t+y}$ : $$f(t,y)=\exp\left[\int\frac 51\,dt+\int\frac{-4y}{-4y}\;dy\right]\Phi\left(\int\frac {dt}{1}+\int\frac {dy}{4y}\right)$$ $$f(t,y)=e^{5t+y}\;\Phi\left(t+\frac {\ln y}4\right)\quad\text{($\Phi$ is an arbitrary function) }$$

that we will rewrite as (taking the exponential of the parameter of $\Phi$) : $$\boxed{\displaystyle f(t,y)=e^{5t+y}\;\Phi\left(y\,e^{4t}\right)}$$ Let's play a little with $\Phi$ setting $\,x:=y\,e^{4t}$ and expanding in Taylor series in $t$ :

  • for $\Phi(x)=1$ we get : $\ \displaystyle e^y+5e^{y}t+\cdots$
  • for $\Phi(x)=e^{-x}$ we get : $\ \displaystyle 1-(4y-5)t+\cdots\ $ (cancelling the $e^y$ term everywhere)
  • for $\Phi(x)=e^{-x}(2x-3)$ we get the pleasant : $$(2y - 3) - \left(8\,y^2 - 30\,y + 15\right)t + \left(16\,y^3 - 112\,y^2 + 165\,y - \frac{75}2\right)t^2+\cdots$$ so that the appropriate $f$ will be : $$\boxed{\displaystyle f(t,y)=e^{5t+y}\;e^{-y\,e^{4t}}\bigl(2\,y\,e^{4t}-3\bigr)}$$
  • $\begingroup$ Thanks @Mhenni (even if I should have taken the time to write down the characteristic system $\,\displaystyle \frac {dt}1=\frac{dy}{-4y}=\frac{df}{(5-4y)f}\ $ and solved it to get $\;y\,e^{4t}=C\;$ and $\;e^{5t+y}=D\;f\;$). $\endgroup$ – Raymond Manzoni Jun 19 '13 at 21:17
  • $\begingroup$ Glad it helped @BobbyOcean ! $\endgroup$ – Raymond Manzoni Jun 24 '13 at 7:42

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