Ayuda con la generación de funciones.

Question

De fondo.

Deje $P_0(y)=2y-3$ y definir de forma recursiva $$P_{n+1}(y)=4y\cdot P_n'(y)+(5-4y)\cdot P_n(y).$$ 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 $\{P_n\}$.

Problema.

En un intento de encontrar la generación de la función que he cometido un terrible error, pero me parece que no puede encontrar lo que mi error. Por favor, muéstrame el error de mis caminos.

Definir el poder formal de la serie, $$f(y,t):= \sum_{n=0}^\infty \frac{P_n(y)}{n!}t^n.$$ 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? $\{P_n\}$ es una secuencia de polinomios???

Yo estaba motivado para intentar generar las funciones de la lectura "Rainville - Funciones Especiales - Página 188" y las derivaciones para los polinomios de Hermite.

Answer 1

Si expande $e^{5t+y}$ en función de $t$ consigue : $$e^{5t+y}=\sum_{n=0}^\infty e^y\frac{(5\,t)^n}{n!}$$ Y su $\ P_n(y)=5^n\,e^{y}\;$ verifica efectivamente $\;P_n(y)=P'_n(y)\;$ mientras $\;P_0(y)=2y-3\,$ no es verificada !

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

que vamos a reescribir como (tomando la exponencial de parámetro de $\Phi$) : $$\boxed{\displaystyle f(t,y)=e^{5t+y}\;\Phi\left(y\,e^{4t}\right)}$$ Vamos a jugar un poco con $\Phi$ $\,x:=y\,e^{4t}$ y la expansión en serie de Taylor en $t$ :

• para$\Phi(x)=1$: $\ \displaystyle e^y+5e^{y}t+\cdots$
• para$\Phi(x)=e^{-x}$: $\ \displaystyle 1-(4y-5)t+\cdots\ $ (cancelación de la $e^y$ plazo en todas partes)
• para $\Phi(x)=e^{-x}(2x-3)$ recibimos la agradable : $$(2y - 3) - \left(8\,y^2 - 30\,y + 15\right)t + \left(16\,y^3 - 112\,y^2 + 165\,y - \frac{75}2\right)t^2+\cdots$$ para que las $f$ será : $$\boxed{\displaystyle f(t,y)=e^{5t+y}\;e^{-y\,e^{4t}}\bigl(2\,y\,e^{4t}-3\bigr)}$$