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How to solve the recurrence relation

$d_n = a \sqrt{n+1}d_{n+1} + b \sqrt{n} d_{n-1}$ (when $n \in \mathbb { Z } ^ { + } )$, and $d_0 := a d_1$ to find the general form of $d_n$?

I was able to solve for the special case when either $a$ or $b$ are zero (dividing both sides by $\sqrt{n!}$), but not in the general case.

Update: It is interesting to note that the $d_n$'s are related to Hermite polynomials (see answers below). $d_n$'s are proportional to the coefficients of a squeezed state of Harmonic oscillator in the basis of Eigenfunctions of the Hamiltonian (That's how I obtained the recurrence relation). The eigenfunctions of Hamiltonian are also themselves related to the Hermite polynomials.

Archisman Panigrahi
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2 Answers2

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Let $d_n=c_n\sqrt{n!}$, then we have $c_n=a(n+1)c_{n+1}+bc_{n-1}$ (when $n>0$) and $c_0=ac_1$.

Then $f(x)=\sum\limits_{n=0}^{\infty}c_n x^n$ satisfies $f(x)=af'(x)+bxf(x)$, solved by $f(x)=c_0 e^{(x-bx^2/2)/a}$.

It remains to expand it into a power series (in $x$). With a known formula in mind, this is easy.

metamorphy
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  • I would advise to re-write generating function $f$ under the form $ce^{(-b/2)(x^2-1/b)^2}$ (by incorporating the residual term $e^{b^2/4}$ into multiplicative constant $c$. – Jean Marie Sep 04 '19 at 22:47
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Let us write the recurrence relationship under the form :

$$b \sqrt{n} d_{n-1} - d_n + a \sqrt{n+1}d_{n+1}=0\tag{1}$$

What will be presented here is not a solution by itself. It is an heuristic way (using linear algebra) to understand that Hermite polynomials (as established in the brilliant solution by @Metamorphy) are involved in the process.

I'll present it for the simple case $a=1$, $b=1$.

Consider the following symmetric tridiagonal $n \times n$ matrix : $$M_n:=\begin{pmatrix}0&\sqrt{\frac12}&&&\\\sqrt{\frac12}&0&\sqrt{\frac22}&&\\&\sqrt{\frac22}&\ddots&\ddots&\\&&\ddots&\ddots&\sqrt{\frac{n-1}2}\\&&&\sqrt{\frac{n-1}2}&0\end{pmatrix}$$

as given in the solution to Proving that all the real roots of Hermite polynomials are in $(-\sqrt{4n+1}, \sqrt{4n+1})$.

Its characteristic polynomial can be shown to be

$$\det(M_n-xI):=\dfrac{(-1)^n}{2^n}H_n(x)\tag{2]$$

where $H_n$ is the $n$-th Hermite polynomial (see proof below).

What is the connection with the question of the OP ?

The recurrence relationship (1) between $d_{k-1},d_{k},d_{k+1}$ for $k=1,\cdots n-2$ can be seen as expressing that (for any fixed $n$) :

Vector $(d_0,d_1,\cdots,d_{n-1})^T$ is an eigenvector of $M_n$ associated with eigenvalue $1/\sqrt{2}$...

One could object that it is not true for the last line. But in fact, one can consider that the recurrence relationship stops to be effective at rank $n$ without "polluting" previous values.

Proof of (2) : Expansion of the characteristic polynomial of $M_n$ along its last column gives :

$$\det(M_n-xI)=-x \det(M_{n-1}-xI) +(-\sqrt{\tfrac{n-1}{2}})^2\det(M_{n-2}-xI) \ \iff$$

$$\chi_n(x)=-x \chi_{n-1}(x)+\tfrac{n-1}{2}\chi_{n-2}(x)$$

$$ \ \text{with} \ \chi_1(x)=-x, \ \chi_2(x)=x^2-\tfrac12.$$

allowing the comparison with the recurrence relationship for Hermite polynomials :

$$H_{n+1} (x)=2xH_n (x)-2nH_{n-1} (x).$$

(end of proof).

Connected as well :

The roots of Hermite polynomials are all real?

https://math.stackexchange.com/q/1629234

Jean Marie
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