The Recursion Relation for larger n | Quantum Science Philippines

## The Recursion Relation for larger n

By Kim S. Ponce, MS Physics I, MSU-IIT

Problem

Given the recursion relation: $a_{n+2} = \frac{2n+1-\varepsilon}{(n+2)(n+1)}a_{n}$. (a) Find $\frac{a_{n+2}}{a_{n}}$  for large $n$. (b) Compare result from (a) to $\xi^{n}e^{\xi^{2}}$.

Solution

Starting with the recursion relation, $\begin{array}{lll} a_{n+2}& = &\frac{2n+1-\varepsilon}{(n+2)(n+1)}a_{n}\end{array}$ $\begin{array}{lll} \frac{a_{n+2}}{a_{n}}& = &\frac{2n+1-\varepsilon}{(n+2)(n+1)}\end{array}$

When $n\rightarrow\infty$, the expression above reduces to $\begin{array}{lll} \frac{a_{n+2}}{a_{n}}& = &\frac{2n}{n^{2}}\\ \\& = &\frac{2}{n}\end{array}$

Now, we compare this to $\xi^{n}e^{\xi^{2}}$ which can be rewritten in the form: $\begin{array}{lll} \xi^{m}e^{\xi^{2}}& = &\sum_{l=0}^{\infty} \frac{\xi^{2l}}{l!}\xi^{m} \end{array}$

since $e^{x} = \sum_{n=0}^{\infty} \frac{x^{n}}{n!}$  and thus, $\begin{array}{lll} \xi^{m}e^{\xi^{2}}& = &\sum_{l=0}^{\infty} \frac{\xi^{2l+m}}{l!} \end{array}$

Finding for $a_{n}$ and $a_{n+2}$ , $a_{n}$ is the coefficient of $\xi^{n} = \frac{1}{l!}$ with $n = 2l+m$. So, $l= \frac{n-m}{2}$. Thus, $a_{n+2} = \frac{1}{(\frac{n+2-m}{2})!}.$

Now, we get the same ratio from (a) that is, $\begin{array}{lll} \frac{a_{n+2}}{a_n}& = &\frac{\frac{1}{(\frac{n+2-m}{2})!}}{\frac{1}{(\frac{n-m}{2})!}}\\ \\& = &\frac{(\frac{n-m}{2})!}{(\frac{n+2-m}{2})!}\\ \\& = &\frac{(\frac{n-m}{2})!}{(\frac{n+2-m}{2})(\frac{n-m}{2})!}\\ \\& = &\frac{1}{\frac{n+2-m}{2}}\end{array}$

If we let $n\rightarrow\infty$, $\begin{array}{lll} \frac{a_{n+2}}{a_n}& = &\frac{2}{n}\end{array}$

which is equal to that of the result from (a).