The Recursion Relation for larger n | Quantum Science Philippines
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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).

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