Recurrence equation question

Question

My question (which has been edited) relates to the following recurrence relation:

$$a_{j+2}=\frac{2 a_{j}}{j}$$

The book which I am reading says that the (approximate) solution is given by:

$$a_{j}=\frac{C}{(j/2)!}$$

(I think there was an assumption of large $j$, too)

Could anyone give me a hand to understand how to arrive at this solution or give me guidelines on how to deal with recurrence relations and convert them to factorials? Thank you so much!

Answer 1

Iterate:

• For even terms, $a_{2n+2}=a_{2n}/n$ hence $a_{2n+2}n!=a_{2n}(n-1)!=\cdots=a_2$ for every $n$, that is, $$a_{2n}=a_2\frac{1}{(n-1)!}.$$
• For odd terms, $a_{2n+1}=2a_{2n-1}/(2n-1)=(4n)a_{2n-1}/(2n)(2n-1)$ hence $a_{2n+1}(2n)!/(4^nn!)=a_{2n-1}(2n-2)!/(4^{n-1}(n-1)!)=\cdots=a_1$ for every $n$, that is, $$a_{2n+1}=a_1\frac{4^{n}n!}{(2n)!}.$$

This yields the exact solution. One sees that $a_j\sim C/(j/2)!$ (whatever the meaning of $i!$ when $i$ is not an integer) is not quite right since $a_{2n}=Cn/n!\gg C/n!$, but with the proper shifts, the correct equivalent is similar to the one in the question.