4
$\begingroup$

If you are given a recurrence relation such that:

$$na_n=2a_{n-2}\implies a_n= \begin{cases} 0 & \text{odd} \,n \\ \frac{2}{n}a_{n-2} & \text{even} \,n \end{cases}$$

My textbook suggests that the series can be simplified by

Putting $n=2m$ (since only even terms appear in this series), we get $$a_{2m}=\frac{2}{2m}a_{2m-2}=\bbox[#AF0]{\frac{1}{m}a_{2m-\color{red}{2}}=\frac{1}{m}\color{red}{\frac{1}{(m-1)}}a_{2m-\color{red}{4}}}=\frac{1}{m!}a_0$$

I understand the final equality as $$\frac{1}{m}\frac{1}{(m-1)}\frac{1}{(m-2)}\frac{1}{(m-3)}\cdots=\frac{1}{m!}$$

But I do not understand the highlighted equality.

Why does the $\color{red}{\cfrac{1}{m-1}}$ appear when $a_{2m-\color{red}{2}}$ is reduced to $a_{2m-\color{red}{4}}$?

$\endgroup$
2
  • $\begingroup$ Use the recurrence for $a_{2m-2}$ and you'll get $$a_{2m-2} = \frac{a_{2m-4}}{m-1}$$ $\endgroup$ – Aritra Das Mar 13 '16 at 6:04
  • $\begingroup$ @Aritra Yes, but I need to know the steps that lead to that equation. Any thoughts? $\endgroup$ – BLAZE Mar 13 '16 at 6:08
4
$\begingroup$

From $$ a_n=\frac{2}{n}\:a_{n-2},\quad n\,\, \text{even} $$ you may just put $n=2m-2$ to obtain $$ a_{2m-\color{red}{2}}=\color{blue}{\frac{2}{2m-2}}\:a_{2m-\color{red}{2-2}} $$ or $$ a_{2m-\color{red}{2}}=\color{blue}{\frac1{m-1}}\:a_{2m-\color{red}{4}}. $$

$\endgroup$
6
  • 1
    $\begingroup$ Thank you for your answer, I understand everything you wrote apart from why "you may just put $n=2m-2$". Can you please elaborate on why $n=2m-2$ is justified as a substitution? For example; Why could I not put $n=2m-3$ or $n=2m-4$ etc? $\endgroup$ – BLAZE Mar 13 '16 at 18:40
  • 1
    $\begingroup$ @BLAZE $n=2m-2$ is justified since the relation is given for any even integer. "Why could I not put $n=2m-3$..." because $n=2m-3$ is not an even integer. We usually write an even integer as $2m$, but I said "you may just put $n=2m-2$ ... " because in this case, we want to obtain the link with $2m-4$... $\endgroup$ – Olivier Oloa Mar 13 '16 at 19:01
  • $\begingroup$ Perfect explanation, thank you very much Olivier for your excellent answer. $\endgroup$ – BLAZE Mar 13 '16 at 19:18
  • $\begingroup$ @BLAZE You are welcome. $\endgroup$ – Olivier Oloa Mar 13 '16 at 19:43
  • $\begingroup$ So I could also put $n=2m-4$ as well? Since that is even right? $\endgroup$ – BLAZE Mar 13 '16 at 19:44
2
$\begingroup$

What has been provided is $$a_{2m} = \frac{1}{m} \, a_{2m-2}$$ From this, then \begin{align} a_{2m} &= \frac{1}{m} \, a_{2(m-1)} \\ &= \frac{1}{m \, (m-1)} \, a_{2(m-2)} \\ &= \frac{1}{m \, (m-1) \, (m-2)} \, a_{2(m-3)} ... \end{align} The pattern is $$a_{2m} = \frac{(m-k)!}{m!} \, a_{2(m-k)},$$ where $0 \leq k \leq m$. When $k=m$ then $a_{2m} = \frac{a_{0}}{m!}$.

$\endgroup$
1
  • 1
    $\begingroup$ Thanks for showing me the pattern of the series; this helped me to understand it a bit better (+1). $\endgroup$ – BLAZE Mar 13 '16 at 19:21

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.