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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 $a_{2m-\color{red}{4}}$?

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  • $\begingroup$ 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
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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 $$ a_{2m-\color{red}{2}}=\color{blue}{\frac1{m-1}}\:a_{2m-\color{red}{4}}. $$

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    $\begingroup$ 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
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    $\begingroup$ @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
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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 $a_{2m} = \frac{a_{0}}{m!}$.

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    $\begingroup$ 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

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