I am trying to figure out how to solve the following equation:

$$a_k - k^2a_{k-1} = 0$$ when $a_0 = 1$ ?


closed as off-topic by zhoraster, user21820, Henrik, Vladhagen, Shailesh Jan 13 '17 at 0:16

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  • 2
    $\begingroup$ 1) Why the name "difference equation" 2) what do you call "solve", have an explicit formula for $a_n$ ? Have you tried to obtain the first terms. In french, we have an expression "Aide toi, le ciel t'aidera" (in substance, "begin by working, then Math SE will help you." $\endgroup$ – Jean Marie Jan 11 '17 at 13:45
  • $\begingroup$ What did you try ? $\endgroup$ – Yves Daoust Jan 11 '17 at 13:53

I guess you want to find an explicit presentation of $a_k$?

So obviously: $$a_k = k^2a_{k-1} = k^2(k-1)^2a_{k-2} = \ldots k^2(k-1)^2(k-2)^2\ldots \cdot 1 \cdot a_0$$

So we get: $$a_k = \prod_{j=1}^k j^2 = \left(\prod_{j=1}^k j\right)^2 = k!^2$$

  • 1
    $\begingroup$ Otherwise known as $k!^2$. $\endgroup$ – user26872 Jan 11 '17 at 13:47
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    $\begingroup$ yeah right, thx :-) $\endgroup$ – Gono Jan 11 '17 at 13:49

The solution by Gono is the one to prefer. Anyway, if you want to tame the equation by means of known methods, you can linearize it with logarithms.



$$\log a_k=\log a_{k-1}+2\log k,$$

and by setting $b_k:=\log a_k$, you get an ordinary linear form

$$b_k=b_{k-1}+2\log k.$$

By recurrence, the solution is

$$b_k=b_0+2\sum_{k=1}^k\log k=b_0+2\log k!$$ and $$a_k=a_0(k!)^2.$$


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