# How to solve the following difference equation? [closed]

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, ShaileshJan 13 '17 at 0:16

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – zhoraster, user21820, Henrik, Vladhagen, Shailesh
If this question can be reworded to fit the rules in the help center, please edit the question.

• 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." – Jean Marie Jan 11 '17 at 13:45
• What did you try ? – 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: $$a_k = \prod_{j=1}^k j^2 = \left(\prod_{j=1}^k j\right)^2 = k!^2$$

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

$$a_k=k^2a_{k-1},$$

becomes

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

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 $$a_k=a_0(k!)^2.$$