Recurrence relation -- simplifying to base case

Question

What would be a closed form relating this recurrence relation $W_n$ to $W_1$?

If $$W_n = \frac{z_n - 4W_{n-1}}{4n}$$

I keep nested fractions, and I'm not sure how to simplify without algebra being all over the page.

This is basically me forming a relationship with my current semester's WAM (average mark per subject) with the previous semester's, so $z_n$ here is just basically the accumulative marks of the current semester over 4 subjects, so it might have a sequence of $z_k$ in the final result.

I hypothesise that

$$W_1 = \frac{1}{4} \left(\sum_{k=2}^{n-1} \: (-1)^k (n-1)! \: z_k \right) + n!\:(-1)^{n+1} \: W_{n}$$ Can someone confirm this?

$z_n$ is just a constant, that doesn't rely on n, as in it's basically saying it's a constant for the 'nth' period. The intent was to find expression for $W_n$ in terms of $z_n, z_{n-1},.. etc$ as well as $W_1$. — Twenty-six colours, Apr 08 '17 at 07:31
Not sure what to input, but I input that relation and it doesn't simplify it in W|A. — Twenty-six colours, Apr 08 '17 at 07:36
@PaulWoch Forget WA to solve the present question (and keep note that the previous commenter's view of mathematics is limited to the blind use of such a CAS). — Did, Apr 08 '17 at 07:59

score 4 · Answer 1 · answered Apr 08 '17 at 07:53

4

By multiplying $$W_n = \frac{z_n - 4W_{n-1}}{4n}$$ by $(-1)^nn!$, one gets $$ (-1)^nn!\cdot W_n = (-1)^n\frac14(n-1)!\cdot z_n + (-1)^{n-1}(n-1)! \cdot W_{n-1} $$ or$$ (-1)^nn!\cdot W_n -(-1)^{n-1}(n-1)! \cdot W_{n-1} = (-1)^n\frac14(n-1)!\cdot z_n $$ then summing from $n=1$ to $n=N$, on the left hand side terms telescope, one has $$ (-1)^N N!\cdot W_N -W_{0}=\sum_{n=1}^N(-1)^n\frac14(n-1)!\cdot z_n $$ giving

$$ W_N=(-1)^N \frac{W_0}{N!}+\frac{(-1)^N}{4N!}\sum_{n=1}^N(-1)^n(n-1)!\cdot z_n. $$

answered Apr 08 '17 at 07:53

Olivier Oloa

120,989

+1. It's similar to 'integrating factor' in differential equations. Fine. – Felix Marin Apr 08 '17 at 15:10
@FelixMarin Yes it is, thank you. – Olivier Oloa Apr 08 '17 at 15:39

Recurrence relation -- simplifying to base case

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