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How do I solve the recurrence relation in terms of $f_0$? $$f_{n+k} = -\frac{f_n}{(n+a+k)(n+b+k)}$$ where $a$ and $k$ are fixed. No idea what to do in this case due to the fact that the difference is bigger than 1 in the $f_i$. Thanks.

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Explain to us what you mean by solving the recurrence. You want to express $f_o$ on a system of equations (possibly nonlinear) in terms of infinite variables? –  Elias Nov 19 '12 at 18:34
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2 Answers 2

It is immediate: put $n=0$.${}{}{}{}{}$

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So there are $k$ undetermined values $f_0$ up to $f_k?$ –  soup Nov 19 '12 at 21:35
    
Put $n=0$. We get $f_k=\frac{f(0)}{(k+a)(k+b)}$. If we prefer the subscript $n$, we get $f_n=\frac{f(0)}{(n+a)(n+b)}$. This is our general formula for $f_n$. Or is $k$ also fixed? –  André Nicolas Nov 19 '12 at 21:40
    
Yeah my guess is $k$ is fixed... –  soup Nov 19 '12 at 21:58
    
Sorry, $k$ is fixed. –  AC21 Nov 20 '12 at 9:53
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Let $n=ku$. Now the equation becomes: $$f_{k(u+1)}=\frac {-f_{ku}}{(ku+a+k)(ku+b+k)}$$

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So $f_n = \frac{(-1)^nf_0}{(a+k)...(a+k+n-1)(b+k)...(b+k+n-1)}$? –  soup Nov 19 '12 at 21:36
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