How to solve recurrence relation

Question

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.

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 Costa, Nov 19 '12 at 18:34

score 1 · Answer 1 · answered Nov 19 '12 at 18:33

1

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

answered Nov 19 '12 at 18:33

André Nicolas

507,029

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 09:53

score 0 · Answer 2 · answered Nov 19 '12 at 18:36

0

Let $n=ku$. Now the equation becomes: $$f_{k(u+1)}=\frac {-f_{ku}}{(ku+a+k)(ku+b+k)}$$

answered Nov 19 '12 at 18:36

Amr

20,030

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

How to solve recurrence relation

2 Answers2