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Jan Pax (Oct 26 2025 at 18:35):How do I solve with initial conditions ?
James Deikun (Oct 26 2025 at 18:52):My best guess is to rewrite it as a "differential equation" in the finite difference calculus ...
Chris Grossack (she/they) (Oct 26 2025 at 23:51):You can also use generating functions -- see the first chapter of this book. I can say more later if you like
Morgan Rogers (he/him) (Oct 27 2025 at 08:50):It looks a lot like Fibonacci. If you want to build up from things you may know, you could try adapting the construction of formulas for those.
Jan Pax (Oct 27 2025 at 09:21):@Chris Grossack (she/they) Dear Chris I'd be much obliged to you if you could fully solve it for me!
Chris Grossack (she/they) (Oct 27 2025 at 14:26):I've spent a lot of time answering questions on math.stackexchange, so it's hard for me to see a question like this without going back to the norms of that site. Especially since that's a better place for a question like this, since it's not really about category theory. So with that in mind, I'll say the same thing I would say there:
What have you tried? Do you have any ideas of your own? Once we have a better idea of exactly where you're struggling we can help you better ^_^
Chris Grossack (she/they) (Oct 27 2025 at 14:31):For example, have you looked at the book I linked, Generatingfunctionology? Your exercise requires a combination of ideas from Example 1.2 (on page 5) and Example 1.3 (on page 8). You'll want to use the method outlined in the section titled The Method on page 8. So by reading the first 10 pages of this book you'll see everything you need in order to solve your problem. But of course, if you have questions about what you've read, I would be happy to help clarify!
Jan Pax (Oct 27 2025 at 19:37):I have tried solving but without any success to apply it to ! I'd be happy seeing a full computation here.
Kevin Carlson (Oct 27 2025 at 21:09):This one is a lot harder than Fibonacci because of the non-constant coefficients: you can see that a quadratic in is certainly not going to suffice, as it would for any non-constant coefficient recurrence relation.
I would suggest you try doing any of the things Chris suggested. It's quite rude to repeatedly ask people to completely solve a problem for you; doing so is of no obvious benefit to the solver, and also likely to be harmful to the asker. If you insist on not trying to learn about the question you're asking, you're probably better off talking about it to robots than to humans.
dusko (Oct 28 2025 at 19:36):People learn a lot from these conversations even without joining. Someone the following solution by recursion. (I hope pasting their ASCII is OK.)
TASK:
compute
a(0) = u
a(1) = v
a(n) = (a(n-2)+a(n-1))/n
recursively.
RECURSION SCHEMA:
using g:B and h:NxB-->B
define f=REC(g,h):N-->B
by
f(0) = g
f(n+1) = h(n,f(n))
CLAIM:
For
B = QxQ
g = <u,v>:B
h(n,x,y) = <y,(x+y)/(n+1)>:NxB-->B
Then f=Rec(g,h):N-->B satisfies
f(0)=<u,v> and
f(n)=<a(n-1), a(n)> for n>0
where a:N—>Q is as above.
PROOF:
Kevin Carlson (Oct 28 2025 at 22:30):I imagine Jan was wondering whether there is an explicit solution, not just whether there indeed exists such a recursively defined sequence. Though good on this mysterious other for thinking to check that!