How would you solve this recurrence relation?
$f_k(a+i b k)=f_{k-1}-f_{k+1}$
$a$ and $b$ are reals and $k$ is an integer in $-\infty$, $+\infty$, $i$ is the imaginary unit.
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2This is called a "recurrence relation". But I like how you named it, it's fancy ^^ – Oussama Boussif May 12 '17 at 20:08
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thank you! I've changed the title accordingly – Andrea May 12 '17 at 20:21
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This is not a recurrence relation in the usual sense, because $f_k$ is defined in terms of both $f_{k-1}$ and $f_{k+1}$, with no "initial conditions." (One obvious solution is $f_k = 0$ for all $k$.) – Fabio Somenzi May 12 '17 at 20:29
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It is a recurrence relation, consider that it defines $f_{k+1}$ in terms of $f_k$ and $f_{k-1}$ – GEdgar May 12 '17 at 20:32
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See Favard's Theorem ... https://en.wikipedia.org/wiki/Favard%27s_theorem – GEdgar May 12 '17 at 20:32
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@GEdgar Thanks for the link. It still seems to require initial conditions. – Fabio Somenzi May 12 '17 at 20:37
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Thank you all. Do you think that there is a chance of finding an analytic solution to this problem? – Andrea May 12 '17 at 21:32
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1For what it's worth, Wolfram Alpha gives a solution in terms of Bessel functions. – dxiv May 13 '17 at 01:34
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It looks like $f_{k-1} $ and $f_{k+1}$ on the right need an argument. Please supply them. Otherwise, they should be constants and this just defines $f_k$ to be a constant. – Ross Millikan May 13 '17 at 02:02
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No, they are not functions. They are constants that are indexed by the index $k$, which runs from $-\infty$ to $+\infty$. Perhaps I should rewrite the equation above as $(a+i b k)f_k=f_{k-1}-f_{k+1}$ to avoid confusion. – Andrea May 13 '17 at 02:06