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Is it possible to define a function such that

$$ fn(0)=a\\fn(1)=b(a-x^0)\\fn(2)=c(b(a-x^0)-x^1)\\fn(3)=d(c(b(a-x^0)-x^1)-x^2)\\ ... $$

I was thinking it would involve something like $fn(n)=a_{n+1}(fn(n-1)-x^{n-1})$ but that breaks where $n=0$.

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  • $\begingroup$ Then define a base case $fn(0) = a$. $\endgroup$
    – peterwhy
    Commented Oct 3, 2022 at 23:21

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Why not? You can say something like $$ f(n) = \cases{a & if $n=0$\cr a_{n+1} (f(n-1) - x^{n-1}) & if $n > 0$\cr}$$

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