Given a recurrence relation: $$ax_n-bx_{n-1}-2x=(xa-b)x_{n-1}-ax_{n-2}-2x$$ where $a=x^3(x^2-3)-(x^2-1) \text{ and } b=x(x^2-3).$
What will be the final polynomial?
What will be the final polynomial for this recurrence relation?
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real-analysis
recurrence-relations
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0'Final polynomial' is rather vague – 2017-01-24
2 Answers
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The given recurrence reduces to:
$$ \require{cancel} a\,x_n\cancel{-b\,x_{n-1}}-\bcancel{2\,x} = a\,x\,x_{n-1}\cancel{-b\,x_{n-1}}-ax_{n-2}-\bcancel{\,2x} \iff a\,x_n = a\,x\,x_{n-1}-ax_{n-2} $$
If $a=0$ then the above is satisfied by any sequence $x_n\,$ otherwise it further reduces to:
$$ x_n = x\,x_{n-1}-x_{n-2} $$
The latter is a linear homogeneous recurrence, which does not depend on $a,b\,$. Given some initial conditions $x_{0,1}$ it can be solved using the standard methods, though the solution will (generally) not be a polynomial in $x\,$.
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So x is a constant? Just get the difference eqiation,
$$ a x_{n-1} - b x_{n-2} -2x=0 $$
Subtract it from the one you have to get rid of the constant. Can you take it fromhere?