Derive 3-term recursion of polynomials from some recursion OR are my polynomials orthogonal
$begingroup$
considering a sequence of polynomials $(Q_n)$ on $mathbb{R}$ given by $$(q_1 - x)Q_0 + p_1 Q_1 =0$$ and for $ngeq 2$ by $$(q_n-x)Q_{n} +p_n Q_{n+1}+sum_{i=0}^{n-1} c_{n,i} Q_i =0$$ where all constants $p_i>0$, $c_{k,i}>0$ and for all $ninmathbb{N}$ the constants satisfy $$sum_{i=0}^{n-1} c_{n,i} + q_n + p_n = 0$$.
I am looking for additional conditions under which these polynomials are orthogonal with respect to some measure or equivalently by Favard's theorem under which additional conditions the aforementionned recursion becomes a recursion of three terms.
Any help or tipps are highly appreciated!
EDIT: Writing stuff down sometimes helps the thinking process. In fact, it is NOT possible to arrive at a orthogonal polynomial, if there is any $c_{k,i}>0$ since then by assuming that there is a measure such that the sequence is orthogonal wrt the SKP <.,.> induced by the measure one arrives at $$<Q_i,Q_i> c_{k,i} = 0$$ st $Q_i=0$ but that is a contradction to being a polynomial of degree $igeq 1$.
recursion orthogonal-polynomials
asked Dec 29 '18 at 8:24
JfischerJfischer
33
$endgroup$
add a comment |
$begingroup$
considering a sequence of polynomials $(Q_n)$ on $mathbb{R}$ given by $$(q_1 - x)Q_0 + p_1 Q_1 =0$$ and for $ngeq 2$ by $$(q_n-x)Q_{n} +p_n Q_{n+1}+sum_{i=0}^{n-1} c_{n,i} Q_i =0$$ where all constants $p_i>0$, $c_{k,i}>0$ and for all $ninmathbb{N}$ the constants satisfy $$sum_{i=0}^{n-1} c_{n,i} + q_n + p_n = 0$$.
I am looking for additional conditions under which these polynomials are orthogonal with respect to some measure or equivalently by Favard's theorem under which additional conditions the aforementionned recursion becomes a recursion of three terms.
Any help or tipps are highly appreciated!
EDIT: Writing stuff down sometimes helps the thinking process. In fact, it is NOT possible to arrive at a orthogonal polynomial, if there is any $c_{k,i}>0$ since then by assuming that there is a measure such that the sequence is orthogonal wrt the SKP <.,.> induced by the measure one arrives at $$<Q_i,Q_i> c_{k,i} = 0$$ st $Q_i=0$ but that is a contradction to being a polynomial of degree $igeq 1$.
recursion orthogonal-polynomials
asked Dec 29 '18 at 8:24
JfischerJfischer
33
$endgroup$
add a comment |
$begingroup$
considering a sequence of polynomials $(Q_n)$ on $mathbb{R}$ given by $$(q_1 - x)Q_0 + p_1 Q_1 =0$$ and for $ngeq 2$ by $$(q_n-x)Q_{n} +p_n Q_{n+1}+sum_{i=0}^{n-1} c_{n,i} Q_i =0$$ where all constants $p_i>0$, $c_{k,i}>0$ and for all $ninmathbb{N}$ the constants satisfy $$sum_{i=0}^{n-1} c_{n,i} + q_n + p_n = 0$$.
I am looking for additional conditions under which these polynomials are orthogonal with respect to some measure or equivalently by Favard's theorem under which additional conditions the aforementionned recursion becomes a recursion of three terms.
Any help or tipps are highly appreciated!
EDIT: Writing stuff down sometimes helps the thinking process. In fact, it is NOT possible to arrive at a orthogonal polynomial, if there is any $c_{k,i}>0$ since then by assuming that there is a measure such that the sequence is orthogonal wrt the SKP <.,.> induced by the measure one arrives at $$<Q_i,Q_i> c_{k,i} = 0$$ st $Q_i=0$ but that is a contradction to being a polynomial of degree $igeq 1$.
recursion orthogonal-polynomials
asked Dec 29 '18 at 8:24
JfischerJfischer
33
$endgroup$
considering a sequence of polynomials $(Q_n)$ on $mathbb{R}$ given by $$(q_1 - x)Q_0 + p_1 Q_1 =0$$ and for $ngeq 2$ by $$(q_n-x)Q_{n} +p_n Q_{n+1}+sum_{i=0}^{n-1} c_{n,i} Q_i =0$$ where all constants $p_i>0$, $c_{k,i}>0$ and for all $ninmathbb{N}$ the constants satisfy $$sum_{i=0}^{n-1} c_{n,i} + q_n + p_n = 0$$.
I am looking for additional conditions under which these polynomials are orthogonal with respect to some measure or equivalently by Favard's theorem under which additional conditions the aforementionned recursion becomes a recursion of three terms.
Any help or tipps are highly appreciated!
EDIT: Writing stuff down sometimes helps the thinking process. In fact, it is NOT possible to arrive at a orthogonal polynomial, if there is any $c_{k,i}>0$ since then by assuming that there is a measure such that the sequence is orthogonal wrt the SKP <.,.> induced by the measure one arrives at $$<Q_i,Q_i> c_{k,i} = 0$$ st $Q_i=0$ but that is a contradction to being a polynomial of degree $igeq 1$.
recursion orthogonal-polynomials
recursion orthogonal-polynomials
asked Dec 29 '18 at 8:24
JfischerJfischer
33
asked Dec 29 '18 at 8:24
JfischerJfischer
33
edited Dec 29 '18 at 9:43
Jfischer
asked Dec 29 '18 at 8:24
JfischerJfischer
33
asked Dec 29 '18 at 8:24
JfischerJfischer
33
asked Dec 29 '18 at 8:24
JfischerJfischer
33
33
add a comment |
add a comment |
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