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I am looking for help on the following, Show that the for odd non-negative integers $m$ there are a polynomials of degree $m$ when $r=1$? This is from the hermite differential equation, have showed the recurrence relation and the indicial equation implies $r=0$ and $r=1$, when the case of $r=1$

I am aware that $a_0$ is free and $a_1=0$, however not sure how to go about that there are polynomials of degree $m$ when $r=1$?

My Hermite equation is $y''+2xy'+2my=0$

My recurrance relation is $a_{n+2}=\frac{2(n+r-m)a_n}{(n+r+2)(n+r+1)} n=0,1...$

My indicial equation is the lowest order term:indicial being $$(r)(r-1){a_0}{x^{-2+r}}+(1+r)(r){a_1}{x^{-1+r}}=0$$ hence

$(r)(r-1))=0$ hence r=0 and r=1, with $y(x)=\sum_{n=0}^{\infty}{a_n}{x^{n+r}}$ and Many thanks in advance.

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I have no idea what you are talking about. I see a differential equation, but I don't see a recurrence relation, or an $r$, or an indicial equation, or any polynomials, or any $a_0$ or $a_1$. Please edit your question into a form where all terms are explained clearly, and then maybe someone will be able to help you. – Gerry Myerson Feb 14 '12 at 0:51
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When $r=1$, and $n=m-1$, the recurrence relation says $a_{m+1}=0$, and then the recurrence tells you $0=a_{m+3}=a_{m+5}=\dots$, that is $a_n=0$ for all even $n\gt m$. From $a_1=0$ and the recurrence you get $a_n=0$ for all odd $n$. So $a_n=0$ for all $n\gt m$, and all you are left with is a polynomial of degree $m$.

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