Series Solutions: Hermite's Equation

Exercise 1:

Find the Hermite Polynomials of order 1 and 3.

Answer.

Recall that the recurrence relations are given by

\begin{displaymath}a_{n+2}= \frac{2(n-k)}{(n+2)(n+1)}a_n=0\mbox{ for all } n=0,1,2,3,\ldots\end{displaymath}

We have to evaluate these coefficients for k=1 and k=3, with initial conditions a0=0, a1=1.

When k=1,

\begin{displaymath}a_3=\frac{2(1-1)}{2\cdot 3} a_1=0.\end{displaymath}

Consequently all odd coefficients other than a1 will be zero. Since a0=0, all even coefficients will be zero, too. Thus

H1(t)=t.

When k=3,

\begin{displaymath}a_3=\frac{2(1-3)}{2\cdot 3} a_1=-\frac{2}{3},\end{displaymath}

and

\begin{displaymath}a_5=\frac{2(3-3)}{4\cdot 5} a_3=0.\end{displaymath}

Consequently all odd coefficients other than a1 and a3 will be zero. Since a0=0, all even coefficients will be zero, too. Thus

\begin{displaymath}H_3(t)=t-\frac{2}{3}t^3.\end{displaymath}

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Helmut Knaust
1998-07-08

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