Series Solutions: Hermite's Equation

Exercise 2:

Find the Hermite Polynomials of order 2, 4 and 6.

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=2, k=4 and k=6, with initial conditions a0=1, a1=0.

When k=2,

\begin{displaymath}a_2=\frac{2(0-2)}{1\cdot 2} a_0=-2,\end{displaymath}

while

\begin{displaymath}a_4=\frac{2(2-2)}{3\cdot 4} a_2=0.\end{displaymath}

Consequently all even coefficients other than a2 will be zero. Since a1=0, all odd coefficients will be zero, too. Thus

H2(t)=1-2t2.

When k=4,

\begin{eqnarray*}a_2&=&\frac{2(0-4)}{1\cdot 2} a_0=-4,\\
a_4&=&\frac{2(2-4)}{3\...
...ft(-4\right)=\frac{4}{3}\\
a_6&=&\frac{2(4-4)}{5\cdot 6} a_4=0.
\end{eqnarray*}


Consequently all even coefficients other than a2 and a4 will be zero. Since a1=0, all odd coefficients will be zero, too. Thus

\begin{displaymath}H_4(t)=1-4t^2+\frac{4}{3}t^4.\end{displaymath}

You can check that

\begin{displaymath}H_6(t)=1 - 6\,{t^2} + 4\,{t^4} - {\frac{8}{15} t^6}.\end{displaymath}

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Helmut Knaust
1998-07-08

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