Find the first $4$ Hermite polynomials using a recursion relation

Question

Given the Probabilists' Hermite differential equation: $$U''-xU'+\lambda U=0\tag{1}$$ A book question asks me to:

Find the first $4$ polynomial solutions (for $\lambda = 0,1,2,3$), each normalised such that the highest power of $x$ has a coefficient of unity.

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So I substituted $$U=\sum_{n=0}^{\lambda}C_nx^n\tag{2}$$ into $(1)$ and equated coefficients of $x^n$ to get the recursion relation: $$C_{n+2}=\frac{(n-\lambda)}{(n+1)(n+2)}C_n$$

So I am looking to find the values of the coefficients $C_0,C_1,C_2$ for $$U=\sum_{n=0}^{3}C_nx^n=C_0+C_1x+C_2x^2+x^3$$ where we are given that $\color{blue}{C_3}=1$.

Using the recursion relation I find that for $(n=0, \lambda=2)$: $$C_2=-C_0$$ and for $(n=1, \lambda=3)$: $$\color{blue}{C_3}=-\frac13C_1\implies C_1=-3$$

This is as far as I can get to in this question.

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The book answer simply states:

Using the recurrence relations, and setting the coefficient of the highest power to $1$, $U_0=1, \,U_1=x,\, U_2=x^2-1\,$ and $U_3=x^3-3x$

I know from this page on Wikipedia that the book answer is right, but the answer is not very helpful to me as I have no idea why $U_0=1$. Does this mean that $C_0=1$, if so how did the author deduce this from the recursion relation? I showed that $C_1$ is equal to $-3$. So why is $U_1\ne-3x$? The same misunderstanding follows for $U_2$ and $U_3$. In fact I don't understand why $U_3$ doesn't contain an $x^2$ term and a constant term. Also, where is the $x$ term in $U_2$ and the constant term in $U_1$?

I have only just started reading about Hermite polynomials so my understanding is very weak (apologies).

Is there any chance someone could explain as simply as possible how to obtain $U_0,U_1,U_2,U_3$?

Thank you.

Answer 1

Two aspects: One addresses the recurrence relation, the other the different polynomial solutions.

• A recurrence relation is not specified if the initial values are missing.

  Here we consider a second-order recurrence relation \begin{align*} (n+1)(n+2)C_{n+2}-(n-\lambda)C_n&=0\qquad\qquad n\geq 0\\ C_0&=???\tag{1}\\ C_1&=???\\ \end{align*} parametrised with $\lambda=0,1,2,\ldots$.

  Please note, in order to fully determine the recurrence relation it's crucial to also determine two initial values. These are usually $C_0$ and $C_1$ from which all other values with the help of the recurrence relation can be calculated.

  The calculation of the initial values is independent from the calculation of the recurrence relation.

We observe due to the special structure of the recurrence relation that once we have determined $C_0$, we can calculate all evenly indexed values, and once we have determined $C_1$, we can calculate all oddly indexed value. \begin{align*} C_0&\qquad\rightarrow\qquad C_2, C_4, C_6, \ldots\\ C_1&\qquad\rightarrow\qquad C_3, C_5, C_7, \ldots\\ \end{align*}

• For each parameter $\lambda=0,1,2,3$ a normalised polynomial $U_\lambda(x)$ with degree $\lambda$ \begin{align*} U_\lambda(x)=\sum_{n=0}^\lambda C_n^{(\lambda)}x^n\\ \end{align*} fulfilling the differential equation $U^{\prime\prime}-xU^\prime+\lambda U=0$ is to determine.

  Since we have to find four different polynomials, it is convenient to use the parameter $\lambda$ as additional marker to make the coefficients distinguishable and write $C_k^{(\lambda)}$.

  The polynomials $U_\lambda$ we are looking for are

  \begin{align*} \lambda=0:\qquad&U_0(x)=1\\ \lambda=1:\qquad&U_1(x)=C_0^{(1)}+x\tag{2}\\ \lambda=2:\qquad&U_2(x)=C_0^{(2)}+C_1^{(2)}x+x^2\\ \lambda=3:\qquad&U_3(x)=C_0^{(3)}+C_1^{(3)}x+C_2^{(3)}x^2+x^3\\ \end{align*}

  Here are four different polynomials $U_\lambda(x)$ with degree $\operatorname{deg}\left(U_\lambda(x)\right)=\lambda$ and coefficient of highest power equal to $1$ since they have to be normalised.

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Calculation of the polynomials:

• Case: $\lambda=0$

Observe, that $U_0(x)=1$ is already given due to the degree requirement of the polynomial and the normalisation requirement. But we have to check that it is a proper solution which fulfills the differential equation.

\begin{align*} \left(U_0(x)\right)^{\prime\prime}&-x\left(U_0(x)\right)^{\prime}+0\cdot U_0(x)=0-x\cdot0+0\cdot 1=0 \end{align*} and we have verified that $$U_0(x)=1$$ is the wanted polynomial for $\lambda=0$.

• Case: $\lambda=1$

In order to determine the missing coefficients of $U_1(x)$ we put it into the differential equation and obtain \begin{align*} 0&=\left(U_1(x)\right)^{\prime\prime}-x\left(U_1(x)\right)^{\prime}+U_1(x)\\ &=0-x\cdot \left(1\right)+1\cdot \left(C_0^{(1)}+x\right)\\ &=C_0^{(1)}\tag{3} \end{align*}

According to (3) we obtain $C_0^{(1)}=0$ and the polynomial $U_1(x)$ is according to (2) \begin{align*} U_1(x)&=x+C_0^{(1)}\\ &=x \end{align*}

• Case: $\lambda=2$

In order to determine the missing coefficients of $U_2(x)$ we put it into the differential equation and obtain \begin{align*} 0&=\left(U_2(x)\right)^{\prime\prime}-x\left(U_2(x)\right)^{\prime}+2U_2(x)\\ &=2-x\left(C_1^{(2)}+2x\right)+2\left(C_0^{(2)}+C_1^{(2)}x+x^2\right)\\ &=C_1^{(2)}x+2\left(C_0^{(2)}+1\right)\tag{4} \end{align*}

Comparing coefficient with equal powers in RHS and LHS of (4) gives \begin{align*} C_1^{(2)}&=0\\ C_0^{(2)}&=-1 \end{align*} and the polynomial $U_2(x)$ is according to (2) \begin{align*} U_2(x)&=x^2+C_1^{(2)}x+C_0^{(2)}\\ &=x^2-1 \end{align*}

• Case: $\lambda=3$

In order to determine the missing coefficients of $U_3(x)$ we put it into the differential equation and obtain \begin{align*} 0&=\left(U_3(x)\right)^{\prime\prime}-x\left(U_3(x)\right)^{\prime}+3U_3(x)\\ &=\left(6x+2C_2^{(3)}\right)-x\left(3x^2+2C_2^{(3)}x+C_1^{(3)}\right) +3\left(C_0^{(3)}+C_1^{(3)}x+C_2^{(3)}x^2+x^3\right)\\ &=C_2^{(3)}x^2+2\left(3+C_1^{(3)}\right)x+2C_2^{(3)}+3C_0^{(3)}\tag{5}\\ \end{align*}

Comparing coefficient with equal powers in RHS and LHS of (4) gives \begin{align*} C_2^{(3)}&=0\\ C_1^{(3)}&=-3\\ C_0^{(3)}&=0 \end{align*} and the polynomial $U_3(x)$ is according to (2) \begin{align*} U_3(x)&=x^3+C_2^{(3)}x^2+C_1^{(3)}x+C_0^{(3)}\\ &=x^3-3x \end{align*}

Notes:

• We observe that the recurrence relation (2) is not that helpful for small values of $\lambda$, since for $\lambda=0,1,2$ the calculation consists of determination of the initial values only. Just in the case $\lambda=3$ we could obtain $C_2^{(3)}$ from the initial value $C_0^{(3)}=0$.

• Nevertheless will the recurrence relation show its strength with increasing $\lambda$. It's also interesting to see that the coefficients depend oddly and evenly indexed from the the initial values.

• The nice fact that evenly indexed polynomials $U_{2\lambda}$ are even functions and oddly indexed polynomials $U_{2\lambda+1}$ are odd functions has to be analysed separately.

Answer 2

You should add a second index to the recurrence formula in order to avoid confusion. $$C_{n+2}^{(\lambda)}=\frac{(n-\lambda)}{(n+1)(n+2)}C_n^{(\lambda)}$$ With this, it is an immediate consequence that in even indexed polynomial $U_0, U_2,\dots$ the odd coefficients $C_{1}^{(\lambda)}, C_{3}^{(\lambda)}$ are zero and analogous for the odd indexed polynomials. So you have the trivial cases

$$U_0(x)=1$$ $$U_1(x)=x$$

For $U_2(x) = C_{0}^{(2)} + x^2$ you get $$1=C_{2}^{(2)} = \frac{(0-2)}{(0+1)(0+2)}C_{0}^{(2)}$$ i.e. $\color{red}{C_{0}^{(2)}}=-1$ and $U_2(x)= x^2-1$.

For $U_3(x) = C_{1}^{(2)}x + x^3$ you get $$1=\color{red}{C_{3}^{(3)}} = \frac{(1-3)}{(1+1)(1+2)}C_{1}^{(3)}=-\frac{1}{3}C_{1}^{(3)}$$

i.e. $\color{red}{C_{1}^{(3)}}=-3$ and $U_2(x)= x^3-3x$.

Answer 3

$$U''-xU'+\lambda U=0$$

This is more a long comment than an answer. If fact I only answer to OP question : "I have no idea why $U_0 =1$. Does this mean that $C_0 =1$ ? " : See below the case $\lambda=0$.

The answer with the method of series and recurrence relations has already been given. No need to repeat it. Of course, it isn't a smart method that is used below.

Case $\lambda=0 \quad\to\quad U''-xU'=0$

The polynomial solution comes from $U'=0 \quad\to\quad U=c=$constant.

This is the highest power of $x$ i.e.: $x^0$ . Taking account of the request of normalisation $c=1$

$$U_0(x)=1$$

Case $\lambda=1$

$U''-xU'+U=0\quad $Polynomial solution : $U=ax+b$

$(ax+b)''-x(ax+b)'+(ax+b)=0 \quad\to\quad \begin{cases} \text{any }a \\ b=0\end{cases} \quad\to\quad U=ax$

$a=1$ to be normalized. $$U_1(x)=x$$

Case $\lambda=2$

$U''-xU'+2U=0\quad$ Polynomial solution $U=ax^2+bx+c$

$2a-x(2ax+b)+2(ax^2+bx+c)=0 \quad\to\quad \begin{cases} \text{any }a \\ b=0\\ c=-a\end{cases} \quad\to\quad U=ax^2-a$

$a=1$ to be normalized. $$U_2(x)=x^2-1$$

Case $\lambda=3$

$U''-xU'+3U=0\quad$ Polynomial solution $U=ax^3+bx^2+cx+d$

$(6ax+2b)-x(3ax^2+2bx+c)+3(ax^3+bx^2+cx+d)=0$

$\begin{cases} \text{any }a \\b=0 \\ c=-3a \\ d=0\end{cases} \quad\to\quad U=ax^3-3a \qquad a=1$ to be normalized. $$U_3(x)=x^3-3x$$

Case $\lambda=4$

$U''-xU'+4U=0\quad$ Polynomial solution $U=ax^4+bx^3+cx^2+dx+E$

$(12ax^2+6bx+2c)-x(4ax^3+3bx^2+2cx+d)+4(ax^4+bx^3+cx^2+dx+E)=0$

$\begin{cases} \text{any }a \\b=0 \\ c=-6a \\ d=0 \\ E=3a\end{cases} \quad\to\quad U=ax^4-6ax^2+3a \qquad a=1$ to be normalized. $$U_4(x)=x^4-6x^2+3$$