Proof. From the definition of \(H_n(x)\), we have \begin {equation} \frac {d^n}{dx^n}\left [e^{-\frac {x^2}{2}}\right ] = (-1)^n\, e^{-\frac {x^2}{2}}\, H_n(x) \end {equation} Differentiating through (4.1), we get \begin {equation} \frac {d^{n+1}}{dx^{n+1}}\left [e^{-\frac {x^2}{2}}\right ] = (-1)^n\left [-xe^{-\frac {x^2}{2}}\, H_n(x) + e^{-\frac {x^2}{2}}\, H_n'(x)\right ] \end {equation} Again using the definition, get \begin {equation} \frac {d^{n+1}}{dx^{n+1}}\left [e^{-\frac {x^2}{2}}\right ] = (-1)^{n+1}\, e^{-\frac {x^2}{2}}\, H_{n+1}(x) \end {equation} from (4.2) and (4.3) get \[H_{n+1}(x) = x\, H_n(x) - H_n'(x)\, , \hspace {0.3cm} n = 0, \, 1, 2, \cdots \cdots \] □

Proof. If \(w(x) = e^{-\frac {x^2}{2}}\), then \begin {equation} w(x - t) = \exp \left (-\frac {x^2}{2} + tx - \frac {t^2}{2}\right ) = e^{-\frac {x^2}{2}}\cdot e^{tx - \frac {t^2}{2}} = w(x)\, \exp \left (tx - \frac {t^2}{2}\right ). \end {equation} Applying Taylor expansion above we get \begin {equation} w(x -t) = \sum ^{\infty }_{n = 0} \frac {(-1)^n}{n!}\, t^n\, \dfrac {d^n}{dx^n}[w(x)] = \sum _{n = 0}^{\infty } \frac {t^n}{n!}\, H_n(x)\, w(x), \end {equation} since \[H_n(x) = (-1)^n\, e^{\frac {x^2}{2}}\, \frac {d^n}{dx^n}[w(x)].\] Then using (4.4) and (4.5), get \[\exp \left (tx - \frac {t^2}{2}\right ) = \sum _{n = 0}^{\infty } \frac {t^n}{n!}\, H_n(x).\] □

Proof. From theorem 4.6.3, \begin {align*} \sum ^{\infty }_{n = 0} \frac {t^n}{n!}\, H_n(x) & = \exp \left (tx - \frac {t^2}{2}\right )\\ & = e^{tx}\, \cdot \, e^{-\frac {t^2}{2}}\\ & = \sum _{k= 0}^{\infty } \frac {(tx)^k}{k!}\,\cdot \, \sum _{r = 0}^{\infty } \frac {\left (\frac {-t^2}{2}\right )^r}{r!}\, , \hspace {0.4cm} \text {Maclaurin expansion}\\ & = \sum _{k,r = 0}^{\infty }\frac {(-1)^r\, x^k\, t^{k + 2r}}{2^r\, k!\, r!} \end {align*}

for a fixed value of \(r\), we get \(t^n\) by setting \(k + 2r = n\), so that \(k = n - 2r\). So for this value of \(r\), the coefficient of \(t^n\) is given by \[ \frac {(-1)^r\, x^{n-2r}}{2^r\, (n-2r)!\, r!}.\] Then the total coefficient of \(t^n\) is obtained by summing over all allowed values of \(r\) since \(k = n-2r\), we have \(n -2r \geq 0\), or \(r \leq \frac {n}{2}\). If \(n = \) even, \(r\) goes from 0 to \(\frac {n}{2}\). If \(n = \) odd, \(r\) goes from 0 to \(\frac {n-1}{2}\).
That is in all cases \(r\) goes from o to \(\left [\frac {n}{2}\right ].\) Thus \[\sum _{r=0}^{\left [\frac {n}{2}\right ]}\frac {(-1)^r\, x^{n - 2r}}{2^r\, (n - 2r)!\, r!} = \frac {1}{n!}\, H_n(x)\] implies \[H_n(x) = \sum _{r=0}^{\left [\frac {n}{2}\right ]}\frac {(-1)^r\, x^{n - 2r}}{2^r\, (n - 2r)!\, r!}.\] □

Proof.

(a).

Differentiating with respect to \(x\), get \[\sum _{n = 0}^{\infty } \frac {t^n}{n!}\, H_n'(x) = t\, \exp \left (tx - \frac {t^2}{2}\right ) = t \,\sum _{n = 0}^{\infty } \frac {t^n}{n!}\, H_n(x).\] Let \(\, s = n + 1\) \[ = \sum _{n = 0}^{\infty }\frac {t^{n + 1}}{n!}\, H_n(x) = \sum _{n = 1}^{\infty } \frac {t^n}{(n - 1)!}\, H_{n-1}(x)\] equating coefficients of \(t^n\), for \(n = 0\), \(H_0'(x) = 0\), and for \(n\geq 1\), \[\frac {H_n'(x)}{n!} = \frac {H_{n-1}(x)}{(n-1)!}\implies H_n'(x) = n\, H_{n-1}(x)\] since \[\frac {n!}{(n-1)!} = \frac {n\, (n-1)!}{(n-1)!} = n.\]

(b).

Differentiating both sides of \[\exp \left (tx - \frac {t^2}{2}\right ) = \sum _{n=0}^{\infty } \frac {t^n}{n!}\, H_n(x)\] with respect to \(t\), get \[(x -t)\, \exp \left (tx - \frac {t^2}{2}\right ) = \sum _{n=0}^{\infty } \frac {n\ t^{n-1}}{n!}\, H_n(x)\] observe that for \(n =0\), \(\frac {n\, t^{n-1}}{n!} = 0\) hence,

\[(x - t) \sum _{n = 0}^{\infty } \frac {t^n}{n!}\, H_n(x) = \sum _{n=1}^{\infty } \frac {t^{n-1}}{(n-1)!}\, H_n(x)\]

\[x\sum _{n=0}^{\infty } \frac {t^n}{n!}\, H_n(x) - \sum _{n=0}^{\infty } \frac {t^{n + 1}}{n!}\, H_n(x) = \sum _{n = 1}^{\infty }\frac {t^{n-1}}{(n-1)!}\, H_n(x)\]

\[x\, \sum _{n=0}^{\infty }\frac {t^n}{n!}\, H_n(x) - \sum _{n=1}^{\infty }\frac {t^{n + 1}}{(n-1)!}\, H_{n-1}(x) = \sum _{n=0}^{\infty } \frac {t^n}{n!}\, H_{n+1}(x)\] comparing coefficients for \(n\geq 1\), get \[\frac {H_n(x)}{n!} - \frac {H_{n-1}(x)}{(n-1)!} = \frac {H_{n+1}(x)}{n!}\, , \] multiply by \(n!\) then \[x\, H_n(x) - n\, H_{n-1}(x) = H_{n+1}(x)\] equating coefficients of \(t^0\), get \[x\, H_0(x) = H_1(x).\]

Proof. Let \[c = \int _{-\infty }^{\infty } e^{-\frac {x^2}{2}}\, H_n(x)\, H_m(x)\, dx.\] Then by definition of \[H_n(x) = (-1)^n\, e^{-\frac {x^2}{2}}\, \frac {d^n}{dx^n}\left (e^{-\frac {x^2}{2}}\right ),\] \begin {align*} c & = (-1)^n\int _{-\infty }^{\infty }H_m(x)\, \frac {d^n}{dx^n}\left (e^{-\frac {x^2}{2}}\right )\, dx\\ & = (-1)^n\,H_m(x)\, \frac {d^{n-1}}{dx^{n-1}}\left (e^{-\frac {x^2}{2}}\right )\int ^{\infty }_{-\infty }\, \frac {d}{dx}H_m(x)\, \cdot \, \frac {d^{n-1}}{dx^{n-1}}\left (e^{-\frac {x^2}{2}}\right )\, dx \end {align*}

by integration by parts. \(u = H_m(x), \, dv = \frac {d^n}{dx^n}\left (e^{-\frac {x^2}{2}}\right )\, dx\) \[c = (-1)^{n+1}\int _{-\infty }^{\infty }\frac {d}{dx}H_m(x)\, \frac {d^{n-1}}{dx^{n-1}}\left (e^{-x^2}\right )\, dx,\] Since \[H_m(x)\, \frac {d^{n-1}}{dx^{n-1}}\left (e^{-\frac {x^2}{2}}\right )\] is a polynomial multiplied by \(e^{-\frac {x^2}{2}}\), and so goes to zero as \(x\rightarrow \infty \).
Integrating by parts \(m-1\) more times, get \begin {equation} c = (-1)^{n+m}\int _{-\infty }^{\infty } \frac {d^m}{dx^m}H_m(x)\, \frac {d^{n-m}}{dx^{n-m}}\left (e^{-\frac {x^2}{2}}\right )\, dx \end {equation} Now, since \(H_m(x)\) is a polynomial of degree \(m\) with leading coefficient 1, \[\frac {d^m}{dx^m}H_m(x) = m!,\] so that \[c = (-1)^{n+m}\int ^{\infty }_{-\infty } \frac {d^{n-m}}{dx^{n-m}}\left (e^{-\frac {x^2}{2}}\right )\, dx = \frac {d^{n-m-1}}{dx^{n-m-1}}\left [e^{-x^2}\right ]^{\infty }_{-\infty } = 0,\] since \(n>m\), and so \(n-m-1 > 0\).
The case \(m<n\) is arrived at in the same way, hence, \[\int _{-\infty }^{\infty }e^{-\frac {x^2}{2}}\, H_m(x)\, H_n(x)\, dx = 0, \hspace {0.3cm} m\neq n.\] When \(m = n\), \begin {align*} c & = \int _{-\infty }^{\infty }\frac {d^n}{dx^n}[H_n(x)]\, e^{-\frac {x^2}{2}}\, dx \hspace {0.4cm} \text {from 4.6}\\ & = n! \int _{-\infty }^{\infty } e^{-\frac {x^2}{2}}\, dx,\,\, \text {since}\, \, H_n(x)\,\, \text {is of degree}\,\, n\,\, \text {with leading coefficient 1}\\ & = 2n! \int _0^{\infty } e^{-\frac {x^2}{2}}\, dx\\ & = 2n!\left (\frac {\sqrt {2}}{2}\,\sqrt {\pi }\right )\\ & = n!\, \sqrt {2\pi }, \hspace {0.3cm} \text {since}\hspace {0.3cm} 2\int _0^{\infty } e^{-t^2}\, dt = \sqrt {\pi },\, \, t^2 = \frac {x^2}{2}. \end {align*} □

Working. Using \[H_n(x) = (-1)^n\, e^{-\frac {x^2}{2}}\, \frac {d^n}{dx^n}\left (e^{-\frac {x^2}{2}}\right )\] we have \begin {align*} C_{2n} & = \frac {1}{(2n)!\, \sqrt {2\pi }}\int _{-\infty }^{\infty } e^{-\frac {x^2}{2}}\, x^{2p}\, H_{2n}(x)\, dx\\\\ & = \frac {(-1)^{2n}}{(2n)!\sqrt {2\pi }}\int _{-\infty }^{\infty } x^{2p}\, \frac {d^{2n}}{dx^{2n}}\left (e^{-\frac {x^2}{2}}\right )\, dx, \hspace {0.3cm} \text {integrating by parts gives}\\\\ & = \frac {(2p)!}{(2n)!\, \sqrt {2\pi }\, (2p-2n)!}\, \int _{-\infty }^{\infty } e^{-\frac {x^2}{2}}\, x^{2p-2n}\, dx\\\\ & = \frac {(2p)!\, 2^{p - n + \frac {1}{2}}}{(2n)!\, \sqrt {2\pi }\, (2p-2n)!}\, \Gamma \left (p-n+\frac {1}{2}\right ) \end {align*} □

Working. \[\exp \left (tx - \frac {t^2}{2}\right ) = \sum _{n=0}^{\infty } \frac {t^n}{n!}\, H_n(x).\] Notice that \[\frac {d^m}{dx^m}\, [H_n(x)]\] is the coefficient of \(\frac {t^n}{n!}\) in the expansion of \[\frac {d^m}{dx^m}\left [\exp \left (tx - \frac {t^2}{2}\right )\right ].\] Now \begin {align*} \frac {d^m}{dx^m}\left [\exp \left (tx - \frac {t^2}{2}\right )\right ] & = t^m\, \exp \left (tx - \frac {t^2}{2}\right )\\ & = t^m\, \sum _{n = 0}^{\infty }\frac {t^n}{n!}\, H_n(x)\\ & = \sum _{n=0}^{\infty } \frac {t^{m + n}}{n!}\, H_n(x). \end {align*}

Let \(r = m+n\) \begin {align*} \frac {d^m}{dx^m}\left [\exp \left (tx - \frac {t^2}{2}\right )\right ] & = \sum _{r=m}^{\infty } \frac {t^r}{(r-m)!}\, H_{r-m}(x)\\ & = \sum _{n=m}^{\infty }\frac {t^n}{(n-m)!}\, H_{n-m}(x)\\ & =\sum _{n=m}^{\infty }\frac {t^n}{n!}\, \cdot \, \frac {n!}{(n-m)!}\, H_{n-m}(x)\, , \end {align*}

so that the coefficient of \(\frac {t^n}{n!}\) is \(\frac {n!}{(n-m)!}\, H_{n-m}(x).\) So \[\frac {d^m}{dx^m}[H_n(x)] = \frac {n!}{(n-m)!}\, H_{n-m}(x).\] □

Working. \begin {align*} \int _{-\infty }^{\infty } x\, e^{-\frac {x^2}{2}}\, H_n(x)\, H_m(x)\, dx & =\int _{-\infty }^{\infty }e^{-\frac {x^2}{2}}[n\, H_{n- 1}(x) + H_{n+1}(x)]\, H_m(x)\, dx\\ & = n\int _{-\infty }^{\infty }e^{-\frac {x^2}{2}}\, H_{n-1}(x)\, H_m(x)\, dx + \int _{-\infty }^{\infty } e^{-\frac {x^2}{2}}\, H_{n+ 1}(x)\, H_m(x)\, dx\\ & = n\, (n-1)!\, \sqrt {2\pi }\, \delta _{(n-1)m} + (n+1)\, \sqrt {2\pi }\, \delta _{(n+1)m}, \end {align*}

since \[\int _{-\infty }^{\infty } e^{-\frac {x^2}{2}}\, H_n(x)\, H_m(x)\, dx = n!\, \sqrt {2\pi }\, \delta _{nm}.\] □