The sequence of functions
\(y_0\), \(y_1\), \(y_2\), \(\ldots\,\) is defined by \(y_0=1\) and, for \(n\ge1\,\),
\[
y_n = (-1)^n \frac {1}{z} \, \frac{\d^{n} z}{\d x^n}
\,,
\]
where \(z= \e^{-x^2}\!\).
Show that
\(\dfrac{\d y_n}{\hspace{-4.7pt}\d x} = 2x y_n -y_{n+1}\,\) for \(n\ge1\,\).
\(y_0 = 1\), \(y_1 = (-1) e^{x^2} \cdot (-2x) \cdot e^{-x^2} = 2x\), \(y_2 = e^{x^2} \frac{\d^2}{\d x^2} \l e^{-x^2}\r = e^{x^2} \frac{\d }{\d x}\l -2xe^{-x^2} \r = e^{x^2} \l -2e^{-x^2}+4x^2e^{-x^2}\r = 4x^2-2\). Therefore \(2xy_1 - 2y_0 = 2x \cdot 2x - 2\cdot1 = 4x^2-2 = y_2\) so our statement is true for \(n=1\). Assume the statement is true for \(n=k\), then
\begin{align*}
&& y_{k+1} &= 2xy_k - 2ky_{k-1} \\
\frac{\d }{\d x}: && \frac{\d y_{k+1}}{\d x} &= 2\frac{\d}{\d x}\l xy_k \r - 2k\frac{\d y_{k-1}}{\d x} \\
\Rightarrow && 2xy_{k+1}-y_{k+2} &= 2y_k+2x \l 2xy_k-y_{k+1}\r - 2k \l 2xy_{k-1}-y_k \r \\
\Rightarrow && y_{k+2} &=2y_k+ 4x \cdot y_{k+1}-(4x^2+2k)y_k+2x \cdot 2k y_{k-1} \\
&&&= 4x \cdot y_{k+1}-(4x^2+2(k+1))y_k+2x \l2xy_k - y_{k+1} \r \\
&&&= 2x \cdot y_{k+1} -2(k+1)y_k
\end{align*}
Therefore since our statement is true for \(n=1\) and if it is true for \(n=k\) it is true for \(n=k=1\), therefore by the principle of mathematical induction it is true for all \(n \geq 1\).
Since \(2x = \frac{y_{n+1}+2ny_{n-1}}{y_n}\) for all \(n\), we must have
\begin{align*}
&& \frac{y_{n+1}+2ny_{n-1}}{y_n} &= \frac{y_{n+2}+2(n+1)y_{n}}{y_{n+1}} \\
\Leftrightarrow && y_{n+1}^2+2ny_{n-1}y_{n+1} &= y_ny_{n+2}+2ny_n^2+2y_n^2 \\
\Leftrightarrow && y_{n+1}^2-y_ny_{n+2} &= 2n(y_n^2-y_{n-1}y_{n+1})+2y_n^2
\end{align*}
Consider the functions \(f_n(x) = y_{n}^2-y_{n-1}y_{n+1}\) then clearly \(f_{n+1} = 2nf_{n} + 2y_n^2 \geq f_{n}\) so to prove \(f_n(x) > 0\) for \(n \geq 1\) it suffices to prove it for \(n = 1\). But \(f_1 = y_1^2 - y_0y_{2} = (2x)^2-(4x^2-2) = 2 > 0\) so we are done.