An operator \(\rm D\) is defined, for any function \(\f\), by
\[
{\rm D}\f(x) = x\frac{\d\f(x)}{\d x}
.\]
The notation \({\rm D}^n\) means that \(\rm D\) is applied \(n\) times; for example
\[
\displaystyle
{\rm D}^2\f(x) = x\frac{\d\ }{\d x}\left( x\frac{\d\f(x)}{\d x} \right)
\,.
\]
Show that, for any constant \(a\), \({\rm D}^2 x^a = a^2 x^a\,\).
Show that if \(\P(x)\) is a polynomial of degree \(r\) (where \(r\ge1\)) then, for any positive integer \(n\), \({\rm D}^n\P(x)\) is also a polynomial of degree \(r\).
Show that if \(n\) and \(m\) are positive integers with \(n < m\), then \({\rm D}^n(1-x)^m\) is divisible by \((1-x)^{m-n}\).
Deduce that, if \(m\) and \(n\) are positive integers with \(n < m\), then
\[
\sum_{r=0}^m (-1)^r \binom m r r^n =0
\, .
\]
[Not on original paper]
Let \(\f_n(x) = D^n(1-x)^n\,\), where \(n\) is a positive integer.
Prove that \(\f_n(1)=(-1)^nn!\, \).
Claim: \({\mathrm D^n}(x^a) =a^n x^a\)
Proof: Induct on \(n\). Base cases we have already seen, so consider \(D^{k+1}(x^a) = D(a^k x^a) = a^{k+1}x^a\) as required.
Claim: \({\mathrm D}\) is linear, ie \({\mathrm D}(f(x) + g(x)) = {\mathrm D}(f(x)) + {\mathrm D}(g(x))\)
Proof:
\begin{align*}
{\mathrm D}(f(x) + g(x)) &= x\frac{\d\ }{\d x}\left(f(x) + g(x) \right) \\
&= x\frac{\d\ }{\d x}f(x) + x\frac{\d\ }{\d x}g(x) \\
&= {\mathrm D}(f(x)) + {\mathrm D}(g(x))
\end{align*}
Claim: If \(p(x)\) is a polynomial degree \(r\) then \({\mathrm D}^n p(x)\) is a polynomial degree \(n\).
Proof: Since \({\mathrm D}\) is linear, it suffices to prove this for a monomial of degree \(n\), but this was already proven in the first question.
Claim: If \(f(x)\) is some polynomial, \({\mathrm D}((1-x)^m f(x))\) is divisible by \((1-x)^{m-1}\)
Proof: \({\mathrm D}((1-x)^mf(x)) = -xm(1-x)^{m-1}f(x) + (1-x)^mxf'(x) = x(1-x)^{m-1}((1-x)f'(x)-xf(x))\) as required.
Therefore repeated application of \({\mathrm D}\) will reduce the factor of \(1-x\) by at most \(1\) each time as required.
\begin{align*}
{\mathrm D}^n(1-x)^m &= {\mathrm D}^n \left ( \sum_{r=0}^m \binom{m}{r}(-1)^r x^r\right) \\
&= \sum_{r=0}^m {\mathrm D}^n \left ( \binom{m}{r}(-1)^r x^r \right ) \\
&= \sum_{r=0}^m\binom{m}{r}(-1)^r r^n x^r
\end{align*}
Since the left-hand side is divisible by \(1-x\), if we substitute \(x = 1\), the sum must be \(0\), i.e., we get the desired result.
On each application of \({\mathrm D}\) to \((1-x)^m f(x)\) we end up with a term in the form \(x(1-x)^{m-1}(x)\) and a term of the form \((1-x)^m\). After the latter term will be annihilated once we evaluate at \(x = 1\) because there will be insufficient applications to remove the factors of \(1-x\). Therefore we only need to focus on the term which does not get annihilated. This term is will be \((-x)^n n \cdot (n-1) \cdots 1\), so \(f_n(1) = (-1)^n n!\) as required.
Alternatively:
\begin{align*}
{\mathrm D}^n((1-x)^n) &= D^{n-1}(-nx(1-x)^{n-1}) \\
&= -n{\mathrm D}^{n-1}(x(1-x)^{n-1}) \\
&= -n{\mathrm D}^{n-1}((x-1+1)(1-x)^{n-1}) \\
&= -n{\mathrm D}^{n-1}(-(1-x)^{n}+(1-x)^{n-1}) \\
&= -n{\mathrm D}^{n-1}(-(1-x)^{n})-n{\mathrm D}^{n-1}((1-x)^{n-1}) \\
\end{align*}
Therefore, when this is evaluated at \(x = 1\), recursively, we will have \(f_n(1) = -nf_{n-1}(1)\), in particular, \(f_n(1) = (-1)^n n!\)