Problem source

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 $a$,  ${\rm D}^2 x^a = a^2 x^a\,$.
\begin{questionparts}
\item 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 $r$.
\item Show that if $n$ and $m$ are  positive integers  with $n < m$, then ${\rm D}^n(1-x)^m$ is divisible by $(1-x)^{m-n}$.
\item Deduce that, if $m$ and $n$ are positive integers with $n < m$, then 
\[
\sum_{r=0}^m (-1)^r \binom m r r^n =0
\, .
\]
\item [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!\, $. 
\end{questionparts}

Solution source

\begin{align*}
{\mathrm D}^2 x^a &= x\frac{\d\ }{\d x}\left(  x\frac{\d}{\d x} \left ( x^a \right) \right) \\
&= x\frac{\d\ }{\d x}\left( ax^a \right) \\
&= a^2 x^a
\end{align*}

\begin{questionparts}
\item 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.

\item 
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.

\item 
\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.

\item 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, $f_n(1) = (-1)^n n!$

\end{questionparts}