Differential Operator Issue

Question

Let us consider the differential operator $H:= x \frac{d}{dx}$ and let us define \begin{equation} \hat{\mathcal{O}_n}= \frac{1}{n!}(H+n)(H+n-1)\cdots(H+2)(H+1). \end{equation}

I proved - by induction - that \begin{equation} \hat{\mathcal{O}_n}\left ( \frac{\log x}{x} \right )=\frac{1}{nx} \end{equation}

I notice that something similar happens with the following: \begin{gather*} \frac{\log(x)}{x^2}; \\ \frac{\log(x)}{x^3}. \end{gather*} In particular: \begin{equation} \hat{\mathcal{O}_n}\left ( \frac{\log x}{x^2} \right )=-\frac{1}{n(n-1)x^2} \end{equation} for $n \ge 2$, and \begin{equation} \hat{\mathcal{O}_n}\left ( \frac{\log x}{x^3} \right )=\frac{2}{n(n-1)(n-2)x^3} \end{equation}

So I think that, generally, the following statement is TRUE.

Statement (Conjecture). Let us consider the differential operator $\hat{\mathcal{O}_n}$, then \begin{equation} \hat{\mathcal{O}_n}\left ( \frac{\log x}{x^m} \right )=(-1)^{m+1}\frac{(m-1)!}{n(n-1)(n-2) \cdots (n-m+1)x^m} \end{equation} with $n \in \mathbb{N}$, $n \ge m$.

Anyone have any idea how it can be proved?

Answer 1

We show the validity of OPs claim by using some operator calculations and deriving binomial identities. We start with some notational conventions. Here we use rising and falling factorial powers \begin{align*} x^{\overline{n}}&=x(x+1)\cdots(x+n-1)\\ x^{\underline{n}}&=x(x-1)\cdots(x-n+1)\\ \end{align*} and the differential operator $D_x=\frac{d}{dx}$. Using this notation the operator $\hat{\mathcal{O}_n}$ can be written $$\hat{\mathcal{O}_n}=\frac{1}{n!}(xD_x+n)^{\underline{n}}$$

We now formulate OPs claim in two steps. At first we transform the operator to a more convenient representation which will be used in the proof below.

The following is valid

\begin{align*} \hat{\mathcal{O}_n}=\frac{1}{n!}(xD_x+n)^{\underline{n}} =\sum_{k=0}^n\binom{n}{k}\frac{1}{k!}x^kD_x^k\qquad &\qquad n\geq 1\tag{1}\\ \end{align*} and then we apply the operator to $\frac{\log x}{x^m}$ \begin{align*} \hat{\mathcal{O}_n}\left(\frac{\log x}{x^m}\right)&=\frac{1}{n!}(xD_x+n)^{\underline{n}}\left(\frac{\log x}{x^m}\right)\\ &=\frac{(-1)^{m+1}}{mx^m}\binom{n}{m}^{-1}\qquad &\qquad 1\leq m\leq n\tag{2} \end{align*}

Approach: The proof is divided into four steps. At first we derive the operator representation (1). Then we apply (1) to $\frac{\log x}{x^m}$ and derive an expression containing a sum with $x^{-m}\log x$ times a binomial expression and another sum with $x^{-m}$ times a binomial expression. In step three we show the first binomial sum cancels and in the last step we simplify the other sum to obtain (2).

Operator methods:

We obtain \begin{align*} \hat{\mathcal{O}_n}&=\frac{1}{n!}(xD_x+n)^{\underline{n}}\tag{3}\\ &=\frac{1}{n!}\sum_{k=0}^n\binom{n}{k}\left(xD_x\right)^{\underline{k}}n^{\underline{n-k}}\\ &=\sum_{k=0}^n\binom{n}{k}\frac{1}{k!}\left(xD_x\right)^{\underline{k}}\tag{4}\\ &=\sum_{k=0}^n\binom{n}{k}\frac{1}{k!}\sum_{j=0}^k(-1)^{k-j}\left[k\atop j\right]\left(xD_x\right)^j\tag{5}\\ &=\sum_{k=0}^n\binom{n}{k}\frac{1}{k!}\sum_{j=0}^k(-1)^{k-j}\left[k\atop j\right]\sum_{l=0}^j{j\brace l}x^lD_x^l\tag{6}\\ &=\sum_{k=0}^n\binom{n}{k}\frac{1}{k!}x^kD_x^k\tag{7} \end{align*} and the claim (1) follows.

Comment:

• In (3) we use the binomial theorem which is also valid for falling factorials. See e.g. exercise 5.37 in Concrete Mathematics by D.E. Knuth et al.

• In (4) we use the representation of falling factorial powers by Stirling numbers of the first kind $\left[n\atop k\right]$ \begin{align*} x^{\underline{n}}=\sum_{k=0}^n(-1)^{n-k}\left[n\atop k\right]x^j \end{align*}

• In (5) we use the representation of $\left(xD_x\right)^n$ by Stirling numbers of the second kind ${n\brace k}$ \begin{align*} \left(xD_x\right)^n=\sum_{k=0}^n(-1)^{n-k}{n\brace k}x^kD_x^k \end{align*}

• In (6) we use that Stirling numbers of first and second kind are binomial inverse pairs. \begin{align*} \sum_{k=0}^n(-1)^{n-k}\left[n\atop k\right]{k\brace l}=[[n=l]] \end{align*}

  with $[[P]]$ the Iverson bracket notation.

Intermezzo:

According to the transformation (7) of the operator $\hat{\mathcal{O}_n}$ we next calculate up the $k$-th derivative of $\frac{\log x}{x^m}$. We do this by applying the general Leibniz rule

\begin{align*} D_x^k\left(\frac{\log x}{x^m}\right)&=\sum_{j=0}^k\binom{k}{j}D_x^j\left(\log x\right)D_x^{k-j}\left(x^{-m}\right)\\ &=\log x D_x^k\left(x^{-m}\right)+\sum_{j=1}^k\binom{k}{j}D_x^j\left(\log x\right)D_x^{k-j}\left(x^{-m}\right) \end{align*}

Since \begin{align*} D_x^k\left(x^{-m}\right)&=(-1)^km^{\overline{k}}x^{-m-k}\\ D_x^k(\log x)&=(-1)^{k-1}(k-1)!x^{-k} \end{align*} we obtain

\begin{align*} D_x^k\left(\frac{\log x}{x^m}\right)&=(-1)^km^{\overline{k}}x^{-m-k}\log x\\ &\qquad -(-1)^kx^{-m-k}\sum_{j=1}^k\binom{k}{j}(j-1)!m^{\overline{k-j}}\qquad\qquad m,k\geq 1 \end{align*}

$$ $$

Applying the operator:

Combining the last result with (7) we obtain after some simplifications \begin{align*} \frac{1}{n!}&(xD_x+n)^{\underline{n}}\left(\frac{\log x}{x^m}\right)\\ &=\sum_{k=0}^n\binom{n}{k}\frac{1}{k!}x^kD_x^k\left(\frac{\log x}{x^m}\right)\\ &=x^{-m}\log x\sum_{k=0}^m\binom{n}{k}(-1)^k\frac{1}{k!}m^{\overline{k}}\\ &\qquad-x^{-m}\sum_{k=1}^n\binom{n}{k}(-1)^k\frac{1}{k!}\sum_{j=1}^k\binom{k}{j}(j-1)!m^{\overline{k-j}}\tag{8} \end{align*}

The last steps consist of simplification of both sums in the last expression (8). In fact this simplification provides some nice binomial identities.

Cancelling first binomial sum:

The following is valid \begin{align*} \sum_{k=0}^n\binom{n}{k}(-1)^k\frac{1}{k!}m^{\overline{k}}=0\qquad\qquad m,n\geq 1\tag{9} \end{align*}

We get \begin{align*} \sum_{k=0}^n\binom{n}{k}(-1)^k\frac{1}{k!}m^{\overline{k}} &=\sum_{k=0}^n(-1)^k\binom{m+k-1}{k}\tag{10}\\ &=\sum_{k=0}^n\binom{n}{k}\binom{-m}{k}\tag{11}\\ &=\binom{n-m}{n}\\ &=[[n=m=0]] \end{align*} and (9) follows.

Comment:

• In (10) we use the upper negation $\binom{-p}{q}=\binom{p+q-1}{q}(-1)^q$.

• In (11) we use Chu-Vandermonde's identity.

Simplifying second binomial sum:

The following is valid \begin{align*} \sum_{k=1}^n\binom{n}{k}\sum_{j=1}^k\frac{1}{j}(-1)^j\binom{-m}{k-j} &=\sum_{j=0}^{n-m}\frac{1}{j+m}(-1)^{j+m}\binom{n-m}{j}\tag{12}\\ &=\frac{(-1)^m}{m}\binom{n}{m}^{-1} \qquad\qquad1\leq m\leq n\tag{13} \end{align*}

We show the identity (12) by application of the Chu-Vandermonde identity. The step from (12) to (13) is a nice identity by its own. We will show it using operator methods.

We obtain \begin{align*} \sum_{k=1}^n&\binom{n}{k}\sum_{j=1}^k\frac{1}{j}(-1)^j\binom{-m}{k-j}\tag{14}\\ &=\sum_{j=1}^n\sum_{k=j}^n\frac{1}{j}(-1)^j\binom{n}{k}\binom{-m}{k-j}\tag{15}\\ &=\sum_{j=1}^n\frac{1}{j}(-1)^{j}\sum_{k=0}^{n-j}\binom{n}{k+j}\binom{-m}{k}\tag{16}\\ &=\sum_{j=m}^n\frac{1}{j}(-1)^{j}\binom{n-m}{n-j}\tag{17}\\ &=\sum_{j=0}^{n-m}\frac{1}{m+j}(-1)^{m+j}\binom{n-m}{j} \end{align*} and (10) follows.

Comment:

• In (14) we exchange the sums.

• In (15) we shift the index $k$ of the inner sum to start from $k=0$.

• In (16) we apply the Chu-Vandermonde identity.

• In (17) we shift the index $j$ to start from $j=0$.

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Note the shift operator $E$ with $E(f(m))=f(m+1)$ and the delta operator $\Delta:=E-1$ are related by \begin{align*} \Delta^n=(E-1)^n=\sum_{j=0}^n\binom{n}{j}(-1)^{n-j}E^j\tag{17} \end{align*}

applying the delta operator to $f(m)=\frac{1}{m}$ results in \begin{align*} \Delta\left(\frac{1}{m}\right)&=\frac{1}{m+1}-\frac{1}{m}=-\frac{1}{m(m+1)}\\ \Delta^2\left(\frac{1}{m}\right)&=-\frac{1}{(m+1)(m+2)}-\frac{1}{m(m+1)}=\frac{2}{m(m+1)(m+2)}\\ &\cdots\\ \Delta^n\left(\frac{1}{m}\right)&=\frac{(-1)^nn!}{m(m+1)\cdots(m+n)}=\frac{(-1)^n}{m}\binom{m+n}{n}^{-1}\tag{18} \end{align*} which can be shown easily by using mathematical induction.

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Applying (17) and (18) is essentially the proof of (11) since \begin{align*} \Delta^n\left(\frac{1}{m}\right)&=\sum_{j=0}^n\binom{n}{j}(-1)^{n-j}E^j\left(\frac{1}{m}\right)\\ &=\sum_{j=0}^n\binom{n}{j}(-1)^{n-j}\frac{1}{m+j}\\ &=\frac{(-1)^m}{m}\binom{m+n}{n}^{-1} \end{align*} Replacing $n$ with $n-m$, cancelling $(-1)^n$ and the claim (11) follows.

Finally putting all together

we obtain from (8), (9) and (13) \begin{align*} \frac{1}{n!}(xD_x+n)^{\underline{n}}\left(\frac{\log x}{x^m}\right) &=\sum_{k=0}^n\binom{n}{k}\frac{1}{k!}x^kD_x^k\left(\frac{\log x}{x^m}\right)\\ &=x^{-m}\log x\sum_{k=0}^m\binom{n}{k}(-1)^k\frac{1}{k!}m^{\overline{k}}\\ &\qquad-x^{-m}\sum_{k=1}^n\binom{n}{k}(-1)^k\frac{1}{k!}\sum_{j=1}^k\binom{k}{j}(j-1)!m^{\overline{k-j}}\\ &=x^{-m}\log x\cdot 0-x^{-m}\cdot\frac{(-1)^m}{m}\binom{n}{m}^{-1}\\ &=\frac{(-1)^{m+1}}{mx^m}\binom{n}{m}^{-1}\qquad \qquad 1\leq m\leq n\\ &\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\Box \end{align*} and the claim (2) follows.