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Let $R[X,\, Y,\, Z]$ denote the ring of the polynomials in noncommuting indeterminates $X,\, Y,\, Z$ over $R$. Define an automorphism $\sigma$ on $R[X,\, Y,\, Z]$ by:

$\sigma[F(X,\, Y,\, Z)]=F(X+1,\, Y,\, Z)$

and a $\sigma$-derivation $\Delta=\sigma-I$:

$\Delta[F(X,\, Y,\, Z)]=F(X+1,\, Y,\, Z)-F(X,\, Y,\, Z)$.

For two polynomials $F$ and $G$ in $R[X,\, Y,\, Z]$ we have (which is easy to show by definition of $\Delta$):

\begin{equation} \Delta[F+G]=\Delta[F]+\Delta[G] \,\,\, (1) \end{equation}

\begin{equation} \Delta[FG]=(\Delta[F])(\sigma[G])+F(\Delta[G]) \,\,\, (2) \end{equation}

and induction gives the Leibniz Formula

\begin{equation} \Delta^{n}[FG]=\sum_{r=0}^{n}\begin{pmatrix}n\\ r \end{pmatrix}(\Delta^{r}[F])(\sigma^{r}\Delta^{n-r}[G]) \end{equation}

I would like to prove this formula by induction. It is easily verified that this is true for $n=1$ and supposing that this is true for $n$ we get for $n+1$

$\Delta^{n+1}[FG]=\Delta(\Delta^{n}[FG])=\Delta\Bigl(\sum_{r=0}^{n}\begin{pmatrix}n\\ r \end{pmatrix}(\Delta^{r}[F])(\sigma^{r}\Delta^{n-r}[G])\Bigr)=\sum_{r=0}^{n}\Bigl(\Delta\begin{pmatrix}n\\ r \end{pmatrix}(\Delta^{r}[F])(\sigma^{r}\Delta^{n-r}[G])\Bigr)=\sum_{r=0}^{n}\Bigl(\begin{pmatrix}n\\ r \end{pmatrix}\Delta(\Delta^{r}[F])(\sigma^{r}\Delta^{n-r}[G])\Bigr) $

Could anyone help me to finish this proof, please? I'm stucked at this point and I can't move forward.

I've done few more steps of these induction.

$\sum_{r=0}^{n}\Bigl(\begin{pmatrix}n\\ r \end{pmatrix}\Delta(\Delta^{r}[F])(\sigma^{r}\Delta^{n-r}[G])=\sum_{r=0}^{n}\begin{pmatrix}n\\ r \end{pmatrix}\Delta^{r+1}[F]\sigma^{r+1}\Delta^{n-r}[G]+\sum_{r=0}^{n}\begin{pmatrix}n\\ r \end{pmatrix}[F]\sigma^{r}\Delta^{n-r-1}[G]=[F]\Delta^{n+1}[G]+\sum_{r=1}^{n}\begin{pmatrix}n\\ r \end{pmatrix}[F]\sigma^{r}\Delta^{n-r-1}[G]+\sum_{r=0}^{n-1}\begin{pmatrix}n\\ r \end{pmatrix}\Delta^{r+1}[F]\sigma^{r+1}\Delta^{n-r}[G]+\Delta^{n+1}[F]\sigma^{n+1}[G] $

And I am stucked again at these point. How can I finish these proof?

Thank you.

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  • $\begingroup$ I suspect it will be easier to prove it by doing $\Delta^{n+1}=\Delta^n\Delta$ rather than $\Delta\Delta^n$. $\endgroup$ Sep 19, 2013 at 17:27
  • $\begingroup$ It looks like you should be able to make some good progress by applying the $\sigma$-derivation property $\Delta(AB) = \Delta(A) \sigma(B) + A\Delta(B)$ in the last expression, along with the fact that $\sigma$ and $\Delta$ commute. I'm guessing that Pascal's rule will finish the job. $\endgroup$ Sep 19, 2013 at 17:31

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